Re: [R] help - Package: stats - function ar.ols

[Ebert,Timothy Aaron]

The data came through fine, the program was a miss. Can you paste the program 
into a ".txt" document like a notepad file and send that? You could also paste 
it into your email IF your email is configured to send text and NOT html.

TIm

-Original Message-
From: R-help  On Behalf Of Pedro Gavronski.
Sent: Friday, February 23, 2024 5:00 AM
To: Rui Barradas 
Cc: r-help@r-project.org; r-help-requ...@r-project.org
Subject: Re: [R] help - Package: stats - function ar.ols

[External Email]

Hello,

Thanks for the reply Rui and for pointing out that I forgot to attach my code. 
Please find attached in this email my code and data.

Thanks in advance.

Best regards, Pedro Gerhardt Gavronski.

On Fri, Feb 23, 2024 at 5:50 AM Rui Barradas  wrote:
>
> Às 16:34 de 22/02/2024, Pedro Gavronski. escreveu:
> > Hello,
> >
> > My name is Pedro and it is nice to meet you all. I am having trouble
> > understanding a message that I receive when use function ar.ols from
> > package stats, it says that "Warning message:
> > In ar.ols(x = dtb[2:6966, ], demean = FALSE, intercept = TRUE,
> > prewhite = TRUE) :
> >model order:  2 singularities in the computation of the
> > projection matrix results are only valid up to model order 1, which
> > I do not know what it means, if someone could clarify it, I would
> > really appreciate it.
> >
> > Attached to this email you will find my code and data I used to run
> > this formula.
> >
> > Thanks in advance.
> >
> > Best regards,  Pedro.
> >
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://st/
> > at.ethz.ch%2Fmailman%2Flistinfo%2Fr-help=05%7C02%7Ctebert%40ufl
> > .edu%7C0afd65836d774adc099408dc349f53c7%7C0d4da0f84a314d76ace60a6233
> > 1e1b84%7C0%7C0%7C638443106144255449%7CUnknown%7CTWFpbGZsb3d8eyJWIjoi
> > MC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C0%7C%7C
> > %7C=12OGc13jk3Lejmcvx5WH8Bko8JSpRxbtW3mzL3OIjyk%3D=0
> > PLEASE do read the posting guide
> > http://www/
> > .r-project.org%2Fposting-guide.html=05%7C02%7Ctebert%40ufl.edu%
> > 7C0afd65836d774adc099408dc349f53c7%7C0d4da0f84a314d76ace60a62331e1b8
> > 4%7C0%7C0%7C638443106144262387%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wL
> > jAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C0%7C%7C%7C
> > data=fAVz0iQ0zS5pmrxrzNeazRrrv%2FyjFdWvwMfgjhibdEA%3D=0
> > and provide commented, minimal, self-contained, reproducible code.
> Hello,
>
> Thanks for the data but the code is missing from the attachment.
> Can you please post your code? In an attachment or directly in the
> e-mail body.
>
> Rui Barradas
>
>
> --
> Este e-mail foi analisado pelo software antivírus AVG para verificar a 
> presença de vírus.
> http://www.a/
> vg.com%2F=05%7C02%7Ctebert%40ufl.edu%7C0afd65836d774adc099408dc34
> 9f53c7%7C0d4da0f84a314d76ace60a62331e1b84%7C0%7C0%7C638443106144266047
> %7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6I
> k1haWwiLCJXVCI6Mn0%3D%7C0%7C%7C%7C=rdqEhXdZ5%2FThzbhuKgspNaBPyhf
> aC%2BWh7MFq4iq%2BxVE%3D=0
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] help - Package: stats - function ar.ols

[Rui Barradas]

Às 16:34 de 22/02/2024, Pedro Gavronski. escreveu:

Hello,

My name is Pedro and it is nice to meet you all. I am having trouble
understanding a message that I receive when use function ar.ols from
package stats, it says that "Warning message:
In ar.ols(x = dtb[2:6966, ], demean = FALSE, intercept = TRUE,
prewhite = TRUE) :
   model order:  2 singularities in the computation of the projection
matrix results are only valid up to model order 1, which I do not know
what it means, if someone could clarify it, I would really appreciate
it.

Attached to this email you will find my code and data I used to run
this formula.

Thanks in advance.

Best regards,  Pedro.


__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Hello,

Thanks for the data but the code is missing from the attachment.
Can you please post your code? In an attachment or directly in the 
e-mail body.


Rui Barradas


--
Este e-mail foi analisado pelo software antivírus AVG para verificar a presença 
de vírus.
www.avg.com

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with spTransform() function and final plot colors

[Poling, William via R-help]

Thank you Ege, I appreciate your response.

I have move this to r-sig-geo.

WHP

Proprietary

-Original Message-
From: Ege Rubak  
Sent: Monday, May 18, 2020 6:41 AM
To: Poling, William ; r-help@r-project.org
Subject: [EXTERNAL] Re: [R] Help with spTransform() function and final plot 
colors

 External Email - Use Caution 

You are more likely to get help with specific problems related to
spTransform() on the dedicated list r-sig-geo.

You should provide a minimal reproducible example. Your code refers to e.g. the 
object `tmp1b`, which we don't have. I think the spTransform() part will work 
with this correction:

xy <- SpatialPointsDataFrame(sample[,2:3], sample[,1,drop=FALSE],
 proj4string=CRS("+proj=longlat
+ellps=WGS84 +datum=WGS84"))

xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84"))

The spatial workflow in R has largely moved to the `sf` package and you are 
probably better off using that in the future if you are new in the game anyway.

Hope this helps,
Ege

On Sun, 2020-05-17 at 09:50 +, Poling, William via R-help wrote:
> #RStudio Version Version 1.2.1335
> sessionInfo()
> # R version 4.0.0 Patched (2020-05-03 r78349)
> #Platform: x86_64-w64-mingw32/x64 (64-bit) #Running under: Windows 10 
> x64 (build 17763)
> 
> Hello. I am running my data through a routine I found that finds 
> clusters of data points based on distance rule.
> 
https://urldefense.proofpoint.com/v2/url?u=https-3A__gis.stackexchange.com_questions_64392_finding-2Dclusters-2Dof-2Dpoints-2Dbased-2Ddistance-2Drule-2Dusing-2Dr=DwIGaQ=wluqKIiwffOpZ6k5sqMWMBOn0vyYnlulRJmmvOXCFpM=j7MrcIQm2xjHa8v-2mTpmTCtKvneM2ExlYvnUWbsByY=sFygsSovnYNWjYOpjSzuSLI1MUrJN8Lih39pDQN98xI=2FCT-JBdWSB66pNvGLTM7Ec5Q7IbXatXBLsQqYT0gng=
 
> 
> 1. I get this error when I get to this point in the routine, see 
> complete routine below?
> xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84")) non finite 
> transformation detected:
> 
> I tried converting columns from numeric to Integer but did not help.
> 
> However, I continue to run the rest of the routine and the final 
> sequence, the plot itself, seems to work Can this error be corrected 
> somehow, despite the fact that it continues to work, just curious what 
> it is I guess"
> 
> 2. However in the plot at the end of the routine the color key appears 
> with the colors but the clusters in the plot are black, see plot call 
> at the end of the routine below?
> 
> Thank you for any advice.
> 
> WHP
> 
> ERROR:
> non finite transformation detected:
>   coords.x1 coords.x2
>  [1,] -119.7339  39.53939 Inf Inf
>  [2,] -119.7665  39.39630 Inf Inf
>  [3,] -119.7794  39.28768 Inf Inf
>  [4,] -121.0234  39.20503 Inf Inf
>  [5,] -122.0047  47.19262 Inf Inf
>  [6,] -122.0135  47.18883 Inf Inf
>  [7,] -122.0379  47.52190 Inf Inf
>  [8,] -122.0578  47.60975 Inf Inf
>  [9,] -122.1330  47.13669 Inf Inf
> [10,] -122.1509  47.55962 Inf Inf
> [11,] -122.1706  47.15546 Inf Inf
> [12,] -122.1846  47.23485 Inf Inf
> [13,] -122.1846  48.15307 Inf Inf
> [14,] -122.1851  47.44870 Inf Inf
> [15,] -122.1954  47.68485 Inf Inf
> [16,] -122.1990  47.51610 Inf Inf
> [17,] -122.2014  47.44772 Inf Inf
> [18,] -122.2025  47.69815 Inf Inf
> [19,] -122.2037  47.67190 Inf Inf
> [20,] -122.2090  47.40378 Inf Inf
> [21,] -122.2108  47.25336 Inf Inf
> [22,] -122.2291  47.63880 Inf Inf
> [23,] -122.2419  47.76870 Inf Inf
> [24,] -122.2722  48.04803 Inf Inf
> [25,] -122.2732  47.87700 Inf Inf
> [26,] -122.2804  47.77620 Inf Inf
> [27,] -122.2839  47.82103 Inf Inf
> [28,] -122.2890  47.86899 Inf Inf
> [29,] -122.2993  47.67306 Inf Inf
> [30,] -122.3180  47.38217 Inf Inf
> [31,] -122.3270  47.40378 Inf Inf
> [32,] -122.3474  47.43884 Inf Inf
> [33,] -122.3484  47.53083 Inf Inf
> [34,] -122.3581  47.27678 Inf Inf
> [35,] -122.3618  47.76735 Inf Inf
> [36,] -122.3700  47.56567 Inf Inf
> [37,] -122.3908  47.54938 Inf Inf
> [38,] -122.4128  47.64622 Inf Inf
> [39,] -122.4293  47.17660 Inf Inf
> [40,] -122.4621  47.44505 Inf Inf
> [41,] -122.4904  47.27460 Inf Inf
> [42,] -122.5515  46.93979 Inf Inf
> [43,] -122.7348  42.37320 Inf Inf
> [44,] -122.7827  47.31059 Inf Inf
> [45,] -122.7987  47.23475 Inf Inf
> [46,] -122.8385  42.35119 Inf Inf
> [47,] -122.8537  42.34495 Inf Inf
> [48,] -122.8904  42.39555 Inf Inf
> [49,] -122.8927  42.33022 Inf Inf
> [50,] -122.9451  47.37574 Inf Inf
> [51,] -122.9594  42.30376 Inf Inf
> [52,] -123.0641  47.16428 Inf Inf
> [53,] -123.3413  42.44117 Inf Inf
> Error in spTransform(xSP, CRSobj, ...) : 
>   failure in points
> 1:2:3:4:5:6:7:8:9:10:11:12:13:14:15:16:17:18:19:20:21:22:23:24:25:26:
> 27:28:29:30:31:32:33:34:35:36:37:38:39:4

Re: [R] Help with spTransform() function and final plot colors

[Ege Rubak]

You are more likely to get help with specific problems related to
spTransform() on the dedicated list r-sig-geo.

You should provide a minimal reproducible example. Your code refers to
e.g. the object `tmp1b`, which we don't have. I think the spTransform()
part will work with this correction:

xy <- SpatialPointsDataFrame(sample[,2:3], sample[,1,drop=FALSE],
 proj4string=CRS("+proj=longlat
+ellps=WGS84 +datum=WGS84"))

xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84"))

The spatial workflow in R has largely moved to the `sf` package and you
are probably better off using that in the future if you are new in the
game anyway.

Hope this helps,
Ege

On Sun, 2020-05-17 at 09:50 +, Poling, William via R-help wrote:
> #RStudio Version Version 1.2.1335 
> sessionInfo() 
> # R version 4.0.0 Patched (2020-05-03 r78349)
> #Platform: x86_64-w64-mingw32/x64 (64-bit)
> #Running under: Windows 10 x64 (build 17763)
> 
> Hello. I am running my data through a routine I found that finds
> clusters of data points based on distance rule.
> 
https://gis.stackexchange.com/questions/64392/finding-clusters-of-points-based-distance-rule-using-r
> 
> 1. I get this error when I get to this point in the routine, see
> complete routine below?
> xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84"))
> non finite transformation detected:
> 
> I tried converting columns from numeric to Integer but did not help.
> 
> However, I continue to run the rest of the routine and the final
> sequence, the plot itself, seems to work
> Can this error be corrected somehow, despite the fact that it
> continues to work, just curious what it is I guess"
> 
> 2. However in the plot at the end of the routine the color key
> appears with the colors but the clusters in the plot are black, see
> plot call at the end of the routine below?
> 
> Thank you for any advice.
> 
> WHP
> 
> ERROR:
> non finite transformation detected:
>   coords.x1 coords.x2
>  [1,] -119.7339  39.53939 Inf Inf
>  [2,] -119.7665  39.39630 Inf Inf
>  [3,] -119.7794  39.28768 Inf Inf
>  [4,] -121.0234  39.20503 Inf Inf
>  [5,] -122.0047  47.19262 Inf Inf
>  [6,] -122.0135  47.18883 Inf Inf
>  [7,] -122.0379  47.52190 Inf Inf
>  [8,] -122.0578  47.60975 Inf Inf
>  [9,] -122.1330  47.13669 Inf Inf
> [10,] -122.1509  47.55962 Inf Inf
> [11,] -122.1706  47.15546 Inf Inf
> [12,] -122.1846  47.23485 Inf Inf
> [13,] -122.1846  48.15307 Inf Inf
> [14,] -122.1851  47.44870 Inf Inf
> [15,] -122.1954  47.68485 Inf Inf
> [16,] -122.1990  47.51610 Inf Inf
> [17,] -122.2014  47.44772 Inf Inf
> [18,] -122.2025  47.69815 Inf Inf
> [19,] -122.2037  47.67190 Inf Inf
> [20,] -122.2090  47.40378 Inf Inf
> [21,] -122.2108  47.25336 Inf Inf
> [22,] -122.2291  47.63880 Inf Inf
> [23,] -122.2419  47.76870 Inf Inf
> [24,] -122.2722  48.04803 Inf Inf
> [25,] -122.2732  47.87700 Inf Inf
> [26,] -122.2804  47.77620 Inf Inf
> [27,] -122.2839  47.82103 Inf Inf
> [28,] -122.2890  47.86899 Inf Inf
> [29,] -122.2993  47.67306 Inf Inf
> [30,] -122.3180  47.38217 Inf Inf
> [31,] -122.3270  47.40378 Inf Inf
> [32,] -122.3474  47.43884 Inf Inf
> [33,] -122.3484  47.53083 Inf Inf
> [34,] -122.3581  47.27678 Inf Inf
> [35,] -122.3618  47.76735 Inf Inf
> [36,] -122.3700  47.56567 Inf Inf
> [37,] -122.3908  47.54938 Inf Inf
> [38,] -122.4128  47.64622 Inf Inf
> [39,] -122.4293  47.17660 Inf Inf
> [40,] -122.4621  47.44505 Inf Inf
> [41,] -122.4904  47.27460 Inf Inf
> [42,] -122.5515  46.93979 Inf Inf
> [43,] -122.7348  42.37320 Inf Inf
> [44,] -122.7827  47.31059 Inf Inf
> [45,] -122.7987  47.23475 Inf Inf
> [46,] -122.8385  42.35119 Inf Inf
> [47,] -122.8537  42.34495 Inf Inf
> [48,] -122.8904  42.39555 Inf Inf
> [49,] -122.8927  42.33022 Inf Inf
> [50,] -122.9451  47.37574 Inf Inf
> [51,] -122.9594  42.30376 Inf Inf
> [52,] -123.0641  47.16428 Inf Inf
> [53,] -123.3413  42.44117 Inf Inf
> Error in spTransform(xSP, CRSobj, ...) : 
>   failure in points
> 1:2:3:4:5:6:7:8:9:10:11:12:13:14:15:16:17:18:19:20:21:22:23:24:25:26:
> 27:28:29:30:31:32:33:34:35:36:37:38:39:40:41:42:43:44:45:46:47:48:49:
> 50:51:52:53
> In addition: Warning message:
> In spTransform(xSP, CRSobj, ...) : 53 projected point(s) not finite
> 
> Here is the data:
> dput(sample)
> structure(list(ID = 1:53, Longitude = c(-119.733899, -119.766493, 
> -119.779416, -121.0234, -122.004736, -122.013456, -122.0379, 
> -122.0578, -122.132971, -122.150901, -122.170608, -122.18462, 
> -122.184639, -122.185079, -122.195398, -122.198994, -122.201356, 
> -122.202507, -122.20371, -122.209047, -122.210797, -122.229095, 
> -122.2419, -122.27216, -122.273164, -122.280355, -122.28389, 
> -122.289043, -122.299261, -122.318009, -122.326987, -122.347382, 
> -122.34844, -122.358115, -122.361839, -122.37003, -122.390815, 
> -122.41282, -122.429323, -122.462136, -122.490417, -122.551483, 
> -122.734847, -122.782736, -122.798669, -122.838498, -122.853683, 
> -122.8904, -122.89271, -122.94511, -122.959407, -123.064087, 
> 

Re: [R] Help with spTransform() function and final plot colors

[Poling, William via R-help]

Hello again.

I also found this discussion on non-finite transformation error, however, I am 
not sure what to look for in the output after I apply my data?

https://stackoverflow.com/questions/14880294/non-finite-transformation-detected-in-sptransform-in-rgdal-r-package


str(sample)
GPS.Points <- sample[,c(2,3)]

n_chunks <- 3 #number of pieces you will break you data into
n.points <- dim(GPS.Points)[1]
breaks <- seq(1,n.points, by=round(n.points/n_chunks))
breaks <- c(breaks, n.points) #make sure to include last points as well

i=1
for (i in 1:(length(breaks)-1)){
  cat('\n','converting points', breaks[i], "to", breaks[i+1])  
  temp.GPS.Points=GPS.Points[breaks[i]:breaks[i+1],]
  temp.GPS.Points.Spatial.Data <- SpatialPoints(temp.GPS.Points, 
proj4string=CRS("+proj=longlat 
+ellps=WGS84"))
  temp.GPS.Points.UTM.Spatial.Data <- spTransform(temp.GPS.Points.Spatial.Data,
  CRS("+proj=utm +south 
+zone=34 
+ellps=WGS84"))
}

Thank you for any advice.

WHP

Proprietary

-Original Message-
From: Poling, William  
Sent: Sunday, May 17, 2020 5:20 AM
To: r-help@r-project.org
Cc: Poling, William 
Subject: RE: Help with spTransform() function and final plot colors

Hello, I have found an additional problem.

I should be getting 3 columns back in the xy@data at some point.
$ ID   : int  1 2 3 4 5 6 7 8 9 10 ...
# $ Clust  : int  1 1 1 1 1 1 1 1 1 1 ... This is always - 1
# $ Clust.1   : int  1 1 1 1 1 1 1 1 1 1 ... This would be the 
cluster number, like 1-3, I am not getting this column back?


Yet the plot runs and indicates three different clusters?



WHP


Proprietary

-Original Message-
From: Poling, William  
Sent: Sunday, May 17, 2020 4:50 AM
To: r-help@r-project.org
Cc: Poling, William 
Subject: Help with spTransform() function and final plot colors

#RStudio Version Version 1.2.1335
sessionInfo()
# R version 4.0.0 Patched (2020-05-03 r78349)
#Platform: x86_64-w64-mingw32/x64 (64-bit) #Running under: Windows 10 x64 
(build 17763)

Hello. I am running my data through a routine I found that finds clusters of 
data points based on distance rule.
https://gis.stackexchange.com/questions/64392/finding-clusters-of-points-based-distance-rule-using-r

1. I get this error when I get to this point in the routine, see complete 
routine below?
xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84")) non finite 
transformation detected:

I tried converting columns from numeric to Integer but did not help.

However, I continue to run the rest of the routine and the final sequence, the 
plot itself, seems to work Can this error be corrected somehow, despite the 
fact that it continues to work, just curious what it is I guess"

2. However in the plot at the end of the routine the color key appears with the 
colors but the clusters in the plot are black, see plot call at the end of the 
routine below?

Thank you for any advice.

WHP

ERROR:
non finite transformation detected:
  coords.x1 coords.x2
 [1,] -119.7339  39.53939 Inf Inf
 [2,] -119.7665  39.39630 Inf Inf
 [3,] -119.7794  39.28768 Inf Inf
 [4,] -121.0234  39.20503 Inf Inf
 [5,] -122.0047  47.19262 Inf Inf
 [6,] -122.0135  47.18883 Inf Inf
 [7,] -122.0379  47.52190 Inf Inf
 [8,] -122.0578  47.60975 Inf Inf
 [9,] -122.1330  47.13669 Inf Inf
[10,] -122.1509  47.55962 Inf Inf
[11,] -122.1706  47.15546 Inf Inf
[12,] -122.1846  47.23485 Inf Inf
[13,] -122.1846  48.15307 Inf Inf
[14,] -122.1851  47.44870 Inf Inf
[15,] -122.1954  47.68485 Inf Inf
[16,] -122.1990  47.51610 Inf Inf
[17,] -122.2014  47.44772 Inf Inf
[18,] -122.2025  47.69815 Inf Inf
[19,] -122.2037  47.67190 Inf Inf
[20,] -122.2090  47.40378 Inf Inf
[21,] -122.2108  47.25336 Inf Inf
[22,] -122.2291  47.63880 Inf Inf
[23,] -122.2419  47.76870 Inf Inf
[24,] -122.2722  48.04803 Inf Inf
[25,] -122.2732  47.87700 Inf Inf
[26,] -122.2804  47.77620 Inf Inf
[27,] -122.2839  47.82103 Inf Inf
[28,] -122.2890  47.86899 Inf Inf
[29,] -122.2993  47.67306 Inf Inf
[30,] -122.3180  47.38217 Inf Inf
[31,] -122.3270  47.40378 Inf Inf
[32,] -122.3474  47.43884 Inf Inf
[33,] -122.3484  47.53083 Inf Inf
[34,] -122.3581  47.27678 Inf Inf
[35,] -122.3618  47.76735 Inf Inf
[36,] -122.3700  47.56567 Inf Inf
[37,] -122.3908  47.54938 Inf Inf
[38,] -122.4128  47.64622 Inf Inf
[39,] -122.4293  47.17660 Inf Inf
[40,] -122.4621  47.44505 Inf Inf
[41,] -122.4904  47.27460 Inf Inf
[42,] -122.5515  46.93979 Inf Inf
[43,] -122.7348  42.37320 Inf Inf
[44,] -122.7827  47.31059 Inf Inf
[45,] -122.7987  47.23475 Inf Inf
[46,] -122.8385  42.35119 Inf Inf
[47,] -122.8537  42.34495 Inf Inf
[48,] -122.8904  42.39555 Inf Inf
[49,] -122.8927  42.33022 Inf Inf
[50,] -122.9451  47.37574 Inf Inf
[51,] -122.9594  42.30376 Inf Inf
[52,] -123.0641  47.16428 Inf Inf
[53,] -123.3413  42.44117 Inf Inf
Error in spTransform(xSP, CRSobj, ...) : 
  failure in points 

Re: [R] Help with spTransform() function and final plot colors

[Poling, William via R-help]

Hello, I have found an additional problem.

I should be getting 3 columns back in the xy@data at some point.
$ ID   : int  1 2 3 4 5 6 7 8 9 10 ...
# $ Clust  : int  1 1 1 1 1 1 1 1 1 1 ... This is always - 1
# $ Clust.1   : int  1 1 1 1 1 1 1 1 1 1 ... This would be the 
cluster number, like 1-3, I am not getting this column back?


Yet the plot runs and indicates three different clusters?



WHP


Proprietary

-Original Message-
From: Poling, William  
Sent: Sunday, May 17, 2020 4:50 AM
To: r-help@r-project.org
Cc: Poling, William 
Subject: Help with spTransform() function and final plot colors

#RStudio Version Version 1.2.1335
sessionInfo()
# R version 4.0.0 Patched (2020-05-03 r78349)
#Platform: x86_64-w64-mingw32/x64 (64-bit) #Running under: Windows 10 x64 
(build 17763)

Hello. I am running my data through a routine I found that finds clusters of 
data points based on distance rule.
https://gis.stackexchange.com/questions/64392/finding-clusters-of-points-based-distance-rule-using-r

1. I get this error when I get to this point in the routine, see complete 
routine below?
xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84")) non finite 
transformation detected:

I tried converting columns from numeric to Integer but did not help.

However, I continue to run the rest of the routine and the final sequence, the 
plot itself, seems to work Can this error be corrected somehow, despite the 
fact that it continues to work, just curious what it is I guess"

2. However in the plot at the end of the routine the color key appears with the 
colors but the clusters in the plot are black, see plot call at the end of the 
routine below?

Thank you for any advice.

WHP

ERROR:
non finite transformation detected:
  coords.x1 coords.x2
 [1,] -119.7339  39.53939 Inf Inf
 [2,] -119.7665  39.39630 Inf Inf
 [3,] -119.7794  39.28768 Inf Inf
 [4,] -121.0234  39.20503 Inf Inf
 [5,] -122.0047  47.19262 Inf Inf
 [6,] -122.0135  47.18883 Inf Inf
 [7,] -122.0379  47.52190 Inf Inf
 [8,] -122.0578  47.60975 Inf Inf
 [9,] -122.1330  47.13669 Inf Inf
[10,] -122.1509  47.55962 Inf Inf
[11,] -122.1706  47.15546 Inf Inf
[12,] -122.1846  47.23485 Inf Inf
[13,] -122.1846  48.15307 Inf Inf
[14,] -122.1851  47.44870 Inf Inf
[15,] -122.1954  47.68485 Inf Inf
[16,] -122.1990  47.51610 Inf Inf
[17,] -122.2014  47.44772 Inf Inf
[18,] -122.2025  47.69815 Inf Inf
[19,] -122.2037  47.67190 Inf Inf
[20,] -122.2090  47.40378 Inf Inf
[21,] -122.2108  47.25336 Inf Inf
[22,] -122.2291  47.63880 Inf Inf
[23,] -122.2419  47.76870 Inf Inf
[24,] -122.2722  48.04803 Inf Inf
[25,] -122.2732  47.87700 Inf Inf
[26,] -122.2804  47.77620 Inf Inf
[27,] -122.2839  47.82103 Inf Inf
[28,] -122.2890  47.86899 Inf Inf
[29,] -122.2993  47.67306 Inf Inf
[30,] -122.3180  47.38217 Inf Inf
[31,] -122.3270  47.40378 Inf Inf
[32,] -122.3474  47.43884 Inf Inf
[33,] -122.3484  47.53083 Inf Inf
[34,] -122.3581  47.27678 Inf Inf
[35,] -122.3618  47.76735 Inf Inf
[36,] -122.3700  47.56567 Inf Inf
[37,] -122.3908  47.54938 Inf Inf
[38,] -122.4128  47.64622 Inf Inf
[39,] -122.4293  47.17660 Inf Inf
[40,] -122.4621  47.44505 Inf Inf
[41,] -122.4904  47.27460 Inf Inf
[42,] -122.5515  46.93979 Inf Inf
[43,] -122.7348  42.37320 Inf Inf
[44,] -122.7827  47.31059 Inf Inf
[45,] -122.7987  47.23475 Inf Inf
[46,] -122.8385  42.35119 Inf Inf
[47,] -122.8537  42.34495 Inf Inf
[48,] -122.8904  42.39555 Inf Inf
[49,] -122.8927  42.33022 Inf Inf
[50,] -122.9451  47.37574 Inf Inf
[51,] -122.9594  42.30376 Inf Inf
[52,] -123.0641  47.16428 Inf Inf
[53,] -123.3413  42.44117 Inf Inf
Error in spTransform(xSP, CRSobj, ...) : 
  failure in points 
1:2:3:4:5:6:7:8:9:10:11:12:13:14:15:16:17:18:19:20:21:22:23:24:25:26:27:28:29:30:31:32:33:34:35:36:37:38:39:40:41:42:43:44:45:46:47:48:49:50:51:52:53
In addition: Warning message:
In spTransform(xSP, CRSobj, ...) : 53 projected point(s) not finite

Here is the data:
dput(sample)
structure(list(ID = 1:53, Longitude = c(-119.733899, -119.766493, -119.779416, 
-121.0234, -122.004736, -122.013456, -122.0379, -122.0578, -122.132971, 
-122.150901, -122.170608, -122.18462, -122.184639, -122.185079, -122.195398, 
-122.198994, -122.201356, -122.202507, -122.20371, -122.209047, -122.210797, 
-122.229095, -122.2419, -122.27216, -122.273164, -122.280355, -122.28389, 
-122.289043, -122.299261, -122.318009, -122.326987, -122.347382, -122.34844, 
-122.358115, -122.361839, -122.37003, -122.390815, -122.41282, -122.429323, 
-122.462136, -122.490417, -122.551483, -122.734847, -122.782736, -122.798669, 
-122.838498, -122.853683, -122.8904, -122.89271, -122.94511, -122.959407, 
-123.064087, -123.341346), Latitude = c(39.53939, 39.396298, 39.287681, 
39.205028, 47.192616, 47.188833, 47.5219, 47.609748, 47.13669, 47.559616, 
47.155455, 47.234849, 48.15307, 47.448697, 47.684854, 47.516104, 47.447723, 
47.698146, 47.6719, 47.403778, 47.253364, 47.638795, 47.768701, 48.048027, 
47.876997, 47.776205, 

[R] Help with spTransform() function and final plot colors

[Poling, William via R-help]

#RStudio Version Version 1.2.1335 
sessionInfo() 
# R version 4.0.0 Patched (2020-05-03 r78349)
#Platform: x86_64-w64-mingw32/x64 (64-bit)
#Running under: Windows 10 x64 (build 17763)

Hello. I am running my data through a routine I found that finds clusters of 
data points based on distance rule.
https://gis.stackexchange.com/questions/64392/finding-clusters-of-points-based-distance-rule-using-r

1. I get this error when I get to this point in the routine, see complete 
routine below?
xy <- spTransform(xy, CRS("+init=epsg:27700 +datum=WGS84"))
non finite transformation detected:

I tried converting columns from numeric to Integer but did not help.

However, I continue to run the rest of the routine and the final sequence, the 
plot itself, seems to work
Can this error be corrected somehow, despite the fact that it continues to 
work, just curious what it is I guess"

2. However in the plot at the end of the routine the color key appears with the 
colors but the clusters in the plot are black, see plot call at the end of the 
routine below?

Thank you for any advice.

WHP

ERROR:
non finite transformation detected:
  coords.x1 coords.x2
 [1,] -119.7339  39.53939 Inf Inf
 [2,] -119.7665  39.39630 Inf Inf
 [3,] -119.7794  39.28768 Inf Inf
 [4,] -121.0234  39.20503 Inf Inf
 [5,] -122.0047  47.19262 Inf Inf
 [6,] -122.0135  47.18883 Inf Inf
 [7,] -122.0379  47.52190 Inf Inf
 [8,] -122.0578  47.60975 Inf Inf
 [9,] -122.1330  47.13669 Inf Inf
[10,] -122.1509  47.55962 Inf Inf
[11,] -122.1706  47.15546 Inf Inf
[12,] -122.1846  47.23485 Inf Inf
[13,] -122.1846  48.15307 Inf Inf
[14,] -122.1851  47.44870 Inf Inf
[15,] -122.1954  47.68485 Inf Inf
[16,] -122.1990  47.51610 Inf Inf
[17,] -122.2014  47.44772 Inf Inf
[18,] -122.2025  47.69815 Inf Inf
[19,] -122.2037  47.67190 Inf Inf
[20,] -122.2090  47.40378 Inf Inf
[21,] -122.2108  47.25336 Inf Inf
[22,] -122.2291  47.63880 Inf Inf
[23,] -122.2419  47.76870 Inf Inf
[24,] -122.2722  48.04803 Inf Inf
[25,] -122.2732  47.87700 Inf Inf
[26,] -122.2804  47.77620 Inf Inf
[27,] -122.2839  47.82103 Inf Inf
[28,] -122.2890  47.86899 Inf Inf
[29,] -122.2993  47.67306 Inf Inf
[30,] -122.3180  47.38217 Inf Inf
[31,] -122.3270  47.40378 Inf Inf
[32,] -122.3474  47.43884 Inf Inf
[33,] -122.3484  47.53083 Inf Inf
[34,] -122.3581  47.27678 Inf Inf
[35,] -122.3618  47.76735 Inf Inf
[36,] -122.3700  47.56567 Inf Inf
[37,] -122.3908  47.54938 Inf Inf
[38,] -122.4128  47.64622 Inf Inf
[39,] -122.4293  47.17660 Inf Inf
[40,] -122.4621  47.44505 Inf Inf
[41,] -122.4904  47.27460 Inf Inf
[42,] -122.5515  46.93979 Inf Inf
[43,] -122.7348  42.37320 Inf Inf
[44,] -122.7827  47.31059 Inf Inf
[45,] -122.7987  47.23475 Inf Inf
[46,] -122.8385  42.35119 Inf Inf
[47,] -122.8537  42.34495 Inf Inf
[48,] -122.8904  42.39555 Inf Inf
[49,] -122.8927  42.33022 Inf Inf
[50,] -122.9451  47.37574 Inf Inf
[51,] -122.9594  42.30376 Inf Inf
[52,] -123.0641  47.16428 Inf Inf
[53,] -123.3413  42.44117 Inf Inf
Error in spTransform(xSP, CRSobj, ...) : 
  failure in points 
1:2:3:4:5:6:7:8:9:10:11:12:13:14:15:16:17:18:19:20:21:22:23:24:25:26:27:28:29:30:31:32:33:34:35:36:37:38:39:40:41:42:43:44:45:46:47:48:49:50:51:52:53
In addition: Warning message:
In spTransform(xSP, CRSobj, ...) : 53 projected point(s) not finite

Here is the data:
dput(sample)
structure(list(ID = 1:53, Longitude = c(-119.733899, -119.766493, 
-119.779416, -121.0234, -122.004736, -122.013456, -122.0379, 
-122.0578, -122.132971, -122.150901, -122.170608, -122.18462, 
-122.184639, -122.185079, -122.195398, -122.198994, -122.201356, 
-122.202507, -122.20371, -122.209047, -122.210797, -122.229095, 
-122.2419, -122.27216, -122.273164, -122.280355, -122.28389, 
-122.289043, -122.299261, -122.318009, -122.326987, -122.347382, 
-122.34844, -122.358115, -122.361839, -122.37003, -122.390815, 
-122.41282, -122.429323, -122.462136, -122.490417, -122.551483, 
-122.734847, -122.782736, -122.798669, -122.838498, -122.853683, 
-122.8904, -122.89271, -122.94511, -122.959407, -123.064087, 
-123.341346), Latitude = c(39.53939, 39.396298, 39.287681, 39.205028, 
47.192616, 47.188833, 47.5219, 47.609748, 47.13669, 47.559616, 
47.155455, 47.234849, 48.15307, 47.448697, 47.684854, 47.516104, 
47.447723, 47.698146, 47.6719, 47.403778, 47.253364, 47.638795, 
47.768701, 48.048027, 47.876997, 47.776205, 47.821029, 47.868987, 
47.673056, 47.382165, 47.403785, 47.438836, 47.530831, 47.276776, 
47.76735, 47.565667, 47.549377, 47.646222, 47.176596, 47.445053, 
47.274599, 46.939789, 42.373195, 47.310595, 47.234748, 42.351189, 
42.344953, 42.395547, 42.33022, 47.375736, 42.303755, 47.164278, 
42.441172)), class = "data.frame", row.names = c(NA, -53L))

Here is the routine:
require(sp)
require(rgdal)
#25*1609.34 = 40233.5  
dis <- 40233.5 #Distance threshold 25miles converted to meters

x <- tmp1b[,c(6)]
head(x)
y <- tmp1b[,c(5)]
head(y)
str(y)

xy <- SpatialPointsDataFrame(matrix(c(x,y), ncol=2), 
data.frame(ID=seq(1:length(x))),
 

Re: [R] Help with gsub function

[Bill Poling]

Good morning Steve. Terrific, so kind of you to follow-up.

I will add that to my ever growing R bag of tips and tricks.

Cheers.

WHP

William H. Poling, Ph.D., MPH | Manager, Revenue Development
Data Intelligence & Analytics
Zelis Healthcare


-Original Message-
From: S Ellison 
Sent: Monday, March 18, 2019 8:32 AM
To: Bill Poling ; r-help (r-help@r-project.org) 

Subject: RE: Help with gsub function

> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")

Just to add something on why this didn't work ...

It looks like you were trying to match a hyphen followed by a number up to 
seven digits. by mistake(?) you gave the digit range as [0-0] so it would 
repmatch a hyphen followed by between none and seven zeroes. When it met "-0" 
it matched that.
And because it was gsub, it replaced what it matched.
If you'd given it the right digit range it would have replaced the whole of the 
number.

If you _really_ wanted to do that kind of thing (control the following 
pattern), you'd have needed something like (untested) gsub("-([0-0]{0,7})", 
"\\1", tb2a$TID)

#The () means 'remember this bit"; the "\\1" means "put the first thing you 
remember here". And it needs to be "\\1" because that becomes "\1" for the grep 
parser.


Steve E


***
This email and any attachments are confidential. Any use...{{dropped:22}}

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function

[S Ellison]

> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")

Just to add something on why this didn't work ...

It looks like you were trying to match a hyphen followed by a number up to 
seven digits. by mistake(?) you gave the digit range as [0-0] so it would 
repmatch a hyphen followed by between none and seven zeroes. When it met "-0" 
it matched that.
And because it was gsub, it replaced what it matched.
If you'd given it the right digit range it would have replaced the whole of the 
number.

If you _really_ wanted to do that kind of thing (control the following 
pattern), you'd have needed something like (untested)
gsub("-([0-0]{0,7})", "\\1", tb2a$TID)

#The () means 'remember this bit"; the "\\1" means "put the first thing you 
remember here". And it needs to be "\\1" because that becomes "\1" for the grep 
parser.


Steve E


***
This email and any attachments are confidential. Any use...{{dropped:8}}

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function

[Bill Poling]

Yep, thank you Jeff, consequence of the first url I landed on asking how to do 
it and rushing off.

All set now.

Appreciate your help.

WHP

From: Jeff Newmiller 
Sent: Friday, March 15, 2019 4:00 PM
To: r-help@r-project.org; Bill Poling ; r-help 
(r-help@r-project.org) 
Subject: Re: [R] Help with gsub function

Your pattern seems ... way overboard? Why not

gsub("-", "", tb2a$TID)

On March 15, 2019 12:45:27 PM PDT, Bill Poling <mailto:bill.pol...@zelis.com> 
wrote:
>Good afternoon.
>
>sessionInfo()
>#R version 3.5.3 (2019-03-11)
>#Platform: x86_64-w64-mingw32/x64 (64-bit)
>#Running under: Windows >= 8 x64 (build 9200)
>
>I am using gsub function to remove a hyphen in a 9 character column of
>values in order to convert it to integer.
>
>Works fine except where the second segment has a leading 0, then it is
>eliminating the 0
>
>Example "73-0700090" becomes " 73700090"
> "77-0633896" becomes "77633896"
>
>Is there a remedy for this?
>
>tb2a$TID2 <-
>
>head(tb2a$TID,n=10)
>[1] "11-1352310" "45-2711804" "35-6001540" "77-0633896" "62-1762545"
>"61-1029768" "73-0700090" "47-0376604" "47-0486026" "38-3833117"
>> head(tb2a$TID2,n=10)
>[1] "111352310" "452711804" "356001540" "77633896" "621762545"
>"611029768" "73700090" "47376604" "47486026" "383833117"
>
>I have googled the problem and have not found a solution.
>
>http://www.endmemo.com/program/R/gsub.php
>http://r.789695.n4.nabble.com/extracting-characters-from-string-td3298971.html
>
>
>Thank you.
>
>WHP
>
>Confidentiality Notice This message is sent from Zelis.
>...{{dropped:13}}
>
>__
>mailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.

--
Sent from my phone. Please excuse my brevity.

Confidentiality Notice This message is sent from Zelis. This transmission may 
contain information which is privileged and confidential and is intended for 
the personal and confidential use of the named recipient only. Such information 
may be protected by applicable State and Federal laws from this disclosure or 
unauthorized use. If the reader of this message is not the intended recipient, 
or the employee or agent responsible for delivering the message to the intended 
recipient, you are hereby notified that any disclosure, review, discussion, 
copying, or taking any action in reliance on the contents of this transmission 
is strictly prohibited. If you have received this transmission in error, please 
contact the sender immediately. Zelis, 2018.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function

[Bill Poling]

Good morning Peter, yes that works fine. My attempt was based on a google 
search that looked promising but was obviously more complicated than it needed 
to be.

Thank you.

WHP

From: Peter Langfelder 
Sent: Friday, March 15, 2019 3:53 PM
To: Bill Poling 
Cc: r-help (r-help@r-project.org) 
Subject: Re: [R] Help with gsub function

If you want to remove just the hyphen, why not do

sub("-", "", tb2a$TID)

sub("-", "", "73-017323")
[1] "73017323"

Am I missing something?

Peter

On Fri, Mar 15, 2019 at 12:46 PM Bill Poling <mailto:bill.pol...@zelis.com> 
wrote:
>
> Good afternoon.
>
> sessionInfo()
> #R version 3.5.3 (2019-03-11)
> #Platform: x86_64-w64-mingw32/x64 (64-bit)
> #Running under: Windows >= 8 x64 (build 9200)
>
> I am using gsub function to remove a hyphen in a 9 character column of values 
> in order to convert it to integer.
>
> Works fine except where the second segment has a leading 0, then it is 
> eliminating the 0
>
> Example "73-0700090" becomes " 73700090"
> "77-0633896" becomes "77633896"
>
> Is there a remedy for this?
>
> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")
>
> head(tb2a$TID,n=10)
> [1] "11-1352310" "45-2711804" "35-6001540" "77-0633896" "62-1762545" 
> "61-1029768" "73-0700090" "47-0376604" "47-0486026" "38-3833117"
> > head(tb2a$TID2,n=10)
> [1] "111352310" "452711804" "356001540" "77633896" "621762545" "611029768" 
> "73700090" "47376604" "47486026" "383833117"
>
> I have googled the problem and have not found a solution.
>
> http://www.endmemo.com/program/R/gsub.php
> http://r.789695.n4.nabble.com/extracting-characters-from-string-td3298971.html
>
>
> Thank you.
>
> WHP
>
> Confidentiality Notice This message is sent from Zelis. ...{{dropped:13}}
>
> __
> mailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Confidentiality Notice This message is sent from Zelis. This transmission may 
contain information which is privileged and confidential and is intended for 
the personal and confidential use of the named recipient only. Such information 
may be protected by applicable State and Federal laws from this disclosure or 
unauthorized use. If the reader of this message is not the intended recipient, 
or the employee or agent responsible for delivering the message to the intended 
recipient, you are hereby notified that any disclosure, review, discussion, 
copying, or taking any action in reliance on the contents of this transmission 
is strictly prohibited. If you have received this transmission in error, please 
contact the sender immediately. Zelis, 2018.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function

[Jeff Newmiller]

Your pattern seems ... way overboard? Why not

gsub("-", "", tb2a$TID)

On March 15, 2019 12:45:27 PM PDT, Bill Poling  wrote:
>Good afternoon.
>
>sessionInfo()
>#R version 3.5.3 (2019-03-11)
>#Platform: x86_64-w64-mingw32/x64 (64-bit)
>#Running under: Windows >= 8 x64 (build 9200)
>
>I am using gsub function to remove a hyphen in a 9 character column of
>values in order to convert it to integer.
>
>Works fine except where the second segment has a leading 0, then it is
>eliminating the 0
>
>Example "73-0700090" becomes " 73700090"
> "77-0633896" becomes "77633896"
>
>Is there a remedy for this?
>
>tb2a$TID2 <- 
>
>head(tb2a$TID,n=10)
>[1] "11-1352310" "45-2711804" "35-6001540" "77-0633896" "62-1762545"
>"61-1029768" "73-0700090" "47-0376604" "47-0486026" "38-3833117"
>> head(tb2a$TID2,n=10)
>[1] "111352310" "452711804" "356001540" "77633896"  "621762545"
>"611029768" "73700090"  "47376604"  "47486026"  "383833117"
>
>I have googled the problem and have not found a solution.
>
>http://www.endmemo.com/program/R/gsub.php
>http://r.789695.n4.nabble.com/extracting-characters-from-string-td3298971.html
>
>
>Thank you.
>
>WHP
>
>Confidentiality Notice This message is sent from Zelis.
>...{{dropped:13}}
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.

-- 
Sent from my phone. Please excuse my brevity.

__
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Re: [R] Help with gsub function

[Peter Langfelder]

If you want to remove just the hyphen, why not do

sub("-", "", tb2a$TID)

sub("-", "", "73-017323")
[1] "73017323"

Am I missing something?

Peter

On Fri, Mar 15, 2019 at 12:46 PM Bill Poling  wrote:
>
> Good afternoon.
>
> sessionInfo()
> #R version 3.5.3 (2019-03-11)
> #Platform: x86_64-w64-mingw32/x64 (64-bit)
> #Running under: Windows >= 8 x64 (build 9200)
>
> I am using gsub function to remove a hyphen in a 9 character column of values 
> in order to convert it to integer.
>
> Works fine except where the second segment has a leading 0, then it is 
> eliminating the 0
>
> Example "73-0700090" becomes " 73700090"
>  "77-0633896" becomes "77633896"
>
> Is there a remedy for this?
>
> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")
>
> head(tb2a$TID,n=10)
>  [1] "11-1352310" "45-2711804" "35-6001540" "77-0633896" "62-1762545" 
> "61-1029768" "73-0700090" "47-0376604" "47-0486026" "38-3833117"
> > head(tb2a$TID2,n=10)
>  [1] "111352310" "452711804" "356001540" "77633896"  "621762545" "611029768" 
> "73700090"  "47376604"  "47486026"  "383833117"
>
> I have googled the problem and have not found a solution.
>
> http://www.endmemo.com/program/R/gsub.php
> http://r.789695.n4.nabble.com/extracting-characters-from-string-td3298971.html
>
>
> Thank you.
>
> WHP
>
> Confidentiality Notice This message is sent from Zelis. ...{{dropped:13}}
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function

[Ivan Krylov]

On Fri, 15 Mar 2019 19:45:27 +
Bill Poling  wrote:

Hello Bill,

> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")

Is the pattern supposed to mean something besides the "-" you want to
remove? For the problem you describe, pattern="-" should be enough. It
should locate all hyphens in the string and replace them with empty
strings, i.e. remove them.

-- 
Best regards,
Ivan

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and provide commented, minimal, self-contained, reproducible code.

[R] Help with gsub function

[Bill Poling]

Good afternoon.

sessionInfo()
#R version 3.5.3 (2019-03-11)
#Platform: x86_64-w64-mingw32/x64 (64-bit)
#Running under: Windows >= 8 x64 (build 9200)

I am using gsub function to remove a hyphen in a 9 character column of values 
in order to convert it to integer.

Works fine except where the second segment has a leading 0, then it is 
eliminating the 0

Example "73-0700090" becomes " 73700090"
 "77-0633896" becomes "77633896"

Is there a remedy for this?

tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")

head(tb2a$TID,n=10)
 [1] "11-1352310" "45-2711804" "35-6001540" "77-0633896" "62-1762545" 
"61-1029768" "73-0700090" "47-0376604" "47-0486026" "38-3833117"
> head(tb2a$TID2,n=10)
 [1] "111352310" "452711804" "356001540" "77633896"  "621762545" "611029768" 
"73700090"  "47376604"  "47486026"  "383833117"

I have googled the problem and have not found a solution.

http://www.endmemo.com/program/R/gsub.php
http://r.789695.n4.nabble.com/extracting-characters-from-string-td3298971.html


Thank you.

WHP

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[R] help with read function in Azure Data Lake Remote environment

[Bill Poling]

Hi,

Locally I am using windows:
R version 3.5.1 (2018-07-02)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

Locally on my laptop using RStudio I normally set my working directory and read 
in my csv files.

setwd("C:/WHP/Appeals")

getwd()#-Check to verify

read.csv("WHP_EditAppealRev_Tmp V1.csv",nrows=1, head=FALSE)#N=61 columns

appdf1 <- fread("WHP_EditAppealRev_Tmp 
V1.csv",select=c(5,9,10,11,14,15,16,17,18,25,31,28,32,42,45,46,47,48,49,53,54,58,59,61),header=TRUE,
 stringsAsFactors=TRUE)

This all works fine locally.

Now I am working in RStudio in a Remote desk top AZURE Data Lake environment 
(DSVM) however the remote file path out there is does not appear to be that 
different.

Remotley I am using windows:
R version 3.5.1 (2018-07-02)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows Server >= 2012 x64 (build 9200)

setwd("C:/Users/sysAdmin/Documents/WHP/NJDemo082018")
getwd()#-Check to verify

The directory gets set but the csv does not get read for some reason?

read.csv("C:/Users/sysAdmin/Documents/WHP/NJDemo082018/TestDemoV1.csv",nrows=1, 
head=FALSE)#N= columns
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
  cannot open file 
'C:/Users/sysAdmin/Documents/WHP/NJDemo082018/TestDemoV1.csv': No such file or 
directory

demo1 <- 
fread("C:/Users/sysAdmin/Documents/WHP/NJDemo082018/TestDemoV1.csv",select=c(1,2,3),header=TRUE,
 stringsAsFactors=TRUE)
Error in fread("C:/Users/sysAdmin/Documents/WHP/NJDemo082018/TestDemoV1.csv",  :
  File 'C:/Users/sysAdmin/Documents/WHP/NJDemo082018/TestDemoV1.csv' does not 
exist; getwd()=='C:/Users/sysAdmin/Documents/WHP/NJDemo082018'.
Include correct full path, or one or more spaces to consider the input a system 
command.

I have tried moving the csv file further forward to the root C: but still no 
luck?

read.csv("C:/TestDemoV1.csv",nrows=1, head=FALSE)#N= columns
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
  cannot open file 'C:/TestDemoV1.csv': No such file or directory


Any suggestions would be appreciated, thank you.

WHP


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and provide commented, minimal, self-contained, reproducible code.

Re: [R] help with recursive function

[William Dunlap via R-help]

recursive_funlp <- function(dataset = dat1, func = funlp2) {
...
if (!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
   else {
df2 <- recursive_funlp() # GIVE SOME ARGUMENTS HERE
return(df2)
}
}

When you recurse into recursve_funip(), give it at least a new dataset
argument.
Otherwise it just processes the entire dat1 again and the recursion never
stops.

Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Thu, Dec 14, 2017 at 9:11 AM, DIGHE, NILESH [AG/2362] <
nilesh.di...@monsanto.com> wrote:

> Eric:  I will try and see if I can figure out the issue by debugging as
> you suggested. I don’t know why my code after stopifnot is not getting
> executed where I like the code to run the funlp2 function when the if
> statement is TRUE but when it is false, I like it to keep running until the
> stopifnot condition is met.
>
> When the stopifnot condition is met, I like to get the output from if
> statement saved.
> Anyway,  I will keep trying.
> Again, Thanks for your help!
> Nilesh
>
> From: Eric Berger [mailto:ericjber...@gmail.com]
> Sent: Thursday, December 14, 2017 10:29 AM
> To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
> Cc: r-help <r-help@r-project.org>
> Subject: Re: [R] help with recursive function
>
> If you are trying to understand why the "stopifnot" condition is met you
> can replace it by something like:
>
> if ( any(dat2$norm_sd >= 1) )
>browser()
>
> This will put you in a debugging session where you can examine your
> variables, e.g.
>
> > dat$norm_sd
>
> HTH,
> Eric
>
>
>
> On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger <ericjber...@gmail.com
> <mailto:ericjber...@gmail.com>> wrote:
> The message is coming from your stopifnot() condition being met.
>
>
> On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] <
> nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
> Hi, I accidently left out few lines of code from the calclp function.
> Updated function is pasted below.
> I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is
> not TRUE“
>
> I would appreciate any help.
> Nilesh
> dput(calclp)
> function (dataset)
> {
> dat1 <- funlp1(dataset)
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
> grep("_",
> names(.)), funs(norm = round(scale(.), 3)))
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
> 1, sd, na.rm = TRUE), 3)
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
> 1, function(x) {
> max(abs(x), na.rm = TRUE)
> }), 3)
> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
> "\\.") %>% spread(treatment, value) %>%
> mutate(outlier = NA)
> stopifnot(!(any(data1$norm_sd >= 1)))
> if (!(any(data1$norm_sd >= 1))) {
> df1 <- dat1
> return(df1)
> }
>    else {
> df2 <- recursive_funlp()
> return(df2)
> }
> }
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
> df3
> }
>
>
> From: DIGHE, NILESH [AG/2362]
> Sent: Thursday, December 14, 2017 9:01 AM
> To: 'Eric Berger' <ericjber...@gmail.com<mailto:ericjber...@gmail.com>>
> Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
> Subject: RE: [R] help with recursive function
>
> Eric:  Thanks for taking time to look into my problem.  Despite of making
> the change you suggested, I am still getting the same error.  I am
> wondering if the logic I am using in the stopifnot and if functions is a
> problem.
> I like the recursive function to stop whenever the norm_sd column has zero
> values that are above or equal to 1. Below is the calclp function after the
> changes you suggested.
> Thanks. Nilesh
>
> dput(calclp)
> function (dataset)
> {
> dat1 <- funlp1(dataset)
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>   

Re: [R] help with recursive function

[DIGHE, NILESH [AG/2362]]

When I run the code without stopifnot, the code takes 5 min to run and then it 
throws an error listed below without producing any results.
Error: node stack overflow
In addition: There were 50 or more warnings (use warnings() to see the first 50)
Error during wrapup: node stack overflow

Thanks.
Nilesh

From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Thursday, December 14, 2017 11:26 AM
To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
Cc: Eric Berger <ericjber...@gmail.com>; r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function

Your code contains the lines
stopifnot(!(any(data1$norm_sd >= 1)))
if(!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}

stop() "throws an error", causing the current function and all functions in the 
call
stack to abort and return nothing.  It does not mean to stop now and return a 
result.
Does the function give the correct results if you just leave out the stopifnot 
line?


Bill Dunlap
TIBCO Software
wdunlap tibco.com<http://tibco.com>

On Thu, Dec 14, 2017 at 9:11 AM, DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
Eric:  I will try and see if I can figure out the issue by debugging as you 
suggested. I don’t know why my code after stopifnot is not getting executed 
where I like the code to run the funlp2 function when the if statement is TRUE 
but when it is false, I like it to keep running until the stopifnot condition 
is met.

When the stopifnot condition is met, I like to get the output from if statement 
saved.
Anyway,  I will keep trying.
Again, Thanks for your help!
Nilesh

From: Eric Berger [mailto:ericjber...@gmail.com<mailto:ericjber...@gmail.com>]
Sent: Thursday, December 14, 2017 10:29 AM
To: DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>>
Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
Subject: Re: [R] help with recursive function

If you are trying to understand why the "stopifnot" condition is met you can 
replace it by something like:

if ( any(dat2$norm_sd >= 1) )
   browser()

This will put you in a debugging session where you can examine your variables, 
e.g.

> dat$norm_sd

HTH,
Eric



On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger 
<ericjber...@gmail.com<mailto:ericjber...@gmail.com><mailto:ericjber...@gmail.com<mailto:ericjber...@gmail.com>>>
 wrote:
The message is coming from your stopifnot() condition being met.


On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com><mailto:nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>>>
 wrote:
Hi, I accidently left out few lines of code from the calclp function.  Updated 
function is pasted below.
I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is not 
TRUE“

I would appreciate any help.
Nilesh
dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
-norm_sd) %>% separate(key, c("field_rep", "treatment"),
"\\.") %>% spread(treatment, 
value) %>% mutate(outlier = NA)
stopifnot(!(any(data1$norm_sd >= 1)))
if (!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
   else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: DIGHE, NILESH [AG/2362]
Sent: Thursday, December 14, 2017 9:01 AM
To: 'Eric Berger' 
<ericjber...@gmail.com<mailto:ericjber...@gmail.com><mailto:ericjber...@gmail.com<mailto:ericjber...@gmail.com>>>
Cc: r-help 
<r-help@r-project.org<mailto:r-help@r-project.org><mailto:r-help@r-project.org<mailto:r-help@r-project.org>>>
Subject: RE: [R] help with recursive function

Eric:  Thanks for taking time to look into my problem.  Despite of making the 
change you suggested, I am still getting the same error.  I am wondering if the 
l

Re: [R] help with recursive function

[William Dunlap via R-help]

Your code contains the lines
stopifnot(!(any(data1$norm_sd >= 1)))
if(!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}

stop() "throws an error", causing the current function and all functions in
the call
stack to abort and return nothing.  It does not mean to stop now and return
a result.
Does the function give the correct results if you just leave out the
stopifnot line?


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Thu, Dec 14, 2017 at 9:11 AM, DIGHE, NILESH [AG/2362] <
nilesh.di...@monsanto.com> wrote:

> Eric:  I will try and see if I can figure out the issue by debugging as
> you suggested. I don’t know why my code after stopifnot is not getting
> executed where I like the code to run the funlp2 function when the if
> statement is TRUE but when it is false, I like it to keep running until the
> stopifnot condition is met.
>
> When the stopifnot condition is met, I like to get the output from if
> statement saved.
> Anyway,  I will keep trying.
> Again, Thanks for your help!
> Nilesh
>
> From: Eric Berger [mailto:ericjber...@gmail.com]
> Sent: Thursday, December 14, 2017 10:29 AM
> To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
> Cc: r-help <r-help@r-project.org>
> Subject: Re: [R] help with recursive function
>
> If you are trying to understand why the "stopifnot" condition is met you
> can replace it by something like:
>
> if ( any(dat2$norm_sd >= 1) )
>browser()
>
> This will put you in a debugging session where you can examine your
> variables, e.g.
>
> > dat$norm_sd
>
> HTH,
> Eric
>
>
>
> On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger <ericjber...@gmail.com
> <mailto:ericjber...@gmail.com>> wrote:
> The message is coming from your stopifnot() condition being met.
>
>
> On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] <
> nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
> Hi, I accidently left out few lines of code from the calclp function.
> Updated function is pasted below.
> I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is
> not TRUE“
>
> I would appreciate any help.
> Nilesh
> dput(calclp)
> function (dataset)
> {
> dat1 <- funlp1(dataset)
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
> grep("_",
> names(.)), funs(norm = round(scale(.), 3)))
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
> 1, sd, na.rm = TRUE), 3)
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
> 1, function(x) {
> max(abs(x), na.rm = TRUE)
> }), 3)
> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
> "\\.") %>% spread(treatment, value) %>%
> mutate(outlier = NA)
> stopifnot(!(any(data1$norm_sd >= 1)))
> if (!(any(data1$norm_sd >= 1))) {
> df1 <- dat1
> return(df1)
> }
>    else {
> df2 <- recursive_funlp()
> return(df2)
> }
> }
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
> df3
> }
>
>
> From: DIGHE, NILESH [AG/2362]
> Sent: Thursday, December 14, 2017 9:01 AM
> To: 'Eric Berger' <ericjber...@gmail.com<mailto:ericjber...@gmail.com>>
> Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
> Subject: RE: [R] help with recursive function
>
> Eric:  Thanks for taking time to look into my problem.  Despite of making
> the change you suggested, I am still getting the same error.  I am
> wondering if the logic I am using in the stopifnot and if functions is a
> problem.
> I like the recursive function to stop whenever the norm_sd column has zero
> values that are above or equal to 1. Below is the calclp function after the
> changes you suggested.
> Thanks. Nilesh
>
> dput(calclp)
> function (dataset)
> {
> dat1 <- funlp1(dataset)
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>

Re: [R] help with recursive function

[DIGHE, NILESH [AG/2362]]

Eric:  I will try and see if I can figure out the issue by debugging as you 
suggested. I don’t know why my code after stopifnot is not getting executed 
where I like the code to run the funlp2 function when the if statement is TRUE 
but when it is false, I like it to keep running until the stopifnot condition 
is met.

When the stopifnot condition is met, I like to get the output from if statement 
saved.
Anyway,  I will keep trying.
Again, Thanks for your help!
Nilesh

From: Eric Berger [mailto:ericjber...@gmail.com]
Sent: Thursday, December 14, 2017 10:29 AM
To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
Cc: r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function

If you are trying to understand why the "stopifnot" condition is met you can 
replace it by something like:

if ( any(dat2$norm_sd >= 1) )
   browser()

This will put you in a debugging session where you can examine your variables, 
e.g.

> dat$norm_sd

HTH,
Eric



On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger 
<ericjber...@gmail.com<mailto:ericjber...@gmail.com>> wrote:
The message is coming from your stopifnot() condition being met.


On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
Hi, I accidently left out few lines of code from the calclp function.  Updated 
function is pasted below.
I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is not 
TRUE“

I would appreciate any help.
Nilesh
dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
-norm_sd) %>% separate(key, c("field_rep", "treatment"),
"\\.") %>% spread(treatment, value) %>% mutate(outlier = 
NA)
stopifnot(!(any(data1$norm_sd >= 1)))
if (!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
   else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: DIGHE, NILESH [AG/2362]
Sent: Thursday, December 14, 2017 9:01 AM
To: 'Eric Berger' <ericjber...@gmail.com<mailto:ericjber...@gmail.com>>
Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
Subject: RE: [R] help with recursive function

Eric:  Thanks for taking time to look into my problem.  Despite of making the 
change you suggested, I am still getting the same error.  I am wondering if the 
logic I am using in the stopifnot and if functions is a problem.
I like the recursive function to stop whenever the norm_sd column has zero 
values that are above or equal to 1. Below is the calclp function after the 
changes you suggested.
Thanks. Nilesh

dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
stopifnot(!(any(dat2$norm_sd >= 1)))
if (!(any(dat2$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: Eric Berger [mailto:ericjber...@gmail.com]
Sent: Thursday, December 14, 2017 8:17 AM
To: DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>>
Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
Subject: Re: [R] help with recursive function

My own typo ... whoops ...

Re: [R] help with recursive function

[Eric Berger]

If you are trying to understand why the "stopifnot" condition is met you
can replace it by something like:

if ( any(dat2$norm_sd >= 1) )
   browser()

This will put you in a debugging session where you can examine your
variables, e.g.

> dat$norm_sd

HTH,
Eric



On Thu, Dec 14, 2017 at 5:33 PM, Eric Berger <ericjber...@gmail.com> wrote:

> The message is coming from your stopifnot() condition being met.
>
>
> On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] <
> nilesh.di...@monsanto.com> wrote:
>
>> Hi, I accidently left out few lines of code from the calclp function.
>> Updated function is pasted below.
>>
>> I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is
>> not TRUE“
>>
>>
>>
>> I would appreciate any help.
>>
>> Nilesh
>>
>> dput(calclp)
>>
>> function (dataset)
>>
>> {
>>
>> dat1 <- funlp1(dataset)
>>
>> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>>
>> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>>
>> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>>
>> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
>> grep("_",
>>
>> names(.)), funs(norm = round(scale(.), 3)))
>>
>> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, sd, na.rm = TRUE), 3)
>>
>> dat2$norm_max <- round(apply(dat2[, grep("lp_norm",
>> names(dat2))],
>>
>> 1, function(x) {
>>
>> max(abs(x), na.rm = TRUE)
>>
>> }), 3)
>>
>> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
>>
>> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
>>
>> "\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
>>
>> stopifnot(!(any(data1$norm_sd >= 1)))
>>
>> if (!(any(data1$norm_sd >= 1))) {
>>
>> df1 <- dat1
>>
>> return(df1)
>>
>> }
>>
>>else {
>>
>> df2 <- recursive_funlp()
>>
>> return(df2)
>>
>> }
>>
>> }
>>
>> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>>
>> df3
>>
>> }
>>
>>
>>
>>
>>
>> *From:* DIGHE, NILESH [AG/2362]
>> *Sent:* Thursday, December 14, 2017 9:01 AM
>> *To:* 'Eric Berger' <ericjber...@gmail.com>
>> *Cc:* r-help <r-help@r-project.org>
>> *Subject:* RE: [R] help with recursive function
>>
>>
>>
>> Eric:  Thanks for taking time to look into my problem.  Despite of making
>> the change you suggested, I am still getting the same error.  I am
>> wondering if the logic I am using in the stopifnot and if functions is a
>> problem.
>>
>> I like the recursive function to stop whenever the norm_sd column has
>> zero values that are above or equal to 1. Below is the calclp function
>> after the changes you suggested.
>>
>> Thanks. Nilesh
>>
>>
>>
>> dput(calclp)
>>
>> function (dataset)
>>
>> {
>>
>> dat1 <- funlp1(dataset)
>>
>> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>>
>> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>>
>> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>>
>> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
>> grep("_",
>>
>> names(.)), funs(norm = round(scale(.), 3)))
>>
>>     dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, sd, na.rm = TRUE), 3)
>>
>> dat2$norm_max <- round(apply(dat2[, grep("lp_norm",
>> names(dat2))],
>>
>> 1, function(x) {
>>
>> max(abs(x), na.rm = TRUE)
>>
>> }), 3)
>>
>> stopifnot(!(any(dat2$norm_sd >= 1)))
>>
>> if (!(any(dat2$norm_sd >= 1))) {
>>
>> df1 <- dat1
>>
>> return(df1)
>>
>> }
>>
>> else {
>>
>> df2 <- r

Re: [R] help with recursive function

[Eric Berger]

The message is coming from your stopifnot() condition being met.


On Thu, Dec 14, 2017 at 5:31 PM, DIGHE, NILESH [AG/2362] <
nilesh.di...@monsanto.com> wrote:

> Hi, I accidently left out few lines of code from the calclp function.
> Updated function is pasted below.
>
> I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is
> not TRUE“
>
>
>
> I would appreciate any help.
>
> Nilesh
>
> dput(calclp)
>
> function (dataset)
>
> {
>
> dat1 <- funlp1(dataset)
>
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>
> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
> grep("_",
>
> names(.)), funs(norm = round(scale(.), 3)))
>
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, sd, na.rm = TRUE), 3)
>
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, function(x) {
>
> max(abs(x), na.rm = TRUE)
>
> }), 3)
>
> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
>
> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
>
> "\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
>
> stopifnot(!(any(data1$norm_sd >= 1)))
>
> if (!(any(data1$norm_sd >= 1))) {
>
> df1 <- dat1
>
> return(df1)
>
> }
>
>else {
>
> df2 <- recursive_funlp()
>
> return(df2)
>
> }
>
> }
>
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>
> df3
>
> }
>
>
>
>
>
> *From:* DIGHE, NILESH [AG/2362]
> *Sent:* Thursday, December 14, 2017 9:01 AM
> *To:* 'Eric Berger' <ericjber...@gmail.com>
> *Cc:* r-help <r-help@r-project.org>
> *Subject:* RE: [R] help with recursive function
>
>
>
> Eric:  Thanks for taking time to look into my problem.  Despite of making
> the change you suggested, I am still getting the same error.  I am
> wondering if the logic I am using in the stopifnot and if functions is a
> problem.
>
> I like the recursive function to stop whenever the norm_sd column has zero
> values that are above or equal to 1. Below is the calclp function after the
> changes you suggested.
>
> Thanks. Nilesh
>
>
>
> dput(calclp)
>
> function (dataset)
>
> {
>
> dat1 <- funlp1(dataset)
>
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>
> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
> grep("_",
>
> names(.)), funs(norm = round(scale(.), 3)))
>
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, sd, na.rm = TRUE), 3)
>
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, function(x) {
>
> max(abs(x), na.rm = TRUE)
>
>     }), 3)
>
> stopifnot(!(any(dat2$norm_sd >= 1)))
>
> if (!(any(dat2$norm_sd >= 1))) {
>
> df1 <- dat1
>
> return(df1)
>
> }
>
> else {
>
> df2 <- recursive_funlp()
>
> return(df2)
>
> }
>
> }
>
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>
> df3
>
> }
>
>
>
>
>
> *From:* Eric Berger [mailto:ericjber...@gmail.com <ericjber...@gmail.com>]
>
> *Sent:* Thursday, December 14, 2017 8:17 AM
> *To:* DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
> *Cc:* r-help <r-help@r-project.org>
> *Subject:* Re: [R] help with recursive function
>
>
>
> My own typo ... whoops ...
>
>
>
> !( any(dat2$norm_sd >= 1 ))
>
>
>
>
>
>
>
> On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger <ericjber...@gmail.com>
> wrote:
>
> You seem to have a typo at this expression (and some others like it)
>
>
>
> Namely, you write
>
>
>
> any(!dat2$norm_sd) >= 1
>
>
>
> when you possibly meant to write
>
>
>
> !

Re: [R] help with recursive function

[DIGHE, NILESH [AG/2362]]

Hi, I accidently left out few lines of code from the calclp function.  Updated 
function is pasted below.
I am still getting the same error “Error: !(any(data1$norm_sd >= 1)) is not 
TRUE“

I would appreciate any help.
Nilesh
dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
-norm_sd) %>% separate(key, c("field_rep", "treatment"),
"\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
stopifnot(!(any(data1$norm_sd >= 1)))
if (!(any(data1$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
   else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: DIGHE, NILESH [AG/2362]
Sent: Thursday, December 14, 2017 9:01 AM
To: 'Eric Berger' <ericjber...@gmail.com>
Cc: r-help <r-help@r-project.org>
Subject: RE: [R] help with recursive function

Eric:  Thanks for taking time to look into my problem.  Despite of making the 
change you suggested, I am still getting the same error.  I am wondering if the 
logic I am using in the stopifnot and if functions is a problem.
I like the recursive function to stop whenever the norm_sd column has zero 
values that are above or equal to 1. Below is the calclp function after the 
changes you suggested.
Thanks. Nilesh

dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
stopifnot(!(any(dat2$norm_sd >= 1)))
if (!(any(dat2$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: Eric Berger [mailto:ericjber...@gmail.com]
Sent: Thursday, December 14, 2017 8:17 AM
To: DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>>
Cc: r-help <r-help@r-project.org<mailto:r-help@r-project.org>>
Subject: Re: [R] help with recursive function

My own typo ... whoops ...

!( any(dat2$norm_sd >= 1 ))



On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger 
<ericjber...@gmail.com<mailto:ericjber...@gmail.com>> wrote:
You seem to have a typo at this expression (and some others like it)

Namely, you write

any(!dat2$norm_sd) >= 1

when you possibly meant to write

!( any(dat2$norm_sd) >= 1 )

i.e. I think your ! seems to be in the wrong place.

HTH,
Eric


On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
Hi, I need some help with running a recursive function. I like to run funlp2 
recursively.
When I try to run recursive function in another function named "calclp" I get 
this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".

I have never built a recursive function before so having trouble executing it 
in this case.  I would appreciate any help or guidance to resolve this issue. 
Please see my data and the three functions that I am using below.
Please note that calclp is the function I am running and the other two 
functions are within this calclp function.

# code:
Test<- calclp(dataset = dat)

# calclp function

calclp<- function (dataset)

{

dat1 <- funlp1(dataset)

recursive_funlp <- function(dataset = dat1, func = funlp2) {

dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%

   

Re: [R] help with recursive function

[DIGHE, NILESH [AG/2362]]

Eric:  Thanks for taking time to look into my problem.  Despite of making the 
change you suggested, I am still getting the same error.  I am wondering if the 
logic I am using in the stopifnot and if functions is a problem.
I like the recursive function to stop whenever the norm_sd column has zero 
values that are above or equal to 1. Below is the calclp function after the 
changes you suggested.
Thanks. Nilesh

dput(calclp)
function (dataset)
{
dat1 <- funlp1(dataset)
recursive_funlp <- function(dataset = dat1, func = funlp2) {
dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",
names(.)), funs(norm = round(scale(.), 3)))
dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, sd, na.rm = TRUE), 3)
dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
1, function(x) {
max(abs(x), na.rm = TRUE)
}), 3)
stopifnot(!(any(dat2$norm_sd >= 1)))
if (!(any(dat2$norm_sd >= 1))) {
df1 <- dat1
return(df1)
}
else {
df2 <- recursive_funlp()
return(df2)
}
}
df3 <- recursive_funlp(dataset = dat1, func = funlp2)
df3
}


From: Eric Berger [mailto:ericjber...@gmail.com]
Sent: Thursday, December 14, 2017 8:17 AM
To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
Cc: r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function

My own typo ... whoops ...

!( any(dat2$norm_sd >= 1 ))



On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger 
<ericjber...@gmail.com<mailto:ericjber...@gmail.com>> wrote:
You seem to have a typo at this expression (and some others like it)

Namely, you write

any(!dat2$norm_sd) >= 1

when you possibly meant to write

!( any(dat2$norm_sd) >= 1 )

i.e. I think your ! seems to be in the wrong place.

HTH,
Eric


On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] 
<nilesh.di...@monsanto.com<mailto:nilesh.di...@monsanto.com>> wrote:
Hi, I need some help with running a recursive function. I like to run funlp2 
recursively.
When I try to run recursive function in another function named "calclp" I get 
this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".

I have never built a recursive function before so having trouble executing it 
in this case.  I would appreciate any help or guidance to resolve this issue. 
Please see my data and the three functions that I am using below.
Please note that calclp is the function I am running and the other two 
functions are within this calclp function.

# code:
Test<- calclp(dataset = dat)

# calclp function

calclp<- function (dataset)

{

dat1 <- funlp1(dataset)

recursive_funlp <- function(dataset = dat1, func = funlp2) {

dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%

mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%

spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",

names(.)), funs(norm = round(scale(.), 3)))

dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, sd, na.rm = TRUE), 3)

dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, function(x) {

max(abs(x), na.rm = TRUE)

}), 3)

stopifnot(any(!dat2$norm_sd) >= 1)

if (any(!dat2$norm_sd) >= 1) {

df1 <- dat1

return(df1)

}

else {

df2 <- recursive_funlp()

return(df2)

}

}

df3 <- recursive_funlp(dataset = dat1, func = funlp2)

df3

}


# funlp1 function

funlp1<- function (dataset)

{

dat2 <- dataset %>% select(field, set, ent_num, rep_num,

lp) %>% unite(uniqueid, set, ent_num, sep = ".") %>%

unite(field_rep, field, rep_num) %>% mutate(field_rep = paste(field_rep,

"lp", sep = ".")) %>% spread(key = field_rep, value = lp) %>%

mutate_at(.vars = grep("_", names(.)), funs(norm = round(scale(.),

3)))

dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, sd, na.rm = TRUE), 3)

dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, function(x) {

max(abs(x), na.rm = TRUE)

}), 3)

data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,

-norm_sd) %>% separate(key, c("field_rep", "treatment"),

"\\.") %>% spr

Re: [R] help with recursive function

[Eric Berger]

My own typo ... whoops ...

!( any(dat2$norm_sd >= 1 ))



On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger  wrote:

> You seem to have a typo at this expression (and some others like it)
>
> Namely, you write
>
> any(!dat2$norm_sd) >= 1
>
> when you possibly meant to write
>
> !( any(dat2$norm_sd) >= 1 )
>
> i.e. I think your ! seems to be in the wrong place.
>
> HTH,
> Eric
>
>
> On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] <
> nilesh.di...@monsanto.com> wrote:
>
>> Hi, I need some help with running a recursive function. I like to run
>> funlp2 recursively.
>> When I try to run recursive function in another function named "calclp" I
>> get this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".
>>
>> I have never built a recursive function before so having trouble
>> executing it in this case.  I would appreciate any help or guidance to
>> resolve this issue. Please see my data and the three functions that I am
>> using below.
>> Please note that calclp is the function I am running and the other two
>> functions are within this calclp function.
>>
>> # code:
>> Test<- calclp(dataset = dat)
>>
>> # calclp function
>>
>> calclp<- function (dataset)
>>
>> {
>>
>> dat1 <- funlp1(dataset)
>>
>> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>>
>> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>>
>> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>>
>> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
>> grep("_",
>>
>> names(.)), funs(norm = round(scale(.), 3)))
>>
>> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, sd, na.rm = TRUE), 3)
>>
>> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, function(x) {
>>
>> max(abs(x), na.rm = TRUE)
>>
>> }), 3)
>>
>> stopifnot(any(!dat2$norm_sd) >= 1)
>>
>> if (any(!dat2$norm_sd) >= 1) {
>>
>> df1 <- dat1
>>
>> return(df1)
>>
>> }
>>
>> else {
>>
>> df2 <- recursive_funlp()
>>
>> return(df2)
>>
>> }
>>
>> }
>>
>> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>>
>> df3
>>
>> }
>>
>>
>> # funlp1 function
>>
>> funlp1<- function (dataset)
>>
>> {
>>
>> dat2 <- dataset %>% select(field, set, ent_num, rep_num,
>>
>> lp) %>% unite(uniqueid, set, ent_num, sep = ".") %>%
>>
>> unite(field_rep, field, rep_num) %>% mutate(field_rep =
>> paste(field_rep,
>>
>> "lp", sep = ".")) %>% spread(key = field_rep, value = lp) %>%
>>
>> mutate_at(.vars = grep("_", names(.)), funs(norm = round(scale(.),
>>
>> 3)))
>>
>> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, sd, na.rm = TRUE), 3)
>>
>> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>>
>> 1, function(x) {
>>
>> max(abs(x), na.rm = TRUE)
>>
>> }), 3)
>>
>> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
>>
>> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
>>
>> "\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
>>
>> df_clean <- with(data1, data1[norm_sd < 1, ])
>>
>> datD <- with(data1, data1[norm_sd >= 1, ])
>>
>> s <- split(datD, datD$uniqueid)
>>
>> sdf <- lapply(s, function(x) {
>>
>> data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,
>>
>> ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),
>>
>> x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))
>>
>> x
>>
>> })
>>
>> sdf2 <- bind_rows(sdf)
>>
>> all_dat <- bind_rows(df_clean, sdf2)
>>
>> all_dat
>>
>> }
>>
>>
>> # funlp2 function
>>
>> funlp2<-function (dataset)
>>
>> {
>>
>> data1 <- dataset
>>
>> df_clean <- with(data1, data1[norm_sd < 1, ])
>>
>> datD <- with(data1, data1[norm_sd >= 1, ])
>>
>> s <- split(datD, datD$uniqueid)
>>
>> sdf <- lapply(s, function(x) {
>>
>> data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,
>>
>> ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),
>>
>> x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))
>>
>> x
>>
>> })
>>
>> sdf2 <- bind_rows(sdf)
>>
>> all_dat <- bind_rows(df_clean, sdf2)
>>
>> all_dat
>>
>> }
>>
>>
>> # dataset
>> dput(dat)
>> structure(list(field = c("LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
>> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", 

Re: [R] help with recursive function

[Eric Berger]

You seem to have a typo at this expression (and some others like it)

Namely, you write

any(!dat2$norm_sd) >= 1

when you possibly meant to write

!( any(dat2$norm_sd) >= 1 )

i.e. I think your ! seems to be in the wrong place.

HTH,
Eric


On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362] <
nilesh.di...@monsanto.com> wrote:

> Hi, I need some help with running a recursive function. I like to run
> funlp2 recursively.
> When I try to run recursive function in another function named "calclp" I
> get this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".
>
> I have never built a recursive function before so having trouble executing
> it in this case.  I would appreciate any help or guidance to resolve this
> issue. Please see my data and the three functions that I am using below.
> Please note that calclp is the function I am running and the other two
> functions are within this calclp function.
>
> # code:
> Test<- calclp(dataset = dat)
>
> # calclp function
>
> calclp<- function (dataset)
>
> {
>
> dat1 <- funlp1(dataset)
>
> recursive_funlp <- function(dataset = dat1, func = funlp2) {
>
> dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%
>
> mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%
>
> spread(key = field_rep, value = lp) %>% mutate_at(.vars =
> grep("_",
>
> names(.)), funs(norm = round(scale(.), 3)))
>
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, sd, na.rm = TRUE), 3)
>
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, function(x) {
>
> max(abs(x), na.rm = TRUE)
>
> }), 3)
>
> stopifnot(any(!dat2$norm_sd) >= 1)
>
> if (any(!dat2$norm_sd) >= 1) {
>
> df1 <- dat1
>
> return(df1)
>
> }
>
> else {
>
> df2 <- recursive_funlp()
>
> return(df2)
>
> }
>
> }
>
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>
> df3
>
> }
>
>
> # funlp1 function
>
> funlp1<- function (dataset)
>
> {
>
> dat2 <- dataset %>% select(field, set, ent_num, rep_num,
>
> lp) %>% unite(uniqueid, set, ent_num, sep = ".") %>%
>
> unite(field_rep, field, rep_num) %>% mutate(field_rep =
> paste(field_rep,
>
> "lp", sep = ".")) %>% spread(key = field_rep, value = lp) %>%
>
> mutate_at(.vars = grep("_", names(.)), funs(norm = round(scale(.),
>
> 3)))
>
> dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, sd, na.rm = TRUE), 3)
>
> dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],
>
> 1, function(x) {
>
> max(abs(x), na.rm = TRUE)
>
> }), 3)
>
> data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,
>
> -norm_sd) %>% separate(key, c("field_rep", "treatment"),
>
> "\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
>
> df_clean <- with(data1, data1[norm_sd < 1, ])
>
> datD <- with(data1, data1[norm_sd >= 1, ])
>
> s <- split(datD, datD$uniqueid)
>
> sdf <- lapply(s, function(x) {
>
> data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,
>
> ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),
>
> x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))
>
> x
>
> })
>
> sdf2 <- bind_rows(sdf)
>
> all_dat <- bind_rows(df_clean, sdf2)
>
> all_dat
>
> }
>
>
> # funlp2 function
>
> funlp2<-function (dataset)
>
> {
>
> data1 <- dataset
>
> df_clean <- with(data1, data1[norm_sd < 1, ])
>
> datD <- with(data1, data1[norm_sd >= 1, ])
>
> s <- split(datD, datD$uniqueid)
>
> sdf <- lapply(s, function(x) {
>
> data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,
>
> ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),
>
> x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))
>
> x
>
> })
>
> sdf2 <- bind_rows(sdf)
>
> all_dat <- bind_rows(df_clean, sdf2)
>
> all_dat
>
> }
>
>
> # dataset
> dput(dat)
> structure(list(field = c("LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
> "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "OL01",
> "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
> "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
> "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
> "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
> "OL01", 

[R] help with recursive function

[DIGHE, NILESH [AG/2362]]

Hi, I need some help with running a recursive function. I like to run funlp2 
recursively.
When I try to run recursive function in another function named "calclp" I get 
this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".

I have never built a recursive function before so having trouble executing it 
in this case.  I would appreciate any help or guidance to resolve this issue. 
Please see my data and the three functions that I am using below.
Please note that calclp is the function I am running and the other two 
functions are within this calclp function.

# code:
Test<- calclp(dataset = dat)

# calclp function

calclp<- function (dataset)

{

dat1 <- funlp1(dataset)

recursive_funlp <- function(dataset = dat1, func = funlp2) {

dat2 <- dataset %>% select(uniqueid, field_rep, lp) %>%

mutate(field_rep = paste(field_rep, "lp", sep = ".")) %>%

spread(key = field_rep, value = lp) %>% mutate_at(.vars = grep("_",

names(.)), funs(norm = round(scale(.), 3)))

dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, sd, na.rm = TRUE), 3)

dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, function(x) {

max(abs(x), na.rm = TRUE)

}), 3)

stopifnot(any(!dat2$norm_sd) >= 1)

if (any(!dat2$norm_sd) >= 1) {

df1 <- dat1

return(df1)

}

else {

df2 <- recursive_funlp()

return(df2)

}

}

df3 <- recursive_funlp(dataset = dat1, func = funlp2)

df3

}


# funlp1 function

funlp1<- function (dataset)

{

dat2 <- dataset %>% select(field, set, ent_num, rep_num,

lp) %>% unite(uniqueid, set, ent_num, sep = ".") %>%

unite(field_rep, field, rep_num) %>% mutate(field_rep = paste(field_rep,

"lp", sep = ".")) %>% spread(key = field_rep, value = lp) %>%

mutate_at(.vars = grep("_", names(.)), funs(norm = round(scale(.),

3)))

dat2$norm_sd <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, sd, na.rm = TRUE), 3)

dat2$norm_max <- round(apply(dat2[, grep("lp_norm", names(dat2))],

1, function(x) {

max(abs(x), na.rm = TRUE)

}), 3)

data1 <- dat2 %>% gather(key, value, -uniqueid, -norm_max,

-norm_sd) %>% separate(key, c("field_rep", "treatment"),

"\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)

df_clean <- with(data1, data1[norm_sd < 1, ])

datD <- with(data1, data1[norm_sd >= 1, ])

s <- split(datD, datD$uniqueid)

sdf <- lapply(s, function(x) {

data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,

ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),

x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))

x

})

sdf2 <- bind_rows(sdf)

all_dat <- bind_rows(df_clean, sdf2)

all_dat

}


# funlp2 function

funlp2<-function (dataset)

{

data1 <- dataset

df_clean <- with(data1, data1[norm_sd < 1, ])

datD <- with(data1, data1[norm_sd >= 1, ])

s <- split(datD, datD$uniqueid)

sdf <- lapply(s, function(x) {

data.frame(x, x$outlier <- ifelse(is.na(x$lp_norm), NA,

ifelse(abs(x$lp_norm) == x$norm_max, "yes", "no")),

x$lp <- with(x, ifelse(outlier == "yes", NA, lp)))

x

})

sdf2 <- bind_rows(sdf)

all_dat <- bind_rows(df_clean, sdf2)

all_dat

}


# dataset
dput(dat)
structure(list(field = c("LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01",
"LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "LM01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01",
"OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "OL01", "SGI1",
"SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1",
"SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1",
"SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1",
"SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1",
"SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", "SGI1", 

Re: [R] Help with optim function in R, please?

[J C Nash]

I haven't tried your code, but I've seen far too many attempts like this where
there is no simple call to the function with starting parameters to see if a
sensible answer is returned. If you get NaN or Inf or ... that isn't sensible,
then you know it is not a good idea to try further.

Beyond this, you don't have gradients. I think I would try nmkb (an updated
Nelder-Mead style code that allows for bounds) from dfoptim package. Note that
you cannot start on the bounds with this code. And it won't be terribly 
efficient, but it
will -- if you ask for intermediate results -- give you some idea about your
function. If everything then working reasonably, you can try L-BFGS-B, possibly
with "better" starting parameters. However, I'm fairly certain you will find
something to fix before then.

It really does pay to be very pedestrian in using optimization software. I've 
been
at this game (and note I wrote 3 of the 5 routines in optim(), which now all 
have
replacements!) and I am constantly reminded in my own work to ALWAYS run some 
simple
tests with the objective function and, if I have them, the gradients.

JN


On 2017-08-06 11:11 AM, fadeh Alanazi wrote:

Hi all,

Many thank in advance for helping me.  I tried to fit Expectation Maximization 
algorithm for mixture data. I must used one of numerical method to maximize my 
function.

I built my code but I do not know how to make the optim function run over a 
different value of the parameters.  That is,

For E-step I need to get the value of mixture weights based on the current 
(initial) values of the parameter of the density. Then, multiple the weight by 
the logliklihood function and maximize it (M-step)
Then, I would like to take the new values of the parameter (from M-step) and 
plug it in the weight, to get a new value of the weight. Then, iterate till 
converges.

I tried the following code, but it does not work.

library(copula)
library(VineCopula)

## to generate the data
set.seed(123)
cp <- mixCopula(list(frankCopula(4),claytonCopula(2)))
cop <- rCopula(100,cp)
x <- pobs(cop) ## this is the data
## my function including optim function
myfunc <- function(data,copula=list(frankCopula(4,dim=2), 
claytonCopula(0.5,dim=2)),maxit=200){
   
   # copula[[1]]@parameters <- par[1]

   # copula[[2]]@parameters <- par[2]
   optim_1 <- function(par, data.=data, copula.=copula){
 copula[[1]]@parameters <- par[1]
 copula[[2]]@parameters <- par[2]
 tau_1 <- par[3]*dCopula(data,copula[[1]],log = F)
 tau_2 <- par[4]*dCopula(data,copula[[2]],log=F)
 Tau_1 <- tau_1 / sum(tau_1 + tau_2)
 Tau_2 <- tau_2 / sum(tau_1 + tau_2)
 up_pi1 <- sum(Tau_1)/ 100
 up_pi2 <- sum(Tau_2) / 100
 # Tau <- c(Tau_1, Tau_2)
 ll <- (-sum(Tau_1*dCopula(data,copula[[1]],log=T)))
 #ll_1 <- -sum(Tau_2*dCopula(data,copula[[2]],log=T))
 

 if (is.finite(ll)) {

   return(ll)
 } else {
   if (is.na(ll)) {
 message(par)
   }
   message(ll)
   return(-3^305)
 }
   }
   
   
   
   xy <- optim(par=c(2,0.5,0.2,0.2),fn=optim_1,method="L-BFGS-B",control = list(maxit= 200, trace=1),lower= c(copula[[1]]@param.lowbnd, copula[[2]]@param.lowbnd,0,0), upper = c(copula[[1]]@param.upbnd, copula[[2]]@param.upbnd,1,1))
   
   return(xy$par)

}

Any help please?

Regards,
Fadhah



[[alternative HTML version deleted]]

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.



__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Help with optim function in R, please?

[fadeh Alanazi]

Hi all, 

Many thank in advance for helping me.  I tried to fit Expectation Maximization 
algorithm for mixture data. I must used one of numerical method to maximize my 
function. 

I built my code but I do not know how to make the optim function run over a 
different value of the parameters.  That is, 

For E-step I need to get the value of mixture weights based on the current 
(initial) values of the parameter of the density. Then, multiple the weight by 
the logliklihood function and maximize it (M-step)
Then, I would like to take the new values of the parameter (from M-step) and 
plug it in the weight, to get a new value of the weight. Then, iterate till 
converges. 

I tried the following code, but it does not work. 

library(copula)
library(VineCopula)

## to generate the data
set.seed(123)
cp <- mixCopula(list(frankCopula(4),claytonCopula(2)))
cop <- rCopula(100,cp)
x <- pobs(cop) ## this is the data
## my function including optim function
myfunc <- function(data,copula=list(frankCopula(4,dim=2), 
claytonCopula(0.5,dim=2)),maxit=200){
  
  # copula[[1]]@parameters <- par[1]
  # copula[[2]]@parameters <- par[2]
  optim_1 <- function(par, data.=data, copula.=copula){
    copula[[1]]@parameters <- par[1]
    copula[[2]]@parameters <- par[2]
    tau_1 <- par[3]*dCopula(data,copula[[1]],log = F)
    tau_2 <- par[4]*dCopula(data,copula[[2]],log=F)
    Tau_1 <- tau_1 / sum(tau_1 + tau_2)
    Tau_2 <- tau_2 / sum(tau_1 + tau_2)
    up_pi1 <- sum(Tau_1)/ 100
    up_pi2 <- sum(Tau_2) / 100
    # Tau <- c(Tau_1, Tau_2)
    ll <- (-sum(Tau_1*dCopula(data,copula[[1]],log=T)))
    #ll_1 <- -sum(Tau_2*dCopula(data,copula[[2]],log=T))
    
   
if (is.finite(ll)) {
  return(ll)
    } else {
  if (is.na(ll)) {
    message(par)
  }
  message(ll)
      return(-3^305)
    }
  }
  
  
  
  xy <- optim(par=c(2,0.5,0.2,0.2),fn=optim_1,method="L-BFGS-B",control = 
list(maxit= 200, trace=1),lower= c(copula[[1]]@param.lowbnd, 
copula[[2]]@param.lowbnd,0,0), upper = c(copula[[1]]@param.upbnd, 
copula[[2]]@param.upbnd,1,1))
  
  return(xy$par)
}

Any help please?

Regards, 
Fadhah 



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Re: [R] Help with improveProb function in Hmisc in R

[Frank Harrell]

Please note that Kirsten is cross-posting to stats.stackexchange.com 
creating extra work for everyone.

-- 

Frank E Harrell Jr  Professor and Chairman  School of Medicine

Department of *Biostatistics*   *Vanderbilt University*


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[R] Help with improveProb function in Hmisc in R

[kirsada]

Please bear with me, I am very new to R.

My question is regarding the use of the improveProb function in the Hmisc
package. I have two logistic models, the only difference being that the
second model contains my novel marker of interest. I am trying to calculate
NRI and IDI to compare models.

I have the PredRisks for both models - PredRisk1 and PredRisk2, and my
outcome is disease 0/1. How do I define this in R in order to run

improveProb(x1, x2, y)?

Many thanks in advance



--
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http://r.789695.n4.nabble.com/Help-with-improveProb-function-in-Hmisc-in-R-tp4713004.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Help with improveProb function in Hmisc in R

[David Winsemius]

On Oct 1, 2015, at 2:26 AM, kirsada wrote:

> Please bear with me, I am very new to R.
> 
> My question is regarding the use of the improveProb function in the Hmisc
> package. I have two logistic models, the only difference being that the
> second model contains my novel marker of interest. I am trying to calculate
> NRI and IDI to compare models.
> 
> I have the PredRisks for both models - PredRisk1 and PredRisk2, and my
> outcome is disease 0/1. How do I define this in R in order to run
> 
> improveProb(x1, x2, y)?

If you are saying that you have two numerical values each of length one, than I 
do not think you can pass those to a function that expects raw data. If your 
"PredRisk" variables are vectors of the same length as y and having a range of 
[0,1] as befits a probability, then what problems are you experiencing?

-- 
David Winsemius
Alameda, CA, USA

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Re: [R] Help with abs function

[Andrés Aragón Martínez]

Hi,

Just do the following:


 tran-c(7.2)
 tgrid-c(7.1,7.4,7.3,7.1,7.3)
 tgrid-tgrid-tran
 tgrid
[1] -0.1  0.2  0.1 -0.1  0.1
 abs(tgrid[tgrid0.1])
[1] 0.2

Andrés



 El 12/06/2015, a las 11:01, Jeff Newmiller jdnew...@dcn.davis.ca.us 
 escribió:
 
 FAQ 7.31
 ---
 Jeff NewmillerThe .   .  Go Live...
 DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
 --- 
 Sent from my phone. Please excuse my brevity.
 
 On June 12, 2015 8:39:48 AM PDT, Nelson, Gary (MISC) 
 gary.nel...@state.ma.us wrote:
 I have come across some odd behavior (to me) using the abs() function
 that I wonder if anyone can explain.
 
 Using  R version 3.2.0, I created a vector of absolute values using 
 the following code:
 
 tran-c(7.2)
 tgrid-c(7.1,7.4,7.3,7.1,7.3)
 dgrid-abs(tgrid-tran)
 dgrid
 [1] 0.1 0.2 0.1 0.1 0.1
 
 When I tried to extract just the rows with values0.1, I get
 
 dgrid[dgrid0.1]
 [1] 0.1 0.2 0.1
 
 There should be only 1 value extracted.
 
 However, if I enter the values by hand
 
 bgrid-c(0.1,0.2,0.1,0.1,0.1)
 bgrid
 [1] 0.1 0.2 0.1 0.1 0.1
 bgrid[bgrid0.1]
 [1] 0.2
 
 The result is correct.  So why is this happening?
 
 I did explore a little bit and found
 
 as.character(dgrid)
 [1] 0.101  0.2
 [3] 0.0996 0.101
 [5] 0.0996
 
 which shows the absolute values of the negative differences  are
 slightly greater than the difference of 0.1
 
 Is this normal behavior for the abs() function?
 
 Thanks,
 
 Gary Nelson
 
  [[alternative HTML version deleted]]
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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Re: [R] Help with abs function

[Jeff Newmiller]

FAQ 7.31
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

On June 12, 2015 8:39:48 AM PDT, Nelson, Gary (MISC) 
gary.nel...@state.ma.us wrote:
I have come across some odd behavior (to me) using the abs() function
that I wonder if anyone can explain.

Using  R version 3.2.0, I created a vector of absolute values using 
the following code:

 tran-c(7.2)
 tgrid-c(7.1,7.4,7.3,7.1,7.3)
 dgrid-abs(tgrid-tran)
 dgrid
[1] 0.1 0.2 0.1 0.1 0.1

When I tried to extract just the rows with values0.1, I get

 dgrid[dgrid0.1]
[1] 0.1 0.2 0.1

There should be only 1 value extracted.

However, if I enter the values by hand

 bgrid-c(0.1,0.2,0.1,0.1,0.1)
 bgrid
[1] 0.1 0.2 0.1 0.1 0.1
 bgrid[bgrid0.1]
[1] 0.2

The result is correct.  So why is this happening?

I did explore a little bit and found

 as.character(dgrid)
[1] 0.101  0.2
[3] 0.0996 0.101
[5] 0.0996

which shows the absolute values of the negative differences  are
slightly greater than the difference of 0.1

Is this normal behavior for the abs() function?

Thanks,

Gary Nelson

   [[alternative HTML version deleted]]

__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Help with abs function

[Sarah Goslee]

Please read R FAQ 7.31.

On Fri, Jun 12, 2015 at 11:39 AM, Nelson, Gary (MISC)
gary.nel...@state.ma.us wrote:
 I have come across some odd behavior (to me) using the abs() function that I 
 wonder if anyone can explain.

 Using  R version 3.2.0, I created a vector of absolute values using  the 
 following code:

 tran-c(7.2)
 tgrid-c(7.1,7.4,7.3,7.1,7.3)
 dgrid-abs(tgrid-tran)
 dgrid
 [1] 0.1 0.2 0.1 0.1 0.1

 When I tried to extract just the rows with values0.1, I get

 dgrid[dgrid0.1]
 [1] 0.1 0.2 0.1

 There should be only 1 value extracted.

 However, if I enter the values by hand

 bgrid-c(0.1,0.2,0.1,0.1,0.1)
 bgrid
 [1] 0.1 0.2 0.1 0.1 0.1
 bgrid[bgrid0.1]
 [1] 0.2

 The result is correct.  So why is this happening?

 I did explore a little bit and found

 as.character(dgrid)
 [1] 0.101  0.2
 [3] 0.0996 0.101
 [5] 0.0996

 which shows the absolute values of the negative differences  are slightly 
 greater than the difference of 0.1

 Is this normal behavior for the abs() function?

 Thanks,

 Gary Nelson
-- 
Sarah Goslee
http://www.functionaldiversity.org

__
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and provide commented, minimal, self-contained, reproducible code.

[R] Help with abs function

[Nelson, Gary (MISC)]

I have come across some odd behavior (to me) using the abs() function that I 
wonder if anyone can explain.

Using  R version 3.2.0, I created a vector of absolute values using  the 
following code:

 tran-c(7.2)
 tgrid-c(7.1,7.4,7.3,7.1,7.3)
 dgrid-abs(tgrid-tran)
 dgrid
[1] 0.1 0.2 0.1 0.1 0.1

When I tried to extract just the rows with values0.1, I get

 dgrid[dgrid0.1]
[1] 0.1 0.2 0.1

There should be only 1 value extracted.

However, if I enter the values by hand

 bgrid-c(0.1,0.2,0.1,0.1,0.1)
 bgrid
[1] 0.1 0.2 0.1 0.1 0.1
 bgrid[bgrid0.1]
[1] 0.2

The result is correct.  So why is this happening?

I did explore a little bit and found

 as.character(dgrid)
[1] 0.101  0.2
[3] 0.0996 0.101
[5] 0.0996

which shows the absolute values of the negative differences  are slightly 
greater than the difference of 0.1

Is this normal behavior for the abs() function?

Thanks,

Gary Nelson

[[alternative HTML version deleted]]

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Re: [R] Help with pedigree() function in kinship2

[Therneau, Terry M., Ph.D.]

Your problem is that PatientID, FatherID, MotherID are factors.  The authors of kinship2 
(myself and Jason) simply never thought of someone doing this.  Yes, that is an oversight. 
 We will correct it by adding some more checks and balances.  For now, turn your id 
variables into character or numeric.


Terry Therneau


On 05/13/2015 05:00 AM, r-help-requ...@r-project.org wrote:

Dear R-help,

I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with
Now, when running

## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)


## data to plot
d - structure(list(FamilyID = c(1, 1, 1, 1, 1, 1, 1,
1, 1), PatientID = structure(c(2L, 3L, 5L, 11L, 12L, 15L,
16L, 17L, 6L), .Label = c( 1,  2,  3,  4,  5,  6,
 7,  9, 10, 11, 13, 14, 18, 20, 23, 24, 25,
27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39,
41, 43, 45, 50, 62, 63, 64, 65, 66, 67, 85,
88), class = factor), FatherID = structure(c(1L, 1L, 6L,
1L, 5L, 6L, 1L, 7L, 6L), .Label = c(0, 1, 10, 11, 13,
2, 23, 27, 28, 3, 33, 34, 35, 38, 5, 62,
64, 66, 9), class = factor), MotherID = structure(c(1L,
1L, 7L, 1L, 14L, 7L, 1L, 5L, 7L), .Label = c(0, 10, 18,
2, 24, 29, 3, 30, 33, 34, 39, 4, 43, 5,
6, 63, 65, 9), class = factor), Sex = structure(c(2L,
1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L), .Label = c(Female, Male), class =
factor),
 AffectionStatus = structure(c(1L, 1L, 2L, 1L, 2L, 2L, 1L,
 2L, 2L), .Label = c(1, 2), class = factor)), .Names =
c(FamilyID,
PatientID, FatherID, MotherID, Sex, AffectionStatus
), row.names = c(NA, 9L), class = data.frame)

## plotting
ped - with(d, pedigree(PatientID, FatherID, MotherID, Sex,  affected =
AffectionStatus, famid = FamilyID))

## Error in pedigree(PatientID, FatherID, MotherID, Sex, affected =
AffectionStatus,  :
##Value of 'dadid' not found in the id list 1/0 1/0 1/2
1/0 1/2

I get an error.  My sessionInfo() is at the end.  I was wondering if
someone could help me to dissect what the cause of this error is and how it
can be fixed.

Thank you very much for your help.

Best regards,
Jorge Velez.-


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Re: [R] help with by function

[DIGHE, NILESH [AG/2362]]

Jean:  Thanks a lot!!  The changes you made to the code gave me what I needed.  
I truly appreciate your time in correcting the code.
Nilesh

From: Adams, Jean [mailto:jvad...@usgs.gov]
Sent: Tuesday, May 12, 2015 2:14 PM
To: DIGHE, NILESH [AG/2362]
Cc: r-help@r-project.org
Subject: Re: [R] help with by function

Nilesh,

I found a couple errors in your code.  First, in your by() statement you have a 
function to operate on the selected subset of data, which you refer to as
 x
but then, in your aov statement you refer to
 data_set
not
 x

Second, your funC() statement is a function of
 trait_names
but to make sure that the name of the variable is included in the formula, I 
changed this to a character variable.

Give the code below a try and see if it works for you.

Jean


library(agricolae)
library(dplyr)

funC - function(trait_names){
  by(data_set, data_set$Isopair, function(x) {
  mod - aov(formula(paste(trait_names, ~ STGgroup*Field + Rep%in%Field)),
data=x)
  out - HSD.test(mod, STGgroup, group=TRUE, console=TRUE)
  dfout - arrange(data.frame(out$groups), desc(trt))
  })
}

funC(Trait1)


On Tue, May 12, 2015 at 1:03 PM, DIGHE, NILESH [AG/2362] 
nilesh.di...@monsanto.commailto:nilesh.di...@monsanto.com wrote:
Hi,
I have an anonymous function called function(x) that will run anova, run 
HSD.test on the model, and then sort the results.  I am passing this anonymous 
function to the by function to get results by Isopair factor which is my 
index variable.  Since I want to run the anova on multiple dependent variables 
including Trait1  Trait2, I am calling the dependent variable, 
trait_names in the model and then passing the by function to another 
function called funC which takes the trait_names as an argument to execute.

After I execute the funC(data_set$trait1), I am getting the results by Isopair 
but the results for the two Isopairs (Isopair-A Isopair-B) are the same which 
I know is not correct.  It looks like the data is not getting split by Isopairs 
and so ALL data is used in anova for both Isopairs.  Any help in modifying 
function, funC or any other ways to achieve the desired outcome will be highly 
appreciated.

Thanks.  Nilesh

R code, data set, and session info is pasted below.
R code:
library(agricolae)
funC- function(trait_names){
  by(data_set, data_set$Isopair,function(x){
  mod- aov(trait_names~ STGgroup*Field + Rep%in%Field, data=data_set)
  out-HSD.test(mod,STGgroup,group=TRUE,console=TRUE)
  dfout- arrange(data.frame(out$groups),desc(trt))
  })
}
Results:
##execute funC function for Trait1  Trait2
funC(data_set$Trait1)


data_set$Isopair: Isopair-A

  trtmeans M

1 STG 776.9167 a

2 Non-STG 779.0833 a

---

data_set$Isopair: Isopair-B

  trtmeans M

1 STG 776.9167 a

2 Non-STG 779.0833 a

Data:
data_set- structure(list(Field = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), .Label = c(LML6, TZL2), class = factor), Isopair = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(Isopair-A, Isopair-B
), class = factor), STGgroup = structure(c(1L, 1L, 1L, 2L,
2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L,
1L, 2L, 2L, 2L), .Label = c(Non-STG, STG), class = factor),
Rep = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c(Rep1,
Rep2, Rep3), class = factor), Trait1 = c(686L, 641L,
642L, 727L, 619L, 562L, 808L, 739L, 744L, 873L, 797L, 868L,
782L, 783L, 675L, 713L, 762L, 641L, 1009L, 995L, 845L, 1186L,
912L, 663L), Trait2 = c(45L, 65L, 70L, 35L, 20L, 80L, 70L,
65L, 70L, 20L, 30L, 35L, 40L, 55L, 35L, 40L, 35L, 40L, 40L,
35L, 25L, 40L, 35L, 25L)), .Names = c(Field, Isopair,
STGgroup, Rep, Trait1, Trait2), class = data.frame, row.names = c(NA,
-24L))

Session info:

R version 3.1.3 (2015-03-09)

Platform: i386-w64-mingw32/i386 (32-bit)

Running under: Windows 7 x64 (build 7601) Service Pack 1



locale:

[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252

[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C

[5] LC_TIME=English_United States.1252



attached base packages:

[1] grid  stats graphics  grDevices utils datasets  methods   base



other attached packages:

 [1] gridExtra_0.9.1 Hmisc_3.16-0Formula_1.2-1   survival_2.38-1 
caret_6.0-41ggplot2_1.0.1

 [7] lattice_0.20-30 MASS_7.3-39 dplyr_0.4.1 agricolae_1.2-1



loaded via a namespace (and not attached):

 [1] acepack_1.3-3.3 assertthat_0.1  boot_1.3-15 
BradleyTerry2_1.0-6

 [5] brglm_0.5-9 car_2.0-25  cluster_2.0.1   coda_0.17-1

 [9] codetools_0.2-10colorspace_1.2-6combinat_0.0-8  DBI_0.3.1

[13] deldir_0.1-9digest_0.6.8foreach_1.4.2   foreign_0.8

[R] help with by function

[DIGHE, NILESH [AG/2362]]

Hi,
I have an anonymous function called function(x) that will run anova, run 
HSD.test on the model, and then sort the results.  I am passing this anonymous 
function to the by function to get results by Isopair factor which is my 
index variable.  Since I want to run the anova on multiple dependent variables 
including Trait1  Trait2, I am calling the dependent variable, 
trait_names in the model and then passing the by function to another 
function called funC which takes the trait_names as an argument to execute.

After I execute the funC(data_set$trait1), I am getting the results by Isopair 
but the results for the two Isopairs (Isopair-A Isopair-B) are the same which 
I know is not correct.  It looks like the data is not getting split by Isopairs 
and so ALL data is used in anova for both Isopairs.  Any help in modifying 
function, funC or any other ways to achieve the desired outcome will be highly 
appreciated.

Thanks.  Nilesh

R code, data set, and session info is pasted below.
R code:
library(agricolae)
funC- function(trait_names){
  by(data_set, data_set$Isopair,function(x){
  mod- aov(trait_names~ STGgroup*Field + Rep%in%Field, data=data_set)
  out-HSD.test(mod,STGgroup,group=TRUE,console=TRUE)
  dfout- arrange(data.frame(out$groups),desc(trt))
  })
}
Results:
##execute funC function for Trait1  Trait2
funC(data_set$Trait1)


data_set$Isopair: Isopair-A

  trtmeans M

1 STG 776.9167 a

2 Non-STG 779.0833 a

---

data_set$Isopair: Isopair-B

  trtmeans M

1 STG 776.9167 a

2 Non-STG 779.0833 a

Data:
data_set- structure(list(Field = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), .Label = c(LML6, TZL2), class = factor), Isopair = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(Isopair-A, Isopair-B
), class = factor), STGgroup = structure(c(1L, 1L, 1L, 2L,
2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L,
1L, 2L, 2L, 2L), .Label = c(Non-STG, STG), class = factor),
Rep = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c(Rep1,
Rep2, Rep3), class = factor), Trait1 = c(686L, 641L,
642L, 727L, 619L, 562L, 808L, 739L, 744L, 873L, 797L, 868L,
782L, 783L, 675L, 713L, 762L, 641L, 1009L, 995L, 845L, 1186L,
912L, 663L), Trait2 = c(45L, 65L, 70L, 35L, 20L, 80L, 70L,
65L, 70L, 20L, 30L, 35L, 40L, 55L, 35L, 40L, 35L, 40L, 40L,
35L, 25L, 40L, 35L, 25L)), .Names = c(Field, Isopair,
STGgroup, Rep, Trait1, Trait2), class = data.frame, row.names = c(NA,
-24L))

Session info:

R version 3.1.3 (2015-03-09)

Platform: i386-w64-mingw32/i386 (32-bit)

Running under: Windows 7 x64 (build 7601) Service Pack 1



locale:

[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252

[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C

[5] LC_TIME=English_United States.1252



attached base packages:

[1] grid  stats graphics  grDevices utils datasets  methods   base



other attached packages:

 [1] gridExtra_0.9.1 Hmisc_3.16-0Formula_1.2-1   survival_2.38-1 
caret_6.0-41ggplot2_1.0.1

 [7] lattice_0.20-30 MASS_7.3-39 dplyr_0.4.1 agricolae_1.2-1



loaded via a namespace (and not attached):

 [1] acepack_1.3-3.3 assertthat_0.1  boot_1.3-15 
BradleyTerry2_1.0-6

 [5] brglm_0.5-9 car_2.0-25  cluster_2.0.1   coda_0.17-1

 [9] codetools_0.2-10colorspace_1.2-6combinat_0.0-8  DBI_0.3.1

[13] deldir_0.1-9digest_0.6.8foreach_1.4.2   foreign_0.8-63

[17] gtable_0.1.2gtools_3.4.1iterators_1.0.7 klaR_0.6-12

[21] latticeExtra_0.6-26 lazyeval_0.1.10 LearnBayes_2.15 lme4_1.1-7

[25] magrittr_1.5Matrix_1.1-5mgcv_1.8-4  minqa_1.2.4

[29] munsell_0.4.2   nlme_3.1-120nloptr_1.0.4nnet_7.3-9

[33] parallel_3.1.3  pbkrtest_0.4-2  plyr_1.8.1  proto_0.3-10

[37] quantreg_5.11   RColorBrewer_1.1-2  Rcpp_0.11.6 reshape2_1.4.1

[41] rpart_4.1-9 scales_0.2.4sp_1.0-17   SparseM_1.6

[45] spdep_0.5-88splines_3.1.3   stringr_0.6.2   tools_3.1.3





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Re: [R] help with by function

[Adams, Jean]

Nilesh,

I found a couple errors in your code.  First, in your by() statement you
have a function to operate on the selected subset of data, which you refer
to as
 x
but then, in your aov statement you refer to
 data_set
not
 x

Second, your funC() statement is a function of
 trait_names
but to make sure that the name of the variable is included in the formula,
I changed this to a character variable.

Give the code below a try and see if it works for you.

Jean


library(agricolae)
library(dplyr)

funC - function(trait_names){
  by(data_set, data_set$Isopair, function(x) {
  mod - aov(formula(paste(trait_names, ~ STGgroup*Field +
Rep%in%Field)),
data=x)
  out - HSD.test(mod, STGgroup, group=TRUE, console=TRUE)
  dfout - arrange(data.frame(out$groups), desc(trt))
  })
}

funC(Trait1)


On Tue, May 12, 2015 at 1:03 PM, DIGHE, NILESH [AG/2362] 
nilesh.di...@monsanto.com wrote:

 Hi,
 I have an anonymous function called function(x) that will run anova, run
 HSD.test on the model, and then sort the results.  I am passing this
 anonymous function to the by function to get results by Isopair factor
 which is my index variable.  Since I want to run the anova on multiple
 dependent variables including Trait1  Trait2, I am calling the
 dependent variable, trait_names in the model and then passing the by
 function to another function called funC which takes the trait_names as
 an argument to execute.

 After I execute the funC(data_set$trait1), I am getting the results by
 Isopair but the results for the two Isopairs (Isopair-A Isopair-B) are the
 same which I know is not correct.  It looks like the data is not getting
 split by Isopairs and so ALL data is used in anova for both Isopairs.  Any
 help in modifying function, funC or any other ways to achieve the desired
 outcome will be highly appreciated.

 Thanks.  Nilesh

 R code, data set, and session info is pasted below.
 R code:
 library(agricolae)
 funC- function(trait_names){
   by(data_set, data_set$Isopair,function(x){
   mod- aov(trait_names~ STGgroup*Field + Rep%in%Field, data=data_set)
   out-HSD.test(mod,STGgroup,group=TRUE,console=TRUE)
   dfout- arrange(data.frame(out$groups),desc(trt))
   })
 }
 Results:
 ##execute funC function for Trait1  Trait2
 funC(data_set$Trait1)


 data_set$Isopair: Isopair-A

   trtmeans M

 1 STG 776.9167 a

 2 Non-STG 779.0833 a

 ---

 data_set$Isopair: Isopair-B

   trtmeans M

 1 STG 776.9167 a

 2 Non-STG 779.0833 a

 Data:
 data_set- structure(list(Field = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L), .Label = c(LML6, TZL2), class = factor), Isopair =
 structure(c(1L,
 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(Isopair-A, Isopair-B
 ), class = factor), STGgroup = structure(c(1L, 1L, 1L, 2L,
 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L,
 1L, 2L, 2L, 2L), .Label = c(Non-STG, STG), class = factor),
 Rep = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label =
 c(Rep1,
 Rep2, Rep3), class = factor), Trait1 = c(686L, 641L,
 642L, 727L, 619L, 562L, 808L, 739L, 744L, 873L, 797L, 868L,
 782L, 783L, 675L, 713L, 762L, 641L, 1009L, 995L, 845L, 1186L,
 912L, 663L), Trait2 = c(45L, 65L, 70L, 35L, 20L, 80L, 70L,
 65L, 70L, 20L, 30L, 35L, 40L, 55L, 35L, 40L, 35L, 40L, 40L,
 35L, 25L, 40L, 35L, 25L)), .Names = c(Field, Isopair,
 STGgroup, Rep, Trait1, Trait2), class = data.frame, row.names =
 c(NA,
 -24L))

 Session info:

 R version 3.1.3 (2015-03-09)

 Platform: i386-w64-mingw32/i386 (32-bit)

 Running under: Windows 7 x64 (build 7601) Service Pack 1



 locale:

 [1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United
 States.1252

 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C

 [5] LC_TIME=English_United States.1252



 attached base packages:

 [1] grid  stats graphics  grDevices utils datasets  methods
  base



 other attached packages:

  [1] gridExtra_0.9.1 Hmisc_3.16-0Formula_1.2-1   survival_2.38-1
 caret_6.0-41ggplot2_1.0.1

  [7] lattice_0.20-30 MASS_7.3-39 dplyr_0.4.1 agricolae_1.2-1



 loaded via a namespace (and not attached):

  [1] acepack_1.3-3.3 assertthat_0.1  boot_1.3-15
  BradleyTerry2_1.0-6

  [5] brglm_0.5-9 car_2.0-25  cluster_2.0.1
  coda_0.17-1

  [9] codetools_0.2-10colorspace_1.2-6combinat_0.0-8  DBI_0.3.1

 [13] deldir_0.1-9digest_0.6.8foreach_1.4.2
  foreign_0.8-63

 [17] gtable_0.1.2gtools_3.4.1iterators_1.0.7
  klaR_0.6-12

 [21] latticeExtra_0.6-26 lazyeval_0.1.10 LearnBayes_2.15 lme4_1.1-7

 [25] magrittr_1.5Matrix_1.1-5mgcv_1.8-4
 minqa_1.2.4

 [29] munsell_0.4.2   nlme_3.1-120nloptr_1.0.4   

[R] Help with pedigree() function in kinship2

[Jorge I Velez]

Dear R-help,

I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with.

Now, when running

## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)


## data to plot
d - structure(list(FamilyID = c(1, 1, 1, 1, 1, 1, 1,
1, 1), PatientID = structure(c(2L, 3L, 5L, 11L, 12L, 15L,
16L, 17L, 6L), .Label = c( 1,  2,  3,  4,  5,  6,
 7,  9, 10, 11, 13, 14, 18, 20, 23, 24, 25,
27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39,
41, 43, 45, 50, 62, 63, 64, 65, 66, 67, 85,
88), class = factor), FatherID = structure(c(1L, 1L, 6L,
1L, 5L, 6L, 1L, 7L, 6L), .Label = c(0, 1, 10, 11, 13,
2, 23, 27, 28, 3, 33, 34, 35, 38, 5, 62,
64, 66, 9), class = factor), MotherID = structure(c(1L,
1L, 7L, 1L, 14L, 7L, 1L, 5L, 7L), .Label = c(0, 10, 18,
2, 24, 29, 3, 30, 33, 34, 39, 4, 43, 5,
6, 63, 65, 9), class = factor), Sex = structure(c(2L,
1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L), .Label = c(Female, Male), class =
factor),
AffectionStatus = structure(c(1L, 1L, 2L, 1L, 2L, 2L, 1L,
2L, 2L), .Label = c(1, 2), class = factor)), .Names =
c(FamilyID,
PatientID, FatherID, MotherID, Sex, AffectionStatus
), row.names = c(NA, 9L), class = data.frame)

## plotting
ped - with(d, pedigree(PatientID, FatherID, MotherID, Sex,  affected =
AffectionStatus, famid = FamilyID))

## Error in pedigree(PatientID, FatherID, MotherID, Sex, affected =
AffectionStatus,  :
##Value of 'dadid' not found in the id list 1/0 1/0 1/2
1/0 1/2

I get an error.  My sessionInfo() is at the end.  I was wondering if
someone could help me to dissect what the cause of this error is and how it
can be fixed.

Thank you very much for your help.

Best regards,
Jorge Velez.-


R sessionInfo()
R version 3.2.0 Patched (2015-05-04 r68320)
Platform: x86_64-apple-darwin13.4.0 (64-bit)
Running under: OS X 10.10.3 (Yosemite)

locale:
[1] en_AU.UTF-8/en_AU.UTF-8/en_AU.UTF-8/C/en_AU.UTF-8/en_AU.UTF-8

attached base packages:
[1] stats graphics  grDevices utils datasets  parallel  compiler
 methods
[9] base

other attached packages:
[1] kinship2_1.6.0 quadprog_1.5-5 Matrix_1.2-0   readxl_0.1.0

loaded via a namespace (and not attached):
[1] Rcpp_0.11.6 grid_3.2.0  lattice_0.20-31

[[alternative HTML version deleted]]

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] help please metro_hastings function

[ARNAB KR MAITY via R-help]

Hi,
I have used this function before successfully. I could help you if you could 
provide your code.
Thanks  Regards,Arnab 
  From: hms Dreams cute_loo...@hotmail.com
 To: r-help@r-project.org r-help@r-project.org 
 Sent: Saturday, February 14, 2015 6:27 AM
 Subject: [R] help please metro_hastings function
   



Hi :)anybody can help me please I'm trying to use Metro_Hastings (

MHadaptive package)the proplem is:  How can I know the covariance matrix( 
prop_sigma ) to enter it in Metro_Hastings: 
 mcmc_r=Metro_Hastings(li_func=baysianlog, pars=c(1,1,1), prop_sigma 
=NULL,par_names=c('alpha','gamma','delta'),data=x ) its gave me an error , I 
must enter the cov matrix but I don't know how to calculate it, somebody told 
me to wrote the function without prop_sigma but its also gave me an error what 
can I do?? Thank you,Sara
                         
    [[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.


  
[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] help please metro_hastings function

[John Kane]

It looks like you posted in HTML and the result are garbbled. ONly post in 
plain text. Also it might help to read one or both of these
https://github.com/hadley/devtools/wiki/Reproducibility
 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

John Kane
Kingston ON Canada


 -Original Message-
 From: cute_loo...@hotmail.com
 Sent: Sat, 14 Feb 2015 15:27:58 +0300
 To: r-help@r-project.org
 Subject: [R] help please metro_hastings function
 
 
 
 
 Hi :)anybody can help me please I'm trying to use Metro_Hastings (
 
 MHadaptive package)the proplem is:  How can I know the covariance matrix(
 prop_sigma ) to enter it in Metro_Hastings:
  mcmc_r=Metro_Hastings(li_func=baysianlog, pars=c(1,1,1), prop_sigma
 =NULL,par_names=c('alpha','gamma','delta'),data=x ) its gave me an error
 , I must enter the cov matrix but I don't know how to calculate it,
 somebody told me to wrote the function without prop_sigma but its also
 gave me an error what can I do?? Thank you,Sara
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks  orcas on your 
desktop!

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] help please metro_hastings function

[hms Dreams]

Hi :)anybody can help me please I'm trying to use Metro_Hastings (

MHadaptive package)the proplem is:  How can I know the covariance matrix( 
prop_sigma ) to enter it in Metro_Hastings: 
 mcmc_r=Metro_Hastings(li_func=baysianlog, pars=c(1,1,1), prop_sigma 
=NULL,par_names=c('alpha','gamma','delta'),data=x ) its gave me an error , I 
must enter the cov matrix but I don't know how to calculate it, somebody told 
me to wrote the function without prop_sigma but its also gave me an error what 
can I do?? Thank you,Sara
  
[[alternative HTML version deleted]]

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with a function [along columns]

[Jim Lemon]

On Mon, 13 Oct 2014 08:54:51 PM Kate Ignatius wrote:
 Just an update to this:
 
 gtal - function(d) {
 alleles - sapply(d, function(.) strsplit(as.character(.), /))
 gt - unlist(lapply(alleles, function(x)
ifelse(identical(x[[1]], vcf[,3])  identical(x[[2]], vcf[,3]),
 'RR', ifelse(identical(x[[1]], vcf[,4])  identical(x[[2]], vcf[,4]), 'AA',
 ifelse(identical(x[[1]], vcf[,3])  identical(x[[2]], vcf[,4]), 'RA',
 ifelse(identical(x[[1]], vcf[,4])  identical(x[[2]],
 vcf[,3]), 'RA', ''))
 }
 
 I've got something working but I'm having trouble with the gt part...
 I'm getting the error: object of type 'closure' is not subsettable.
 The vcf is my original file that I want to match with so not sure
 whether this a problem.
 

Hi Kate,
Unless you have passed vcf to your function, it is unlikely to recognize 
it. As you are working with genome data, I suspect that there is a 
function named vcf somewhere in the parent environment and you 
can't subset a function.

Jim

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Help with a function [along columns]

[Kate Ignatius]

Hi all,

I need help with a function.  I'm trying to write a function to apply
to varying number of columns in a lot of files - hence the function...
but I'm getting stuck.  Here it is:

gt- function(x) {
alleles - sapply(x, function(.) strsplit(as.character(.), /))
gt - apply(x, function(.) ifelse(x[1] == vcf[3]  x[2] == vcf[3], 'RR',
ifelse(x[1] == vcf[4]  x[2] == vcf[4], 'AA',
ifelse(x[1] == vcf[3]  x[2] == vcf[4], 'RA',
ifelse(x[1] == vcf[4]  x[2] == vcf[3], 'RA', '')
}

I have different sized family genetic files and at the end of the day
I want to see whether the alleles of each person in the family match
the ref and/or the alt and if so, give AA, RA or RR.

Like so:

REF ALT Sample_1 GT_1 Sample_2 GT_2
A G A/A RR A/G RA
T G G/G AA T/T RR
A T T/T AA A/A RR
G A G/A RA G/G RR
G A G/G RR G/A RA
T C C/C AA C/C AA
T C C/C AA C/C AA
C T C/T RA T/T AA
G A A/A AA A/A AA
T G T/G RA G/G AA


Is there an easy way to do this?

Thanks!

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with a function [along columns]

[Kate Ignatius]

Just an update to this:

gtal - function(d) {
alleles - sapply(d, function(.) strsplit(as.character(.), /))
gt - unlist(lapply(alleles, function(x)
   ifelse(identical(x[[1]], vcf[,3])  identical(x[[2]], vcf[,3]), 'RR',
   ifelse(identical(x[[1]], vcf[,4])  identical(x[[2]], vcf[,4]), 'AA',
   ifelse(identical(x[[1]], vcf[,3])  identical(x[[2]], vcf[,4]), 'RA',
   ifelse(identical(x[[1]], vcf[,4])  identical(x[[2]],
vcf[,3]), 'RA', ''))
}

I've got something working but I'm having trouble with the gt part...
I'm getting the error: object of type 'closure' is not subsettable.
The vcf is my original file that I want to match with so not sure
whether this a problem.

On Mon, Oct 13, 2014 at 4:46 PM, Kate Ignatius kate.ignat...@gmail.com wrote:
 Hi all,

 I need help with a function.  I'm trying to write a function to apply
 to varying number of columns in a lot of files - hence the function...
 but I'm getting stuck.  Here it is:

 gt- function(x) {
 alleles - sapply(x, function(.) strsplit(as.character(.), /))
 gt - apply(x, function(.) ifelse(x[1] == vcf[3]  x[2] == vcf[3], 'RR',
 ifelse(x[1] == vcf[4]  x[2] == vcf[4], 'AA',
 ifelse(x[1] == vcf[3]  x[2] == vcf[4], 'RA',
 ifelse(x[1] == vcf[4]  x[2] == vcf[3], 'RA', '')
 }

 I have different sized family genetic files and at the end of the day
 I want to see whether the alleles of each person in the family match
 the ref and/or the alt and if so, give AA, RA or RR.

 Like so:

 REF ALT Sample_1 GT_1 Sample_2 GT_2
 A G A/A RR A/G RA
 T G G/G AA T/T RR
 A T T/T AA A/A RR
 G A G/A RA G/G RR
 G A G/G RR G/A RA
 T C C/C AA C/C AA
 T C C/C AA C/C AA
 C T C/T RA T/T AA
 G A A/A AA A/A AA
 T G T/G RA G/G AA


 Is there an easy way to do this?

 Thanks!

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Help with Cast Function

[Burhan ul haq]

Hi,

This is the input data frame:

###
df.1 = read.table(header=T,text=
id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
RT_no.drug
1 1 female 3.7 3.7 3.9 488 236 371
2 2 female 6.4 7.3 7.9 607 376 349
3 3 female 4.6 7.4 7.3 643 226 412
4 4 male 6.4 7.8 8.2 684 206 252
5 5 female 4.9 5.2 7.0 593 262 439
6 6 male 5.4 6.6 7.2 492 230 464
7 7 male 7.9 7.9 8.9 690 259 327
8 8 male 4.1 5.9 4.5 486 230 305
9 9 female 5.2 6.2 7.2 686 273 327
10 10 female 6.2 7.4 7.8 645 240 498
  )
###

This is the desired output:
###
id gender drug WMC RT
1 1 female alcohol 3.7 488
2 2 female alcohol 6.4 607
3 3 female alcohol 4.6 643
4 4 male alcohol 6.4 684
5 5 female alcohol 4.9 593
6 6 male alcohol 5.4 492
7 7 male alcohol 7.9 690
8 8 male alcohol 4.1 486
9 9 female alcohol 5.2 686
10 10 female alcohol 6.2 645
11 1 female caffeine 3.7 236
12 2 female caffeine 7.3 376
###

I know some melt and cast magic is required. But I was unable to sort it
myself.

Here are the dput versions:

Input Data Frame
###
 dput(df.1)
structure(list(id = 1:10, gender = structure(c(1L, 1L, 1L, 2L,
1L, 2L, 2L, 2L, 1L, 1L), .Label = c(female, male), class = factor),
WMC_alcohol = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2,
6.2), WMC_caffeine = c(3.7, 7.3, 7.4, 7.8, 5.2, 6.6, 7.9,
5.9, 6.2, 7.4), WMC_no.drug = c(3.9, 7.9, 7.3, 8.2, 7, 7.2,
8.9, 4.5, 7.2, 7.8), RT_alcohol = c(488L, 607L, 643L, 684L,
593L, 492L, 690L, 486L, 686L, 645L), RT_caffeine = c(236L,
376L, 226L, 206L, 262L, 230L, 259L, 230L, 273L, 240L), RT_no.drug =
c(371L,
349L, 412L, 252L, 439L, 464L, 327L, 305L, 327L, 498L)), .Names =
c(id,
gender, WMC_alcohol, WMC_caffeine, WMC_no.drug, RT_alcohol,
RT_caffeine, RT_no.drug), class = data.frame, row.names = c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10))


Output Data Frame
###
 dput(df.output)
structure(list(id = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
1L, 2L), gender = structure(c(1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L,
1L, 1L, 1L, 1L), .Label = c(female, male), class = factor),
drug = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 2L), .Label = c(alcohol, caffeine), class = factor),
WMC = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2, 6.2,
3.7, 7.3), RT = c(488L, 607L, 643L, 684L, 593L, 492L, 690L,
486L, 686L, 645L, 236L, 376L)), .Names = c(id, gender,
drug, WMC, RT), class = data.frame, row.names = c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12))


Cheers !

[[alternative HTML version deleted]]

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Re: [R] Help with Cast Function

[Carl Witthoft]

I see your desired output has rather fewer data than the input data frame.  
Instead of making us pore over a bunch of numbers, can you explain exactly
what filtering you wish to do to get the specific subset 
of {male/female} {alcohol/caffeine} you're trying to get?



BHM wrote
 Hi,
 
 This is the input data frame:
 
 ###
 df.1 = read.table(header=T,text=
 id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
 RT_no.drug
 1 1 female 3.7 3.7 3.9 488 236 371
 2 2 female 6.4 7.3 7.9 607 376 349
 3 3 female 4.6 7.4 7.3 643 226 412
 4 4 male 6.4 7.8 8.2 684 206 252
 5 5 female 4.9 5.2 7.0 593 262 439
 6 6 male 5.4 6.6 7.2 492 230 464
 7 7 male 7.9 7.9 8.9 690 259 327
 8 8 male 4.1 5.9 4.5 486 230 305
 9 9 female 5.2 6.2 7.2 686 273 327
 10 10 female 6.2 7.4 7.8 645 240 498
   )
 ###
 
 This is the desired output:
 ###
 id gender drug WMC RT
 1 1 female alcohol 3.7 488
 2 2 female alcohol 6.4 607
 3 3 female alcohol 4.6 643
 4 4 male alcohol 6.4 684
 5 5 female alcohol 4.9 593
 6 6 male alcohol 5.4 492
 7 7 male alcohol 7.9 690
 8 8 male alcohol 4.1 486
 9 9 female alcohol 5.2 686
 10 10 female alcohol 6.2 645
 11 1 female caffeine 3.7 236
 12 2 female caffeine 7.3 376
 ###
 
 I know some melt and cast magic is required. But I was unable to sort it
 myself.
 
 Here are the dput versions:
 
 Input Data Frame
 ###
 dput(df.1)
 structure(list(id = 1:10, gender = structure(c(1L, 1L, 1L, 2L,
 1L, 2L, 2L, 2L, 1L, 1L), .Label = c(female, male), class = factor),
 WMC_alcohol = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2,
 6.2), WMC_caffeine = c(3.7, 7.3, 7.4, 7.8, 5.2, 6.6, 7.9,
 5.9, 6.2, 7.4), WMC_no.drug = c(3.9, 7.9, 7.3, 8.2, 7, 7.2,
 8.9, 4.5, 7.2, 7.8), RT_alcohol = c(488L, 607L, 643L, 684L,
 593L, 492L, 690L, 486L, 686L, 645L), RT_caffeine = c(236L,
 376L, 226L, 206L, 262L, 230L, 259L, 230L, 273L, 240L), RT_no.drug =
 c(371L,
 349L, 412L, 252L, 439L, 464L, 327L, 305L, 327L, 498L)), .Names =
 c(id,
 gender, WMC_alcohol, WMC_caffeine, WMC_no.drug, RT_alcohol,
 RT_caffeine, RT_no.drug), class = data.frame, row.names = c(1,
 2, 3, 4, 5, 6, 7, 8, 9, 10))
 
 
 Output Data Frame
 ###
 dput(df.output)
 structure(list(id = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
 1L, 2L), gender = structure(c(1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L,
 1L, 1L, 1L, 1L), .Label = c(female, male), class = factor),
 drug = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 2L, 2L), .Label = c(alcohol, caffeine), class = factor),
 WMC = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2, 6.2,
 3.7, 7.3), RT = c(488L, 607L, 643L, 684L, 593L, 492L, 690L,
 486L, 686L, 645L, 236L, 376L)), .Names = c(id, gender,
 drug, WMC, RT), class = data.frame, row.names = c(1,
 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12))
 
 
 Cheers !
 
   [[alternative HTML version deleted]]
 
 __

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 PLEASE do read the posting guide
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Re: [R] Help with Cast Function

[David Winsemius]

On Nov 29, 2013, at 9:42 AM, Burhan ul haq wrote:

 Hi,
 
 This is the input data frame:
 
 ###
 df.1 = read.table(header=T,text=
 id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
 RT_no.drug
 1 1 female 3.7 3.7 3.9 488 236 371
 2 2 female 6.4 7.3 7.9 607 376 349
 3 3 female 4.6 7.4 7.3 643 226 412
 4 4 male 6.4 7.8 8.2 684 206 252
 5 5 female 4.9 5.2 7.0 593 262 439
 6 6 male 5.4 6.6 7.2 492 230 464
 7 7 male 7.9 7.9 8.9 690 259 327
 8 8 male 4.1 5.9 4.5 486 230 305
 9 9 female 5.2 6.2 7.2 686 273 327
 10 10 female 6.2 7.4 7.8 645 240 498
  )
 ###
 
 This is the desired output:
 ###
 id gender drug WMC RT
 1 1 female alcohol 3.7 488
 2 2 female alcohol 6.4 607
 3 3 female alcohol 4.6 643
 4 4 male alcohol 6.4 684
 5 5 female alcohol 4.9 593
 6 6 male alcohol 5.4 492
 7 7 male alcohol 7.9 690
 8 8 male alcohol 4.1 486
 9 9 female alcohol 5.2 686
 10 10 female alcohol 6.2 645
 11 1 female caffeine 3.7 236
 12 2 female caffeine 7.3 376
 ###
 
 I know some melt and cast magic is required. But I was unable to sort it
 myself.
 
# this is base::reshape

reshape(df.1, idvar=1:2, sep=_, direction=long, varying=names(df.1)[3:8])


 Here are the dput versions:
 
 Input Data Frame
 ###
 dput(df.1)
 structure(list(id = 1:10, gender = structure(c(1L, 1L, 1L, 2L,
 1L, 2L, 2L, 2L, 1L, 1L), .Label = c(female, male), class = factor),
WMC_alcohol = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2,
6.2), WMC_caffeine = c(3.7, 7.3, 7.4, 7.8, 5.2, 6.6, 7.9,
5.9, 6.2, 7.4), WMC_no.drug = c(3.9, 7.9, 7.3, 8.2, 7, 7.2,
8.9, 4.5, 7.2, 7.8), RT_alcohol = c(488L, 607L, 643L, 684L,
593L, 492L, 690L, 486L, 686L, 645L), RT_caffeine = c(236L,
376L, 226L, 206L, 262L, 230L, 259L, 230L, 273L, 240L), RT_no.drug =
 c(371L,
349L, 412L, 252L, 439L, 464L, 327L, 305L, 327L, 498L)), .Names =
 c(id,
 gender, WMC_alcohol, WMC_caffeine, WMC_no.drug, RT_alcohol,
 RT_caffeine, RT_no.drug), class = data.frame, row.names = c(1,
 2, 3, 4, 5, 6, 7, 8, 9, 10))
 
 
 Output Data Frame
 ###
 dput(df.output)
 structure(list(id = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
 1L, 2L), gender = structure(c(1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L,
 1L, 1L, 1L, 1L), .Label = c(female, male), class = factor),
drug = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 2L), .Label = c(alcohol, caffeine), class = factor),
WMC = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2, 6.2,
3.7, 7.3), RT = c(488L, 607L, 643L, 684L, 593L, 492L, 690L,
486L, 686L, 645L, 236L, 376L)), .Names = c(id, gender,
 drug, WMC, RT), class = data.frame, row.names = c(1,
 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12))
 
 
 Cheers !
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

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Re: [R] Help with Cast Function

[Burhan ul haq]

Hi,


First, a big thanks to all those who replied.


I am including all the replies in one email for easier reference later:

# Input from David
# 
  reshape(df.1, idvar=1:2, sep=_, direction=long,
varying=names(df.1)[3:8])
# 


# Input from Dennis
# 
  dfr1 - reshape(df.1, idvar = c(id, gender), v.names = c(WMC, RT),
timevar = type, times = c(alcohol, caffeine,
no.drug),
varying = list(3:5, 6:8), direction = long)
  rownames(dfr1) - NULL
  dfr
# 


# Input from Arun
# 
  library(reshape2)
  library(plyr)
  join_all(lapply(c(WMC,RT),function(x)
transform(melt(df.1[,c(1:2,grep(x,names(df.1)))],id.vars=c(id,gender),var=drug),drug=gsub(.*\\_,,drug))),by=c
 (id,gender,drug))
# 


Cheers !


On Sat, Nov 30, 2013 at 1:20 AM, David Winsemius dwinsem...@comcast.netwrote:


 On Nov 29, 2013, at 9:42 AM, Burhan ul haq wrote:

  Hi,
 
  This is the input data frame:
 
  ###
  df.1 = read.table(header=T,text=
  id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
  RT_no.drug
  1 1 female 3.7 3.7 3.9 488 236 371
  2 2 female 6.4 7.3 7.9 607 376 349
  3 3 female 4.6 7.4 7.3 643 226 412
  4 4 male 6.4 7.8 8.2 684 206 252
  5 5 female 4.9 5.2 7.0 593 262 439
  6 6 male 5.4 6.6 7.2 492 230 464
  7 7 male 7.9 7.9 8.9 690 259 327
  8 8 male 4.1 5.9 4.5 486 230 305
  9 9 female 5.2 6.2 7.2 686 273 327
  10 10 female 6.2 7.4 7.8 645 240 498
   )
  ###
 
  This is the desired output:
  ###
  id gender drug WMC RT
  1 1 female alcohol 3.7 488
  2 2 female alcohol 6.4 607
  3 3 female alcohol 4.6 643
  4 4 male alcohol 6.4 684
  5 5 female alcohol 4.9 593
  6 6 male alcohol 5.4 492
  7 7 male alcohol 7.9 690
  8 8 male alcohol 4.1 486
  9 9 female alcohol 5.2 686
  10 10 female alcohol 6.2 645
  11 1 female caffeine 3.7 236
  12 2 female caffeine 7.3 376
  ###
 
  I know some melt and cast magic is required. But I was unable to sort it
  myself.
 
 # this is base::reshape

 reshape(df.1, idvar=1:2, sep=_, direction=long,
 varying=names(df.1)[3:8])


  Here are the dput versions:
 
  Input Data Frame
  ###
  dput(df.1)
  structure(list(id = 1:10, gender = structure(c(1L, 1L, 1L, 2L,
  1L, 2L, 2L, 2L, 1L, 1L), .Label = c(female, male), class = factor),
 WMC_alcohol = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2,
 6.2), WMC_caffeine = c(3.7, 7.3, 7.4, 7.8, 5.2, 6.6, 7.9,
 5.9, 6.2, 7.4), WMC_no.drug = c(3.9, 7.9, 7.3, 8.2, 7, 7.2,
 8.9, 4.5, 7.2, 7.8), RT_alcohol = c(488L, 607L, 643L, 684L,
 593L, 492L, 690L, 486L, 686L, 645L), RT_caffeine = c(236L,
 376L, 226L, 206L, 262L, 230L, 259L, 230L, 273L, 240L), RT_no.drug =
  c(371L,
 349L, 412L, 252L, 439L, 464L, 327L, 305L, 327L, 498L)), .Names =
  c(id,
  gender, WMC_alcohol, WMC_caffeine, WMC_no.drug, RT_alcohol,
  RT_caffeine, RT_no.drug), class = data.frame, row.names = c(1,
  2, 3, 4, 5, 6, 7, 8, 9, 10))
 
 
  Output Data Frame
  ###
  dput(df.output)
  structure(list(id = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
  1L, 2L), gender = structure(c(1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L,
  1L, 1L, 1L, 1L), .Label = c(female, male), class = factor),
 drug = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 2L, 2L), .Label = c(alcohol, caffeine), class = factor),
 WMC = c(3.7, 6.4, 4.6, 6.4, 4.9, 5.4, 7.9, 4.1, 5.2, 6.2,
 3.7, 7.3), RT = c(488L, 607L, 643L, 684L, 593L, 492L, 690L,
 486L, 686L, 645L, 236L, 376L)), .Names = c(id, gender,
  drug, WMC, RT), class = data.frame, row.names = c(1,
  2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12))
 
 
  Cheers !
 
[[alternative HTML version deleted]]
 
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  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.

 David Winsemius
 Alameda, CA, USA



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Re: [R] Help for a function

[Rui Barradas]

   2  1488  4038  12   4
48 2009-07-01   2   2  1587  4090  15   6
49 2009-07-03   5   2  1587  4568  15   6
50 2009-07-06   5   2  2485  5298  19   7

...
# I transform the date in numeric
t1-derdata[[DATE]]
t - as.numeric(t1)
t
#i add the numeric date to the dataset
Mydata-data.frame(derdata,t)
Mydata
# I select each country
ARGENTINA- subset(Mydata,select = c(DATE,t,Canada))
names(ARGENTINA)[3] - Symptomes
ARGENTINA

# I create two functions because i want to create two variables by country 
:Incubation and CONTAGIEUX

# 1. Here is the incubation function (a person getting a virus  ,incub the 
virus for a certain time : 2 days)
incub - function(x){
   x$Incubation - 0
   x[order(x$t),]
   x$Incubation[1] - x$Symptomes[1]
   if(nrow(x)  1)
 x$Incubation[2] - sum(x$Symptomes[1:2])
   for(i in seq_len(nrow(x))[-(1:2)])
 x$Incubation[i] - sum(x$Symptomes[i - (0:1)])
   x
}

2. Contagion function (after incubing the virus, the subject can spread now the 
virus for the 7 coming days)

contag - function(x){
   x$CONTAGIEUX - 0
   for(i in 1:min(nrow(x), 7))
 x$CONTAGIEUX[i] - sum(x$Symptomes[1:i], na.rm = TRUE)
   for (i in seq_len(nrow(x))[-(1:7)]) {
 x$CONTAGIEUX[i] - x$Symptomes[i] + x$CONTAGIEUX[i-1] -
   x$Symptomes[i-7]
   }
   x
}

# and So i obtain by country what i need.
incub_ARGENTINA -incub(ARGENTINA)
incub_ARGENTINA
contag_ARGENTINA -contag(ARGENTINA)
contag_ARGENTINA
derdata_ARGENTINA -merge(contag_ARGENTINA, incub_ARGENTINA)
derdata_ARGENTINA



Now my problem is that i want to update those functions  incub  and contag  so 
that i can easily change the incubation and contagion period.
  they will not be fixe.
the incubation period can be 2 or 1 or 3 or 4 or 5 and so.
also the contagion period can be 4 or 5 or 6 or 8 or 9,..
And i want to not select the country before launch the the two funcions (incub  
and contag)
so that i have all the datasets for all country in the same dataset

Maurice










  De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au
Cc : anoumou teko_maur...@yahoo.fr; r-help@r-project.org
Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function


Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the
loop it never exits.
And

  res1[j] -(a*h)
  res2 -sum( res1[j])

is equivalent to

  res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:

On 12/05/2012 01:01 AM, anoumou wrote:

Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont
understant the
problem.


Hi anoumou,
Your function provides almost no indication of what two of its five
arguments are supposed to be. If, with a certain degree of optimistic
inference, we suppose x to be a data frame organized as shown below
the function. t, i and CONTAGIEUX must be the appropriately
named columns of the data frame. This leaves two columns, Symptomes
and Incubation. Say we flip a coin to decide which of these to
assign to r, and with all of our degrees of freedom gone, we assume
that the other is h. We are forced to the conclusion that a is a
nuisance argument, added to throw us off the scent.

Peering within the function, we notice that the date format is wrong,
braces are unmatched and that our data frame is alphabetically ordered
by name of country to no purpose whatsoever. The best I can do here is
to make the function potentially able to do something if you can work
out what to do with it.

Lambda-function (x,date1,r,h,a) {
   ndate1 - as.Date(date1, %Y-%m-%d)
   t1 - as.numeric(ndate1)
   x[order(x$i),]
   xt -x[,t]
   xi -x[,i]
   CONTAGIEUX -x[,CONTAGIEUX]
   while ( t1  min(xt) ){
for (i in 1:length(xi) ){
 for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
  res1[j] -(a*h)
  res2 -sum( res1[j])
 }
}
   lambda[i] - r*res2
   }
   x-data.frame(x,lambda)
   x
}

Jim

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and provide commented, minimal, self-contained, reproducible code.


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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Help with dataEllipse function

[Jana Makedonska]

Hi Everyone,

I am working with the R function dataEllipse. I plot the 95% confidence
ellipses for several different samples in the same plot and I color-code
the ellipse of each sample, but I do not know how to specify a different
line pattern for each ellipse. I can only modify the pattern for all
ellipses with the lty argument. Any help will be highly appreciated.

Thanks in advance!
Jana

-- 


Jana Makedonska,
B.Sc. Biology, Universite Paul Sabatier Toulouse III
M.Sc. Paleontology, Paleobiology and Phylogeny, Universite de Montpellier II
Ph.D. candidate in Physical Anthropology and Part-time lecturer
Department of Anthropology
College of Arts  Sciences
State University of New York at Albany
1400 Washington Avenue
1 Albany, NY
Office phone: 518-442-4699
http://electricsongs.academia.edu/JanaMakedonska

[[alternative HTML version deleted]]

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with dataEllipse function

[John Fox]

Dear Jana,

The lty argument to dataEllipse() (in the car package) isn't vectorized. It 
could be, and I'll add that as a feature request. Actually, lty isn't an 
explicit argument to dataEllipse(); it's simply passed through to the lines() 
function, which draws the ellipses.

You should be able to do what you want by adding the ellipses one at a time to 
your plot (see the argument add in ?dataEllipse) or by using the coordinates of 
the ellipses, returned by dataEllipse(), to a customized graph.

I hope that this helps,
 John


John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

On Thu, 25 Apr 2013 20:00:20 -0400
 Jana Makedonska jmakedon...@gmail.com wrote:
 Hi Everyone,
 
 I am working with the R function dataEllipse. I plot the 95% confidence
 ellipses for several different samples in the same plot and I color-code
 the ellipse of each sample, but I do not know how to specify a different
 line pattern for each ellipse. I can only modify the pattern for all
 ellipses with the lty argument. Any help will be highly appreciated.
 
 Thanks in advance!
 Jana
 
 -- 
 
 
 Jana Makedonska,
 B.Sc. Biology, Universite Paul Sabatier Toulouse III
 M.Sc. Paleontology, Paleobiology and Phylogeny, Universite de Montpellier II
 Ph.D. candidate in Physical Anthropology and Part-time lecturer
 Department of Anthropology
 College of Arts  Sciences
 State University of New York at Albany
 1400 Washington Avenue
 1 Albany, NY
 Office phone: 518-442-4699
 http://electricsongs.academia.edu/JanaMakedonska
 
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with dataEllipse function

[Michael Friendly]

On 4/25/2013 8:00 PM, Jana Makedonska wrote:

Hi Everyone,

I am working with the R function dataEllipse. I plot the 95% confidence
ellipses for several different samples in the same plot and I color-code
the ellipse of each sample, but I do not know how to specify a different
line pattern for each ellipse. I can only modify the pattern for all
ellipses with the lty argument. Any help will be highly appreciated.


lty is not an argument of car::dataEllipse, but is passed via ...
So, when you use the groups= argument, only a single value gets used.
You would have to modify the function to allow what you want.



--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.  Chair, Quantitative Methods
York University  Voice: 416 736-2100 x66249 Fax: 416 736-5814
4700 Keele StreetWeb:   http://www.datavis.ca
Toronto, ONT  M3J 1P3 CANADA

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[R] Help with merge function

[Catarina Ferreira]

Dear all,

I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.

Dataframe x1

State_prov Shape_name   bob2009   bob 2010   bob2011
Nova ScotiaAnnapolis 0  0  1
Nova ScotiaAntigonish0  0  0
Nova ScotiaGly   NA   NA NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009
 0Nova ScotiaAnnapolis 0
0  10
 1Nova ScotiaAntigonish0
0  1
 2Nova ScotiaGly   0
0  1

So when I do

x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

it should do the trick. The thing is that it works for the columns (it adds
all the new columns not common to both dataframes), but it also adds the
rows. This is what I get (x3):

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
 0Nova ScotiaAnnapolis 0
0  10NA
 NA  Nova ScotiaAnnapolis NA   NA  NA
1
 1Nova ScotiaAntigonish0
0  1   NA
NA  Nova ScotiaAntigonishNA   NA  NA
0
 2Nova ScotiaGly   0
0  1   NA
NA  Nova ScotiaGly   NA   NA
NA NA

What I want to get is a true merge, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
 0Nova ScotiaAnnapolis 0
0  101
 1Nova ScotiaAntigonish0
0  1   0
 2Nova ScotiaGly   0
0  1   NA

Can anybody please help me to understand what I'm doing wrong.
Any help will be much appreciated!!


-- 
Catarina C. Ferreira, PhD

[[alternative HTML version deleted]]

__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with merge function

[Rui Barradas]

Hello,

The following seems to do the trick.



x1 -
structure(list(State_prov = c(Nova Scotia, Nova Scotia, Nova Scotia
), Shape_name = c(Annapolis, Antigonish, Gly), bob2009 = c(0L,
0L, NA), bob2010 = c(0L, 0L, NA), bob2011 = c(1L, 0L, NA)), .Names = 
c(State_prov,
Shape_name, bob2009, bob2010, bob2011), class = data.frame, 
row.names = c(NA,

-3L))

x2 -
structure(list(FID = 0:2, State_prov = c(Nova Scotia, Nova Scotia,
Nova Scotia), Shape_name = c(Annapolis, Antigonish, Gly
), bob2009 = c(0L, 0L, 0L), bob2010 = c(0L, 0L, 0L), coy2009 = c(10L,
1L, 1L)), .Names = c(FID, State_prov, Shape_name, bob2009,
bob2010, coy2009), class = data.frame, row.names = c(NA,
-3L))

x3  - merge(x1, x2, all.y = TRUE)



Note also that since by = intersect(names(x1), names(x2)), you really 
don't need it, it's the default behavior.


Hope this helps,

Rui Barradas

Em 26-04-2013 18:10, Catarina Ferreira escreveu:

Dear all,

I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.

Dataframe x1

State_prov Shape_name   bob2009   bob 2010   bob2011
Nova ScotiaAnnapolis 0  0  1
Nova ScotiaAntigonish0  0  0
Nova ScotiaGly   NA   NA NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009
  0Nova ScotiaAnnapolis 0
0  10
  1Nova ScotiaAntigonish0
0  1
  2Nova ScotiaGly   0
0  1

So when I do

x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

it should do the trick. The thing is that it works for the columns (it adds
all the new columns not common to both dataframes), but it also adds the
rows. This is what I get (x3):

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
  0Nova ScotiaAnnapolis 0
0  10NA
  NA  Nova ScotiaAnnapolis NA   NA  NA
 1
  1Nova ScotiaAntigonish0
0  1   NA
NA  Nova ScotiaAntigonishNA   NA  NA
 0
  2Nova ScotiaGly   0
0  1   NA
NA  Nova ScotiaGly   NA   NA
NA NA

What I want to get is a true merge, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
  0Nova ScotiaAnnapolis 0
0  101
  1Nova ScotiaAntigonish0
0  1   0
  2Nova ScotiaGly   0
0  1   NA

Can anybody please help me to understand what I'm doing wrong.
Any help will be much appreciated!!




__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with merge function

[Catarina Ferreira]

Hello, Thank you for your help. However the dataframes I gave you were only
examples, the actual dataframes are very big. Does this mean I have to
write every range of data for each variable??


On Fri, Apr 26, 2013 at 2:25 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 The following seems to do the trick.



 x1 -
 structure(list(State_prov = c(Nova Scotia, Nova Scotia, Nova Scotia
 ), Shape_name = c(Annapolis, Antigonish, Gly), bob2009 = c(0L,
 0L, NA), bob2010 = c(0L, 0L, NA), bob2011 = c(1L, 0L, NA)), .Names =
 c(State_prov,
 Shape_name, bob2009, bob2010, bob2011), class = data.frame,
 row.names = c(NA,
 -3L))

 x2 -
 structure(list(FID = 0:2, State_prov = c(Nova Scotia, Nova Scotia,
 Nova Scotia), Shape_name = c(Annapolis, Antigonish, Gly
 ), bob2009 = c(0L, 0L, 0L), bob2010 = c(0L, 0L, 0L), coy2009 = c(10L,
 1L, 1L)), .Names = c(FID, State_prov, Shape_name, bob2009,
 bob2010, coy2009), class = data.frame, row.names = c(NA,
 -3L))

 x3  - merge(x1, x2, all.y = TRUE)



 Note also that since by = intersect(names(x1), names(x2)), you really
 don't need it, it's the default behavior.

 Hope this helps,

 Rui Barradas

 Em 26-04-2013 18:10, Catarina Ferreira escreveu:

  Dear all,

 I'm trying to merge 2 dataframes, but I'm not being entirely successful
 and
 I can't understand why.

 Dataframe x1

 State_prov Shape_name   bob2009   bob 2010   bob2011
 Nova ScotiaAnnapolis 0  0  1
 Nova ScotiaAntigonish0  0  0
 Nova ScotiaGly   NA   NA NA

 Dataframe x2 - has 2 rows and 193 variables, contains one important
 field which is FID that is a link to a shapefile (this is not in x1) and
 shares common columns with x1, like this:

 FID State_prov Shape_name   bob2009   bob 2010  coy 2009
   0Nova ScotiaAnnapolis 0
 0  10
   1Nova ScotiaAntigonish0
 0  1
   2Nova ScotiaGly   0
 0  1

 So when I do

 x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

 it should do the trick. The thing is that it works for the columns (it
 adds
 all the new columns not common to both dataframes), but it also adds the
 rows. This is what I get (x3):

 FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
   0Nova ScotiaAnnapolis 0
 0  10NA
   NA  Nova ScotiaAnnapolis NA   NA  NA
  1
   1Nova ScotiaAntigonish0
 0  1   NA
 NA  Nova ScotiaAntigonishNA   NA  NA
  0
   2Nova ScotiaGly   0
 0  1   NA
 NA  Nova ScotiaGly   NA   NA
 NA NA

 What I want to get is a true merge, like this:

 FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
   0Nova ScotiaAnnapolis 0
 0  101
   1Nova ScotiaAntigonish0
 0  1   0
   2Nova ScotiaGly   0
 0  1   NA

 Can anybody please help me to understand what I'm doing wrong.
 Any help will be much appreciated!!





-- 
Catarina C. Ferreira, PhD
Post-doctoral Research Fellow
Department of Biology
Trent University
Peterborough, ON Canada
URL: http://www.researcherid.com/rid/A-3898-2011

[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with merge function

[arun]

Hi,

The format is bit messed up. 
So, not sure this is what you wanted.

x1- read.table(text=State_prov,Shape_name,bob2009,bob2010,bob2011
Nova Scotia,Annapolis,0,0,1
Nova Scotia,Antigonish,0,0,0
Nova Scotia,Gly,NA,NA,NA
,sep=,,header=TRUE,stringsAsFactors=FALSE)

x2- read.table(text=
FID,State_prov,Shape_name,bob2009,bob2010,coy2009
0,Nova Scotia,Annapolis,0,0,10
1,Nova Scotia,Antigonish,0,0,1
2,Nova Scotia,Gly,0,0,1
,sep=,,header=TRUE,stringsAsFactors=FALSE)
 merge(x1,x2,all=TRUE)
#   State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1
#4 Nova Scotia    Gly  NA  NA  NA  NA  NA





- Original Message -
From: Catarina Ferreira catferre...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Friday, April 26, 2013 1:10 PM
Subject: [R] Help with merge function

Dear all,

I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.

Dataframe x1

State_prov     Shape_name   bob2009   bob 2010   bob2011
Nova Scotia    Annapolis         0                  0              1
Nova Scotia    Antigonish        0                  0              0
Nova Scotia    Gly                   NA               NA             NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009
0        Nova Scotia    Annapolis         0
0              10
1        Nova Scotia    Antigonish        0
0              1
2        Nova Scotia    Gly                   0
0              1

So when I do

x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

it should do the trick. The thing is that it works for the columns (it adds
all the new columns not common to both dataframes), but it also adds the
rows. This is what I get (x3):

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009   bob2011
0        Nova Scotia    Annapolis         0
0              10            NA
NA      Nova Scotia    Annapolis         NA               NA          NA
            1
1        Nova Scotia    Antigonish        0
0              1               NA
NA      Nova Scotia    Antigonish        NA               NA          NA
            0
2        Nova Scotia    Gly                   0
0              1               NA
NA      Nova Scotia    Gly                   NA               NA
NA             NA

What I want to get is a true merge, like this:

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009   bob2011
0        Nova Scotia    Annapolis         0
0              10            1
1        Nova Scotia    Antigonish        0
0              1               0
2        Nova Scotia    Gly                   0
0              1               NA

Can anybody please help me to understand what I'm doing wrong.
Any help will be much appreciated!!


-- 
Catarina C. Ferreira, PhD

    [[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with merge function

[Rui Barradas]

Hello,

I don't understand the question, what range? I've just changed the 'all' 
argument to 'all.y', without doing anything special to the variables.

Can you explain what you mean?

Rui Barradas


Em 26-04-2013 19:30, Catarina Ferreira escreveu:

Hello, Thank you for your help. However the dataframes I gave you were only
examples, the actual dataframes are very big. Does this mean I have to
write every range of data for each variable??


On Fri, Apr 26, 2013 at 2:25 PM, Rui Barradas ruipbarra...@sapo.pt wrote:


Hello,

The following seems to do the trick.



x1 -
structure(list(State_prov = c(Nova Scotia, Nova Scotia, Nova Scotia
), Shape_name = c(Annapolis, Antigonish, Gly), bob2009 = c(0L,
0L, NA), bob2010 = c(0L, 0L, NA), bob2011 = c(1L, 0L, NA)), .Names =
c(State_prov,
Shape_name, bob2009, bob2010, bob2011), class = data.frame,
row.names = c(NA,
-3L))

x2 -
structure(list(FID = 0:2, State_prov = c(Nova Scotia, Nova Scotia,
Nova Scotia), Shape_name = c(Annapolis, Antigonish, Gly
), bob2009 = c(0L, 0L, 0L), bob2010 = c(0L, 0L, 0L), coy2009 = c(10L,
1L, 1L)), .Names = c(FID, State_prov, Shape_name, bob2009,
bob2010, coy2009), class = data.frame, row.names = c(NA,
-3L))

x3  - merge(x1, x2, all.y = TRUE)



Note also that since by = intersect(names(x1), names(x2)), you really
don't need it, it's the default behavior.

Hope this helps,

Rui Barradas

Em 26-04-2013 18:10, Catarina Ferreira escreveu:

  Dear all,


I'm trying to merge 2 dataframes, but I'm not being entirely successful
and
I can't understand why.

Dataframe x1

State_prov Shape_name   bob2009   bob 2010   bob2011
Nova ScotiaAnnapolis 0  0  1
Nova ScotiaAntigonish0  0  0
Nova ScotiaGly   NA   NA NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009
   0Nova ScotiaAnnapolis 0
0  10
   1Nova ScotiaAntigonish0
0  1
   2Nova ScotiaGly   0
0  1

So when I do

x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

it should do the trick. The thing is that it works for the columns (it
adds
all the new columns not common to both dataframes), but it also adds the
rows. This is what I get (x3):

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
   0Nova ScotiaAnnapolis 0
0  10NA
   NA  Nova ScotiaAnnapolis NA   NA  NA
  1
   1Nova ScotiaAntigonish0
0  1   NA
NA  Nova ScotiaAntigonishNA   NA  NA
  0
   2Nova ScotiaGly   0
0  1   NA
NA  Nova ScotiaGly   NA   NA
NA NA

What I want to get is a true merge, like this:

FID State_prov Shape_name   bob2009   bob 2010  coy 2009   bob2011
   0Nova ScotiaAnnapolis 0
0  101
   1Nova ScotiaAntigonish0
0  1   0
   2Nova ScotiaGly   0
0  1   NA

Can anybody please help me to understand what I'm doing wrong.
Any help will be much appreciated!!








__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with merge function

[arun]

Hi,
From the output you wanted, it looks like:
library(plyr)
join(x1,x2,type=right)
#Joining by: State_prov, Shape_name, bob2009, bob2010
 #  State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1
 merge(x1,x2,all.y=TRUE)
#   State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1

A.K.







From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Friday, April 26, 2013 2:23 PM
Subject: Re: [R] Help with merge function



Hello,

I didn't realize that the format had been changed after I sent the email. I'm 
sending you the original mail in attach in a word with the correct format, 
since I don't think your answer is the one I'm looking for, likely due to the 
erroneous format.

Thank you again for your help.




On Fri, Apr 26, 2013 at 2:11 PM, arun smartpink...@yahoo.com wrote:

Hi,

The format is bit messed up. 
So, not sure this is what you wanted.

x1- read.table(text=State_prov,Shape_name,bob2009,bob2010,bob2011
Nova Scotia,Annapolis,0,0,1
Nova Scotia,Antigonish,0,0,0
Nova Scotia,Gly,NA,NA,NA
,sep=,,header=TRUE,stringsAsFactors=FALSE)

x2- read.table(text=
FID,State_prov,Shape_name,bob2009,bob2010,coy2009
0,Nova Scotia,Annapolis,0,0,10
1,Nova Scotia,Antigonish,0,0,1
2,Nova Scotia,Gly,0,0,1
,sep=,,header=TRUE,stringsAsFactors=FALSE)
 merge(x1,x2,all=TRUE)
#   State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1
#4 Nova Scotia    Gly  NA  NA  NA  NA  NA






- Original Message -
From: Catarina Ferreira catferre...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, April 26, 2013 1:10 PM
Subject: [R] Help with merge function

Dear all,

I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.

Dataframe x1

State_prov     Shape_name   bob2009   bob 2010   bob2011
Nova Scotia    Annapolis         0                  0              1
Nova Scotia    Antigonish        0                  0              0
Nova Scotia    Gly                   NA               NA             NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009
0        Nova Scotia    Annapolis         0
0              10
1        Nova Scotia    Antigonish        0
0              1
2        Nova Scotia    Gly                   0
0              1

So when I do

x3  - merge(x1, x2, by=intersect(names(x1), names(x2)), all=TRUE)

it should do the trick. The thing is that it works for the columns (it adds
all the new columns not common to both dataframes), but it also adds the
rows. This is what I get (x3):

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009   bob2011
0        Nova Scotia    Annapolis         0
0              10            NA
NA      Nova Scotia    Annapolis         NA               NA          NA
            1
1        Nova Scotia    Antigonish        0
0              1               NA
NA      Nova Scotia    Antigonish        NA               NA          NA
            0
2        Nova Scotia    Gly                   0
0              1               NA
NA      Nova Scotia    Gly                   NA               NA
NA             NA

What I want to get is a true merge, like this:

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009   bob2011
0        Nova Scotia    Annapolis         0
0              10            1
1        Nova Scotia    Antigonish        0
0              1               0
2        Nova Scotia    Gly                   0
0              1               NA

Can anybody please help me to understand what I'm doing wrong.
Any help will be much appreciated!!


--
Catarina C. Ferreira, PhD

    [[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.




-- 
Catarina C. Ferreira, PhD
Post-doctoral Research Fellow
Department of Biology
Trent University
Peterborough, ON Canada
URL: http://www.researcherid.com/rid/A-3898-2011

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do

Re: [R] Help with merge function

[arun]

Hi,

Check whether this works.

Lines1-readLines(NS_update.txt)
x1-read.table(text=gsub('\',,Lines1),sep=,,header=TRUE,stringsAsFactors=FALSE)
 x2- read.table(data.txt,sep=,header=TRUE,stringsAsFactors=FALSE,fill=TRUE)
 dim(x2)
#[1] 34577   189
library(plyr) 
 res- join(x1,x2,type=right)
#Joining by: State_Prov, Shape_name, bob2009, bob2010, red2009, red2010, 
coy2009, coy2010, lyn2009, lyn2010
 dim(res)
#[1] 34577   193

 res2- merge(x1,x2,all.y=TRUE)
 dim(res2)
#[1] 34577   193
A.K.






From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Friday, April 26, 2013 4:20 PM
Subject: Re: [R] Help with merge function



here they are. As you see the NS_update is data for only 1 province and I want 
it to add this data to the bigger file (data), merging the common columns and 
adding the new columns. But what it is doing is duplicating the rows in the 
bigger file that correspond to NS_update, as well as creating the new columns 
(this is ok).




On Fri, Apr 26, 2013 at 4:16 PM, arun smartpink...@yahoo.com wrote:

You can send the files. 








From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, April 26, 2013 4:15 PM

Subject: Re: [R] Help with merge function



is it ok if I send you the files, it's probably better for you to understand 
me. It didn't work on my files.




On Fri, Apr 26, 2013 at 4:12 PM, arun smartpink...@yahoo.com wrote:

Hi,

 I am not sure what is the problem.  I used the datasets your provided x1 
and x2.  I got the result that was shown in the output your desired.
Are you saying that this didn't worked in your original dataset or the one 
your provided?  In that case, could you dput(dataset,20)?









From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, April 26, 2013 4:01 PM

Subject: Re: [R] Help with merge function



Thank you. It still isn't working. Thank you in any case.




On Fri, Apr 26, 2013 at 2:31 PM, arun smartpink...@yahoo.com wrote:

Hi,
From the output you wanted, it looks like:
library(plyr)
join(x1,x2,type=right)
#Joining by: State_prov, Shape_name, bob2009, bob2010

 #  State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1
 merge(x1,x2,all.y=TRUE)

#   State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1

A.K.








From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, April 26, 2013 2:23 PM
Subject: Re: [R] Help with merge function




Hello,

I didn't realize that the format had been changed after I sent the email. 
I'm sending you the original mail in attach in a word with the correct 
format, since I don't think your answer is the one I'm looking for, likely 
due to the erroneous format.

Thank you again for your help.




On Fri, Apr 26, 2013 at 2:11 PM, arun smartpink...@yahoo.com wrote:

Hi,

The format is bit messed up. 
So, not sure this is what you wanted.

x1- read.table(text=State_prov,Shape_name,bob2009,bob2010,bob2011
Nova Scotia,Annapolis,0,0,1
Nova Scotia,Antigonish,0,0,0
Nova Scotia,Gly,NA,NA,NA
,sep=,,header=TRUE,stringsAsFactors=FALSE)

x2- read.table(text=
FID,State_prov,Shape_name,bob2009,bob2010,coy2009
0,Nova Scotia,Annapolis,0,0,10
1,Nova Scotia,Antigonish,0,0,1
2,Nova Scotia,Gly,0,0,1
,sep=,,header=TRUE,stringsAsFactors=FALSE)
 merge(x1,x2,all=TRUE)
#   State_prov Shape_name bob2009 bob2010 bob2011 FID coy2009
#1 Nova Scotia  Annapolis   0   0   1   0  10
#2 Nova Scotia Antigonish   0   0   0   1   1
#3 Nova Scotia    Gly   0   0  NA   2   1
#4 Nova Scotia    Gly  NA  NA  NA  NA  NA






- Original Message -
From: Catarina Ferreira catferre...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, April 26, 2013 1:10 PM
Subject: [R] Help with merge function

Dear all,

I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.

Dataframe x1

State_prov     Shape_name   bob2009   bob 2010   bob2011
Nova Scotia    Annapolis         0                  0              1
Nova Scotia    Antigonish        0                  0              0
Nova Scotia    Gly                   NA               NA             NA

Dataframe x2 - has 2 rows and 193 variables, contains one important
field which is FID that is a link to a shapefile (this is not in x1) and
shares common columns with x1, like this:

FID     State_prov     Shape_name   bob2009   bob 2010  coy 2009
0        Nova

[R] Help with lmRob function

[Mark Chaffin]

Hi,

I am fairly new to R and have encountered an issue with the lmRob function that 
I have been unable to resolve. I am trying to run a robust regression using the 
lmRob function which runs successfully, but the results are rather strange. I'm 
not sure it's important, but my model has 3 dichotomous categorical variables 
and 2 continuous variables in it. When I look at a summary of my model, 1 of my 
continuous variables and 1 of my categorical variables both have an Estimate 
value of 1. My other 2 categorical variables and 1 other continuous variables 
all have Estimate values of 0. These results seem strange to me. When I run a 
simple rlm function the model runs fine and produces the results I'd expect to 
see.

Another potentially useful piece of information is that when I run a model with 
just the two variables that have an Estimate value of 1 (one categorical and 
one continuous), I get the following error message:

1: In lmRob.fit.compute(x, y, x1.idx = x1.idx, nrep = nrep, robust.control = 
robust.control,  :
  Sum(psi.weight(wi)) less than 1e-06 in lmRob.ucovcoef.  during initial 
estimation.
2: In lmRob.fit.compute(x, y, x1.idx = x1.idx, nrep = nrep, robust.control = 
robust.control,  :
  Sum(psi.weight(wi)) less than 1e-06 in lmRob.ucovcoef. during final scale 
estimation.

I do not understand what this error message means and have been unable to 
resolve it. I tried several different permutations of my five variables in 
varying combinations and sometimes the models work and sometimes I get this 
error message. There seems to be no pattern as to when the error message occurs 
and when it does not.

If anyone has encountered this or has any help I'd appreciate it. Thanks.

[[alternative HTML version deleted]]

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[R] help with simple function

[Miguel Eduardo Delgado Burbano]

hello all

I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function

sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error message

Error: attempt to apply non-function

alternatively I wrote this
 sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))]

but this error message appear

[1] NA
Warning message:
NAs introduced by coercion

some suggestion to solve this?? thank you so much for your help

Miguel

[[alternative HTML version deleted]]

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Re: [R] help with simple function

[Berend Hasselman]

On 17-03-2013, at 16:47, Miguel Eduardo Delgado Burbano 
mdelgadoburb...@gmail.com wrote:

 hello all
 
 I am writing a quite simple script to study dental wear patterns in humans
 and I wrote this function
 
 sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
 error message
 
 Error: attempt to apply non-function
 

Write 4*( instead of 4( assuming you want to multiply.

 alternatively I wrote this
 sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))]
 
 but this error message appear
 
 [1] NA
 Warning message:
 NAs introduced by coercion
 

[ is an indexing operator. You are trying to index the number 4. Which is 
impossible.

Berend


 some suggestion to solve this?? thank you so much for your help
 
 Miguel
 
   [[alternative HTML version deleted]]
 
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Re: [R] help with simple function

[arun]

Hi,

 Y1- 1:4
 Y2- 5:8

sqrt(var(Y1)+var(Y2)^2)-4*((var(Y1)*(var(Y2)-cov(Y1,Y2)^2)))
#[1] 9.515593
A.K.

- Original Message -
From: Miguel Eduardo Delgado Burbano mdelgadoburb...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Sunday, March 17, 2013 11:47 AM
Subject: [R] help with simple function

hello all

I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function

sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error message

Error: attempt to apply non-function

alternatively I wrote this
sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))]

but this error message appear

[1] NA
Warning message:
NAs introduced by coercion

some suggestion to solve this?? thank you so much for your help

Miguel

    [[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.


__
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Re: [R] help with simple function

[Rui Barradas]

Hello,

You are missing a '*' in your first try.
(As for the second, in R parenthesis are round, not [])

sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2))

This written as a function becomes

fun - function(Y1, Y2)
sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2))

x1 - rnorm(10)
x2 - rnorm(10)
fun(x1, x2)


Hope this helps,

Rui Barradas

Em 17-03-2013 15:47, Miguel Eduardo Delgado Burbano escreveu:

hello all

I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function

sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error message

Error: attempt to apply non-function

alternatively I wrote this
  sqrt(var(Y1)+var(Y2)^2)-4[(var(Y1)*(var(Y2)-cov(Y1,Y2)^2))]

but this error message appear

[1] NA
Warning message:
NAs introduced by coercion

some suggestion to solve this?? thank you so much for your help

Miguel

[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.



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Re: [R] Help with a function and text

[Eliano]

Thanks for that. Eliano

2013/3/7 David Winsemius [via R] ml-node+s789695n4660565...@n4.nabble.com


 On Mar 6, 2013, at 3:44 PM, Eliano wrote:

  Thanks. Btw are you able to help with my issue? Thanks, Eliano

 I'm sorry,  I was too busy answering the question from 'Eliano' over
 on StackOverflow. I didn't have time to address this one.

 (Please do note that cross-posting questions to Rhelp is contrary to
 advice in the Posting Guide.)

 You might also do further searching in the Archives with the search terms:
 substitute text eval
  and perhaps narrow it down further with contributors named:
 grothendeick, dunlap, ligges, venables

 
  Sent from my iPhone
 
  On 6 Mar 2013, at 23:41, David Winsemius [via R] 
  [hidden email] http://user/SendEmail.jtp?type=nodenode=4660565i=0
 wrote:
 
 
  On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:
 
  Hi, can I understand why this message was rejected ?
  Thanks,
  Eliano
 
  First hit on a Markmail search:
 
 
 http://markmail.org/message/5xog3ayx4amprsdx?q=list:org%2Er-project%2Er-help+nabble+rejected
 
  --
  David.
 
 
  Sent from my iPhone
 
  On 6 Mar 2013, at 19:18, Eliano [hidden
 email]/user/SendEmail.jtp?type=nodenode=4660547i=0
   wrote:
 
  Hi everyone,
 
  I am writing some code to generate a function. I am passing that code
 to
  a
  dataset which i'm importing in R, e.g.
  Test=read.table('C:/test.txt', header=F, sep='\t', na.strings='NA',
  dec='.',
  strip.white=TRUE)
  Test
 
  V1
  (if(nclusters0){OptmizationInputs[3,3]*beta[1]}else{0})+
  (if(nclusters1){OptmizationInputs[3,3]*beta[1]}else{0})+
  V1 has inside a code for a function.
 
  I'm having problems with 2 things:
 
  1 - I need to take out from V1 all  that appears in the text, i tried
 a
  replace but did not work.
  Test=replace(Test,'  ', ' ')  , did not work.
 
  2 - Writing a function like this :
 
  nlog=function(par)
   {
   beta=par[1:n]
   Measure=Test[1]  # would this read the text?
   return(Measure)
   }
 
  So i need to use that code inside the function as above.
  Any suggestion on how you would do this?
 
  Kind Regards,
  Eliano
 
 
 
  --
  View this message in context:
 
 http://r.789695.n4.nabble.com/Help-with-a-function-and-text-tp4660523.html
  Sent from the R help mailing list archive at Nabble.com.
 
 
  __
  [hidden email] /user/SendEmail.jtp?type=nodenode=4660547i=1 mailing
  list
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  PLEASE do read the posting guide
  http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
  David Winsemius
  Alameda, CA, USA
 
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Re: [R] Help with a function and text

[Eliano Marques]

Hi, can I understand why this message was rejected ?
Thanks,
Eliano

Sent from my iPhone

On 6 Mar 2013, at 19:18, Eliano eliano.m.marq...@gmail.com wrote:

 Hi everyone,

 I am writing some code to generate a function. I am passing that code to a
 dataset which i'm importing in R, e.g.
 Test=read.table('C:/test.txt', header=F, sep='\t', na.strings='NA', dec='.',
 strip.white=TRUE)
 Test

 V1
 (if(nclusters0){OptmizationInputs[3,3]*beta[1]}else{0})+
 (if(nclusters1){OptmizationInputs[3,3]*beta[1]}else{0})+
 V1 has inside a code for a function.

 I'm having problems with 2 things:

 1 - I need to take out from V1 all  that appears in the text, i tried a
 replace but did not work.
 Test=replace(Test,'  ', ' ')  , did not work.

 2 - Writing a function like this :

 nlog=function(par)
{
beta=par[1:n]
Measure=Test[1]  # would this read the text?
return(Measure)
}

 So i need to use that code inside the function as above.
 Any suggestion on how you would do this?

 Kind Regards,
 Eliano



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-with-a-function-and-text-tp4660523.html
 Sent from the R help mailing list archive at Nabble.com.


__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Help with a function and text

[David Winsemius]

On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:

 Hi, can I understand why this message was rejected ?
 Thanks,
 Eliano

First hit on a Markmail search:

http://markmail.org/message/5xog3ayx4amprsdx?q=list:org%2Er-project%2Er-help+nabble+rejected

-- 
David.

 
 Sent from my iPhone
 
 On 6 Mar 2013, at 19:18, Eliano eliano.m.marq...@gmail.com wrote:
 
 Hi everyone,
 
 I am writing some code to generate a function. I am passing that code to a
 dataset which i'm importing in R, e.g.
 Test=read.table('C:/test.txt', header=F, sep='\t', na.strings='NA', dec='.',
 strip.white=TRUE)
 Test
 
 V1
 (if(nclusters0){OptmizationInputs[3,3]*beta[1]}else{0})+
 (if(nclusters1){OptmizationInputs[3,3]*beta[1]}else{0})+
 V1 has inside a code for a function.
 
 I'm having problems with 2 things:
 
 1 - I need to take out from V1 all  that appears in the text, i tried a
 replace but did not work.
 Test=replace(Test,'  ', ' ')  , did not work.
 
 2 - Writing a function like this :
 
 nlog=function(par)
   {
   beta=par[1:n]
   Measure=Test[1]  # would this read the text?
   return(Measure)
   }
 
 So i need to use that code inside the function as above.
 Any suggestion on how you would do this?
 
 Kind Regards,
 Eliano
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-with-a-function-and-text-tp4660523.html
 Sent from the R help mailing list archive at Nabble.com.
 
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

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Re: [R] Help with a function and text

[Eliano]

Thanks. Btw are you able to help with my issue? Thanks, Eliano

Sent from my iPhone

On 6 Mar 2013, at 23:41, David Winsemius [via R] 
ml-node+s789695n4660547...@n4.nabble.com wrote:


On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:

 Hi, can I understand why this message was rejected ?
 Thanks,
 Eliano

First hit on a Markmail search:

http://markmail.org/message/5xog3ayx4amprsdx?q=list:org%2Er-project%2Er-help+nabble+rejected

-- 
David.


 Sent from my iPhone

 On 6 Mar 2013, at 19:18, Eliano [hidden 
 email]/user/SendEmail.jtp?type=nodenode=4660547i=0
wrote:

 Hi everyone,

 I am writing some code to generate a function. I am passing that code to
a
 dataset which i'm importing in R, e.g.
 Test=read.table('C:/test.txt', header=F, sep='\t', na.strings='NA',
dec='.',
 strip.white=TRUE)
 Test

 V1
 (if(nclusters0){OptmizationInputs[3,3]*beta[1]}else{0})+
 (if(nclusters1){OptmizationInputs[3,3]*beta[1]}else{0})+
 V1 has inside a code for a function.

 I'm having problems with 2 things:

 1 - I need to take out from V1 all  that appears in the text, i tried a
 replace but did not work.
 Test=replace(Test,'  ', ' ')  , did not work.

 2 - Writing a function like this :

 nlog=function(par)
   {
   beta=par[1:n]
   Measure=Test[1]  # would this read the text?
   return(Measure)
   }

 So i need to use that code inside the function as above.
 Any suggestion on how you would do this?

 Kind Regards,
 Eliano



 --
 View this message in context:
http://r.789695.n4.nabble.com/Help-with-a-function-and-text-tp4660523.html
 Sent from the R help mailing list archive at Nabble.com.


 __
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 PLEASE do read the posting guide
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 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Help with a function and text

[David Winsemius]

On Mar 6, 2013, at 3:44 PM, Eliano wrote:

 Thanks. Btw are you able to help with my issue? Thanks, Eliano

I'm sorry,  I was too busy answering the question from 'Eliano' over on 
StackOverflow. I didn't have time to address this one.

(Please do note that cross-posting questions to Rhelp is contrary to advice in 
the Posting Guide.)

You might also do further searching in the Archives with the search terms: 
substitute text eval 
 and perhaps narrow it down further with contributors named: grothendeick, 
dunlap, ligges, venables

 
 Sent from my iPhone
 
 On 6 Mar 2013, at 23:41, David Winsemius [via R] 
 ml-node+s789695n4660547...@n4.nabble.com wrote:
 
 
 On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:
 
 Hi, can I understand why this message was rejected ?
 Thanks,
 Eliano
 
 First hit on a Markmail search:
 
 http://markmail.org/message/5xog3ayx4amprsdx?q=list:org%2Er-project%2Er-help+nabble+rejected
 
 -- 
 David.
 
 
 Sent from my iPhone
 
 On 6 Mar 2013, at 19:18, Eliano [hidden 
 email]/user/SendEmail.jtp?type=nodenode=4660547i=0
 wrote:
 
 Hi everyone,
 
 I am writing some code to generate a function. I am passing that code to
 a
 dataset which i'm importing in R, e.g.
 Test=read.table('C:/test.txt', header=F, sep='\t', na.strings='NA',
 dec='.',
 strip.white=TRUE)
 Test
 
 V1
 (if(nclusters0){OptmizationInputs[3,3]*beta[1]}else{0})+
 (if(nclusters1){OptmizationInputs[3,3]*beta[1]}else{0})+
 V1 has inside a code for a function.
 
 I'm having problems with 2 things:
 
 1 - I need to take out from V1 all  that appears in the text, i tried a
 replace but did not work.
 Test=replace(Test,'  ', ' ')  , did not work.
 
 2 - Writing a function like this :
 
 nlog=function(par)
  {
  beta=par[1:n]
  Measure=Test[1]  # would this read the text?
  return(Measure)
  }
 
 So i need to use that code inside the function as above.
 Any suggestion on how you would do this?
 
 Kind Regards,
 Eliano
 
 
 
 --
 View this message in context:
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 Alameda, CA, USA
 
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Re: [R] Help for a function

[Rui Barradas]

Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the 
loop it never exits.

And

res1[j] -(a*h)
res2 -sum( res1[j])

is equivalent to

res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:

On 12/05/2012 01:01 AM, anoumou wrote:

Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont 
understant the

problem.


Hi anoumou,
Your function provides almost no indication of what two of its five 
arguments are supposed to be. If, with a certain degree of optimistic 
inference, we suppose x to be a data frame organized as shown below 
the function. t, i and CONTAGIEUX must be the appropriately 
named columns of the data frame. This leaves two columns, Symptomes 
and Incubation. Say we flip a coin to decide which of these to 
assign to r, and with all of our degrees of freedom gone, we assume 
that the other is h. We are forced to the conclusion that a is a 
nuisance argument, added to throw us off the scent.


Peering within the function, we notice that the date format is wrong, 
braces are unmatched and that our data frame is alphabetically ordered 
by name of country to no purpose whatsoever. The best I can do here is 
to make the function potentially able to do something if you can work 
out what to do with it.


Lambda-function (x,date1,r,h,a) {
 ndate1 - as.Date(date1, %Y-%m-%d)
 t1 - as.numeric(ndate1)
 x[order(x$i),]
 xt -x[,t]
 xi -x[,i]
 CONTAGIEUX -x[,CONTAGIEUX]
 while ( t1  min(xt) ){
  for (i in 1:length(xi) ){
   for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] -(a*h)
res2 -sum( res1[j])
   }
  }
 lambda[i] - r*res2
 }
 x-data.frame(x,lambda)
 x
}

Jim

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PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html

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Re: [R] Help for a function

[teko maurice]

Thanks you all

 
Maurice TEKO

 
Biostatisticien,Doctorant.

Université de Liège
Département des Sciences et Gestion de l'environnement
185 avenue de Longwy
6700 Arlon (Belgique)
Mail :teko_maur...@yahoo.fr
:anoumou.tekoahate...@ulg.ac.be
 http://www.ulg.ac.be
  




 De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au 

Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function

Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the loop it 
never exits.
And

    res1[j] -(a*h)
    res2 -sum( res1[j])

is equivalent to

    res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:
 On 12/05/2012 01:01 AM, anoumou wrote:
 Hello all,
 I need a help.
 I am modeling a disease and a create a R function like that:
 ...
 But i do not get the results,i try by all means but i d'ont understant the
 problem.
 
 Hi anoumou,
 Your function provides almost no indication of what two of its five arguments 
 are supposed to be. If, with a certain degree of optimistic inference, we 
 suppose x to be a data frame organized as shown below the function. t, 
 i and CONTAGIEUX must be the appropriately named columns of the data 
 frame. This leaves two columns, Symptomes and Incubation. Say we flip a 
 coin to decide which of these to assign to r, and with all of our degrees 
 of freedom gone, we assume that the other is h. We are forced to the 
 conclusion that a is a nuisance argument, added to throw us off the scent.
 
 Peering within the function, we notice that the date format is wrong, braces 
 are unmatched and that our data frame is alphabetically ordered by name of 
 country to no purpose whatsoever. The best I can do here is to make the 
 function potentially able to do something if you can work out what to do with 
 it.
 
 Lambda-function (x,date1,r,h,a) {
  ndate1 - as.Date(date1, %Y-%m-%d)
  t1 - as.numeric(ndate1)
  x[order(x$i),]
  xt -x[,t]
  xi -x[,i]
  CONTAGIEUX -x[,CONTAGIEUX]
  while ( t1  min(xt) ){
   for (i in 1:length(xi) ){
    for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
     res1[j] -(a*h)
     res2 -sum( res1[j])
    }
   }
  lambda[i] - r*res2
  }
  x-data.frame(x,lambda)
  x
 }
 
 Jim
 
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[[alternative HTML version deleted]]

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Re: [R] Help for a function

[teko maurice]

Rui ,
I discover all this when running and rerunning my function one and thousand 
times and per line.
Now it's perfect, and work so well.
Thanks.






À : Rui Barradas ruipbarra...@sapo.pt; Jim Lemon j...@bitwrit.com.au
Cc : r-help@r-project.org r-help@r-project.org 
Envoyé le : Mercredi 5 décembre 2012 16h45
Objet : Re: [R] Help for a function


Thanks you all
 
Maurice TEKO

 
Biostatisticien,Doctorant.

Université de Liège
Département des Sciences et Gestion de l'environnement
185 avenue de Longwy
6700 Arlon (Belgique)

:anoumou.tekoahate...@ulg.ac.be
 http://www.ulg.ac.be
  




 De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au 

Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function

Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the loop it 
never
 exits.
And

    res1[j] -(a*h)
    res2 -sum( res1[j])

is equivalent to

    res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:
 On 12/05/2012 01:01 AM, anoumou wrote:
 Hello all,
 I need a help.
 I am modeling a disease and a create a R function like that:
 ...
 But i do not get the results,i try by all means but i d'ont understant the
 problem.
 
 Hi anoumou,
 Your function provides almost no indication of what two of its five arguments 
 are supposed to be. If, with a certain degree of optimistic inference, we 
 suppose x to be a data frame organized as shown below the function. t, 
 i and CONTAGIEUX must be the appropriately named columns of the data 
 frame. This leaves two columns,
 Symptomes and Incubation. Say we flip a coin to decide which of these to 
assign to r, and with all of our degrees of freedom gone, we assume that the 
other is h. We are forced to the conclusion that a is a nuisance argument, 
added to throw us off the scent.
 
 Peering within the function, we notice that the date format is wrong, braces 
 are unmatched and that our data frame is alphabetically ordered by name of 
 country to no purpose whatsoever. The best I can do here is to make the 
 function potentially able to do something if you can work out what to do with 
 it.
 
 Lambda-function (x,date1,r,h,a) {
  ndate1 - as.Date(date1, %Y-%m-%d)
  t1 - as.numeric(ndate1)
  x[order(x$i),]
  xt -x[,t]
  xi -x[,i]
  CONTAGIEUX -x[,CONTAGIEUX]
  while ( t1  min(xt) ){
   for (i in 1:length(xi)
 ){
    for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
     res1[j] -(a*h)
     res2 -sum( res1[j])
    }
   }
  lambda[i] - r*res2
  }
  x-data.frame(x,lambda)
  x
 }
 
 Jim
 
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[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

[R] Help for a function

[anoumou]

Hello all,
I need a help.
I am modeling a disease and a create a R function like that:

Lambda-function (x,date1,r,h,a){
  ndate1 - as.Date(date1, %d/%m/%Y)
  t1 - as.numeric(ndate1)
  x[order(x$i),]
  t -x[,t]
  i -x[,i]
  CONTAGIEUX -x[,CONTAGIEUX]
  while ( t1  min(t) ){
  for (i in 1:length(i) ){
{for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
  res1[j] -(a*h)
  res2 -sum( res1[j])
}
 }
  lambda[i] - r*res2
  }
  }
x-data.frame(x,lambda)
x
}

on such data :

DATEi   Symptomes   t   Incubation  CONTAGIEUX
1   2009-04-29   Canada 13  14363   13  13
2   2009-05-01   Israel 2   14365   2   2
3   2009-05-09  argentina   1   14373   1   1
5   2009-05-09  australia   1   14373   1   1
6   2009-05-10  australia   1   14374   2   2
7   2009-04-29  Austria 1   14363   1   1
8   2009-04-30  Austria 1   14364   2   2
9   2009-05-01  Austria 1   14365   2   3
10  2009-05-02  Austria 1   14366   2   4
11  2009-05-03  Austria 1   14367   2   5
17  2009-05-09  Austria 1   14373   2   7
18  2009-05-10  Austria 1   14374   2   7
19  2009-05-08  brasil  4   14372   4   4
20  2009-05-09  brazil  6   14373   6   6
21  2009-05-10  brazil  6   14374   12  12
22  2009-05-02  canada  51  14366   51  51
23  2009-05-03  canada  85  14367   136 136
24  2009-05-04  canada  101 14368   186 237
31  2009-04-27  Canada  6   14361   6   6
32  2009-04-28  Canada  6   14362   6   6
33  2009-04-30  Canada  19  14364   25  25
34  2009-05-01  Canada  34  14365   53  59
35  2009-05-01  China,HongKong, SAR 1   14365   1   1
36  2009-05-02  China,HongKong, SAR 1   14366   2   2
37  2009-05-03  China,HongKong, SAR 1   14367   2   3
38  2009-05-04  China,HongKong, SAR 1   14368   2   4
44  2009-05-10  China,HongKong, SAR 1   14374   2   7
45  2009-05-04  colombia1   14368   1   1
46  2009-05-05  colombia1   14369   2   2
47  2009-05-06  colombia

But i do not get the results,i try by all means but i d'ont understant the
problem.
Thanks for your help.



--
View this message in context: 
http://r.789695.n4.nabble.com/Help-for-a-function-tp4652054.html
Sent from the R help mailing list archive at Nabble.com.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help for a function

[Jeremy Miles]

What are you expecting?

What do you get?

What is the problem?

J

On 4 December 2012 06:01, anoumou teko_maur...@yahoo.fr wrote:
 Hello all,
 I need a help.
 I am modeling a disease and a create a R function like that:

 Lambda-function (x,date1,r,h,a){
   ndate1 - as.Date(date1, %d/%m/%Y)
   t1 - as.numeric(ndate1)
   x[order(x$i),]
   t -x[,t]
   i -x[,i]
   CONTAGIEUX -x[,CONTAGIEUX]
   while ( t1  min(t) ){
   for (i in 1:length(i) ){
 {for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
   res1[j] -(a*h)
   res2 -sum( res1[j])
 }
  }
   lambda[i] - r*res2
   }
   }
 x-data.frame(x,lambda)
 x
 }

 on such data :

 DATEi   Symptomes   t   Incubation  CONTAGIEUX
 1   2009-04-29   Canada 13  14363   13  13
 2   2009-05-01   Israel 2   14365   2   2
 3   2009-05-09  argentina   1   14373   1   1
 5   2009-05-09  australia   1   14373   1   1
 6   2009-05-10  australia   1   14374   2   2
 7   2009-04-29  Austria 1   14363   1   1
 8   2009-04-30  Austria 1   14364   2   2
 9   2009-05-01  Austria 1   14365   2   3
 10  2009-05-02  Austria 1   14366   2   4
 11  2009-05-03  Austria 1   14367   2   5
 17  2009-05-09  Austria 1   14373   2   7
 18  2009-05-10  Austria 1   14374   2   7
 19  2009-05-08  brasil  4   14372   4   4
 20  2009-05-09  brazil  6   14373   6   6
 21  2009-05-10  brazil  6   14374   12  12
 22  2009-05-02  canada  51  14366   51  51
 23  2009-05-03  canada  85  14367   136 136
 24  2009-05-04  canada  101 14368   186 237
 31  2009-04-27  Canada  6   14361   6   6
 32  2009-04-28  Canada  6   14362   6   6
 33  2009-04-30  Canada  19  14364   25  25
 34  2009-05-01  Canada  34  14365   53  59
 35  2009-05-01  China,HongKong, SAR 1   14365   1   1
 36  2009-05-02  China,HongKong, SAR 1   14366   2   2
 37  2009-05-03  China,HongKong, SAR 1   14367   2   3
 38  2009-05-04  China,HongKong, SAR 1   14368   2   4
 44  2009-05-10  China,HongKong, SAR 1   14374   2   7
 45  2009-05-04  colombia1   14368   1   1
 46  2009-05-05  colombia1   14369   2   2
 47  2009-05-06  colombia

 But i do not get the results,i try by all means but i d'ont understant the
 problem.
 Thanks for your help.



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-for-a-function-tp4652054.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help for a function

[Jim Lemon]

On 12/05/2012 01:01 AM, anoumou wrote:

Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont understant the
problem.


Hi anoumou,
Your function provides almost no indication of what two of its five 
arguments are supposed to be. If, with a certain degree of optimistic 
inference, we suppose x to be a data frame organized as shown below 
the function. t, i and CONTAGIEUX must be the appropriately named 
columns of the data frame. This leaves two columns, Symptomes and 
Incubation. Say we flip a coin to decide which of these to assign to 
r, and with all of our degrees of freedom gone, we assume that the 
other is h. We are forced to the conclusion that a is a nuisance 
argument, added to throw us off the scent.


Peering within the function, we notice that the date format is wrong, 
braces are unmatched and that our data frame is alphabetically ordered 
by name of country to no purpose whatsoever. The best I can do here is 
to make the function potentially able to do something if you can work 
out what to do with it.


Lambda-function (x,date1,r,h,a) {
 ndate1 - as.Date(date1, %Y-%m-%d)
 t1 - as.numeric(ndate1)
 x[order(x$i),]
 xt -x[,t]
 xi -x[,i]
 CONTAGIEUX -x[,CONTAGIEUX]
 while ( t1  min(xt) ){
  for (i in 1:length(xi) ){
   for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] -(a*h)
res2 -sum( res1[j])
   }
  }
 lambda[i] - r*res2
 }
 x-data.frame(x,lambda)
 x
}

Jim

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with predict function in glm

[Rui Barradas]

Hello,

Why mail a question just to me? Post to the list and the odds of getting 
more answers (and better) are bigger.
As for your question, the problem is in the call to glm, you don't need 
the prefix 'train$' in the formula, the argument 'data' solves that and 
when predicting R will look for the columns with names in the formula 
and is unable to find columns called train$Outcome and train$Weight in 
the new data.frame 'test'. Corrected:


mylogit - glm(Outcome ~ Weight, data=train, family = binomial(logit))
predictions - predict(mylogit, newdata = test, type= response)


Hope this helps,

Rui Barradas
Em 26-11-2012 01:42, somnath bandyopadhyay escreveu:


Hi,
I am trying some basic logistic regression analysis using glm. I just have one dependent 
variable (Outcome) which is binary in nature and one independent variable (Weight). I fit 
a model using a training data set (train) which has 85 observations and try to apply it 
on an independent dataset (test) which has 55 observations. When I apply the predict 
function on the fitted model for the new dataset, I get the following warning 
Warning message: 'newdata' had 55 rows but variable(s) found have 85 rows and 
the predict works on the training observations and not on the test observations.

Following is he session info, code and the training and test datasets I am 
using.

What am I doing wrong? Any help would be greatly appreciated.

Thanks,
S.


train - read.table(train_data.txt, header=T, row.names=1, sep=\t)
test- read.table(test_data.txt, header=T, row.names=1, sep=\t)
mylogit - glm(train$Outcome ~ train$Weight, data=train, family = 
binomial(logit))
predictions - predict(mylogit, newdata = test, type= response)

Warning message:
'newdata' had 55 rows but variable(s) found have 85 rows



sessionInfo()

R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-mingw32/x64 (64-bit)

locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 
LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.1252

attached base packages:
[1] stats graphics grDevices utils datasets methods base





train

Outcome Weight
AB256939_21 0 0.331
AB257076_21 0 0.308
AB257079_21 0 0.453
AB415508_21 0 0.303
AB700497_21 0 0.354
AB904508_21 0 0.336
AC048719_21 0 0.420
AC185939_21 0 0.249
AC185940_21 0 1.525
AC445840_21 0 0.261
E7490523_21 0 0.269
E7490524_21 0 0.213
E7659579_21 0 0.360
E7661528_21 0 0.271
E7781094_21 0 0.156
E7781095_21 0 0.221
E7781096_21 0 0.098
E7969081_21 0 0.430
E8117594_21 0 0.321
E8133295_21 0 0.166
E8161578_22 0 0.269
E8483037_21 0 0.162
E8559720_21 0 0.226
L1065550_18 0 0.396
L1065607_17 0 0.541
L1065944_24 0 0.131
L1066017_20 0 0.421
L1069261_12 0 0.357
L1069262_14 0 0.309
L1069263_27 0 0.283
L1069297_24 0 0.620
L1081528_21 0 0.561
L1084066_21 0 0.564
L1086090_21 0 0.649
L1104280_17 0 0.181
L362_22 0 0.199
L1118063_15 0 0.369
L1133550_21 0 0.302
L1144201_14 0 0.249
L1155023_7 0 0.257
L1158386_21 0 0.470
L1163051_4 0 0.446
...
...
...



test

Weight
AB256870_21 0.364
AB256873_21 0.329
AB415518_21 0.219
AB460669_21 0.481
AB609036_21 0.313
AB609038_21 0.196
AB700495_21 0.402
AB700498_21 0.343
AC112834_21 0.372
AC185937_21 0.270
AC269527_21 0.285
E7352023_21 0.358
E7661554_21 0.471
E7750502_21 0.437
E7845183_21 0.232
E7854155_21 0.474
E7854156_21 0.121
E7924877_21 0.312
E7969079_21 0.423
E8139256_21 0.329
E8161577_22 1.060
E8161580_21 0.157
E8364473_21 0.227
E8364474_21 0.069
L1065940_14 0.256
L1065946_10 0.184
L1066018_25 0.282
L1069260_15 1.094









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Re: [R] Help with predict function in glm

[genome1976]

That did it!



Thanks so much as always.





I emailed the question to you because I think you are an R expert based 
on all the suggestions, feedback and codes I have received from you in 
the past. 
Yes, I do look for answers in the open 
forum but when it comes to a question for which the answers out there 
are not very clear I prefer to ask somebody whom I can trust. I hope I 
don’t bother you with my questions.
Thanks once again.




Cheers!
-Som.

Date: Mon, 26 Nov 2012 03:33:43 -0800
From: ml-node+s789695n4650829...@n4.nabble.com
To: genome1...@hotmail.com
Subject: Re: Help with predict function in glm



Hello,


Why mail a question just to me? Post to the list and the odds of getting 

more answers (and better) are bigger.

As for your question, the problem is in the call to glm, you don't need 

the prefix 'train$' in the formula, the argument 'data' solves that and 

when predicting R will look for the columns with names in the formula 

and is unable to find columns called train$Outcome and train$Weight in 

the new data.frame 'test'. Corrected:


mylogit - glm(Outcome ~ Weight, data=train, family = binomial(logit))

predictions - predict(mylogit, newdata = test, type= response)



Hope this helps,


Rui Barradas

Em 26-11-2012 01:42, somnath bandyopadhyay escreveu:



 Hi,

 I am trying some basic logistic regression analysis using glm. I just have 
 one dependent variable (Outcome) which is binary in nature and one 
 independent variable (Weight). I fit a model using a training data set 
 (train) which has 85 observations and try to apply it on an independent 
 dataset (test) which has 55 observations. When I apply the predict function 
 on the fitted model for the new dataset, I get the following warning Warning 
 message: 'newdata' had 55 rows but variable(s) found have 85 rows and the 
 predict works on the training observations and not on the test observations.



 Following is he session info, code and the training and test datasets I am 
 using.



 What am I doing wrong? Any help would be greatly appreciated.



 Thanks,

 S.



 train - read.table(train_data.txt, header=T, row.names=1, sep=\t)

 test- read.table(test_data.txt, header=T, row.names=1, sep=\t)

 mylogit - glm(train$Outcome ~ train$Weight, data=train, family = 
 binomial(logit))

 predictions - predict(mylogit, newdata = test, type= response)

 Warning message:

 'newdata' had 55 rows but variable(s) found have 85 rows





 sessionInfo()

 R version 2.15.0 (2012-03-30)

 Platform: x86_64-pc-mingw32/x64 (64-bit)



 locale:

 [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 
 LC_MONETARY=English_United States.1252

 [4] LC_NUMERIC=C LC_TIME=English_United States.1252



 attached base packages:

 [1] stats graphics grDevices utils datasets methods base









 train

 Outcome Weight

 AB256939_21 0 0.331

 AB257076_21 0 0.308

 AB257079_21 0 0.453

 AB415508_21 0 0.303

 AB700497_21 0 0.354

 AB904508_21 0 0.336

 AC048719_21 0 0.420

 AC185939_21 0 0.249

 AC185940_21 0 1.525

 AC445840_21 0 0.261

 E7490523_21 0 0.269

 E7490524_21 0 0.213

 E7659579_21 0 0.360

 E7661528_21 0 0.271

 E7781094_21 0 0.156

 E7781095_21 0 0.221

 E7781096_21 0 0.098

 E7969081_21 0 0.430

 E8117594_21 0 0.321

 E8133295_21 0 0.166

 E8161578_22 0 0.269

 E8483037_21 0 0.162

 E8559720_21 0 0.226

 L1065550_18 0 0.396

 L1065607_17 0 0.541

 L1065944_24 0 0.131

 L1066017_20 0 0.421

 L1069261_12 0 0.357

 L1069262_14 0 0.309

 L1069263_27 0 0.283

 L1069297_24 0 0.620

 L1081528_21 0 0.561

 L1084066_21 0 0.564

 L1086090_21 0 0.649

 L1104280_17 0 0.181

 L362_22 0 0.199

 L1118063_15 0 0.369

 L1133550_21 0 0.302

 L1144201_14 0 0.249

 L1155023_7 0 0.257

 L1158386_21 0 0.470

 L1163051_4 0 0.446

 ...

 ...

 ...





 test

 Weight

 AB256870_21 0.364

 AB256873_21 0.329

 AB415518_21 0.219

 AB460669_21 0.481

 AB609036_21 0.313

 AB609038_21 0.196

 AB700495_21 0.402

 AB700498_21 0.343

 AC112834_21 0.372

 AC185937_21 0.270

 AC269527_21 0.285

 E7352023_21 0.358

 E7661554_21 0.471

 E7750502_21 0.437

 E7845183_21 0.232

 E7854155_21 0.474

 E7854156_21 0.121

 E7924877_21 0.312

 E7969079_21 0.423

 E8139256_21 0.329

 E8161577_22 1.060

 E8161580_21 0.157

 E8364473_21 0.227

 E8364474_21 0.069

 L1065940_14 0.256

 L1065946_10 0.184

 L1066018_25 0.282

 L1069260_15 1.094

 

 











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If you reply to this email, your message will be added to the 
discussion below:

[R] Help with glht function and binary data!

[Jessica McLachlan]

Hi, 

I carried out an experiment, using a repeated measures design, in which I
broadcast 6 playbacks to 15 nests. I am trying to run a posthoc pairwise
comparison on a binomial GLMM with missing data points to determine which
playback treatments differ. 

The response variable (Response) is binary, there is one categorical factor
(Treatment) and Nest is included as a random factor. I have run a binomial
GLMM using the lme4 package and found the effect of Treatment to be highly
significant. I followed this up using the glht function in the multcomp
package to look at the pairwise comparisons. 

However, it is giving me strange results for some of the comparisons,
producing p-values of 1 for comparisons between what graphically (plot of
mean and std error/boxplots) appear to be the most different treatments. 
For my other binomial GLMM, where I look at proportion rather than binary
data, the glht function yields the expected results without problems. 

Can anyone help? Is the glht function not appropriate for binary data? Or
have I got something wrong?

Thank you!

Jessica


My data and code are below.

# My data:
Nest-c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,8,9,9,9,9,9,9,10,10,10,10,10,10,11,11,11,11,11,11,12,12,12,12,12,12,13,13,13,13,13,14,14,14,14,15,15,15,15,15)
Treatment-factor(c(CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CRS,CL,CS,SL,SS,CRL,CL,CS,SL,SS,CRL,CS,SL,SS,CRL,CRS,CS,SL,SS))
Response-c(0,0,1,1,1,1,0,0,1,1,1,1,0,0,1,1,1,1,0,1,1,1,1,1,0,0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,0,1,1,1,0,0,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,0,1,1,1,0,0,1,1,1)
binomialGLMM-data.frame(Nest,Treatment,Response)


# Binomial GLMM

library(lme4)
lmm1-lmer(Response~Treatment+(1|factor(Nest)),data=binomialGLMM,family=binomial)
lmm2-lmer(Response~1+(1|factor(Nest)),data=binomialGLMM,family=binomial)
anova(lmm1,lmm2)
Data: binomialGLMM
Models:
lmm2: Response ~ 1 + (1 | factor(Nest))
lmm1: Response ~ Treatment + (1 | factor(Nest))
 Df AIC BIC  logLik  Chisq Chi Df Pr(Chisq)
lmm2  2 109.405 114.314 -52.703 
lmm1  7  57.013  74.193 -21.506 62.392  5  3.889e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


# Post-hoc pairwise comparison 

library(multcomp)
posthoc-glht(lmm1, linfct = mcp(Treatment = Tukey))
summary(posthoc)

Simultaneous Tests for General Linear Hypotheses

Multiple Comparisons of Means: Tukey Contrasts


Fit: glmer(formula = Response ~ Treatment + (1 | factor(Nest)), data =
binomialGLMM, 
family = binomial)

Linear Hypotheses:
 Estimate Std. Error z value Pr(|z|)   
!!! *CRL - CL == 0  -2.088e+01  4.581e+03  -0.005  1.0*   
!!! *CRS - CL == 0  -2.247e+01  4.581e+03  -0.005  1.0*   
CS - CL == 0   -1.652e+01  4.581e+03  -0.004  1.0   
SL - CL == 09.148e-07  6.276e+03   0.000  1.0   
SS - CL == 0   -1.735e+01  4.581e+03  -0.004  1.0   
CRS - CRL == 0 -1.595e+00  1.348e+00  -1.183  0.79608   
CS - CRL == 0   4.359e+00  1.359e+00   3.206  0.01107 * 
!!! *SL - CRL == 0   2.088e+01  4.290e+03   0.005  1.0 *  
SS - CRL == 0   3.531e+00  1.072e+00   3.294  0.00790 **
CS - CRS == 0   5.953e+00  1.700e+00   3.502  0.00382 **
!!! *SL - CRS == 0   2.247e+01  4.290e+03   0.005  1.0*   
SS - CRS == 0   5.125e+00  1.480e+00   3.464  0.00421 **
SL - CS == 01.652e+01  4.290e+03   0.004  1.0   
SS - CS == 0   -8.279e-01  1.465e+00  -0.565  0.98998   
SS - SL == 0   -1.735e+01  4.290e+03  -0.004  1.0   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
(Adjusted p values reported -- single-step method)




!!! on the left indicates unexpected p-values.

 



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Re: [R] help using optim function

[Ben Bolker]

tmuman mumantariq at gmail.com writes:

  Hi, am very new to R and I've written an optim function, but can't
 get it to work
 
 least.squares.fitter-function(start.params,gr,
  low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
  {
 result-optim(par=start.params,
  method=c('Nelder-Mead'),
   fn=least.squares.fit,
lower=low.constraints,
   upper=high.constraints,data=data,scale=scale,ploton=ploton)
return(result)  
}

 least.squares.fitter(c(2,2),c(0,0),c(Inf,Inf),ricker.one.step,dat,'linear')
 
 where least.squares.fit is the function returning the difference between
 model data points and predicted points (residuals, residuals^2 and SSR),
 which i am trying to minimise
 
 and ricker.one.step is the function that generated the predictions
 
 I've been trying to figure out whats wrong for hours but can't.
 

  You haven't given us much to work with here -- have you
read the posting guide (URL at the bottom of every message)?
It's not reproducible ( http://tinyurl.com/reproducible-000 ),
and you haven't told us what's going wrong.  Are you getting
error messages?  What are they?  Warnings?  No errors but
you have reason to believe the answer is wrong?

A few comments:
*   'scale' and 'ploton' are not arguments to optim(),
so they'd better be accepted by your least.squares.fit() function
* the Nelder-Mead algorithm doesn't accept constraints (lower/upper)
  (see the nloptr package for a Nelder-Mead implementation that does)
* except for the constraints, you might be better off with ?nls
if all you want to do is least-squares fitting; nlsLM in the
minpack.lm package does least-squares fitting with constraints

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and provide commented, minimal, self-contained, reproducible code.

[R] help using optim function

[tmuman]

Hi, am very new to R and I've written an optim function, but can't get it to
work

least.squares.fitter-function(start.params,gr,low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
 {
  
result-optim(par=start.params,method=c('Nelder-Mead'),fn=least.squares.fit,lower=low.constraints,upper=high.constraints,data=data,scale=scale,ploton=ploton)
   return(result)   
 
 }
least.squares.fitter(c(2,2),c(0,0),c(Inf,Inf),ricker.one.step,dat,'linear')

where least.squares.fit is the function returning the difference between
model data points and predicted points (residuals, residuals^2 and SSR),
which i am trying to minimise

and ricker.one.step is the function that generated the predictions

I've been trying to figure out whats wrong for hours but can't.

Many thanks,

Tariq



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[R] help on plotfit function in package nlstools

[Yi Li]

I encountered this problem when I was learning nonlinear regression with R :
there is this function plotfit in package nlstools,which displays
a superimposed plot of the dependent variable versus one the
independent variables together with the fitted model.,but when I
execute the example given in the help page, it keeps saying that :

 plotfit(nls1, smooth = TRUE)
Error in form[[3]] : object of type 'symbol' is not subsettable

can anybody tell me what this means and how to fix it ?

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Re: [R] help with rle function on paired data

[raishilpa]

 hello,
I have a question regarding the function WebCorpus..When I am using this
function its showing error: couldnt find the function...I have downloaded
the tm package...Can you help me in this regard? Also I want to know how we
can data/information about a product or topic from google,facebook etc
using R...I have find a command

corpus - WebCorpus(GoogleFinanceSource(AAPL))

but this is not working and showing error couldnt find the function
GoogleFinanceSource...

please help me ..I will be thankful to you

thanking you in anticipation




Shilpa Rai
MSc.(2011-2013)
Applied Statistics and Informatics
Indian Institute of Technology,Bombay


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Re: [R] help with rle function on paired data

[William Dunlap]

   ldiff - function(x)c(TRUE, x[-1]!=x[-length(x)]) # logical diff function
   d[ldiff(d$DataValue), ]
ElpsdTime DataValue
  1 0 1
  490 2
  6   150 3
  7   180 2
  8   210 3

ldiff does the first half of what rle does.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Steve E.
 Sent: Friday, June 08, 2012 2:33 PM
 To: r-help@r-project.org
 Subject: [R] help with rle function on paired data
 
 Dear R Community - I hope you might be able to provide some guidance
 regarding the use of the rle function. I have a set of time-series data
 where a measured value is recorded every 30 seconds after the start of an
 experiment. Many of the measured values repeat and I am interested only in
 the values when there is a change. If I turn the measured values into a
 vector, the rle function works perfectly for this but I need also the
 corresponding time of the value and I am not sure how to use rle on paired
 data. Below is a brief example to help explain the problem. I thank you in
 advance for any assistance you might be able to provide. Regards, Steve
 
 Original dataset:
 
 ElpsdTime, DataValue
 0, 1
 30, 1
 60, 1
 90, 2
 120, 2
 150, 3
 180, 2
 210, 3
 240, 3
 .
 .
 
 Desired dataset:
 
 ElpTime   DataValue
 0, 1
 90, 2
 150, 3
 180, 2
 210, 3
 .
 .
 
 
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 paired-data-tp4632856.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] Help with permutation function from Turner et al. 2010 (Ecology)

[Uwe Ligges]

On 09.06.2012 03:57, Ashley Stasko wrote:

Hello,


I'm using R code that includes a residual permutation that was written as a 
supplement to the paper:


Turner et al. 2010. A general hypothesis-testing framework for stable isotopes 
ratios in ecological studies. Ecology 91:2227-2233.
The supplemental code is available at: 
http://www.esapubs.org/archive/ecol/E091/157/suppl-1.htm


When I execute the function, no warnings are given (seems fine). But the resulting output 
matrix is simply a matrix of repeating rows (i.e., it appears as if the function is 
selecting the same random sample every time). The function is as follows 
(please keep in mind that many of the internal functions were written by Turner et al. as 
well and may not be familiar):


line.1-c(0,traj.result(traj.size,traj.dir,traj.shape))


permute-function(y,x2,g,o,p) {
   p.table-line.1
   yhat-predict(x2)
   res-resid(x2)
   line-nrow(y)

   for(i in 1:p){
 line.rand-sample(line,replace=FALSE)
 res.temp-cbind(line.rand,res)
 z-(order(line.rand))
 res.temp2-as.matrix(res.temp[z,])
 res.p-res.temp2[,-1]  # Rows of residuals are now randomized
 y.rand-yhat+res.p

 lm.new-lm(y.rand~o*g,x=T,model=T)
 ls-data.frame(expand.grid(o=levels(o),g=levels(g)))
 ls[]-lapply(ls, factor)
 means.r-predict(lm.new,ls)

 traj.r-arrayspecs(means.r,g,o)
 traj.size.r-trajsize(traj.r)
 traj.dir.r-trajorient(traj.r,ang.cor=T)
 traj.shape.r-trajshape(traj.r)
 result-traj.result(traj.size.r,traj.dir.r,traj.shape.r)
 p.table-rbind(p.table,c(i,result))
   }
   p.table
}



But when I try to execute the function the rows in the output matrix repeat the 
same numbers:



test3-permute(Y,lm.2f.red,group,obs,99)
test3

 [,1] [,2]  [,3]  [,4]
p.table0 3.612322 0.9910104 0.3998703
  1 3.612322 0.9910104 0.3998703
  2 3.612322 0.9910104 0.3998703
  3 3.612322 0.9910104 0.3998703
  4 3.612322 0.9910104 0.3998703
  5 3.612322 0.9910104 0.3998703
  ..


Can anyone see where there may be a problem with the permutation code? I'm 
relatively new to functions, and especially to writing permutation functions, 
so I'm having difficulty finding the problem.




If the results are suspicious (we cannot reproduce easily since you have 
not provided reproducible code as the posting guide asks you), why not 
ask the authors of the code? They hopefully know better what they aimed 
to get than readers of this list.


Best,
Uwe Ligges




Thanks,


Ashley






Ashley Stasko
M.Sc. Candidate
Cooperative Freshwater Ecology Unit
Living With Lakes Centre
Laurentian University
Sudbury, ON
Tel: 705-675-1151 ext. 2209
Email: ax_sta...@laurentian.ca

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Re: [R] Help with permutation function from Turner et al. 2010 (Ecology)

[mlcollyer]

Dear Ashley,

I am one of the co-authors for these scripts.  I just downloaded the example
files from the ESA archives and ran in naively on some data and it worked
fine.  My guess is that the problem lies in a portion of the script that you
did not provide.  If I had to be more certain with my guess, I bet your 'y'
data are the predicted values from a linear model instead of the raw data. 
Therefore, when the permutation procedure tries to shuffle residuals, there
are no residuals to shuffle.

If this is not the case, please email me and I can look at the rest of your
adapted script.  My new email address is michael (dot) collyer (at) wku
(dot) edu.  

By the way, one regret that we had after publishing these R scripts in the
ESA archives was to not provide a link for updates.  ESA is going to allow
us to make an amendment.  Look for this change in the near future.  We are
in the process of streamlining the original functions a bit and plan to make
these updates available.

Sincerely,
Michael Collyer
Western Kentucky University, Department of Biology




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[R] help with rle function on paired data

[Steve E.]

Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repeat and I am interested only in
the values when there is a change. If I turn the measured values into a
vector, the rle function works perfectly for this but I need also the
corresponding time of the value and I am not sure how to use rle on paired
data. Below is a brief example to help explain the problem. I thank you in
advance for any assistance you might be able to provide. Regards, Steve

Original dataset:

ElpsdTime, DataValue
0, 1
30, 1
60, 1
90, 2
120, 2
150, 3
180, 2
210, 3
240, 3
.
.

Desired dataset:

ElpTime DataValue
0, 1
90, 2
150, 3
180, 2
210, 3
.
.


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Re: [R] help with rle function on paired data

[Peter Ehlers]

On 2012-06-08 14:33, Steve E. wrote:

Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repeat and I am interested only in
the values when there is a change. If I turn the measured values into a
vector, the rle function works perfectly for this but I need also the
corresponding time of the value and I am not sure how to use rle on paired
data. Below is a brief example to help explain the problem. I thank you in
advance for any assistance you might be able to provide. Regards, Steve

Original dataset:

ElpsdTime, DataValue
0, 1
30, 1
60, 1
90, 2
120, 2
150, 3
180, 2
210, 3
240, 3
.
.

Desired dataset:

ElpTime DataValue
0, 1
90, 2
150, 3
180, 2
210, 3
.
.


Let's call your dataframe 'dat'. Here's a pretty brute-force way:

  z - rle(dat[, 2])
  zV - z$values
  zL - z$lengths
  idx - c(0, head(cumsum(zL), -1)) + 1
  newdat - data.frame(time = dat[idx, 1], value = zV)
  newdat

Peter Ehlers

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[R] Help with permutation function from Turner et al. 2010 (Ecology)

[Ashley Stasko]

Hello,


I'm using R code that includes a residual permutation that was written as a 
supplement to the paper:


Turner et al. 2010. A general hypothesis-testing framework for stable isotopes 
ratios in ecological studies. Ecology 91:2227-2233.
The supplemental code is available at: 
http://www.esapubs.org/archive/ecol/E091/157/suppl-1.htm


When I execute the function, no warnings are given (seems fine). But the 
resulting output matrix is simply a matrix of repeating rows (i.e., it appears 
as if the function is selecting the same random sample every time). The 
function is as follows (please keep in mind that many of the internal functions 
were written by Turner et al. as well and may not be familiar):


line.1-c(0,traj.result(traj.size,traj.dir,traj.shape))


permute-function(y,x2,g,o,p) { 
  p.table-line.1
  yhat-predict(x2)
  res-resid(x2)
  line-nrow(y)
  
  for(i in 1:p){
line.rand-sample(line,replace=FALSE)
res.temp-cbind(line.rand,res)
z-(order(line.rand))
res.temp2-as.matrix(res.temp[z,])
res.p-res.temp2[,-1]  # Rows of residuals are now randomized
y.rand-yhat+res.p

lm.new-lm(y.rand~o*g,x=T,model=T) 
ls-data.frame(expand.grid(o=levels(o),g=levels(g))) 
ls[]-lapply(ls, factor)
means.r-predict(lm.new,ls)

traj.r-arrayspecs(means.r,g,o)
traj.size.r-trajsize(traj.r)
traj.dir.r-trajorient(traj.r,ang.cor=T)
traj.shape.r-trajshape(traj.r)
result-traj.result(traj.size.r,traj.dir.r,traj.shape.r)
p.table-rbind(p.table,c(i,result))
  }
  p.table
}



But when I try to execute the function the rows in the output matrix repeat the 
same numbers:


 test3-permute(Y,lm.2f.red,group,obs,99)
 test3
[,1] [,2]  [,3]  [,4]
p.table0 3.612322 0.9910104 0.3998703
 1 3.612322 0.9910104 0.3998703
 2 3.612322 0.9910104 0.3998703
 3 3.612322 0.9910104 0.3998703
 4 3.612322 0.9910104 0.3998703
 5 3.612322 0.9910104 0.3998703 
 ..


Can anyone see where there may be a problem with the permutation code? I'm 
relatively new to functions, and especially to writing permutation functions, 
so I'm having difficulty finding the problem.


Thanks,


Ashley






Ashley Stasko
M.Sc. Candidate
Cooperative Freshwater Ecology Unit
Living With Lakes Centre
Laurentian University
Sudbury, ON
Tel: 705-675-1151 ext. 2209
Email: ax_sta...@laurentian.ca

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[R] Help for numericDeriv function

[Cengiz Zopluoğlu]

Hi,

I am stuck on something for a couple days, I am almost about to give up.
This looks simple, but I can't figure out. I hope I can get some help here.

I am trying to do some symbolic and numerical derivations. Let me explain
the problem. Let's say, I have a matrix as follows:

 load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)

 load
  [,1] [,2]
[1,]30
[2,]14
[3,]13

I will create objects for each element of the matrix, and assign the matrix
elements to these objects. These objects will be also an element of another
matrix.

l - matrix(nrow=nrow(load),ncol=ncol(load))
for(i in 1:nrow(load)) {
for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}

 l
 [,1]  [,2]
[1,] l11 l12
[2,] l21 l22
[3,] l31 l32

 l11
[1] 3
 l12
[1] 0
 l21
[1] 1
 l22
[1] 4
 l31
[1] 1
 l32
[1] 3

Let's say, I need to take the derivative of  (l11*l21+l12*l22) with respect
to l11,l12,l21,l22,l31,l32.

 numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))
[1] 3
attr(,gradient)
 [,1] [,2] [,3] [,4] [,5] [,6]
[1,]143000

Everything is fine up to this point. I need what I have. But, I don't want
to manipulate the elements in quote() manually within the numericDeriv
function. For instance, I want to call these elements from the matrix l
as below:

 a - paste(
+ paste(l[1,1],l[2,1],sep=*),
+ paste(l[1,2],l[1,2],sep=*),
+ sep=+
+ )
 a
[1] l11*l21+l12*l12

Now, a is a character. Do you have any idea about how to use the
character value l11*l21+l12*l12 assigned to object a within the
numericDeriv function. So, I can get the same result as above without
manipulating the expression in quote() manually.

Thanks




-- 

Cengiz Zopluoglu

[[alternative HTML version deleted]]

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Re: [R] Help for numericDeriv function

[Cengiz Zopluoğlu]

I missed a couple line of codes in the previous e-mail. Here is whole code
again:

load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)

l - matrix(nrow=nrow(load),ncol=ncol(load))
for(i in 1:nrow(load)) {
for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}

for(i in 1:nrow(load)){
for(j in 1:ncol(load)){ assign(paste(l,i,j,sep=),load[i,j]) }}

numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))

a - paste(
paste(l[1,1],l[2,1],sep=*),
paste(l[1,2],l[1,2],sep=*),
sep=+
)

numericDeriv(???a???, c(t(l),recursive=TRUE))

Thanks

On Fri, May 18, 2012 at 1:25 PM, Cengiz Zopluoğlu cen@gmail.com wrote:

 Hi,

 I am stuck on something for a couple days, I am almost about to give up.
 This looks simple, but I can't figure out. I hope I can get some help here.

 I am trying to do some symbolic and numerical derivations. Let me explain
 the problem. Let's say, I have a matrix as follows:

  load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)
 
  load
   [,1] [,2]
 [1,]30
 [2,]14
 [3,]13

 I will create objects for each element of the matrix, and assign the
 matrix elements to these objects. These objects will be also an element of
 another matrix.

 l - matrix(nrow=nrow(load),ncol=ncol(load))
 for(i in 1:nrow(load)) {
 for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}

  l
  [,1]  [,2]
 [1,] l11 l12
 [2,] l21 l22
 [3,] l31 l32

  l11
 [1] 3
  l12
 [1] 0
  l21
 [1] 1
  l22
 [1] 4
  l31
 [1] 1
  l32
 [1] 3

 Let's say, I need to take the derivative of  (l11*l21+l12*l22) with
 respect to l11,l12,l21,l22,l31,l32.

  numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))
 [1] 3
 attr(,gradient)
  [,1] [,2] [,3] [,4] [,5] [,6]
 [1,]143000

 Everything is fine up to this point. I need what I have. But, I don't want
 to manipulate the elements in quote() manually within the numericDeriv
 function. For instance, I want to call these elements from the matrix l
 as below:

  a - paste(
 + paste(l[1,1],l[2,1],sep=*),
 + paste(l[1,2],l[1,2],sep=*),
 + sep=+
 + )
  a
 [1] l11*l21+l12*l12

 Now, a is a character. Do you have any idea about how to use the
 character value l11*l21+l12*l12 assigned to object a within the
 numericDeriv function. So, I can get the same result as above without
 manipulating the expression in quote() manually.

 Thanks




 --

 Cengiz Zopluoglu







-- 

Cengiz Zopluoglu

[[alternative HTML version deleted]]

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Re: [R] Help for numericDeriv function

[William Dunlap]

You could use parse(text=a)[[1]]:
numericDeriv(parse(text=a)[[1]], c(t(l),recursive=TRUE))
  [1] 3
  attr(,gradient)
   [,1] [,2] [,3] [,4] [,5] [,6]
  [1,]103000

or construct your expression directly:
   lnames - array(lapply(l, as.name), dim=dim(l))
   aa - call(+, call(*, lnames[[1,1]], lnames[[2,1]]), call(*, 
lnames[[1,2]], lnames[[1,2]]))
   numericDeriv(aa, c(t(l),recursive=TRUE))
  [1] 3
  attr(,gradient)
   [,1] [,2] [,3] [,4] [,5] [,6]
  [1,]103000

The results are identical.  The latter method is a bit more work to set up but
works no matter what the names in the expression look like.  The former
fails if the names contain spaces or other things that are not allowed in
R names.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Cengiz Zopluoglu
 Sent: Friday, May 18, 2012 11:33 AM
 To: r-help@r-project.org
 Subject: Re: [R] Help for numericDeriv function
 
 I missed a couple line of codes in the previous e-mail. Here is whole code
 again:
 
 load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)
 
 l - matrix(nrow=nrow(load),ncol=ncol(load))
 for(i in 1:nrow(load)) {
 for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}
 
 for(i in 1:nrow(load)){
 for(j in 1:ncol(load)){ assign(paste(l,i,j,sep=),load[i,j]) }}
 
 numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))
 
 a - paste(
 paste(l[1,1],l[2,1],sep=*),
 paste(l[1,2],l[1,2],sep=*),
 sep=+
 )
 
 numericDeriv(???a???, c(t(l),recursive=TRUE))
 
 Thanks
 
 On Fri, May 18, 2012 at 1:25 PM, Cengiz ZopluoÄŸlu cen@gmail.com wrote:
 
  Hi,
 
  I am stuck on something for a couple days, I am almost about to give up.
  This looks simple, but I can't figure out. I hope I can get some help here.
 
  I am trying to do some symbolic and numerical derivations. Let me explain
  the problem. Let's say, I have a matrix as follows:
 
   load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)
  
   load
[,1] [,2]
  [1,]30
  [2,]14
  [3,]13
 
  I will create objects for each element of the matrix, and assign the
  matrix elements to these objects. These objects will be also an element of
  another matrix.
 
  l - matrix(nrow=nrow(load),ncol=ncol(load))
  for(i in 1:nrow(load)) {
  for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}
 
   l
   [,1]  [,2]
  [1,] l11 l12
  [2,] l21 l22
  [3,] l31 l32
 
   l11
  [1] 3
   l12
  [1] 0
   l21
  [1] 1
   l22
  [1] 4
   l31
  [1] 1
   l32
  [1] 3
 
  Let's say, I need to take the derivative of  (l11*l21+l12*l22) with
  respect to l11,l12,l21,l22,l31,l32.
 
   numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))
  [1] 3
  attr(,gradient)
   [,1] [,2] [,3] [,4] [,5] [,6]
  [1,]143000
 
  Everything is fine up to this point. I need what I have. But, I don't want
  to manipulate the elements in quote() manually within the numericDeriv
  function. For instance, I want to call these elements from the matrix l
  as below:
 
   a - paste(
  + paste(l[1,1],l[2,1],sep=*),
  + paste(l[1,2],l[1,2],sep=*),
  + sep=+
  + )
   a
  [1] l11*l21+l12*l12
 
  Now, a is a character. Do you have any idea about how to use the
  character value l11*l21+l12*l12 assigned to object a within the
  numericDeriv function. So, I can get the same result as above without
  manipulating the expression in quote() manually.
 
  Thanks
 
 
 
 
  --
 
  Cengiz Zopluoglu
 
 
 
 
 
 
 
 --
 
 Cengiz Zopluoglu
 
   [[alternative HTML version deleted]]

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Re: [R] Help with V function in igraph

[Gábor Csárdi]

Something weird must be going on in your s641_social object. Can you
just simply check that the vertex names look OK with
'V(s641_social)$name'?

If they look good, then can you send me the s641_social object in
private? (Or part of it, assuming a part is enough to reproduce the
problem.)

Best,
Gabor

On Sat, May 12, 2012 at 8:07 PM, bmccowan bjmcco...@ucdavis.edu wrote:
 I am using the code below to output some network measures:

 central_social - data.frame(V(s641_social)$name, indegree_social,
 outdegree_social, incloseness_social, outcloseness_social,
 betweenness_social, eigen_social)

 and I get the following error:


 Error in Re(z) : non-numeric argument to function

 I know this has to do with V but I cannot figure out what is wrong.
 s641-social is a graph object and the vertices do have a name attribute.

 What can I do to fix?

 Thanks

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-with-V-function-in-igraph-tp4629767.html
 Sent from the R help mailing list archive at Nabble.com.

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Gabor Csardi csa...@rmki.kfki.hu     MTA KFKI RMKI

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Re: [R] Help with V function in igraph

[bmccowan]

Thank for your response. It is oddly working now.

Thanks again,

Brenda



On 5/14/12 10:36 AM, Gábor Csárdi-2 [via R]
ml-node+s789695n4629973...@n4.nabble.com wrote:

 Something weird must be going on in your s641_social object. Can you just
 simply check that the vertex names look OK with 'V(s641_social)$name'? If they
 look good, then can you send me the s641_social object in private? (Or part of
 it, assuming a part is enough to reproduce the problem.) Best, Gabor On Sat,
 May 12, 2012 at 8:07 PM, bmccowan [hidden email]
 /user/SendEmail.jtp?type=nodenode=4629973i=0  wrote:
  I am using the code below to output some network measures:  
 central_social - data.frame(V(s641_social)$name, indegree_social, 
 outdegree_social, incloseness_social, outcloseness_social, 
 betweenness_social, eigen_social)   and I get the following error:   
 Error in Re(z) : non-numeric argument to function   I know this has to do
 with V but I cannot figure out what is wrong.  s641-social is a graph object
 and the vertices do have a name attribute.   What can I do to fix?  
 Thanks   --  View this message in context:
 http://r.789695.n4.nabble.com/Help-with-V-function-in-igraph-tp4629767.html
 Sent from the R help mailing list archive at Nabble.com.  
 __  [hidden email]
 /user/SendEmail.jtp?type=nodenode=4629973i=1  mailing list 
 https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting
 guide http://www.R-project.org/posting-guide.html and provide commented,
 minimal, self-contained, reproducible code.
 -- Gabor Csardi [hidden email]
 /user/SendEmail.jtp?type=nodenode=4629973i=2      MTA KFKI RMKI
 __ [hidden email]
 /user/SendEmail.jtp?type=nodenode=4629973i=3  mailing list
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 http://www.R-project.org/posting-guide.htmland provide commented, minimal,
 self-contained, reproducible code.
 
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[R] Help with V function in igraph

[bmccowan]

I am using the code below to output some network measures:

central_social - data.frame(V(s641_social)$name, indegree_social,
outdegree_social, incloseness_social, outcloseness_social,
betweenness_social, eigen_social)

and I get the following error:


Error in Re(z) : non-numeric argument to function

I know this has to do with V but I cannot figure out what is wrong.
s641-social is a graph object and the vertices do have a name attribute.

What can I do to fix?

Thanks

--
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http://r.789695.n4.nabble.com/Help-with-V-function-in-igraph-tp4629767.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Help with gsub function or a similar function

[Sarah Goslee]

You don't provide a reproducible example, or even str(), but I'd guess you need 
to match ^15 instead of just 15.

Sarah

On Apr 5, 2012, at 10:38 PM, ieatnapalm era...@tulane.edu wrote:

 Hey, sorry if this has been addressed before, but I'm really new to R and
 having trouble with the gsub function. I need a way to make this function
 exclude certain values from being substituted:
 ie my data looks something like (15:.0234,10:.0157) and I'm trying to
 replace the leading 15 with something else - but of course it replaces the
 second 15 with something else too. If there's a way to exclude anything with
 a leading decimal point or something like that, that would do the trick.
 Thanks yall.
 
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[R] Help with gsub function or a similar function

[ieatnapalm]

Hey, sorry if this has been addressed before, but I'm really new to R and
having trouble with the gsub function. I need a way to make this function
exclude certain values from being substituted:
ie my data looks something like (15:.0234,10:.0157) and I'm trying to
replace the leading 15 with something else - but of course it replaces the
second 15 with something else too. If there's a way to exclude anything with
a leading decimal point or something like that, that would do the trick.
Thanks yall.

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View this message in context: 
http://r.789695.n4.nabble.com/Help-with-gsub-function-or-a-similar-function-tp4536584p4536584.html
Sent from the R help mailing list archive at Nabble.com.

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help with gsub function or a similar function

[David Winsemius]

On Apr 5, 2012, at 10:38 PM, ieatnapalm wrote:

Hey, sorry if this has been addressed before, but I'm really new to  
R and
having trouble with the gsub function. I need a way to make this  
function

exclude certain values from being substituted:
ie my data looks something like (15:.0234,10:.0157) and I'm trying to
replace the leading 15 with something else - but of course it  
replaces the
second 15 with something else too. If there's a way to exclude  
anything with
a leading decimal point or something like that, that would do the  
trick.

Thanks yall.



Cross-posting to SO and rhelp in quick succession is deprecated. If  
you hadn't gotten an answer on SO after a day or so, then fine, go a  
head cross-post but in point of fact you have already gotten three  
answer on SO so you are just wasting rhelp brain-bandwidth.



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David Winsemius, MD
West Hartford, CT

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