date:20130118


Hello,

Maybe instead of a loop, you could try

lapply(frames, function(y) plot(y$hour1))

Hope this helps,

Rui Barradas

Em 18-01-2013 09:16, condor escreveu:

So by hand the command would be

par(mfrow=c(1,2))
plot(frames$'1'hour1)
plot(frames$'2'hour1)

But in my case there are far more than 2 days, so I want to use a loop.
Suppose I have 10 plots
par(mfrow=c(2,5))
for(i in 1:10){
plot( /what should be put here??/)
}



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[R] scaling of nonbinROC penalties

2013-01-18 Thread Jonathan Williams

Dear R Helpers

I am having difficulty understanding how to use the penalty matrix for the 
nomROC function in package 'nonbinROC'.

The documentation says that the values of the penalty matrix code the 
penalty function L[i,j] in which 0 = L[i,j] = 1 for
j  i. It gives an example that if we have an ordered response with 4 
categories, then we might wish to penalise larger misclassifications more - so 
there is (for example) 0 penalty for correct classifications, 0.25 penalty for 
misclassifying by one category, 0.5 penalty for misclassifying by two 
categories and 1.0 penalty for misclassifying by 3 categories. I wanted to use 
this sort of penalty - but with equal distances between the 4 categories (0, 
1/3, 2/3, 1). But, I found that if I simply re-scale the penalty matrix, while 
maintaining equal distances between categories, then the estimate of overall 
accuracy increases. In effect I can achieve any value for accuracy - including 
unity - by re-scaling the penalty matrix. So, I'd like to ask what, if any are 
the contraints on the scaling process?

Here is a working code that illustrates my difficulty:-

set.seed(1); gldstd=round(runif(200)*4); gldstd[gldstd==0]=4; table(gldstd)
pred1=gldstd*rnorm(200, mean=1, sd=2/gldstd); boxplot(pred1~gldstd)
library(nonbinROC); gldstd=ordered(gldstd)
ordered_penalty = matrix(c(0,0,0,0,1/3,0,0,0,2/3,1/3,0,0,1,2/3,1/3,0), nrow = 4)
constant_penalty = matrix(c(0,0,0,0,1,0,0,0,1,1,0,0,1,1,1,0), nrow = 4)

# first using the constant_penalty (default) matrix
ordROC(gldstd,pred1, penalty=constant_penalty)

# now re-scaling the constant_penalty matrix by 1/2
ordROC(gldstd,pred1, penalty=constant_penalty/2)

# now using an ordered penalty matrix
ordROC(gldstd,pred1,penalty=ordered_penalty)

# now re-scaling the ordered_penalty matrix by 1/2
ordROC(gldstd,pred1,penalty=ordered_penalty/2)

Here is the resulting output:-
 # first using the constant_penalty (default) matrix
 ordROC(gldstd,pred1, penalty=constant_penalty)
$`Pairwise Accuracy`
Pair  Estimate Standard.Error
1 1 vs 2 0.6420973 0.05667060
2 1 vs 3 0.7083172 0.05293275
3 1 vs 4 0.8500507 0.04149018
4 2 vs 3 0.6081169 0.05438183
5 2 vs 4 0.7950680 0.04736247
6 3 vs 4 0.7025974 0.05371362

$`Penalty Matrix`
  1 2 3 4
1 0 1 1 1
2 0 0 1 1
3 0 0 0 1
4 0 0 0 0

$`Overall Accuracy`
   Estimate Standard.Error
1 0.7074935 0.02764736

 # now re-scaling the constant_penalty matrix by 1/2
 ordROC(gldstd,pred1, penalty=constant_penalty/2)
$`Pairwise Accuracy`
Pair  Estimate Standard.Error
1 1 vs 2 0.6420973 0.05667060
2 1 vs 3 0.7083172 0.05293275
3 1 vs 4 0.8500507 0.04149018
4 2 vs 3 0.6081169 0.05438183
5 2 vs 4 0.7950680 0.04736247
6 3 vs 4 0.7025974 0.05371362

$`Penalty Matrix`
  1   2   3   4
1 0 0.5 0.5 0.5
2 0 0.0 0.5 0.5
3 0 0.0 0.0 0.5
4 0 0.0 0.0 0.0

$`Overall Accuracy`
   Estimate Standard.Error
1 0.8537467 0.01382368   === larger estimate of overall accuracy 
(cf 0.707 for original constant_penalty matrix)

 # now using an ordered penalty matrix
 ordROC(gldstd,pred1,penalty=ordered_penalty)
$`Pairwise Accuracy`
Pair  Estimate Standard.Error
1 1 vs 2 0.6420973 0.05667060
2 1 vs 3 0.7083172 0.05293275
3 1 vs 4 0.8500507 0.04149018
4 2 vs 3 0.6081169 0.05438183
5 2 vs 4 0.7950680 0.04736247
6 3 vs 4 0.7025974 0.05371362

$`Penalty Matrix`
  1 2 3 4
1 0 0.333 0.667 1.000
2 0 0.000 0.333 0.667
3 0 0.000 0.000 0.333
4 0 0.000 0.000 0.000

$`Overall Accuracy`
   Estimate Standard.Error
1 0.8616933 0.01622384

 # now re-scaling the ordered_penalty matrix by 1/2
 ordROC(gldstd,pred1,penalty=ordered_penalty/2)
$`Pairwise Accuracy`
Pair  Estimate Standard.Error
1 1 vs 2 0.6420973 0.05667060
2 1 vs 3 0.7083172 0.05293275
3 1 vs 4 0.8500507 0.04149018
4 2 vs 3 0.6081169 0.05438183
5 2 vs 4 0.7950680 0.04736247
6 3 vs 4 0.7025974 0.05371362

$`Penalty Matrix`
  1 2 3 4
1 0 0.167 0.333 0.500
2 0 0.000 0.167 0.333
3 0 0.000 0.000 0.167
4 0 0.000 0.000 0.000

$`Overall Accuracy`
   Estimate Standard.Error
1 0.9308467 0.00811192 === larger estimate of overall accuracy (cf 
0.862 for original ordered_penalty matrix)

I can see that penalising differences between categories less might produce 
better overall accuracy. However, if I use a constant penalty matrix with all 
off-diagonal values very close to zero, then the overall accuracy approaches 1. 
It seems counter-intuitive to me that the estimate of overall accuracy for an 
ordinal gold standard should depend on the absolute values of the penalty 
matrix.

So, I would like to ask, (a) should the penalty matrix always contain at least 
one penalty with a value of 1 and/or (b) should there be any other constraint 
on the sum of penalties in the matrix (e.g. should the matrix sum to some

[R] eliminate double entries in data.frame

2013-01-18 Thread Mat

hello together,

i want to eliminate double entries in a data.frame. 
I have a data.frame, like this one:

 1  2
A   Albert1800
B   Albert1800
C   Jack2000
D   Mike  2200

As you can see, Albert is two times available.
I want a solution like this one:

 1  2
A   Albert1800
B   Jack2000
C   Mike  2200

How can i eliminate this double entries?

Thanks.

Mat?



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[R] Classification by Standard Deviation of Lognormal with Weight

2013-01-18 Thread Fiona

Hi,

We got a actuarial question which cannot be solved in Excel, so we are
wondering if R can help us on it.

As the sample table below, variable X has 50 different values and the
weighted Y has a lognormal distribution.

We want to make X into four or five classes, based on the standard
deviation ó in lognormal of the weighted Y.

X Y Weight of Y
01  500  2
01  600  1
02  350  3
03  400  3
03  550  2
04  200  4
...   ... ...

If it is doable in R, can you please give us some suggestions?

Thank you very much.

Best regards,
Fiona

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Re: [R] reading multiple key=value pairs per line

HI,

May be this helps:
Lines1-readLines(textConnection('key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks'))
res-read.table(text=gsub(key{0,1}\\d,,gsub([\,],,Lines1)),sep==,header=FALSE,stringsAsFactors=F)[-1]


 names(res)- paste(substr(Lines1,1,3),1:3,sep=)
 res
#  key1 key2    key3
#1   23   67 hello there
#2    7   22 how are you
#3    2   77 nice day thanks
A.K.



- Original Message -
From: Frank Singleton b17fly...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Thursday, January 17, 2013 11:21 PM
Subject: [R] reading multiple key=value pairs per line

Hi,

Thanks for a great environmentfor statistical  computing :-)

I have some input data in a file (input_kvpairs.csv) of the form

key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks

Now in my head I wish it was of the form (input.csv)

#key1, key2, key3
23,    67,   hello there
7,     22,   how are you
2,     77,   nice day, thanks

so I could do

data - read.csv(input.csv, header=TRUE)

where the header column names are derived from the key names dynamically,
and I could access the data using normal data$key1 or data$key2 mechanism.

I guess I could just pre process the file first  using python etc to create
a CSV file with column header derived from key names, and values derived from
key values, but I am interested to see how experienced R folks would handle this
inside R.

Thanks,

Frank

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Re: [R] reading multiple key=value pairs per line

Hi,

Sorry, there was a mistake.  I didn't notice comma in key3
.Lines1-readLines(textConnection('key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks'))
res1-read.table(text=gsub(key{0,1}\\d,,gsub([\],,Lines1)),sep==,header=FALSE,stringsAsFactors=F)[-1]
 res1[,1:2]-sapply(res1[,1:2],function(x) as.numeric(gsub(\\, $,,x)))
names(res1)- paste(substr(Lines1,1,3),1:3,sep=)
res1
#  key1 key2 key3
#1   23   67  hello there
#2    7   22  how are you
#3    2   77 nice day, thanks
 str(res1)
#'data.frame':    3 obs. of  3 variables:
# $ key1: num  23 7 2
# $ key2: num  67 22 77
# $ key3: chr  hello there how are you nice day, thanks
A.K.



- Original Message -
From: Frank Singleton b17fly...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Thursday, January 17, 2013 11:21 PM
Subject: [R] reading multiple key=value pairs per line

Hi,

Thanks for a great environmentfor statistical  computing :-)

I have some input data in a file (input_kvpairs.csv) of the form

key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks

Now in my head I wish it was of the form (input.csv)

#key1, key2, key3
23,    67,   hello there
7,     22,   how are you
2,     77,   nice day, thanks

so I could do

data - read.csv(input.csv, header=TRUE)

where the header column names are derived from the key names dynamically,
and I could access the data using normal data$key1 or data$key2 mechanism.

I guess I could just pre process the file first  using python etc to create
a CSV file with column header derived from key names, and values derived from
key values, but I am interested to see how experienced R folks would handle this
inside R.

Thanks,

Frank

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Re: [R] function approx interpolation of time series data sets


Hello,

Both Gabor's and my way work and produce the same results:


#-- Gabor
library(zoo)
library(chron)

data1 - 
01:23:40  5
01:23:45 10
01:23:50 12
01:23:55  7

data2 - 
01:23:42
01:23:47
01:23:51
01:23:54
01:23:58
01:23:59

data1zoo - read.zoo(text=data1, FUN=times)
data2zoo - read.zoo(text=data2, FUN=times)
na.approx(data1zoo, xout = time(data2zoo))

#-- Rui

data1 - read.table(text = 
01:23:40 5
01:23:45 10
01:23:50 12
01:23:55 7
)

data2 - read.table(text = 
01:23:42
01:23:47
01:23:51
01:23:54
01:23:58
01:23:59
)

approx(as.POSIXct(data1$V1, format = %H:%M:%S),
y = data1$V2,
xout = as.POSIXct(data2$V1, format = %H:%M:%S))


Note that the last two values of data2 are outside the range of data1, 
and therefore the interpolation functions return nothing 
(zoo::na.approx) or NA (stats::approx)


Hope this helps,

Rui Barradas

Em 18-01-2013 13:51, e-letter escreveu:

On 18/01/2013, Gabor Grothendieck ggrothendi...@gmail.com wrote:

On Fri, Jan 18, 2013 at 7:31 AM, e-letter inp...@gmail.com wrote:

On 16/01/2013, Rui Barradas ruipbarra...@sapo.pt wrote:

Hello,

Like this?


data1 - read.table(text = 
01:23:40 5
01:23:45 10
01:23:50 12
01:23:55 7
)

data2 - read.table(text = 
01:23:42
01:23:47
01:23:51
01:23:54
)

approx(as.POSIXct(data1$V1, format = %H:%M:%S), y = data1$V2, xout =
as.POSIXct(data2$V1, format = %H:%M:%S))



Thanks. I have later realised that if data frames are unequal (i.e.
data2edit

01:23:42
01:23:47
01:23:51
01:23:54
01:23:58
01:23:59

the result for 'y' is

$y
[1] NA NA NA NA NA NA

Similar error occurs with the 'zoo' package.



its not clear precisely what you tried with zoo but have a look at this:


data1 - 

+  01:23:40 5
+  01:23:45 10
+  01:23:50 12
+  01:23:55 7


data2 - 

+  01:23:42
+  01:23:47
+  01:23:51
+  01:23:54


library(zoo)
library(chron)
z1 - read.zoo(text = data1, FUN = times)
z2 - read.zoo(text = data2, FUN = times)
na.approx(z1, xout = time(z2))

01:23:42 01:23:47 01:23:51 01:23:54
  7.0 10.8 11.0  8.0



Latest error:


data1

 V1 V2
1 01:23:40  5
2 01:23:45 10
3 01:23:50 12
4 01:23:55  7


data2

 V1
1 01:23:42
2 01:23:47
3 01:23:51
4 01:23:54
5 01:23:58
6 01:23:59


data1zoo-read.zoo(text=data1,FUN=times)

Error in textConnection(text) : invalid 'text' argument


data2zoo-read.zoo(text=data2,FUN=times)

Error in textConnection(text) : invalid 'text' argument

Then I tried to create objects differently:

data1zoo-read.zoo('test1.txt',FUN=times)

data1zoo

01:23:40 01:23:45 01:23:50 01:23:55
5   10   127

data2zoo-read.zoo('test2.txt',FUN=times)

data2zoo


01:23:42
01:23:47
01:23:51
01:23:54
01:23:58
01:23:59


data3-(na.approx(merge(data1zoo,data2zoo),time(data1zoo)))

Error in na.approx.default(object, x = x, xout = xout, na.rm = FALSE,  :
   x and index must have the same length



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Re: [R] eliminate double entries in data.frame

2013-01-18 Thread Mat

thx, works perfectly :-)

Mat



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Re: [R] plotting from dataframes

2013-01-18 Thread Jeff Newmiller

You need to (re-) read the Introduction to R document that comes with R, 
particularly about indexing lists.

Briefly, there are three ways: integer, string, and approximate string 
indexing. You seem to be stuck now using approximate string indexing with the $ 
operator. Integer indexing is more appropriate for your loop.

for(i in 1:10){
plot( frames[[ i ]]$hour1 )
}

---
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.

condor radonniko...@hotmail.nl wrote:

So by hand the command would be 

par(mfrow=c(1,2))
plot(frames$'1'hour1)
plot(frames$'2'hour1)

But in my case there are far more than 2 days, so I want to use a loop.
Suppose I have 10 plots
par(mfrow=c(2,5))
for(i in 1:10){
plot( /what should be put here??/)
}



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[R] sem package, suppress warnings

2013-01-18 Thread Dustin Fife

Hi,

I'm using the sem package under conditions where I know beforehand that
several models will be problematic. Because of that, I want to suppress
warnings. I've used the following code

sem(mod, S = as.matrix(dataset), N = 1000, maxiter = 1,
warn = FALSE)

But I still get warnings. I've also tried adding check.analytic = FALSE,
debug = FALSE, and that doesn't seem to do it either. Any ideas why it's
not working?

It may be important to note that the command above is wrapped within a
function and I'm using try() to avoid errors.

Thanks in advance!


-- 
Dustin Fife
PhD Student
Quantitative Psychology
University of Oklahoma

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Re: [R] lattice: loess smooths based on y-axis values

2013-01-18 Thread Bert Gunter

Yes ... that's much better. RTFM, Bert.

-- Bert

On Fri, Jan 18, 2013 at 7:57 AM, Raeanne Miller
raeanne.mil...@sams.ac.uk wrote:
 Hi Bert,

 Thanks for your answer - I've done a good bit of research in to panel 
 functions this afternoon, and they're starting to seem a *little* less 
 intimidating. I didn't know there was a panel.lines, so that's very helpful!

 Turns out that panel.loess has an argument 'horizontal', which when set to 
 'TRUE' bases the smooth on the y axis, as opposed to the x. So by using a 
 panel function:

 panel.groups = function(x, y, ..., lty) {
   panel.xyplot(x, y, ...)
   panel.loess(x,y, horizontal=TRUE, lty=lty,...)
}

 in my call to xyplot, it seems I can get the smooth the way I want it, with 
 Depth on the y axis and abundance on the x axis.

 Hope that is useful for anyone else interested - and Bert, thanks again for 
 the swift reply!

 Raeanne

 -Original Message-
 From: Bert Gunter [mailto:gunter.ber...@gene.com]
 Sent: 18 January 2013 15:08
 To: Raeanne Miller
 Cc: r-help@r-project.org
 Subject: Re: [R] lattice: loess smooths based on y-axis values

 Reanne:


 UNTESTED:

 I would reverse the xyplot call as xyplot(x~y,...) to get all axes set up 
 properly and then just write an explicit panel function using
 loess() and predict.loess the correct way, e.g. as loess(y~x) to get the 
 (x,y) curve pairs to be plotted  via panel.lines(y,x). You will have to 
 make sure the pairs are ordered in decreasing x.

 This presumes familiarity with loess() and lattice panel functions, of 
 course. Deepayan's book is a comprehensive resource for the latter.
 Murrell's also contains a less comprehensive but probably adequate overview. 
 -- If you need anything at all, that is.

 -- Bert



 On Fri, Jan 18, 2013 at 3:36 AM, Raeanne Miller raeanne.mil...@sams.ac.uk 
 wrote:
 Hi there,

 I'm using the lattice package to create an xy plot of abundance vs. depth 
 for 5 stages of barnacle larvae from 5 species. Each panel of the plot 
 represents a different stage, while different loess smoothers within each 
 panel should represent different species.

 However, I would like depth to be on the y-axis and abundance to be on the 
 x-axis, because this is more intuitive as an oceanographer. The smoothing 
 curves would then represent the abundance profiles with depth of the larvae. 
 I have tried just switching depth and abundance in the plot formula/coding, 
 but this means that the loess smoother is based on averaging depth values 
 for each abundance, which means that the curves give multiple possible 
 abundances at each depth level. Rather, I would like the loess smooth to 
 still be based on 'averaging' abundance at each depth.

 Essentially, I would like a mirror image of the plot from this code:

 ord-order(Zoo_nocyp$Depth_m, decreasing=TRUE)
 xyplot(log10Ab[ord]~factor(-1*Depth_m[ord])|StageF[ord],
  data=Zoo_nocyp[Zoo_nocyp$log10Ab!=0 ,],
  type=c(p,smooth),
  par.settings=list(strip.background=list(col=white), 
 fontzize=list(text=16,points=10), lty = 0),
  space=0, border=NA,
  col=c(blue,red,green,purple,orange),
  stack=TRUE, groups=Zoo_nocyp$SpeciesF,
  key=
list(title=Species, 
 cex.title=1,text=list(c(BB,BC,CH,SB,VS)),space=right,
 rectangles=list(size=2, 
 border=white,col=c(blue,red,green,purple,orange))),
  xlab=depth (m), ylab=log10 abundance,
  layout=c(2,3))

 which is different to what you get when you switch log10Ab and Depth_m.

 Is there any way to specify the smoothers to be based on sequential y-axis 
 values, rather than x? Or to get the mirror image of a plot?

 I can provide a dataset offline, for those interested.

 Thanks very much,

 Raeanne


 The Scottish Association for Marine Science (SAMS) is registered in Scotland 
 as a Company Limited by Guarantee (SC009292) and is a registered charity 
 (9206). SAMS has an actively trading wholly owned subsidiary company: SAMS 
 Research Services Ltd a Limited Company (SC224404). All Companies in the 
 group are registered in Scotland and share a registered office at Scottish 
 Marine Institute, Oban Argyll PA37 1QA. The content of this message may 
 contain personal views which are not the views of SAMS unless specifically 
 stated. Please note that all email traffic is monitored for purposes of 
 security and spam filtering. As such individual emails may be examined in 
 more detail.

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



 --

 Bert Gunter
 Genentech Nonclinical Biostatistics

 Internal Contact Info:
 Phone: 467-7374
 Website:

[R] How to re-project ease( Equal Area Scalable Earth) grid with a ~25 km cylindrical projection to WGS84 0.25 degree?

2013-01-18 Thread Jonsson

I have nc files  for global soil moisture,here is one file 
https://echange-fichiers.inra.fr/get?k=f9DDllPKdUKs5ZNQwfq

from the metadata ,the projection is cylindrical and the resolution is 25
km(it is based on authalic sphere based on International 1924 ellipsoid).As
I want to compare with other data, I have to make them identical.
- my other data are in WGS84 with 0.25*0.25 degree resolution and extent
-180, 180, -90, 90.
So I want to re-project  the data I have here
 from 
EASE grid, cylindrical,25 km ,1383 pixel  586 lines
 to  
WGS84,0.25*0.25 degree,1440 pixel 720 lines:   
 
f=open.ncdf(C:\\Users\\aalyaari\\Desktop\\SM_RE01_MIR_CLF31D_20100812T00_20100812T235959_246_001_7.DBL)
 
A =
get.var.ncdf(nc=f,varid=Soil_Moisture,verbose=TRUE)
I tried this:

d
-raster(C:\\Users\\aalyaari\\Desktop\\SM_RE01_MIR_CLF31D_20100812T00_20100812T235959_246_001_7.nc,
varname = Soil_Moisture) 
d:
class   : RasterLayer 
dimensions  : 586, 1383, 810438  (nrow, ncol, ncell)
resolution  : 0.2603037, 0.2916659  (x, y)
extent  : -180, 180, -85.4581, 85.4581  (xmin, xmax, ymin, ymax)
  coord. ref. : +proj=longlat +datum=WGS84 
names   : Retrieved.soil.moisture.value 
   zvar: Soil_Moisture 
   a - spTransform(d, CRS (+proj=longlat +ellps=WGS84 +datum=WGS84
+no_defs))
but I got this error:
  
 Error in function (classes, fdef, mtable)  : 
  unable to find an inherited method for function ‘spTransform’
for signature ‘RasterLayer, CRS’



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[R] longitudinal study

2013-01-18 Thread bibek sharma

Hello R user,
I have a data set from a longitudinal study ( sample below) where
subjects are followed over time. Second column (status) contains info
about if subject is dead or still in the  study and third column is
time measured in the week. Here is what I need: if status is not dead
or unknown take the last week, if status is dead or unknown I need to
have corresponding week.

Desired resulst:

1   no  7
2   yes 5
3   Unknown 4

Sample data
id  status  week
1   no  1
1   no  2
1   no  3
1   no  4
1   no  5
1   no  6
1   no  7
2   no  1
2   no  2
2   no  3
2   no  4
2   yes 5
2   yes 6
2   na  7
2   na  8
2   na  9
3   no  1
3   no  2
3   no  3
3   Unknown 4
3   unknown 5
3   na  6
3   na  7
3   na  8

Any suggestion is much appreciated!
Thank you.
Bibek

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Re: [R] longitudinal study

2013-01-18 Thread Rich Shepard


On Fri, 18 Jan 2013, bibek sharma wrote:


I have a data set from a longitudinal study ( sample below) where subjects
are followed over time. Second column (status) contains info about if
subject is dead or still in the study and third column is time measured in
the week. Here is what I need: if status is not dead or unknown take the
last week, if status is dead or unknown I need to have corresponding week.

Desired resulst:

1   no  7
2   yes 5
3   Unknown 4


  Looks like a survival analysis situation. I know there are R packages for
this.

Rich

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Re: [R] How to convert a string to the column it represents in a dataframe, with a reproducible example

2013-01-18 Thread mtb954

Many thanks to everyone who chimed in on this one. I really appreciate the
time you took to help me. Especially Dave W. who made me question what
exactly I was after.

Dan N.'s solution does exactly what I want, and it helped me learn about
the eval() and parse() functions too.

Thanks again, and all the best.

Mark Na


On Thu, Jan 17, 2013 at 5:15 PM, Nordlund, Dan (DSHS/RDA) 
nord...@dshs.wa.gov wrote:

  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of mtb...@gmail.com
  Sent: Thursday, January 17, 2013 2:27 PM
  To: David Winsemius; r-help@r-project.org
  Subject: Re: [R] How to convert a string to the column it represents in
  a dataframe, with a reproducible example
 
  Hi David,
 
  I would like to have two objects, one containing the values in a column
  and
  the other containing the column's name.
 
  Of course, that's easy to do manually, but I don't want to have to type
  out
  the name of the column more than once (thus, below, I have typed it
  once in
  quotes, and I am trying to find a programatic way to create the other
  object, without typing the column name again).
 
  Thank you for your help.
 
  Mark Na
 

 Something like this

 eval(parse(text=y))

 could be what you want. But even if it is, I am not sure it is what you
 should want.  Without more context, it is hard to say.

 Hope this is at least somewhat helpful,

 Dan

 Daniel J. Nordlund
 Washington State Department of Social and Health Services
 Planning, Performance, and Accountability
 Research and Data Analysis Division
 Olympia, WA 98504-5204


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[R] columns called X rename Y

2013-01-18 Thread Sebastian Kruk

I have a data. frame to which you want to change the names to some of their
columns.

For example:

 seba - data.frame ('constant' = 3, 'numbers' = 1: 10, 'letters' = LETTERS
[1:10], otros = 2:11)

List their names:

 names (Seba)
[1] constant numbers letters

I want to rename c the column called constant and b the column
called numbers.

I do not know in which order appear names.

How could do it?

Thanks in advance,

Sebastian.

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Re: [R] read tab delimited file from a certain line

2013-01-18 Thread Henrik Bengtsson

Christof,

I've added support for this to the R.filesets package.  In your case,
then all you need to do is:

library(R.filesets)
dlf - readDataFrame(filename, skip=^year)

No need to specify any other arguments - they're all automagically
inferred - and the default is stringsAsFactors=FALSE.

This is in R.filesets v2.0.0 which I still haven't published on CRAN.
In the meanwhile, you can install it via:

source(http://aroma-project.org/hbLite.R;)
hbLite(R.filesets)

/Henrik

On Thu, Jan 17, 2013 at 1:34 AM, Christof Kluß ckl...@email.uni-kiel.de wrote:
 Hello

 thank you for the fast and helpful answer! Now the following works fine for
 me

 x - readLines(filename)

 i - grep(^year, x)
 dlf - read.table(textConnection(x[i:length(x)]),
header = T, stringsAsFactors=F,sep=\t)

 Greetings
 Christof


 Am 16-01-2013 16:55, schrieb Rui Barradas:

 Hello,

 Read the file using readLines, then grep ^year. You can then use a
 textConnection to read.table:

 x - readLines(con = textConnection(
 informations (unknown count of lines)
 ... and at some point the table
 --
 year month mday value
 2013 1 16 0 ))

 # This is it
 i - grep(^year, x)
 read.table(textConnection(x[i:length(x)]), header = TRUE)


 Hope this helps,

 Rui Barradas

 Em 16-01-2013 14:17, Christof Kluß escreveu:

 Hi

 I would like to read table data from a text-files with extra
 informations in the header (of unknown line count). Example:

 informations (unknown count of lines)
 ... and at some point the table
 --
 year month mday value
 2013 1 16 0
 ...

 If it was an excel file I could use something like read.xls(...,
 pattern=year) But it is a simple tab seperated text-file. Is there
 an easy way to read only the table? (Without dirty things ;))

 Thx
 Christof

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Re: [R] lattice: loess smooths based on y-axis values

2013-01-18 Thread Raeanne Miller

Hi Bert,

Thanks for your answer - I've done a good bit of research in to panel functions 
this afternoon, and they're starting to seem a *little* less intimidating. I 
didn't know there was a panel.lines, so that's very helpful!

Turns out that panel.loess has an argument 'horizontal', which when set to 
'TRUE' bases the smooth on the y axis, as opposed to the x. So by using a panel 
function:

panel.groups = function(x, y, ..., lty) {
  panel.xyplot(x, y, ...)
  panel.loess(x,y, horizontal=TRUE, lty=lty,...)
   }

in my call to xyplot, it seems I can get the smooth the way I want it, with 
Depth on the y axis and abundance on the x axis.

Hope that is useful for anyone else interested - and Bert, thanks again for the 
swift reply!

Raeanne

-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: 18 January 2013 15:08
To: Raeanne Miller
Cc: r-help@r-project.org
Subject: Re: [R] lattice: loess smooths based on y-axis values

Reanne:


UNTESTED:

I would reverse the xyplot call as xyplot(x~y,...) to get all axes set up 
properly and then just write an explicit panel function using
loess() and predict.loess the correct way, e.g. as loess(y~x) to get the 
(x,y) curve pairs to be plotted  via panel.lines(y,x). You will have to 
make sure the pairs are ordered in decreasing x.

This presumes familiarity with loess() and lattice panel functions, of course. 
Deepayan's book is a comprehensive resource for the latter.
Murrell's also contains a less comprehensive but probably adequate overview. -- 
If you need anything at all, that is.

-- Bert



On Fri, Jan 18, 2013 at 3:36 AM, Raeanne Miller raeanne.mil...@sams.ac.uk 
wrote:
 Hi there,

 I'm using the lattice package to create an xy plot of abundance vs. depth for 
 5 stages of barnacle larvae from 5 species. Each panel of the plot represents 
 a different stage, while different loess smoothers within each panel should 
 represent different species.

 However, I would like depth to be on the y-axis and abundance to be on the 
 x-axis, because this is more intuitive as an oceanographer. The smoothing 
 curves would then represent the abundance profiles with depth of the larvae. 
 I have tried just switching depth and abundance in the plot formula/coding, 
 but this means that the loess smoother is based on averaging depth values for 
 each abundance, which means that the curves give multiple possible abundances 
 at each depth level. Rather, I would like the loess smooth to still be based 
 on 'averaging' abundance at each depth.

 Essentially, I would like a mirror image of the plot from this code:

 ord-order(Zoo_nocyp$Depth_m, decreasing=TRUE)
 xyplot(log10Ab[ord]~factor(-1*Depth_m[ord])|StageF[ord],
  data=Zoo_nocyp[Zoo_nocyp$log10Ab!=0 ,],
  type=c(p,smooth),
  par.settings=list(strip.background=list(col=white), 
 fontzize=list(text=16,points=10), lty = 0),
  space=0, border=NA,
  col=c(blue,red,green,purple,orange),
  stack=TRUE, groups=Zoo_nocyp$SpeciesF,
  key=
list(title=Species, 
 cex.title=1,text=list(c(BB,BC,CH,SB,VS)),space=right,
 rectangles=list(size=2, 
 border=white,col=c(blue,red,green,purple,orange))),
  xlab=depth (m), ylab=log10 abundance,
  layout=c(2,3))

 which is different to what you get when you switch log10Ab and Depth_m.

 Is there any way to specify the smoothers to be based on sequential y-axis 
 values, rather than x? Or to get the mirror image of a plot?

 I can provide a dataset offline, for those interested.

 Thanks very much,

 Raeanne


 The Scottish Association for Marine Science (SAMS) is registered in Scotland 
 as a Company Limited by Guarantee (SC009292) and is a registered charity 
 (9206). SAMS has an actively trading wholly owned subsidiary company: SAMS 
 Research Services Ltd a Limited Company (SC224404). All Companies in the 
 group are registered in Scotland and share a registered office at Scottish 
 Marine Institute, Oban Argyll PA37 1QA. The content of this message may 
 contain personal views which are not the views of SAMS unless specifically 
 stated. Please note that all email traffic is monitored for purposes of 
 security and spam filtering. As such individual emails may be examined in 
 more detail.

 [[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.



--

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
The Scottish Association for Marine Science (SAMS) is registered in Scotland as 
a Company Limited by Guarantee (SC009292) and is a registered

Re: [R] plotting from dataframes

Hi,
May be this helps:
frames-list(data.frame(c1=1:3,day1=17,hour1=c(10,11,6)),data.frame(c1=6:7,day1=19,hour1=8),data.frame(c1=8:10,day1=21,hour1=c(11,15,18)),data.frame(c1=12:13,day1=23,hour1=7))
par(mfrow=c(2,2))
lapply(seq_along(frames),function(i) plot(frames[[i]][,3]))
A.K.




- Original Message -
From: condor radonniko...@hotmail.nl
To: r-help@r-project.org
Cc: 
Sent: Friday, January 18, 2013 4:16 AM
Subject: Re: [R] plotting from dataframes

So by hand the command would be 

par(mfrow=c(1,2))
plot(frames$'1'hour1)
plot(frames$'2'hour1)

But in my case there are far more than 2 days, so I want to use a loop.
Suppose I have 10 plots
par(mfrow=c(2,5))
for(i in 1:10){
plot( /what should be put here??/)
}



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[R] Thermoisoplethendiagramm

2013-01-18 Thread Dominic Roye

Hello,

I would like to generate isolines of temperature for the 24 hours by the 12
month. Something like hier:
http://www.diercke.de/bilder/omeda/800/12351E.jpg.

On the x-axis are the month and on the y-axis the 24 hours, than as contour
lines the temperature.

My data are:

'data.frame':   288 obs. of  3 variables:
 $ Group.1: num  0 1 2 3 4 5 6 7 8 9 ...
 $ Group.2: num  1 1 1 1 1 1 1 1 1 1 ...
 $ x  : num  6.99 6.66 6.36 6.1 5.94 ...


Group 1: hour
Group 2: month
x: temperature ºC

Have anybody an idea? I have tried without success the
command filled.contour().

Many thanks!!

Best regards,

Dominic

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Re: [R] longitudinal study

HI,

May be this helps:
dat1-read.table(text=
id status week
1 no 1
1 no 2
1 no 3
1 no 4
1 no 5
1 no 6
1 no 7
2 no 1
2 no 2
2 no 3
2 no 4
2 yes 5
2 yes 6
2 na 7
2 na 8
2 na 9
3 no 1
3 no 2
3 no 3
3 Unknown 4
3 unknown 5
3 na 6
3 na 7
3 na 8 
,sep=,header=TRUE,stringsAsFactors=FALSE,na.strings=na)
 dat2-dat1[complete.cases(dat1),]
 res-do.call(rbind,lapply(split(dat2,dat2$id),function(x) 
rbind(tail(x[all(x[,2]==no)],1),head(x[x[,2]==yes|x[,2]==Unknown,],1
 res
#  id  status week
#1  1  no    7
#2  2 yes    5
#3  3 Unknown    4
A.K.





- Original Message -
From: Rich Shepard rshep...@appl-ecosys.com
To: R help r-help@r-project.org
Cc: 
Sent: Friday, January 18, 2013 12:18 PM
Subject: Re: [R] longitudinal study

On Fri, 18 Jan 2013, bibek sharma wrote:

 I have a data set from a longitudinal study ( sample below) where subjects
 are followed over time. Second column (status) contains info about if
 subject is dead or still in the study and third column is time measured in
 the week. Here is what I need: if status is not dead or unknown take the
 last week, if status is dead or unknown I need to have corresponding week.

 Desired resulst:

 1    no    7
 2    yes    5
 3    Unknown    4

   Looks like a survival analysis situation. I know there are R packages for
this.

Rich

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[R] Hclust tree to Figtree w/ branch lengths

2013-01-18 Thread Julian Banchier

Hi,

I'm doing hierarchical clustering, and want to export my dendrogram to a
tree-viewing/editing software. I can do this by converting the data to
Newick format (hc2Newick in ctc package), but I can't get branch lengths to
show in the resulting phylogram. I figured it might help to convert my
hclust object into a phylo object (as.phylo in ape package), but the
following lines give me this Error in UseMethod(as.phylo) : no applicable
method for 'as.phylo' applied to an object of class character:

data = read.table(C:/path/data.txt)
d - dist(data, method = euclidean)
library(ape)
fit - as.phylo(hclust(d, method=ward)
library(ctc)
write.table(as.phylo(hc2Newick(fit),
file=C:/path/output.txt,row.names=FALSE,col.names=FALSE))

Would anyone have an idea about how to make this work?

Thanks so much in advance!
Julian



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Re: [R] Thermoisoplethendiagramm


On Jan 18, 2013, at 9:49 AM, Dominic Roye wrote:

 Hello,
 
 I would like to generate isolines of temperature for the 24 hours by the 12
 month. Something like hier:
 http://www.diercke.de/bilder/omeda/800/12351E.jpg.
 
 On the x-axis are the month and on the y-axis the 24 hours, than as contour
 lines the temperature.
 
 My data are:
 
 'data.frame':   288 obs. of  3 variables:
 $ Group.1: num  0 1 2 3 4 5 6 7 8 9 ...
 $ Group.2: num  1 1 1 1 1 1 1 1 1 1 ...
 $ x  : num  6.99 6.66 6.36 6.1 5.94 ...
 


Perhaps:

require(lattice)
levelplot(x ~ Group.2 + Group.1, data =dfrm)


 
 Group 1: hour
 Group 2: month
 x: temperature ºC
 
 Have anybody an idea? I have tried without success the
 command filled.contour().

I believe filled contour would have required the z argument to be a matrix, 
i.e. you owuld haveneeded to recast the data in wide format.

-- 

David Winsemius
Alameda, CA, USA

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[R] A smart way to use $ in data frame

2013-01-18 Thread Yuan, Rebecca

Hello all,

I have a data frame dataa:

newdate newstate newid newbalance newaccounts
1 31DEC2001AR 1 1170   61
2 31DEC2001VA 2  4565   54
3 31DEC2001WA 3 2726   35
4 31DEC2001AR 3 2700   35

The following gives me the balance of state AR:

dataa$newbalance[data$newstate == 'AR']
1170
2700

Now, I have another different data frame datab, it is very similar to data, 
except that the name of the columns are different, and the order of the columns 
are different:

oldstate olddate oldbalance oldid oldaccounts
1 AR   31DEC20121234 7  40
2 WA 31DEC2012 3  30
3 VA   31DEC20122345 5  23
3 AR   31DEC20125673 5  23

datab$oldbalance[datab$oldstate== 'AR' ]
1234
5673

Could I have a way to quote

data$balance[data$state == 'AR']

in general, where balance=oldbalance, state=oldstate when data=dataa, and 
balance = newbalance, state = newstate when data=datab ?

Thanks very much!

Cheers,

Rebecca

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Re: [R] columns called X rename Y


Hello,

Try the following.


names(seba)[grep(numbers, names(seba))] - b
names(seba)[grep(constant, names(seba))] - c

names(seba)


Hope this helps,

Rui Barradas

Em 18-01-2013 18:14, Sebastian Kruk escreveu:

I have a data. frame to which you want to change the names to some of their
columns.

For example:


seba - data.frame ('constant' = 3, 'numbers' = 1: 10, 'letters' = LETTERS

[1:10], otros = 2:11)

List their names:


names (Seba)

[1] constant numbers letters

I want to rename c the column called constant and b the column
called numbers.

I do not know in which order appear names.

How could do it?

Thanks in advance,

Sebastian.

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Re: [R] A smart way to use $ in data frame

2013-01-18 Thread Duncan Murdoch


On 18/01/2013 2:40 PM, Yuan, Rebecca wrote:

Hello all,

I have a data frame dataa:

newdate newstate newid newbalance newaccounts
1 31DEC2001AR 1 1170   61
2 31DEC2001VA 2  4565   54
3 31DEC2001WA 3 2726   35
4 31DEC2001AR 3 2700   35

The following gives me the balance of state AR:

dataa$newbalance[data$newstate == 'AR']
1170
2700

Now, I have another different data frame datab, it is very similar to data, 
except that the name of the columns are different, and the order of the columns 
are different:

oldstate olddate oldbalance oldid oldaccounts
1 AR   31DEC20121234 7  40
2 WA 31DEC2012 3  30
3 VA   31DEC20122345 5  23
3 AR   31DEC20125673 5  23

datab$oldbalance[datab$oldstate== 'AR' ]
1234
5673

Could I have a way to quote

data$balance[data$state == 'AR']

in general, where balance=oldbalance, state=oldstate when data=dataa, and 
balance = newbalance, state = newstate when data=datab ?


Yes, you could use the name of the dataframe to get the column names, e.g.

state - c(dataa=oldstate, datab=newstate)
balance - c(dataa=oldbalance, datab=newbalance)
dfname - dataa
df - get(dfname)
df[ df[,state[dfname]] == 'AR', balance[dfname]]

but that is really, really unreadable code.  You would be much better 
off to name the columns consistently in the first place.


Duncan Murdoch

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[R] tables package: How to remove column headings and alignment issues

2013-01-18 Thread Marius Hofert

Dear expeRts,

Here is a minimal example with the latest version of 'tables' (questions below):

require(tables)

saveopts - table_options(toprule=\\toprule, midrule=\\midrule, 
bottomrule=\\bottomrule,
  titlerule=\\cmidrule(lr), 
rowlabeljustification=r)#, justification=r)

## data.frame
x - expand.grid(beta=as.factor(c(95, 99, 99.9)),
m=as.factor(c(64, 256)),
p=as.factor(c(8, 64, 512)),
f=as.factor(c(A, B)),
gamma=as.factor(c(0.25, 0.5)))
x - cbind(x, value=1:nrow(x))

## tabular
(tab - tabular(Heading($p$) * Justify(c) * p *
Heading($m$) * Justify(c) * m *
Heading($\\gamma$) * Justify(c) * gamma
~
Heading() * f * Heading() * beta *
Heading() * value * Heading() * identity, data=x))
latex(tab)


Questions:

1) How can the column labels f and beta be removed? I tried to use 
Heading() to remove the labels, but some of them (actually all except 
identity) still appear.

2) The above example leads to a tabular with columns aligned like this: 
rrrcc. However, if put in a .tex document, the table entries (1:9) are 
filled with \phantom{0}, so that the numbers do not appear centered (rather 
right-aligned). In contrast, the row labels (given in the first three columns) 
appear centered although they should be right-aligned.

3) Additionally using justification=r in table_options() provides 
right-justification of the entries, however, they are now all wrapped in 
\multicolumn{1}{c}{...}. Why?

Cheers,

Marius



Here is a .tex wrapper for convenience:

\documentclass{article}

\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[american]{babel}

\usepackage{tabularx}
\usepackage{booktabs}


\begin{document}
  % put code here
\end{document}

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Re: [R] columns called X rename Y

2013-01-18 Thread Jose Iparraguirre

Simply

 colnames(seba)[1] - c
 colnames(seba)[2] - b

Regards,

Jose


From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of 
Sebastian Kruk [residuo.so...@gmail.com]
Sent: 18 January 2013 18:14
To: R-help
Subject: [R] columns called X rename Y

I have a data. frame to which you want to change the names to some of their
columns.

For example:

 seba - data.frame ('constant' = 3, 'numbers' = 1: 10, 'letters' = LETTERS
[1:10], otros = 2:11)

List their names:

 names (Seba)
[1] constant numbers letters

I want to rename c the column called constant and b the column
called numbers.

I do not know in which order appear names.

How could do it?

Thanks in advance,

Sebastian.

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Re: [R] read tab delimited file from a certain line


On Jan 18, 2013, at 10:26 AM, Henrik Bengtsson wrote:

 Christof,
 
 I've added support for this to the R.filesets package.  In your case,
 then all you need to do is:
 
 library(R.filesets)
 dlf - readDataFrame(filename, skip=^year)
 
 No need to specify any other arguments - they're all automagically
 inferred - and the default is stringsAsFactors=FALSE.
 
 This is in R.filesets v2.0.0 which I still haven't published on CRAN.
 In the meanwhile, you can install it via:
 
 source(http://aroma-project.org/hbLite.R;)
 hbLite(R.filesets)

Mac GUI users should exit to a Terminal session running R --vanilla  before 
attempting this. (Probably a safer practice in general, but sometimes we get 
get lazy because the GUI Package Installer is so handy.)

-- 
David.
 
 /Henrik
 
 On Thu, Jan 17, 2013 at 1:34 AM, Christof Kluß ckl...@email.uni-kiel.de 
 wrote:
 Hello
 
 thank you for the fast and helpful answer! Now the following works fine for
 me
 
 x - readLines(filename)
 
 i - grep(^year, x)
 dlf - read.table(textConnection(x[i:length(x)]),
   header = T, stringsAsFactors=F,sep=\t)
 
 Greetings
 Christof
 
 
 Am 16-01-2013 16:55, schrieb Rui Barradas:
 
 Hello,
 
 Read the file using readLines, then grep ^year. You can then use a
 textConnection to read.table:
 
 x - readLines(con = textConnection(
 informations (unknown count of lines)
 ... and at some point the table
 --
 year month mday value
 2013 1 16 0 ))
 
 # This is it
 i - grep(^year, x)
 read.table(textConnection(x[i:length(x)]), header = TRUE)
 
 
 Hope this helps,
 
 Rui Barradas
 
 Em 16-01-2013 14:17, Christof Kluß escreveu:
 
 Hi
 
 I would like to read table data from a text-files with extra
 informations in the header (of unknown line count). Example:
 
 informations (unknown count of lines)
 ... and at some point the table
 --
 year month mday value
 2013 1 16 0
 ...
 
 If it was an excel file I could use something like read.xls(...,
 pattern=year) But it is a simple tab seperated text-file. Is there
 an easy way to read only the table? (Without dirty things ;))
 
 Thx
 Christof
 
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[R] select rows with identical columns from a data frame

2013-01-18 Thread Sam Steingold

I have a data frame with several columns.
I want to select the rows with no NAs (as with complete.cases)
and all columns identical.
E.g., for

--8---cut here---start-8---
 f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
 f
   a  b  c
1  1  1  1
2 NA NA NA
3 NA  3  5
4  4 40 40
--8---cut here---end---8---

I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
row because there all 3 columns are the same and none is NA.

thanks!

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://memri.org http://mideasttruth.com
http://honestreporting.com http://pmw.org.il http://iris.org.il
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[R] read.csv returns no lines available for input

2013-01-18 Thread Collins, Stephen


Hello,

I'm trying to read a file rows at a time, so as to not read the entire file 
into memory.  When reading the connections and readLines help, and R help 
archive, it seems this should be possible with read.csv and a file connection, 
making use of the nrows argument.

From certain posts, it seemed that read.csv should return character(0) when 
the end of file is reached, and there are no more rows to read.  Instead, I 
get an error there are no lines available for input.

What is the proper way to check the end-of-file condition with read.csv, such 
that I could break a while loop reading the data in?

#example, make a test file
con - file(test.csv,wt)
cat(a,b,c\n, 1,2,3\n, 4,5,6\n, 7,6,5\n, 4,3,2\n, 3,2,1\n,file=con)
unlink(con)

#show the file is valid
con - file(test.csv,rt)
read.csv(con,header=T)
unlink(con)

#show that readLines ends with character(0), like expected
con - file(test.csv,rt)
readLines(con,n=10)
readLines(con,n=10)
unlink(con)

#show that read.csv end with error
con - file(test.csv,rt)
read.csv(con,header=T,nrows=10)
read.csv(con,header=F,nrows=10)
unlink(con)



Sincerely,

Stephen


 sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: i386-pc-mingw32/i386 (32-bit)

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

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

loaded via a namespace (and not attached):
[1] tools_2.15.0


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Re: [R] select rows with identical columns from a data frame

2013-01-18 Thread Sam Steingold

I can do
  Reduce(==,f[complete.cases(f),])
but that creates an intermediate data frame which I would love to avoid
(to save memory).

 * Sam Steingold f...@tah.bet [2013-01-18 15:53:21 -0500]:

 I have a data frame with several columns.
 I want to select the rows with no NAs (as with complete.cases)
 and all columns identical.
 E.g., for

 f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
 f
a  b  c
 1  1  1  1
 2 NA NA NA
 3 NA  3  5
 4  4 40 40

 I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
 row because there all 3 columns are the same and none is NA.

 thanks!

-- 
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Re: [R] Working with regular expression


Hello,

This one seems to work.

gsub([[:alpha:]_]*(.*), \\1, Text)


Hope this helps,

Rui Barradas

Em 18-01-2013 20:11, Christofer Bogaso escreveu:

Hello again,

I was trying to extract the date element from a character vector using
Regular expression. Below is a sample what I was trying to do:


Text - c(asdf May 09 2009, dsaf_01-01-2001)
gsub([a-zA-Z_], , Text)

[1]   09 2009  01-01-2001


I basically have a long text and the task is to extract the date
element. My above strategy failed because I can not extract the month
for the first element.

Can somebody help me out?

Thanks and regards,

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Re: [R] select rows with identical columns from a data frame


Hello,

Try the following.

complete.cases(f)  apply(f, 1, function(x) all(x == x[1]))


Hope this helps,

Rui Barradas

Em 18-01-2013 20:53, Sam Steingold escreveu:

I have a data frame with several columns.
I want to select the rows with no NAs (as with complete.cases)
and all columns identical.
E.g., for

--8---cut here---start-8---

f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
f

a  b  c
1  1  1  1
2 NA NA NA
3 NA  3  5
4  4 40 40
--8---cut here---end---8---

I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
row because there all 3 columns are the same and none is NA.

thanks!



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Re: [R] Working with regular expression

2013-01-18 Thread Christofer Bogaso

Thanks Rui for your help.

Could you also explain the underlying logic please?

Thanks and regards,


On Sat, Jan 19, 2013 at 2:43 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
 gsub([[:alpha:]_]*(.*), \\1, Text)

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Re: [R] reading multiple key=value pairs per line

2013-01-18 Thread Greg Snow

You could use the strapply function from the gsubfn package to extract the
data from strings.  This will return a list that you could use with
do.call(rbind(

The stringr package may have something similar or an alternative (but I am
less familiar with that package).


On Thu, Jan 17, 2013 at 9:21 PM, Frank Singleton b17fly...@gmail.comwrote:

 Hi,

 Thanks for a great environmentfor statistical  computing :-)

 I have some input data in a file (input_kvpairs.csv) of the form

 key1=23, key2=67, key3=hello there
 key1=7, key2=22, key3=how are you
 key1=2, key2=77, key3=nice day, thanks

 Now in my head I wish it was of the form (input.csv)

 #key1, key2, key3
 23,67,   hello there
 7, 22,   how are you
 2, 77,   nice day, thanks

 so I could do

 data - read.csv(input.csv, header=TRUE)

 where the header column names are derived from the key names dynamically,
 and I could access the data using normal data$key1 or data$key2 mechanism.

 I guess I could just pre process the file first  using python etc to create
 a CSV file with column header derived from key names, and values derived
 from
 key values, but I am interested to see how experienced R folks would
 handle this
 inside R.

 Thanks,

 Frank

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Re: [R] select rows with identical columns from a data frame


On Jan 18, 2013, at 1:02 PM, Rui Barradas wrote:

 Hello,
 
 Try the following.
 
 complete.cases(f)  apply(f, 1, function(x) all(x == x[1]))
 
 
 Hope this helps,
 
 Rui Barradas
 
 Em 18-01-2013 20:53, Sam Steingold escreveu:
 I have a data frame with several columns.
 I want to select the rows with no NAs (as with complete.cases)
 and all columns identical.
 E.g., for
 
 --8---cut here---start-8---
 f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
 f
a  b  c
 1  1  1  1
 2 NA NA NA
 3 NA  3  5
 4  4 40 40
 --8---cut here---end---8---

 f[ which( rowSums(f==f[[1]]) == length(f) ), ]
  a b c
1 1 1 1

 
 I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
 row because there all 3 columns are the same and none is NA.
 
 thanks!
 

David Winsemius
Alameda, CA, USA

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Re: [R] select rows with identical columns from a data frame

2013-01-18 Thread William Dunlap

Here are two related approaches to your problem.  The first uses
a logical vector, keep, to say which rows to keep.  The second
uses an integer vector, it can be considerably faster when the columns
are not well correlated with one another (so the number of desired
rows is small proportion of the input rows).

f1 - function (x) 
{
# sieve with logical 'keep' vector
stopifnot(is.data.frame(x), ncol(x)  1)
keep - x[[1]] == x[[2]]
for (i in seq_len(ncol(x))[-(1:2)]) {
keep - keep  x[[i - 1]] == x[[i]]
}
!is.na(keep)  keep
}

f2 - function (x) 
{
# sieve with integer 'keep' vector
stopifnot(is.data.frame(x), ncol(x)  1)
keep - which(x[[1]] == x[[2]])
for (i in seq_len(ncol(x))[-(1:2)]) {
keep - keep[which(x[[i - 1]][keep] == x[[i]][keep])]
}
seq_len(nrow(x)) %in% keep
}

E.g., for a 10 million by 10 data.frame I get:

 x - data.frame(lapply(structure(1:10,names=letters[1:10]), 
 function(i)sample(c(NA,1,1,1,2,2,2,3), replace=TRUE, size=1e7)))
 system.time(v1 - f1(x))
   user  system elapsed 
   4.040.164.19 
 system.time(v2 - f2(x))
   user  system elapsed 
   0.800.000.79 
 identical(v1, v2)
[1] TRUE
 head(x[v1,])
  a b c d e f g h i j
4811  2 2 2 2 2 2 2 2 2 2
41706 1 1 1 1 1 1 1 1 1 1
56633 1 1 1 1 1 1 1 1 1 1
70859 1 1 1 1 1 1 1 1 1 1
83848 1 1 1 1 1 1 1 1 1 1
84767 1 1 1 1 1 1 1 1 1 1


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 Sam Steingold
 Sent: Friday, January 18, 2013 12:53 PM
 To: r-help@r-project.org
 Subject: [R] select rows with identical columns from a data frame
 
 I have a data frame with several columns.
 I want to select the rows with no NAs (as with complete.cases)
 and all columns identical.
 E.g., for
 
 --8---cut here---start-8---
  f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
  f
a  b  c
 1  1  1  1
 2 NA NA NA
 3 NA  3  5
 4  4 40 40
 --8---cut here---end---8---
 
 I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
 row because there all 3 columns are the same and none is NA.
 
 thanks!
 
 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 
 11.0.11103000
 http://www.childpsy.net/ http://memri.org http://mideasttruth.com
 http://honestreporting.com http://pmw.org.il http://iris.org.il
 All extremists should be taken out and shot.
 
 __
 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
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Re: [R] A smart way to use $ in data frame

2013-01-18 Thread Greg Snow

The important thing to understand is that $ is a shortcut for [[ and you
are moving into the realm where a shortcut is the longest distance between
2 points (see fortune(312)).

So your code can be something like:

state - 'oldstate'
balance - 'oldbalance'
dataa[[balance]][ dataa[[state]]=='AR' ]

You may also benefit from learning to use tools like with and subset
(though subset has its own complications when used inside of other
functions) or grep and match to find the columns of interest.


On Fri, Jan 18, 2013 at 12:40 PM, Yuan, Rebecca 
rebecca.y...@bankofamerica.com wrote:

 Hello all,

 I have a data frame dataa:

 newdate newstate newid newbalance newaccounts
 1 31DEC2001AR 1 1170   61
 2 31DEC2001VA 2  4565   54
 3 31DEC2001WA 3 2726   35
 4 31DEC2001AR 3 2700   35

 The following gives me the balance of state AR:

 dataa$newbalance[data$newstate == 'AR']
 1170
 2700

 Now, I have another different data frame datab, it is very similar to
 data, except that the name of the columns are different, and the order of
 the columns are different:

 oldstate olddate oldbalance oldid oldaccounts
 1 AR   31DEC20121234 7  40
 2 WA 31DEC2012 3  30
 3 VA   31DEC20122345 5  23
 3 AR   31DEC20125673 5  23

 datab$oldbalance[datab$oldstate== 'AR' ]
 1234
 5673

 Could I have a way to quote

 data$balance[data$state == 'AR']

 in general, where balance=oldbalance, state=oldstate when data=dataa, and
 balance = newbalance, state = newstate when data=datab ?

 Thanks very much!

 Cheers,

 Rebecca

 --
 This message, and any attachments, is for the intended...{{dropped:19}}

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Re: [R] select rows with identical columns from a data frame

 apply(f,1,function(x) all(duplicated(x)|duplicated(x,fromLast=TRUE)!is.na(x)))

#[1]  TRUE FALSE FALSE FALSE


A.K.



- Original Message -
From: Sam Steingold s...@gnu.org
To: r-help@r-project.org
Cc: 
Sent: Friday, January 18, 2013 3:53 PM
Subject: [R] select rows with identical columns from a data frame

I have a data frame with several columns.
I want to select the rows with no NAs (as with complete.cases)
and all columns identical.
E.g., for

--8---cut here---start-8---
 f - data.frame(a=c(1,NA,NA,4),b=c(1,NA,3,40),c=c(1,NA,5,40))
 f
   a  b  c
1  1  1  1
2 NA NA NA
3 NA  3  5
4  4 40 40
--8---cut here---end---8---

I want the vector TRUE,FALSE,FALSE,FALSE selecting just the first
row because there all 3 columns are the same and none is NA.

thanks!

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://memri.org http://mideasttruth.com
http://honestreporting.com http://pmw.org.il http://iris.org.il
All extremists should be taken out and shot.

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[R] Error in mer_finalize(ans) : Downdated X'X is not positive definite, 8.

2013-01-18 Thread Yoav Avneon

Dear All,
I have conducted an experiment in order to examine predation pressure in the
surroundings of potential wildlife road-crossing structures. I have
documented predation occurrence (binary…) in these structures and calculated
several possible explanatory variables describing the spatial heterogeneity
in several scales. At the landscape scale I have calculated the percentage
of different land-uses (7) in buffers around the structures with changing
radii (100m, 250m, 500m, 1000m and 2000m). For each radii I have generated
a set of all possible models from the given 7 variables related to this
radii.
I have tried to account for random effects. A spatial random effect – the
structure itself as a repeated measurement, by adding the term
(1|Road_Structure). A temporal random effect – the observation session (is
it correct to say that this examines possible learning in time ?), by adding
the term (1|Session.ord).
I have generated 4 sets for each radii: 1) with both random effects, 2+3)
only one (each) random effect, and 4) none.
I have tried to model the relationships using glm, glmer and lmer.
The problem is that in the full model (all 7 variables) of some sets with
random effects I get the following error:
*Error in mer_finalize(ans) : Downdated X'X is not positive definite, 8.*

The only record of this error I have found in the internet is for data with
missing observation. This doesn't comply with my data set. I think it
might be a singularity problem but I don't know how to check it, and as a
result, I obviously don't know how to fix it.
I have attached example R scripts (for the radius 2000) and the data table.
Any help is highly appreciated !
Thanks and all the best,
--
Yoav Avneon
Department of Life Sciences
The Spatial Ecology Lab
Ben Gurion University of the Negev
P.O.B. 653 Beer-Sheva 84105
ISRAEL

+972-52-3391918 (mobile)
+972-8-6461350 (lab)
---
Yoav Yeled Teva

ANP_multi_landscape_No_RE_Buffer_2000m.R
http://r.789695.n4.nabble.com/file/n4656015/ANP_multi_landscape_No_RE_Buffer_2000m.R

ANP_multi_landscape_Plus_RE_Str_Session_Buffer_2000m.R
http://r.789695.n4.nabble.com/file/n4656015/ANP_multi_landscape_Plus_RE_Str_Session_Buffer_2000m.R

ANP_multi_landscape_Plus_RE_Str_Buffer_2000m.R
http://r.789695.n4.nabble.com/file/n4656015/ANP_multi_landscape_Plus_RE_Str_Buffer_2000m.R

ANP_multi_landscape_Plus_RE_Session_Buffer_2000m.R
http://r.789695.n4.nabble.com/file/n4656015/ANP_multi_landscape_Plus_RE_Session_Buffer_2000m.R

Raw_data_4_logit_reg_Only_str_point_all_Landscape_vars.txt
http://r.789695.n4.nabble.com/file/n4656015/Raw_data_4_logit_reg_Only_str_point_all_Landscape_vars.txt

--
View this message in context:
http://r.789695.n4.nabble.com/Error-in-mer-finalize-ans-Downdated-X-X-is-not-positive-definite-8-tp4656015.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] Working with regular expression

2013-01-18 Thread David Alston

Greetings!

 I hope you don't mind, Rui Barradas, but I'd like to explain the
regex.  Parsing it was a fun exercise!

 Here's the regex broken into two parts..

[[:alpha:]_]*   = match zero or more alphabet or underscore characters
(.*)= match zero or more characters and add them to \1
pattern buffer


Going character by character through the date string asdf May 09 2009

asdf matches the first part
May 09 2009 matches the second part and is stored in the \1 pattern buffer


The gsub command -  gsub([[:alpha:]_]*(.*), \\1, Text)   -
replaces the entire string (because this regex matches the entire
string.. they all begin with a sequence of alphabet and/or underscore
characters and the .* pattern at the end matches the rest of the
line) with the contents of the \1 pattern buffer and stores it in the
variable Text.


 If the length of the string prepended to the date is consistent
another possible solution would be -   gsub(.{5}(.*), \\1, Text)
- which would strip off the first five characters  (.{5} matches
five any characters).


--David Alston
Without rules there  is no game for it is by the rules the game is defined.
  --SOv

On Fri, Jan 18, 2013 at 3:05 PM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:
 Thanks Rui for your help.

 Could you also explain the underlying logic please?

 Thanks and regards,


 On Sat, Jan 19, 2013 at 2:43 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
 gsub([[:alpha:]_]*(.*), \\1, Text)

 __
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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
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Re: [R] Working with regular expression

Hi, 
Not sure what format you wanted the dates: 
gsub(^\\w+ ,,gsub([_], ,Text)) 
#[1] May 09 2009 01-01-2001 

#Another way is: 
gsub(^\\w+ |\\w+_,,Text) 
#[1] May 09 2009 01-01-2001 
res- gsub(^\\w+ |\\w+_,,Text) 
res1-c(as.Date(res[grep( ,res)],format=%b %d %Y), as.Date(res[-grep( 
,res)],format=%m-%d-%Y)) 
res1 
#[1] 2009-05-09 2001-01-01 
A.K. 



- Original Message -
From: Christofer Bogaso bogaso.christo...@gmail.com
To: r-help r-help@r-project.org
Cc: 
Sent: Friday, January 18, 2013 3:11 PM
Subject: [R] Working with regular expression

Hello again,

I was trying to extract the date element from a character vector using
Regular expression. Below is a sample what I was trying to do:

 Text - c(asdf May 09 2009, dsaf_01-01-2001)
 gsub([a-zA-Z_], , Text)
[1]   09 2009  01-01-2001


I basically have a long text and the task is to extract the date
element. My above strategy failed because I can not extract the month
for the first element.

Can somebody help me out?

Thanks and regards,

__
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Re: [R] columns called X rename Y

2013-01-18 Thread Jeff Newmiller

If you have a lot of names to change, or you need to do it regularly, you can 
use something like:

renames - read.table(text=
oldname newname
constant c
numbers b
, header=TRUE, as.is=TRUE )
names(seba)[match(renames$oldname, names(seba))] - renames$newnames

---
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.

Rui Barradas ruipbarra...@sapo.pt wrote:

Hello,

Try the following.


names(seba)[grep(numbers, names(seba))] - b
names(seba)[grep(constant, names(seba))] - c

names(seba)


Hope this helps,

Rui Barradas

Em 18-01-2013 18:14, Sebastian Kruk escreveu:
 I have a data. frame to which you want to change the names to some of
their
 columns.

 For example:

 seba - data.frame ('constant' = 3, 'numbers' = 1: 10, 'letters' =
LETTERS
 [1:10], otros = 2:11)

 List their names:

 names (Seba)
 [1] constant numbers letters

 I want to rename c the column called constant and b the column
 called numbers.

 I do not know in which order appear names.

 How could do it?

 Thanks in advance,

 Sebastian.

  [[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.


__
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[R] Object created within a function disappears after the function is run

2013-01-18 Thread mtb954

Dear R-helpers,

I have run the code below which I expected to make an object called dd1,
but that object does not exist.

So, in summary, my problem is that my function is meant to make an object
(dd1), and it does indeed make that object (I know that the last line of
the function prints it out) but then, after the function has run, the
object has disappeared.

It's late on a Friday so I may be overlooking something obvious, but I'd
appreciate your help if you can see what I'm doing wrong.

Many thanks,

Mark Na



rm(list=ls())

#LOAD package, data, and keep clean copies of data
library(reshape2); library(MuMIn);library(AICcmodavg)
birdmatrix-read.csv(birdmatrix.csv,stringsAsFactors=FALSE);
birdmatrixclean-birdmatrix
habitatmatrix-read.csv(habitatmatrix.csv,stringsAsFactors=FALSE);
habitatmatrixclean-habitatmatrix
traits-read.csv(traits.csv,stringsAsFactors=FALSE); traitsclean-traits
birdhabitat-merge(birdmatrix,habitatmatrix,by=SARBASINYEAR) #merge bird
and habitat data
data-birdhabitat; data$SAR-substr(data$SARBASINYEAR,1,3) #convenience;
create SAR variable

#make a FUNCTION to combine variables into models
dredgeit-function(lm1){
dd1-dredge(lm1,eval=FALSE) #create all possible combinations
dd1-gsub(formula = ,,dd1,fixed=TRUE) #delete characters
dd1-gsub( + 1,,dd1,fixed=TRUE) #delete characters
dd1#inspect model formulae
}

lm1-lm(data$BiRich.o~log(data$HaArea,10)+log(data$HaPeri,10)) #saturated
submodel
dredgeit(lm1)

[[alternative HTML version deleted]]

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

Re: [R] Object created within a function disappears after the function is run

2013-01-18 Thread mtb954

Hi Sarah, that works perfectly, thank you!

Mark Na


On Fri, Jan 18, 2013 at 5:23 PM, Sarah Goslee sarah.gos...@gmail.comwrote:

 Hi,

 You need to assign the result of the function to an object:

 dd1 - dredgeit(lm1)

 On Fri, Jan 18, 2013 at 6:17 PM,  mtb...@gmail.com wrote:
  Dear R-helpers,
 
  I have run the code below which I expected to make an object called dd1,
  but that object does not exist.
 
  So, in summary, my problem is that my function is meant to make an object
  (dd1), and it does indeed make that object (I know that the last line of
  the function prints it out) but then, after the function has run, the
  object has disappeared.
 
  It's late on a Friday so I may be overlooking something obvious, but I'd
  appreciate your help if you can see what I'm doing wrong.
 
  Many thanks,
 
  Mark Na
 
 
 
  rm(list=ls())
 
  #LOAD package, data, and keep clean copies of data
  library(reshape2); library(MuMIn);library(AICcmodavg)
  birdmatrix-read.csv(birdmatrix.csv,stringsAsFactors=FALSE);
  birdmatrixclean-birdmatrix
  habitatmatrix-read.csv(habitatmatrix.csv,stringsAsFactors=FALSE);
  habitatmatrixclean-habitatmatrix
  traits-read.csv(traits.csv,stringsAsFactors=FALSE);
 traitsclean-traits
  birdhabitat-merge(birdmatrix,habitatmatrix,by=SARBASINYEAR) #merge
 bird
  and habitat data
  data-birdhabitat; data$SAR-substr(data$SARBASINYEAR,1,3) #convenience;
  create SAR variable
 
  #make a FUNCTION to combine variables into models
  dredgeit-function(lm1){
  dd1-dredge(lm1,eval=FALSE) #create all possible combinations
  dd1-gsub(formula = ,,dd1,fixed=TRUE) #delete characters
  dd1-gsub( + 1,,dd1,fixed=TRUE) #delete characters
  dd1#inspect model formulae
  }
 
  lm1-lm(data$BiRich.o~log(data$HaArea,10)+log(data$HaPeri,10)) #saturated
  submodel
  dredgeit(lm1)
 
 --
 Sarah Goslee
 http://www.functionaldiversity.org


[[alternative HTML version deleted]]

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

Re: [R] Working with regular expression


Hello,

Right, thanks for the explanation, it saved me time.

Rui Barradas

Em 18-01-2013 22:50, David Alston escreveu:

Greetings!

  I hope you don't mind, Rui Barradas, but I'd like to explain the
regex.  Parsing it was a fun exercise!

  Here's the regex broken into two parts..

[[:alpha:]_]*   = match zero or more alphabet or underscore characters
(.*)= match zero or more characters and add them to \1
pattern buffer


Going character by character through the date string asdf May 09 2009

asdf matches the first part
May 09 2009 matches the second part and is stored in the \1 pattern buffer


The gsub command -  gsub([[:alpha:]_]*(.*), \\1, Text)   -
replaces the entire string (because this regex matches the entire
string.. they all begin with a sequence of alphabet and/or underscore
characters and the .* pattern at the end matches the rest of the
line) with the contents of the \1 pattern buffer and stores it in the
variable Text.


  If the length of the string prepended to the date is consistent
another possible solution would be -   gsub(.{5}(.*), \\1, Text)
- which would strip off the first five characters  (.{5} matches
five any characters).


--David Alston
Without rules there  is no game for it is by the rules the game is defined.
   --SOv

On Fri, Jan 18, 2013 at 3:05 PM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:

Thanks Rui for your help.

Could you also explain the underlying logic please?

Thanks and regards,


On Sat, Jan 19, 2013 at 2:43 AM, Rui Barradas ruipbarra...@sapo.pt wrote:

gsub([[:alpha:]_]*(.*), \\1, Text)


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


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

Re: [R] columns called X rename Y


On Jan 18, 2013, at 3:13 PM, Jeff Newmiller wrote:

 If you have a lot of names to change, or you need to do it regularly, you can 
 use something like:
 
 renames - read.table(text=
 oldname newname
 constant c
 numbers b
 , header=TRUE, as.is=TRUE )
 names(seba)[match(renames$oldname, names(seba))] - renames$newnames

Thnks Jeff;

I like it, plan to add it to my import scripts, ... although the 'newname' 
column is misspelled.

If you have a set of target variable and a diverse set of input variables that 
need to be mapped to a constant set of names and a re thinking of incorporating 
in a function, it might be better with:

names(newdat)[ match(renames[['oldname']], names(newdat))] - 
renames[['newsource3']]
save(renames, file=name_map_data.rda)

 
 ---
 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.
 
 Rui Barradas ruipbarra...@sapo.pt wrote:
 
 Hello,
 
 Try the following.
 
 
 names(seba)[grep(numbers, names(seba))] - b
 names(seba)[grep(constant, names(seba))] - c
 
 names(seba)

I would think that it might be dangerous if there were multiple column names 
that contained either numbers or data.


 
 
 Hope this helps,
 
 Rui Barradas
 
 Em 18-01-2013 18:14, Sebastian Kruk escreveu:
 I have a data. frame to which you want to change the names to some of
 their
 columns.
 
 For example:
 
 seba - data.frame ('constant' = 3, 'numbers' = 1: 10, 'letters' =
 LETTERS
 [1:10], otros = 2:11)
 
 List their names:
 
 names (Seba)
 [1] constant numbers letters
 
 I want to rename c the column called constant and b the column
 called numbers.
 
 I do not know in which order appear names.
 
 How could do it?
 
 Thanks in advance,
 
 Sebastian.
 
 [[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
 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
 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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[R] Tinn-R and R problem

2013-01-18 Thread Roslina Zakaria

Dear r-users,
 
Actually, this is not a big problem, however it is a bit annoying.  Everytime I 
want to use Tinn-R and R. I have to do this step first before I can excute a 
set of R codes:
 
R-- configure -- temporarily (current session)
 
if not it will give this message:
 
 .trPaths - paste(paste(Sys.getenv('APPDATA'), '\\Tinn-R\\tmp\\', sep=''), 
 c('', 'search.txt', 'objects.txt', 'file.r', 'selection.r', 'block.r', 
 'lines.r'), sep=''
 
 source('C:/Users/Owner/AppData/Roaming/Tinn-R/custom/Rconfigure_default.r')
 
Hope somebody can explain to me and help me to solve this problem?
 
R version 2.12.1 (2010-12-16)
 
Thank you so much for your help.
[[alternative HTML version deleted]]

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[R] an apply question

2013-01-18 Thread m p

Hello,
It should be easu but I cannot figure out how to use apply function. I am
trying to replace negative values in an array with these values + 24.
Would appreciate help. Thanks,
Mark

shours - apply(fhours, function(x){if (x  0) x - x+24})
Error in match.fun(FUN) : argument FUN is missing, with no default

[[alternative HTML version deleted]]

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

Re: [R] an apply question

2013-01-18 Thread Jorge I Velez

Hi m p,

You can use either

apply(fhours, 2, function(x) ifelse(x  0, x + 24, x)

or

shours - fhours
shours[shours  0 ] - shours[shours  0 ] + 24
shours


HTH,
Jorge.-


On Sat, Jan 19, 2013 at 4:17 PM, m p  wrote:

 Hello,
 It should be easu but I cannot figure out how to use apply function. I am
 trying to replace negative values in an array with these values + 24.
 Would appreciate help. Thanks,
 Mark

 shours - apply(fhours, function(x){if (x  0) x - x+24})
 Error in match.fun(FUN) : argument FUN is missing, with no default

 [[alternative HTML version deleted]]

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[[alternative HTML version deleted]]

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Re: [R] reading multiple key=value pairs per line

2013-01-18 Thread Frank Singleton


Thanks Arun,Petr,Greg and Christophe,

I have some possibilities to consider now :-)

Cheers / Frank

On 01/18/2013 08:37 AM, arun wrote:

Hi,

Sorry, there was a mistake.  I didn't notice comma in key3
.Lines1-readLines(textConnection('key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks'))
res1-read.table(text=gsub(key{0,1}\\d,,gsub([\],,Lines1)),sep==,header=FALSE,stringsAsFactors=F)[-1]
  res1[,1:2]-sapply(res1[,1:2],function(x) as.numeric(gsub(\\, $,,x)))
names(res1)- paste(substr(Lines1,1,3),1:3,sep=)
res1
#  key1 key2 key3
#1   23   67  hello there
#27   22  how are you
#32   77 nice day, thanks
  str(res1)
#'data.frame':3 obs. of  3 variables:
# $ key1: num  23 7 2
# $ key2: num  67 22 77
# $ key3: chr  hello there how are you nice day, thanks
A.K.



- Original Message -
From: Frank Singleton b17fly...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Thursday, January 17, 2013 11:21 PM
Subject: [R] reading multiple key=value pairs per line

Hi,

Thanks for a great environmentfor statistical  computing :-)

I have some input data in a file (input_kvpairs.csv) of the form

key1=23, key2=67, key3=hello there
key1=7, key2=22, key3=how are you
key1=2, key2=77, key3=nice day, thanks

Now in my head I wish it was of the form (input.csv)

#key1, key2, key3
23,67,   hello there
7, 22,   how are you
2, 77,   nice day, thanks

so I could do

data - read.csv(input.csv, header=TRUE)

where the header column names are derived from the key names dynamically,
and I could access the data using normal data$key1 or data$key2 mechanism.

I guess I could just pre process the file first  using python etc to create
a CSV file with column header derived from key names, and values derived from
key values, but I am interested to see how experienced R folks would handle this
inside R.

Thanks,

Frank

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

Re: [R] random effects model

2013-01-18 Thread rex2013

Hi AK

I got the gg plots , thanks. All of a sudden I just got it. I didn't even
update packages. Not sure what worked.

Just a basic confusion, to find out if in the association between HiBP and
Obese/Overweight status, to quantify if obese males and obese females run a
different risk of becoming hypertensive, will the gee output of

BP.gee - gee(HiBP~Obese*
Sex ,data=BPsub3,id=CODEA,family=binomial,
corstr=exchangeable,na.action=na.omit)

suffice?

Cheers.





 
On Thu, Jan 17, 2013 at 2:01 PM, arun kirshna [via R] 
ml-node+s789695n4655802...@n4.nabble.com wrote:

 HI,

 In the same link at the bottom of the page,

 

 All is well now after updating all packages with the following:
 update.packages()

 It may or may not solve your problem.

 I got your attachments.  You should post those questions in ([hidden
 email] http://user/SendEmail.jtp?type=nodenode=4655802i=0).  I
 suggest you to read lme4 book (http://lme4.r-forge.r-project.org/lMMwR/)
 #lrgprt.pdf

 A.K.







 - Original Message -
 From: rex2013 [hidden 
 email]http://user/SendEmail.jtp?type=nodenode=4655802i=1

 To: [hidden email] http://user/SendEmail.jtp?type=nodenode=4655802i=2
 Cc:
 Sent: Wednesday, January 16, 2013 5:06 AM
 Subject: Re: [R] random effects model

 Hi

 I tried removing the missing values and installing plyr. Still error
 message appears with ggplot2

 Btw, did you get the attachments with my earlier mail?

 Ta.

 On Wed, Jan 16, 2013 at 3:16 AM, arun kirshna [via R] 
 [hidden email] http://user/SendEmail.jtp?type=nodenode=4655802i=3
 wrote:

 
 
  Hi,
  Check these links:
  http://comments.gmane.org/gmane.comp.lang.r.ggplot2/6527
  https://groups.google.com/forum/#!msg/ggplot2/nfVjxL0DXnY/5zf50zCeZuMJ
  A.K.
 
  
  From: Usha Gurunathan [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=0
 
  To: arun [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=1
 
  Cc: R help [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=2
 
  Sent: Tuesday, January 15, 2013 6:31 AM
  Subject: Re: [R] random effects model
 
 
  Hi AK
 
  Got an error message with
  library(ggplot2) 
 
 ggplot(BP.stack1,aes(x=factor(HiBP),fill=Obese))+geom_bar(position=fill)
  Error in rename(x, .base_to_ggplot, warn_missing = FALSE) :  could not
 find
  function revalue 
 
 ggplot(BP.stack1,aes(x=factor(HiBP),fill=Overweight))+geom_bar(position=fill)

  Error in rename(x, .base_to_ggplot, warn_missing = FALSE) :  could not
 find
  function revalue
  I got the dot plot, thanks for that.
 
  I have attached some plots, not sure how to interpret, they had unusual
  patterns.Is it because of missing data? I tried removing the missing
 data
  too. They still appeared the same. Do I need to transform the data?
 
 
  Thanks in advance.
 
 
 
 
 
  On Tue, Jan 15, 2013 at 8:54 AM, arun [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=3
  wrote:
 
  HI,
 
  
  
  BP_2b-read.csv(BP_2b.csv,sep=\t)
  BP_2bNM-na.omit(BP_2b)
  
  BP.stack3 -
 
 reshape(BP_2bNM,idvar=CODEA,timevar=time,sep=,varying=list(c(Obese14,Obese21),c(Overweight14,Overweight21),c(hibp14,hibp21)),v.names=c(Obese,Overweight,HiBP),times=factor(c(1,2)),direction=long)

 
  library(car)
  BP.stack3$Obese- recode(BP.stack3$Obese,1='Obese';0='Not Obese')
  BP.stack3$Overweight-
 recode(BP.stack3$Overweight,1='Overweight';0='Not
  Overweight')
  
  library(ggplot2)
 
 ggplot(BP.stack3,aes(x=factor(HiBP),fill=Obese))+geom_bar(position=fill)
 
 
 ggplot(BP.stack3,aes(x=factor(HiBP),fill=Overweight))+geom_bar(position=fill)

 
  
  You could try lmer() from lme4.
  library(lme4)
  fm1-lmer(HiBP~time+(1|CODEA), family=binomial,data=BP.stack3) #check
  codes, not sure
  print(dotplot(ranef(fm1,post=TRUE),
scales = list(x = list(relation = free)))[[1]])
  qmt1- qqmath(ranef(fm1, postVar=TRUE))
  print(qmt1[[1]])
  
  
  A.K.
  
  
  
  
  
  
  From: Usha Gurunathan [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=4
 
  To: arun [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=5
 
  Cc: R help [hidden email]
 http://user/SendEmail.jtp?type=nodenode=4655612i=6
 
  Sent: Monday, January 14, 2013 6:32 AM
  
  Subject: Re: [R] random effects model
  
  
  Hi AK
  
  I have been trying to create some plots. All being categorical
 variables,
  I am not getting any luck with plots. The few ones that have worked are
  below:
  
  barchart(~table(HiBP)|Obese,data=BP.sub3) ## BP.sub3 is the stacked
 data
  without missing values
  
  barchart(~table(HiBP)|Overweight,data=BP.sub3)
  
 
 plot(jitter(hibp14,factor=2)~jitter(Obese14,factor=2),col=gray,cex=0.7,
  data=Copy.of.BP_2)  ## Copy.of.BP_2 is the original wide format
  
  ## not producing any good plots with mixed models as well.
  summary(lme.3 - lme(HiBP~time, data=BP.sub3,random=~1|CODEA,
  na.action=na.omit))
  anova(lme.3)
  head(ranef(lme.3))
  print(plot(ranef(lme.3))) ##
  
  Thanks

Re: [R] an apply question