Re: [R] yaxp problem for more irregular time series in one plot

[Poersching]

lanc...@fns.uniba.sk schrieb:
 Good day,

 I'm trying to get more time series in one plot. As there are bigger
 differences in values of variables I need logaritmic y axis.

 The code I use is the following:

 nvz_3_data - read.csv('/home/tomas/R_outputs/nvz_3.csv')
 date - (nvz_3_data$date)
 NO3 - (nvz_3_data$NO3)
 NH4 - (nvz_3_data$NH4)
 date_p - as.POSIXct(date, CET)
 par(mfrow=c(2,1), ylog = TRUE, yaxp = c(0.01, 100, 3))
 plot(date_p, NO3, log = y, type = l, col = darkred, main = NVZ-1,
 xlab = time, ylab = NO3- )
 lines(date_p, NH4, col = darkblue, lty = dotted)
 plot(date_p, NH4, log = y, type = l, col = darkblue, main = NVZ-1,
 xlab = time, ylab = NH4+ )

 So, as I anderstood, extreme (max and min) values on the y axis are
 conntrolled byt the yaxp, but it is ignored on the plot, and the NH4
 values are out of the plot (see the attached picture). Do somebody know
 what I am doing wrong?

 Many thanks in advance

 Tomas

   

 

 

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Hey Thomas,

the yaxp command isn't ignored. The problem is the boundary of your
y-data. Your plot is adjusted to the NO3 data and because the higher
values of your NH4 you won't be able to see it on the plot. A solution
lies in implementing the ylim command in your plot.
Here is your changed code:

nvz_3_data - read.csv('/home/tomas/R_outputs/nvz_3.csv')
date - (nvz_3_data$date)
NO3 - (nvz_3_data$NO3)
NH4 - (nvz_3_data$NH4)
maxy-max(NO3,NH4)   ## the maximum value of your data
miny-min(NO3[NO30],NH4[NH40]) ## the minimum value of your data
which are  0 (because: log)
date_p - as.POSIXct(date, CET)
par(mfrow=c(2,1), ylog = TRUE, yaxp = c(0.01, 100, 3))
plot(date_p, NO3,log = y, ylim=c(miny,maxy), type = l, col = darkred,
 main = NVZ-1, xlab = time, ylab = NO3- )
lines(date_p, NH4, col = darkblue, lty = dotted)
plot(date_p, NH4, log = y, type = l, col = darkblue, main = NVZ-1,
xlab = time, ylab = NH4+ )

I hope, I was able to help you.

Regards,
Christian


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Re: [R] graphs

[Poersching]

Mary A. Marion schrieb:
 Hello,

 I am plotting two distributions and want to draw a vertical line at
 the critical point 149.
 How can I stop it from going further up than the norm(140,15) curve?

 x-seq(75,225,0.1)
 plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
 abline(v = 149, col = black)
 curve(dnorm(x,mean=150, sd=15),from=75, to=225, col='orange', add=TRUE)

 Thank you.

 Sincerely,
 Mary A. Marion

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Hey,
in your case you shouldn't use the abline command. Instead try segment like:
segments(149, 0, 149, dnorm(149,mean=140, sd=15))

Regards,
Christian Porsche



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Re: [R] graphs

[Radha Krishna]

Hi Mary,

One can use arrows too...

Here is the code :

x-seq(75,225,0.1)
plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
*arrows(149,0,149,dnorm(149,140,15),length=0)
*par(new=T)
plot(x,dnorm(x,mean=150, sd=15), type='l', col='orange',axes=F)


Regards
Radha


On Sun, Jul 26, 2009 at 5:09 AM, Mary A. Marion mms...@comcast.net wrote:

 Hello,

 I am plotting two distributions and want to draw a vertical line at the
 critical point 149.
 How can I stop it from going further up than the norm(140,15) curve?

 x-seq(75,225,0.1)
 plot(x,dnorm(x,mean=140, sd=15), type='l', col='navy')
 abline(v = 149, col = black)
 curve(dnorm(x,mean=150, sd=15),from=75, to=225, col='orange', add=TRUE)

 Thank you.

 Sincerely,
 Mary A. Marion

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Re: [R] Sweave, cacheSweave, and data frame

[Shige Song]

Dear Gabor,

Thanks for the suggestion. I am writing a research paper using Sweave, not
building a R package. If i understand correctly, the --no-vignettes option
does not really help in my case.

Shige

On Sun, Jul 26, 2009 at 11:56 AM, Gabor Grothendieck 
ggrothendi...@gmail.com wrote:

 Another thing you can do to save time is to use the --no-vignettes switch
 when you build the package.

 On Sat, Jul 25, 2009 at 11:05 PM, Shige Songshiges...@gmail.com wrote:
  Dear All,
 
  I have been using Sweave (mainly via the Sweave.sh script) and really
 like
  it. I am working a paper (using Sweave, of course) which includes several
  time-consuming computations, and it gets tedious to re-compile the whoel
  thing every time I made changes. Then I discover the cacheSweave
 package,
  which seems the right solution to my problem. I only have on problem.
 Here
  is what I did:
 
  --
  results=hide,echo=false=
  library(foreign)
  library(Zelig)
  library(memisc)
  options(digits=4)
  @
 
  echo=false=
  d - read.dta(~/project/abortion/data/data_transition_wide.dta)
  @
 
  ...
 
  --
 
  It can be compiled using Sweave.sh foo.Rnw, but when I tried Sweave.sh
 -c
  foo.Rnw, I got error message:
   ...Processing code chunks ...
  1 : term hide
  Error in data.frame(chunk = options$label, chunkprefix = chunkprefix,  :
   arguments imply differing number of rows: 0, 1
 
  I thought it might be an imcompatibiltiy problem between the cacheSweave
 and
  the foreign packages, but the problem was still there when I tried to
 read
  the data in using read.table function.
 
  Any ideas? Many thanks.
 
  Best,
  Shige
 
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Re: [R] Sweave, cacheSweave, and data frame

[Gabor Grothendieck]

Right. Its to avoid making the vignettes when you are mainly interested
in testing out a package.

On Sun, Jul 26, 2009 at 3:07 AM, Shige Songshiges...@gmail.com wrote:
 Dear Gabor,

 Thanks for the suggestion. I am writing a research paper using Sweave, not
 building a R package. If i understand correctly, the --no-vignettes option
 does not really help in my case.

 Shige

 On Sun, Jul 26, 2009 at 11:56 AM, Gabor Grothendieck
 ggrothendi...@gmail.com wrote:

 Another thing you can do to save time is to use the --no-vignettes switch
 when you build the package.

 On Sat, Jul 25, 2009 at 11:05 PM, Shige Songshiges...@gmail.com wrote:
  Dear All,
 
  I have been using Sweave (mainly via the Sweave.sh script) and really
  like
  it. I am working a paper (using Sweave, of course) which includes
  several
  time-consuming computations, and it gets tedious to re-compile the whoel
  thing every time I made changes. Then I discover the cacheSweave
  package,
  which seems the right solution to my problem. I only have on problem.
  Here
  is what I did:
 
  --
  results=hide,echo=false=
  library(foreign)
  library(Zelig)
  library(memisc)
  options(digits=4)
  @
 
  echo=false=
  d - read.dta(~/project/abortion/data/data_transition_wide.dta)
  @
 
  ...
 
  --
 
  It can be compiled using Sweave.sh foo.Rnw, but when I tried
  Sweave.sh -c
  foo.Rnw, I got error message:
   ...Processing code chunks ...
  1 : term hide
  Error in data.frame(chunk = options$label, chunkprefix = chunkprefix,  :
   arguments imply differing number of rows: 0, 1
 
  I thought it might be an imcompatibiltiy problem between the cacheSweave
  and
  the foreign packages, but the problem was still there when I tried to
  read
  the data in using read.table function.
 
  Any ideas? Many thanks.
 
  Best,
  Shige
 
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Re: [R] Determine the dimension-names of an element in an array in R

[Sauvik De]

Hi Gabor:
Many thanks for your prompt reply!
The code is fine. But I need it in more general form as I had mentioned that
I need to input any 0 to find its dimension-names.

Actually, I was using sapply to calculate correlation and this idea was
required in the middle of correlation calculation.
I am providing the way I tried my calculation.

a= c(A1,A2,A3,A4,A5)
b= c(B1,B2,B3)
c= c(C1,C2,C3,C4)
d= c(D1,D2)
e= c(E1,E2,E3,E4,E5,E6,E7,E8)

DataArray_1 = array(c(rnorm(240)),dim=c(length(a),length(b),
length(d),length(e)),dimnames=list(a,b,d,e))
DataArray_2 = array(c(rnorm(320)), dim=c(length(a),length(c),
length(d),length(e)),dimnames=list(a,c,d,e))

#Defining an empty array which will contain the correlation values (output
array)
Correl = array(NA, dim=c(length(a),length(b),
length(c),length(d)),dimnames=list(a,b,c,d))

#Calculating Correlation between attributes b  c over values of e
Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
use=pairwise.complete.obs))

This is where I get stuck.
In the above, d is acting as an element in the Correl array. Hence I need
to get the dimension-names for d.

#The first element of Correl will be:
cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)

So my problem boils down to extracting the dim-names in terms of element(d)
and not in terms of Correl (that I have mentioned as ... in the above
code)

My sincere thanks for your valuable time  suggestions.

Many Thanks  Kind Regards,
Sauvik


On Sun, Jul 26, 2009 at 5:26 AM, Gabor Grothendieck ggrothendi...@gmail.com
 wrote:

 Try this:

  ix - c(1, 3, 4, 2)
  mapply([, dimnames(mydatastructure), ix)
 [1] S1 T3 U4 V2


 On Sat, Jul 25, 2009 at 5:12 PM, Sauvik Desauvik.s...@gmail.com wrote:
  Hi:
  How can I extract the dimension-names of a pre-defined element in a
  multidimensional array in R ?
 
  A toy example is provided below:
  I have a 4-dimensional array with each dimension having certain length.
 In
  the below example, mydatastructure explains the structure of my data.
 
  mydatastructure = array(0,
 dim=c(length(b),length(z),length(x),length(d)),
  dimnames=list(b,z,x,d))
 
  where,
  b=c(S1,S2,S3,S4,S5)
  z=c(T1,T2, T3)
  x=c(U1,U2,U3,U4)
  d=c(V1,V2)
 
  Clearly, mydatastructure contains many 0's.
  Now how can I get the dimension-names of any particular 0 ?
  That is, my input should be a particular 0 in the array mydatastructure
  (Suppose this 0 corresponds to S1,T3,U4  V2 in the array). Then my
 output
  should be S1,T3,U4  V2.
 
  The function dimnames didn't help me with the solution.
  Any idea will greatly be appreciated.
 
  Thanks for your time!
 
  Kind Regards,
  Sauvik
 
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Re: [R] Sweave, cacheSweave, and data frame

[Dieter Menne]

Shige Song wrote:
 
 I have been using Sweave (mainly via the Sweave.sh script) and really like
 it. I am working a paper (using Sweave, of course) which includes several
 
 

To check what's wrong, please post a reproducible minimal example. It is
difficult to guess if this has to do with the data frame or something else
from your post.

Dieter


-- 
View this message in context: 
http://www.nabble.com/Sweave%2C-cacheSweave%2C-and-data-frame-tp24663644p24665116.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Sweave, cacheSweave, and data frame

[Duncan Murdoch]

Shige Song wrote:

Dear All,

I have been using Sweave (mainly via the Sweave.sh script) and really like
it. I am working a paper (using Sweave, of course) which includes several
time-consuming computations, and it gets tedious to re-compile the whoel
thing every time I made changes. Then I discover the cacheSweave package,
which seems the right solution to my problem. I only have on problem. Here
is what I did:
  


I can't help with your question, but another approach to do what you 
want is to have one script that does the time consuming calculations and 
saves the results (using save()) to a file, then have your Sweave 
document load() the object as necessary.  You could then also move the 
read.dta() line to this other script, and load() that data, simplifying 
your Sweave document.


Duncan Murdoch

--
results=hide,echo=false=
library(foreign)
library(Zelig)
library(memisc)
options(digits=4)
@

echo=false=
d - read.dta(~/project/abortion/data/data_transition_wide.dta)
@

...

--

It can be compiled using Sweave.sh foo.Rnw, but when I tried Sweave.sh -c
foo.Rnw, I got error message:
 ...Processing code chunks ...
1 : term hide
Error in data.frame(chunk = options$label, chunkprefix = chunkprefix,  :
 arguments imply differing number of rows: 0, 1

I thought it might be an imcompatibiltiy problem between the cacheSweave and
the foreign packages, but the problem was still there when I tried to read
the data in using read.table function.

Any ideas? Many thanks.

Best,
Shige

[[alternative HTML version deleted]]

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Re: [R] Determine the dimension-names of an element in an array in R

[Poersching]

Sauvik De schrieb:
 Hi Gabor:
 Many thanks for your prompt reply!
 The code is fine. But I need it in more general form as I had mentioned that
 I need to input any 0 to find its dimension-names.

 Actually, I was using sapply to calculate correlation and this idea was
 required in the middle of correlation calculation.
 I am providing the way I tried my calculation.

 a= c(A1,A2,A3,A4,A5)
 b= c(B1,B2,B3)
 c= c(C1,C2,C3,C4)
 d= c(D1,D2)
 e= c(E1,E2,E3,E4,E5,E6,E7,E8)

 DataArray_1 = array(c(rnorm(240)),dim=c(length(a),length(b),
 length(d),length(e)),dimnames=list(a,b,d,e))
 DataArray_2 = array(c(rnorm(320)), dim=c(length(a),length(c),
 length(d),length(e)),dimnames=list(a,c,d,e))

 #Defining an empty array which will contain the correlation values (output
 array)
 Correl = array(NA, dim=c(length(a),length(b),
 length(c),length(d)),dimnames=list(a,b,c,d))

 #Calculating Correlation between attributes b  c over values of e
 Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
 use=pairwise.complete.obs))

 This is where I get stuck.
 In the above, d is acting as an element in the Correl array. Hence I need
 to get the dimension-names for d.

 #The first element of Correl will be:
 cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)

 So my problem boils down to extracting the dim-names in terms of element(d)
 and not in terms of Correl (that I have mentioned as ... in the above
 code)

 My sincere thanks for your valuable time  suggestions.

 Many Thanks  Kind Regards,
 Sauvik


 On Sun, Jul 26, 2009 at 5:26 AM, Gabor Grothendieck ggrothendi...@gmail.com
   
 wrote:
 

   
 Try this:

 
 ix - c(1, 3, 4, 2)
 mapply([, dimnames(mydatastructure), ix)
   
 [1] S1 T3 U4 V2


 On Sat, Jul 25, 2009 at 5:12 PM, Sauvik Desauvik.s...@gmail.com wrote:
 
 Hi:
 How can I extract the dimension-names of a pre-defined element in a
 multidimensional array in R ?

 A toy example is provided below:
 I have a 4-dimensional array with each dimension having certain length.
   
 In
 
 the below example, mydatastructure explains the structure of my data.

 mydatastructure = array(0,
   
 dim=c(length(b),length(z),length(x),length(d)),
 
 dimnames=list(b,z,x,d))

 where,
 b=c(S1,S2,S3,S4,S5)
 z=c(T1,T2, T3)
 x=c(U1,U2,U3,U4)
 d=c(V1,V2)

 Clearly, mydatastructure contains many 0's.
 Now how can I get the dimension-names of any particular 0 ?
 That is, my input should be a particular 0 in the array mydatastructure
 (Suppose this 0 corresponds to S1,T3,U4  V2 in the array). Then my
   
 output
 
 should be S1,T3,U4  V2.

 The function dimnames didn't help me with the solution.
 Any idea will greatly be appreciated.

 Thanks for your time!

 Kind Regards,
 Sauvik

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

   

   [[alternative HTML version deleted]]

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Hey,

I have spend some time to write a function, which should fulfill your
needs.
so i hope ;-)

findIndex-function(data,element) {
  ld-length(data)
  el-which(is.element(data,element))
  lel-length(el)
  ndim-length(dim(data))
  ind-array(,dim=c(lel,ndim),dimnames=list(el,1:ndim))
  precomma-
  tempdata-data
  tempel-el
  for (j in 1:lel) {
data-tempdata
el-tempel
ld-length(data)
for (i in ndim:1) {
  ratio-el[j]/(ld/dim(data)[i])
  if (ratio-trunc(ratio)0) {
ind[j,i]-trunc(ratio)+1
  } else {
ind[j,i]-trunc(ratio)
  }
  if (length(dim(data))1) {
k-1
while (k=1  k=(i-1)) {
  precomma-paste(precomma,,,sep=)
  k-k+1
}
   
data-as.array(eval(parse(text=paste(data[,precomma,ind[j,i],],sep=
precomma-
ld-length(data)
el[j]-which(is.element(data,element))
  }
}
  }
  return(ind)
}

Regards,
Christian Porsche

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Re: [R] zoo plot: yearly marks on X-Axis

[Poersching]

stvienna wiener schrieb:
 Hi all,

 I am plotting a financial time series, but I need a more detailed X-Axis.

 Example:
 x - zoo(rnorm(1:6000), as.Date(1992-11-11)+c(1:6000))
 plot(x)

 The X-Axis is labeled 1995, 2000 and 2005.
 I would need either 1995, 1997, etc. or maybe yearly


 I used google first, then look at ?plot.zoo but could't get it working.


 Regards,
 Steve

 __
 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.

   
Hey,

try something like the following:

plot(x, y, xaxt=n)
axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), years),
labels = FALSE, tcl = -0.2)
axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), 5 years),
labels = TRUE, las=3, tcl = -0.2)

Regards,
Christian

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[R] Is there an R implementation for the Barnard's exact test (a substitute for fisher.test) ?

[Tal Galili]

Hello R help members. I came across today with an article on Barnard's exact
test (http://www.cytel.com/Papers/twobinomials.pdf), that is supposed to
give a more powerful fisher.test - Because it doesn't assume that we know
the row and column totals are in advance. Any pointers to such a function ?
Thanks, Tal




-- 
--


My contact information:
Tal Galili
Phone number: 972-50-3373767
FaceBook: Tal Galili
My Blogs:
http://www.r-statistics.com/
http://www.talgalili.com
http://www.biostatistics.co.il

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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] Determine the dimension-names of an element in an array in R

[Sauvik De]

Hi:
Lots of thanks for your valuable time!

But I am not sure how you would like to use the function in this situation.

As I had mentioned that the first element of my output array should be like:

cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)

in my below code.

and

the output array of correlation I wish to get using sapply as follows:

Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
use=pairwise.complete.obs))

So it would be of great help if you could kindly specify how to utilise your
function findIndex in ...

Apologies for all this!

Thanks  Regards,
Sauvik


On Sun, Jul 26, 2009 at 3:54 PM, Poerschingpoerschin...@web.de wrote:
 Sauvik De schrieb:

 Hi Gabor:
 Many thanks for your prompt reply!
 The code is fine. But I need it in more general form as I had mentioned
that
 I need to input any 0 to find its dimension-names.

 Actually, I was using sapply to calculate correlation and this idea was
 required in the middle of correlation calculation.
 I am providing the way I tried my calculation.

 a= c(A1,A2,A3,A4,A5)
 b= c(B1,B2,B3)
 c= c(C1,C2,C3,C4)
 d= c(D1,D2)
 e= c(E1,E2,E3,E4,E5,E6,E7,E8)

 DataArray_1 = array(c(rnorm(240)),dim=c(length(a),length(b),
 length(d),length(e)),dimnames=list(a,b,d,e))
 DataArray_2 = array(c(rnorm(320)), dim=c(length(a),length(c),
 length(d),length(e)),dimnames=list(a,c,d,e))

 #Defining an empty array which will contain the correlation values (output
 array)
 Correl = array(NA, dim=c(length(a),length(b),
 length(c),length(d)),dimnames=list(a,b,c,d))

 #Calculating Correlation between attributes b  c over values of e
 Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
 use=pairwise.complete.obs))

 This is where I get stuck.
 In the above, d is acting as an element in the Correl array. Hence I
need
 to get the dimension-names for d.

 #The first element of Correl will be:

cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)

 So my problem boils down to extracting the dim-names in terms of
element(d)
 and not in terms of Correl (that I have mentioned as ... in the above
 code)

 My sincere thanks for your valuable time  suggestions.

 Many Thanks  Kind Regards,
 Sauvik


 On Sun, Jul 26, 2009 at 5:26 AM, Gabor Grothendieck 
ggrothendi...@gmail.com


 wrote:




 Try this:



 ix - c(1, 3, 4, 2)
 mapply([, dimnames(mydatastructure), ix)


 [1] S1 T3 U4 V2


 On Sat, Jul 25, 2009 at 5:12 PM, Sauvik Desauvik.s...@gmail.com wrote:


 Hi:
 How can I extract the dimension-names of a pre-defined element in a
 multidimensional array in R ?

 A toy example is provided below:
 I have a 4-dimensional array with each dimension having certain length.


 In


 the below example, mydatastructure explains the structure of my data.

 mydatastructure = array(0,


 dim=c(length(b),length(z),length(x),length(d)),


 dimnames=list(b,z,x,d))

 where,
 b=c(S1,S2,S3,S4,S5)
 z=c(T1,T2, T3)
 x=c(U1,U2,U3,U4)
 d=c(V1,V2)

 Clearly, mydatastructure contains many 0's.
 Now how can I get the dimension-names of any particular 0 ?
 That is, my input should be a particular 0 in the array mydatastructure
 (Suppose this 0 corresponds to S1,T3,U4  V2 in the array). Then my


 output


 should be S1,T3,U4  V2.

 The function dimnames didn't help me with the solution.
 Any idea will greatly be appreciated.

 Thanks for your time!

 Kind Regards,
 Sauvik

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 http://www.R-project.org/posting-guide.html


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



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 Hey,

 I have spend some time to write a function, which should fulfill your
needs.
 so i hope ;-)

 findIndex-function(data,element) {
   ld-length(data)
   el-which(is.element(data,element))
   lel-length(el)
   ndim-length(dim(data))
   ind-array(,dim=c(lel,ndim),dimnames=list(el,1:ndim))
   precomma-
   tempdata-data
   tempel-el
   for (j in 1:lel) {
 data-tempdata
 el-tempel
 ld-length(data)
 for (i in ndim:1) {
   ratio-el[j]/(ld/dim(data)[i])
   if (ratio-trunc(ratio)0) {
 ind[j,i]-trunc(ratio)+1
   } else {
 ind[j,i]-trunc(ratio)
   }
   if (length(dim(data))1) {
 k-1
 while (k=1  

Re: [R] How to add 95% confidence intervals in the calibration plot?

[Frank E Harrell Jr]

zhu yao wrote:

Dear experts:

I am a newbie to R. Recently, I try to make prediction models with R and the
Design library.
I have read Prof. Harrell's excellent book. But I did not quite understand.
I have two problems about the validation and calibration of prediction
models:
1. Can someone explain the results outputted by the validate() function? How
to get 95% of c-value of validate?


validate does not provide that confidence interval, unfortunately.


2. How to add 95% ci in the calibration plot?


That is not provided except for survival models.

Next time please include your code so we can see what model you are using.

Thanks
Frank



Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center

[[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.




--
Frank E Harrell Jr   Professor and Chair   School of Medicine
 Department of Biostatistics   Vanderbilt University

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Re: [R] How to add 95% confidence intervals in the calibration plot?

[zhu yao]

Thanks for your reply.

Actually, I'm confused about the results in the article Postoperative
nomogram for survival of patients with retroperitoneal sarcoma treated with
curative intent
http://annonc.oxfordjournals.org/cgi/content/abstract/mdp298v1

It stated as:

nomogram model
The Cox model was used as the basis for the nomogram
(Table 2). Figure 2 depicts the final nomogram and portrays the
association between each variable and survival based on the
scoring system derived from this analysis. The concordance
index (discrimination) after internal validation with 200
bootstrapping resamples was 0.73 (95% CI 0.71–0.75).
Similarly, Figure 3 illustrates the calibration of the nomogram
before and after internal validation with bootstrapping samples.
Calibration was excellent with observed outcomes always
within 95% CI of the predicted survival probability.

Figure 3 is
http://i3.6.cn/cvbnm/a9/c8/8b/c01aad248a0b4ae6ef677600614bd4fa.jpg


2009/7/26 Frank E Harrell Jr f.harr...@vanderbilt.edu

 zhu yao wrote:

 Dear experts:

 I am a newbie to R. Recently, I try to make prediction models with R and
the
 Design library.
 I have read Prof. Harrell's excellent book. But I did not quite
understand.
 I have two problems about the validation and calibration of prediction
 models:
 1. Can someone explain the results outputted by the validate() function?
How
 to get 95% of c-value of validate?

 validate does not provide that confidence interval, unfortunately.

 2. How to add 95% ci in the calibration plot?

 That is not provided except for survival models.

 Next time please include your code so we can see what model you are using.

 Thanks
 Frank


 Yao Zhu
 Department of Urology
 Fudan University Shanghai Cancer Center

[[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.



 --
 Frank E Harrell Jr   Professor and Chair   School of Medicine
 Department of Biostatistics   Vanderbilt University

[[alternative HTML version deleted]]

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

Re: [R] How to add 95% confidence intervals in the calibration plot?

[Frank E Harrell Jr]

zhu yao wrote:

Thanks for your reply.

Actually, I'm confused about the results in the article Postoperative 
nomogram for survival of patients with retroperitoneal sarcoma treated 
with curative intent

http://annonc.oxfordjournals.org/cgi/content/abstract/mdp298v1

It stated as:

nomogram model
The Cox model was used as the basis for the nomogram
(Table 2). Figure 2 depicts the final nomogram and portrays the
association between each variable and survival based on the
scoring system derived from this analysis. The concordance
index (discrimination) after internal validation with 200
bootstrapping resamples was 0.73 (95% CI 0.71–0.75).
Similarly, Figure 3 illustrates the calibration of the nomogram
before and after internal validation with bootstrapping samples.
Calibration was excellent with observed outcomes always
within 95% CI of the predicted survival probability.


Figure 3 is provided by the Design package without modification.  As I 
stated before it does provide those CIs for survival models.  I guess 
that the CI for the c-index was obtained without bootstrap validation 
using the Hmisc package's rcorr.cens function (and Dxy=2*(C-.5)) or by 
using an approximate bootstrap analysis they programmed.


Note that in the abstract the authors wrongly used the confidence 
intervals in Fig 3 to conclude excellent validation of the model.  Their 
conclusion can arise from just having large confidence intervals.


Frank



Figure 3 is
http://i3.6.cn/cvbnm/a9/c8/8b/c01aad248a0b4ae6ef677600614bd4fa.jpg


2009/7/26 Frank E Harrell Jr f.harr...@vanderbilt.edu 
mailto:f.harr...@vanderbilt.edu

 
  zhu yao wrote:
 
  Dear experts:
 
  I am a newbie to R. Recently, I try to make prediction models with R 
and the

  Design library.
  I have read Prof. Harrell's excellent book. But I did not quite 
understand.

  I have two problems about the validation and calibration of prediction
  models:
  1. Can someone explain the results outputted by the validate() 
function? How

  to get 95% of c-value of validate?
 
  validate does not provide that confidence interval, unfortunately.
 
  2. How to add 95% ci in the calibration plot?
 
  That is not provided except for survival models.
 
  Next time please include your code so we can see what model you are 
using.

 
  Thanks
  Frank
 
 
  Yao Zhu
  Department of Urology
  Fudan University Shanghai Cancer Center
 
 [[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailto: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.

[R] R: R: Is there a way to extract some fields data fromHTML pages through any R function ?

[mauede]

It works if the web file adress is of the type:  http://;.
It does not work if the web file adress is of the type:  'ftp://;.
 outFile - 
 read.xls(ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls;)
Error in xls2csv(xls, sheet, verbose = verbose, ..., perl = perl) : 
  Unable to read xls file 
'ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls'.
Error in file.exists(tfn) : invalid 'file' argument

But the file does exists as shown in the following:
 download.file(ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls,outFile;)
trying URL 'ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls'
ftp data connection made, file length 2563072 bytes
opened URL
downloaded 2.4 Mb

Can the two steps (download + read.xls) be performed with one command line  
only ?

Thank you,
Maura



-Messaggio originale-
Da: r-help-boun...@r-project.org per conto di Daniel Nordlund
Inviato: lun 06/07/2009 20.45
A: r-h...@stat.math.ethz.ch
Oggetto: Re: [R] R: R: Is there a way to extract some fields data fromHTML  
pages through any R function ?
 
 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of mau...@alice.it
 Sent: Sunday, July 05, 2009 11:28 PM
 To: Martin Morgan
 Cc: r-h...@stat.math.ethz.ch
 Subject: [R] R: R: Is there a way to extract some fields data 
 from HTML pages through any R function ?
 
 It helps. But it is overly sophisticated.
 I have already downloaded and used the Excel file containing 
 the validated stuff.
 
 Since there are R commands to download gzip as well as FASTA 
 files, I wonder whether it is possible to 
 automatically download the Excel file from 
 http://mirecords.umn.edu/miRecords/download.php
 Actually the latter may not be the actual file URL because it 
 is necessary to click on the word here to download the file.
 
 Thank you,
 Maura
 
Maura,

I haven't seen a response to your question (however, I just may have missed
it, or you mave have received an off-line response).  I went to the URL
above and found that the Excel file is at

http://mirecords.umn.edu/miRecords/download_data.php?v=1

I think you could use the read.xls() function from the gdata package to get
the file, something like this

library(gdata)
df - read.xls(http://mirecords.umn.edu/miRecords/download_data.php?v=1;)

Hope this is helpful,

Dan

Daniel Nordlund
Bothell, WA USA

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[[alternative HTML version deleted]]

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

Re: [R] R: R: Is there a way to extract some fields data fromHTML pages through any R function ?

[Gabor Grothendieck]

I will send you offline an enhancement for read.xls that accepts
ftp connections.

On Sun, Jul 26, 2009 at 11:32 AM, mau...@alice.it wrote:
 It works if the web file adress is of the type:  http://;.
 It does not work if the web file adress is of the type:  'ftp://;.
 outFile - 
 read.xls(ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls;)
 Error in xls2csv(xls, sheet, verbose = verbose, ..., perl = perl) :
  Unable to read xls file 
 'ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls'.
 Error in file.exists(tfn) : invalid 'file' argument

 But the file does exists as shown in the following:
 download.file(ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls,outFile;)
 trying URL 'ftp://ftp.sanger.ac.uk/pub/mirbase/sequences/CURRENT/miRNA.xls'
 ftp data connection made, file length 2563072 bytes
 opened URL
 downloaded 2.4 Mb

 Can the two steps (download + read.xls) be performed with one command line  
 only ?

 Thank you,
 Maura



 -Messaggio originale-
 Da: r-help-boun...@r-project.org per conto di Daniel Nordlund
 Inviato: lun 06/07/2009 20.45
 A: r-h...@stat.math.ethz.ch
 Oggetto: Re: [R] R: R: Is there a way to extract some fields data fromHTML    
   pages through any R function ?

 -Original Message-
 From: r-help-boun...@r-project.org
 [mailto:r-help-boun...@r-project.org] On Behalf Of mau...@alice.it
 Sent: Sunday, July 05, 2009 11:28 PM
 To: Martin Morgan
 Cc: r-h...@stat.math.ethz.ch
 Subject: [R] R: R: Is there a way to extract some fields data
 from HTML pages through any R function ?

 It helps. But it is overly sophisticated.
 I have already downloaded and used the Excel file containing
 the validated stuff.

 Since there are R commands to download gzip as well as FASTA
 files, I wonder whether it is possible to
 automatically download the Excel file from
 http://mirecords.umn.edu/miRecords/download.php
 Actually the latter may not be the actual file URL because it
 is necessary to click on the word here to download the file.

 Thank you,
 Maura

 Maura,

 I haven't seen a response to your question (however, I just may have missed
 it, or you mave have received an off-line response).  I went to the URL
 above and found that the Excel file is at

 http://mirecords.umn.edu/miRecords/download_data.php?v=1

 I think you could use the read.xls() function from the gdata package to get
 the file, something like this

 library(gdata)
 df - read.xls(http://mirecords.umn.edu/miRecords/download_data.php?v=1;)

 Hope this is helpful,

 Dan

 Daniel Nordlund
 Bothell, WA USA

 __
 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.



        [[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.


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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] Collinearity in Linear Multiple Regression

[John P. Burkett]

Stephan Kolassa wrote:

Hi Alex,

I personally have had more success with the (more complicated) 
collinearity diagnostics proposed by Belsley, Kuh  Welsch in their book 
Regression Diagnostics than with Variance Inflation Factors. See also:


Belsley, D. A. A Guide to using the collinearity diagnostics. 
Computational Economics, 1991, 4, 33-50


However, I know of no R package that implements these diagnostics. 
Anyway, it's not hard to do so oneself.


R code utilizing singular value decomposition and variance decomposition 
proportions along lines proposed by D. Belsley, Conditioning Diagnostics 
(1991) is available at http://www.uri.edu/artsci/ecn/burkett/scivdp.R


-John



Good luck!
Stephan


Alex Roy schrieb:

Dear all,
  How can I test for collinearity in the predictor 
data set

for multiple linear regression.

Thanks

Alex

[[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
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.




--
John P. Burkett
Department of Economics
University of Rhode Island
Kingston, RI 02881-0808
USA

phone (401) 874-9195

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Re: [R] suggestion for paired t-tests

[Ben Bolker]

jacktanner wrote:
 
 There's a funny inconsistency in how t.test handles paired=T or paired=F.
 If x
 and y parameters are lists, paired=F works, but paired=T doesn't.
 
 lg=read.csv(my.csv)
 a = subset(lg, condition==a)[score]
 b = subset(lg, condition==b)[score]
 t.test(a,b)
 t.test(a,b, paired=TRUE)
 Error in `[.data.frame`(y, yok) : undefined columns selected
 
 But this works
 a=a[,1]
 b=b[,1]
 t.test(a,b, paired=TRUE)
 ...
 
 

It's sort of an accident that this works for the unpaired case.
You can follow what happens via debug(stats:::t.test.default) ...
there is some code

 if (paired) 
xok - yok - complete.cases(x, y)
else {
yok - !is.na(y)
xok - !is.na(x)
}

if paired is FALSE, !is.na(y) and !is.na(x) happen to convert x and y
into matrices, whence they can be used for the rest of the computations.
If paired is TRUE, x and y remain data frames.

  Bottom line: a data frame with a single column in it really isn't the
same as a vector ...
-- 
View this message in context: 
http://www.nabble.com/suggestion-for-paired-t-tests-tp24651851p24668046.html
Sent from the R help mailing list archive at Nabble.com.

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

[R] moving text labels in plot

[Luis Iván Ortiz Valencia]

Hi R users

I need to specify some parameter input in plot code to move Y text label to
left.

plot(temp, develo_rate, xlab = expression(paste(Temperature (C^o,))),
ylab = expression(paste(Development rate (d^-1,))),las=1,pch=19,
xlim=c(0,32),ylim=c(0,0.03),xaxs = i, yaxs = i)

Plot result is added.

any help?

Ivan
attachment: scater_temp_desen_rate.png__
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[R] Non-Linear Regression with two Predictors

[Berlinerfee]

Hello there,

I am using nls the first time for a non-linear regression with a 
logistic growth function:

startparam - c(alpha=3e+07,beta=4000,gamma=2)
fit - nls(dataset$V2~(( alpha / ( 1 + exp( beta - gamma * dataset$V1 ) 
) ) ),data=dataset,start=startparam)


Everything works fine and i get good results. Now I would like to 
improve the results using my DUMMY Variable (dataset$V6) the runs half 
of the time 0 and then 1. This is my new nls:

startparam - c(alpha=3e+07,beta=4000,gamma=2,delta=100)
fit - nls(dataset$V2~(( alpha / ( 1 + exp( beta - gamma * dataset$V1 ) 
) ) + (dataset$V6*dataset$V1*delta) ),data=dataset,start=startparam)


I get Singular Gradient Matrice. May anyone give me the right nls 
function for this problem??


Regards

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[R] problems hist() and density

[Jan Teichmann]

Hello,

I have a problem with the hist() function and showing densities. The
densities sum to 50 and not to 1! I use R version 2.9.1 (2009-06-26) and
I load the seqinR library.

My data is the following vector:
[1] 0.140 0.200 0.220 0.2828283 0.160 0.160
0.360
[8] 0.160 0.220 0.260 0.200 0.300 0.220
0.2342342
[15] 0.180 0.220 0.160 0.230 0.200 0.220
0.240
[22] 0.200 0.220 0.220 0.260 0.200 0.160
0.220
[29] 0.2342342 0.200 0.220 0.200 0.200 0.140
0.180
[36] 0.220 0.160 0.160 0.140 0.220 0.200
0.2871287
[43] 0.290 0.200 0.1836735 0.200 0.200 0.290
0.240
[50] 0.220 0.280 0.200 0.2745098 0.220 0.230
0.180
[57] 0.230 0.180 0.260 0.220 0.222 0.220
0.260
[64] 0.220 0.220 0.260 0.220 0.200 0.220

I use the following command:
tmp - hist(data, freq=FALSE, plot=FALSE)

and that's the result:
$breaks
 [1] 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36

$counts
 [1] 10  4 15 19  8  5  2  5  0  0  1

$intensities
 [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
 [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

$density
 [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
 [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

$mids
 [1] 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35

$xname
[1] data

$equidist
[1] TRUE

attr(,class)
[1] histogram
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[R] smoothScatter problems

[Jeroen van der Ham]

Hello,

I'm having some trouble getting a good result for a smoothScatter plot.
I have some data that I want to log-plot, but when I use smoothScatter
the result is not correct.
The problem seems to be that with the log=x argument smoothScatter
calculates the bins linearly, so the plot will be skewed towards the right.

See for example:

file(http://dckd.nl/~jeroen/drop/example.rdata;)
smoothScatter(d,log=x)
smoothScatter(log(d$x),d$y)

I could also use the latter way to produce the result, but I would like
to use the original units on the x-axis, not the log units.

I also want to use these results in a publication, but the PDFs saved
from these results are not very nice. With Acrobat there are lots of
blocks, and with other viewers there are many white lines through the
blue colours.

Thanks,
Jeroen.

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Re: [R] problems hist() and density

[jim holtman]

It is the 'area' under the curve that sums to zero.  Look at what the
difference is between the 'breaks' (0.02).  multiply this by 50 and
you get 1.

On Sun, Jul 26, 2009 at 7:43 AM, Jan
Teichmannjan.teichm...@googlemail.com wrote:
 Hello,

 I have a problem with the hist() function and showing densities. The
 densities sum to 50 and not to 1! I use R version 2.9.1 (2009-06-26) and
 I load the seqinR library.

 My data is the following vector:
 [1] 0.140 0.200 0.220 0.2828283 0.160 0.160
 0.360
 [8] 0.160 0.220 0.260 0.200 0.300 0.220
 0.2342342
 [15] 0.180 0.220 0.160 0.230 0.200 0.220
 0.240
 [22] 0.200 0.220 0.220 0.260 0.200 0.160
 0.220
 [29] 0.2342342 0.200 0.220 0.200 0.200 0.140
 0.180
 [36] 0.220 0.160 0.160 0.140 0.220 0.200
 0.2871287
 [43] 0.290 0.200 0.1836735 0.200 0.200 0.290
 0.240
 [50] 0.220 0.280 0.200 0.2745098 0.220 0.230
 0.180
 [57] 0.230 0.180 0.260 0.220 0.222 0.220
 0.260
 [64] 0.220 0.220 0.260 0.220 0.200 0.220

 I use the following command:
 tmp - hist(data, freq=FALSE, plot=FALSE)

 and that's the result:
 $breaks
  [1] 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36

 $counts
  [1] 10  4 15 19  8  5  2  5  0  0  1

 $intensities
  [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
  [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

 $density
  [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
  [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

 $mids
  [1] 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35

 $xname
 [1] data

 $equidist
 [1] TRUE

 attr(,class)
 [1] histogram

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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] problems hist() and density

[jim holtman]

sums to one  I should have said.

On Sun, Jul 26, 2009 at 7:43 AM, Jan
Teichmannjan.teichm...@googlemail.com wrote:
 Hello,

 I have a problem with the hist() function and showing densities. The
 densities sum to 50 and not to 1! I use R version 2.9.1 (2009-06-26) and
 I load the seqinR library.

 My data is the following vector:
 [1] 0.140 0.200 0.220 0.2828283 0.160 0.160
 0.360
 [8] 0.160 0.220 0.260 0.200 0.300 0.220
 0.2342342
 [15] 0.180 0.220 0.160 0.230 0.200 0.220
 0.240
 [22] 0.200 0.220 0.220 0.260 0.200 0.160
 0.220
 [29] 0.2342342 0.200 0.220 0.200 0.200 0.140
 0.180
 [36] 0.220 0.160 0.160 0.140 0.220 0.200
 0.2871287
 [43] 0.290 0.200 0.1836735 0.200 0.200 0.290
 0.240
 [50] 0.220 0.280 0.200 0.2745098 0.220 0.230
 0.180
 [57] 0.230 0.180 0.260 0.220 0.222 0.220
 0.260
 [64] 0.220 0.220 0.260 0.220 0.200 0.220

 I use the following command:
 tmp - hist(data, freq=FALSE, plot=FALSE)

 and that's the result:
 $breaks
  [1] 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36

 $counts
  [1] 10  4 15 19  8  5  2  5  0  0  1

 $intensities
  [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
  [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

 $density
  [1]  7.2463754  2.8985507 10.8695652 13.7681159  5.7971014  3.6231884
  [7]  1.4492754  3.6231884  0.000  0.000  0.7246377

 $mids
  [1] 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35

 $xname
 [1] data

 $equidist
 [1] TRUE

 attr(,class)
 [1] histogram

 __
 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
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-- 
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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] moving text labels in plot

[jim holtman]

Try this:

par(mar=c(4,6,2,1))
plot(0, xlab = expression(paste(Temperature (C^o,))),
   ylab =  ,las=1,pch=19,
   xlim=c(0,32),ylim=c(0,0.03),xaxs = i, yaxs = i)
mtext(expression(paste(Development rate (d^-1,))), 2, line=4)

2009/7/26 Luis Iván Ortiz Valencia liov2...@gmail.com:
 Hi R users

 I need to specify some parameter input in plot code to move Y text label to
 left.

 plot(temp, develo_rate, xlab = expression(paste(Temperature (C^o,))),
    ylab = expression(paste(Development rate (d^-1,))),las=1,pch=19,
    xlim=c(0,32),ylim=c(0,0.03),xaxs = i, yaxs = i)

 Plot result is added.

 any help?

 Ivan

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-- 
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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] Determine the dimension-names of an element in an array in R

[Poersching]

Sauvik De schrieb:
 Hi:
 Lots of thanks for your valuable time!

 But I am not sure how you would like to use the function in this
 situation.

 As I had mentioned that the first element of my output array should be
 like:

 cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)

 in my below code.

 and

 the output array of correlation I wish to get using sapply as follows:

 Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
 use=pairwise.complete.obs))

 So it would be of great help if you could kindly specify how to
 utilise your function findIndex in ...

 Apologies for all this!

 Thanks  Regards,
 Sauvik

Hey,
sorry, I haven't understood your problem last time, but now this
solution should solve your problem, so I hope. :-)
It's only a for to loop, but an apply function may work too. I will
think about this, but for now...  ;-)

la-length(a)
lb-length(b)
lc-length(c)
ld-length(d)
for (ia in 1:la) {
  for (ib in 1:lb) {
for (ic in 1:lc) {
  for (id in 1:ld) {
Correl[ia,ib,ic,id]-cor(
 DataArray_1[dimnames(Correl)[[1]][ia],
 dimnames(Correl)[[2]][ib],
 dimnames(Correl)[[4]][id],]
 ,
 DataArray_2[dimnames(Correl)[[1]][ia],
  dimnames(Correl)[[3]][ic],
  dimnames(Correl)[[4]][id],]
 ,
 use=pairwise.complete.obs)
  }
}
  }
}
## with function findIndex you can find the dimensions with
## i.e. cor values greater 0.5 or smaller -0.5, like:
findIndex(Correl,Correl[Correl0.5])
findIndex(Correl,Correl[Correl(-0.5)])

I have changed the code of the function findIndex in line which
contents: el[j]-which(is.element(data,element[j]))

Rigards,
Christian


 On Sun, Jul 26, 2009 at 3:54 PM, Poerschingpoerschin...@web.de
 mailto:poerschin...@web.de wrote:
  Sauvik De schrieb:
 
  Hi Gabor:
  Many thanks for your prompt reply!
  The code is fine. But I need it in more general form as I had
 mentioned that
  I need to input any 0 to find its dimension-names.
 
  Actually, I was using sapply to calculate correlation and this
 idea was
  required in the middle of correlation calculation.
  I am providing the way I tried my calculation.
 
  a= c(A1,A2,A3,A4,A5)
  b= c(B1,B2,B3)
  c= c(C1,C2,C3,C4)
  d= c(D1,D2)
  e= c(E1,E2,E3,E4,E5,E6,E7,E8)
 
  DataArray_1 = array(c(rnorm(240)),dim=c(length(a),length(b),
  length(d),length(e)),dimnames=list(a,b,d,e))
  DataArray_2 = array(c(rnorm(320)), dim=c(length(a),length(c),
  length(d),length(e)),dimnames=list(a,c,d,e))
 
  #Defining an empty array which will contain the correlation values
 (output
  array)
  Correl = array(NA, dim=c(length(a),length(b),
  length(c),length(d)),dimnames=list(a,b,c,d))
 
  #Calculating Correlation between attributes b  c over values of e
  Correl = sapply(Correl,function(d)
 cor(DataArray_1[...],DataArray_2[...],
  use=pairwise.complete.obs))
 
  This is where I get stuck.
  In the above, d is acting as an element in the Correl array. Hence
 I need
  to get the dimension-names for d.
 
  #The first element of Correl will be:
 
 cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use=pairwise.complete.obs)
 
  So my problem boils down to extracting the dim-names in terms of
 element(d)
  and not in terms of Correl (that I have mentioned as ... in the above
  code)
 
  My sincere thanks for your valuable time  suggestions.
 
  Many Thanks  Kind Regards,
  Sauvik
 
 
  On Sun, Jul 26, 2009 at 5:26 AM, Gabor Grothendieck
 ggrothendi...@gmail.com mailto:ggrothendi...@gmail.com
   
 
  wrote:
 
 
   
 
  Try this:
 
 
 
  ix - c(1, 3, 4, 2)
  mapply([, dimnames(mydatastructure), ix)
   
 
  [1] S1 T3 U4 V2
 
 
  On Sat, Jul 25, 2009 at 5:12 PM, Sauvik Desauvik.s...@gmail.com
 mailto:sauvik.s...@gmail.com wrote:
 
 
  Hi:
  How can I extract the dimension-names of a pre-defined element in a
  multidimensional array in R ?
 
  A toy example is provided below:
  I have a 4-dimensional array with each dimension having certain length.
   
 
  In
 
 
  the below example, mydatastructure explains the structure of my data.
 
  mydatastructure = array(0,
   
 
  dim=c(length(b),length(z),length(x),length(d)),
 
 
  dimnames=list(b,z,x,d))
 
  where,
  b=c(S1,S2,S3,S4,S5)
  z=c(T1,T2, T3)
  x=c(U1,U2,U3,U4)
  d=c(V1,V2)
 
  Clearly, mydatastructure contains many 0's.
  Now how can I get the dimension-names of any particular 0 ?
  That is, my input should be a particular 0 in the array
 mydatastructure
  (Suppose this 0 corresponds to S1,T3,U4  V2 in the array). Then my
   
 
  output
 
 
  should be S1,T3,U4  V2.
 
  The function dimnames didn't help me with the solution.
  Any idea will greatly be appreciated.
 
  Thanks for 

[R] obtain names of variables and data from glm object

[Jacob Wegelin]

Suppose we have some glm object such as:

myglm - glm( y ~ x, data=DAT)

Is there an elegant way--or the right way within the R way of thinking--to
obtain the names of the response variable, the predictor variables, and the
dataset, as character strings?

For instance, suppose the right way was to use the (currently fictitious)
functions theresponse(), thepredictors(), and theDataSet().  Then I would be
able to write a function that obtains the names and subsequently pastes text
along the following lines:

theResponse - theresponse( myglm )

theFirstPredictor - thepredictors( myglm )[1]

theDataSet - theDataSet(myglm)

title(main= paste(theResponse,  is the response and , theFirstPredictor, 
is the only predictor)

In reality, I can of course extract

 formula(myglm)
y ~ x

but I see no elegant way to extract the names of the predictor and response
from this object. The deparse() function doesn't quite solve this problem:

 deparse(formula(myglm))
[1] y ~ x
 deparse(formula(myglm)[2])
[1] y()
 deparse(formula(myglm)[3])
[1] x()

Ideally the elegant method would, in this example, return the character
strings x, y, and DAT.

Thanks for any insights.

Jake

Jacob A. Wegelin
Assistant Professor
Department of Biostatistics
Virginia Commonwealth University
730 East Broad Street Room 3006
P. O. Box 980032
Richmond VA 23298-0032
U.S.A.
E-mail: jwege...@vcu.edu
URL: http://www.people.vcu.edu/~jwegelin

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

Re: [R] obtain names of variables and data from glm object

[Gabor Grothendieck]

Try this:

g - glm(demand ~ Time, BOD, family = gaussian)
all.vars(formula(g))

The result will be a character vector whose 1st component is
the name of the response and whose subsequent components
are the names of the predictor variables.

On Sun, Jul 26, 2009 at 3:14 PM, Jacob Wegelinjacob.wege...@gmail.com wrote:
 Suppose we have some glm object such as:

 myglm - glm( y ~ x, data=DAT)

 Is there an elegant way--or the right way within the R way of thinking--to
 obtain the names of the response variable, the predictor variables, and the
 dataset, as character strings?

 For instance, suppose the right way was to use the (currently fictitious)
 functions theresponse(), thepredictors(), and theDataSet().  Then I would be
 able to write a function that obtains the names and subsequently pastes text
 along the following lines:

 theResponse - theresponse( myglm )

 theFirstPredictor - thepredictors( myglm )[1]

 theDataSet - theDataSet(myglm)

 title(main= paste(theResponse,  is the response and , theFirstPredictor, 
 is the only predictor)

 In reality, I can of course extract

 formula(myglm)
 y ~ x

 but I see no elegant way to extract the names of the predictor and response
 from this object. The deparse() function doesn't quite solve this problem:

 deparse(formula(myglm))
 [1] y ~ x
 deparse(formula(myglm)[2])
 [1] y()
 deparse(formula(myglm)[3])
 [1] x()

 Ideally the elegant method would, in this example, return the character
 strings x, y, and DAT.

 Thanks for any insights.

 Jake

 Jacob A. Wegelin
 Assistant Professor
 Department of Biostatistics
 Virginia Commonwealth University
 730 East Broad Street Room 3006
 P. O. Box 980032
 Richmond VA 23298-0032
 U.S.A.
 E-mail: jwege...@vcu.edu
 URL: http://www.people.vcu.edu/~jwegelin

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[R] splitting multiple data in one column into multiple rows with one entry per column

[Felix Müller-Sarnowski]

Dear R colleagues,

I annotated a list of single nuclotide polymorphiosms (SNP) with the
corresponding genes using biomaRt. The result is the following
data.frame (pasted from R):

snp ensembl_gene_id
1  rs8032583
2  rs1071600 ENSG0101605
3  rs13406898 ENSG0167165
4  rs7030479 ENSG0107249
5  rs1244414 ENSG0165629
6  rs1005636 ENSG0230681
7  rs927913 ENSG0151655;ENSG0227546
8  rs4832680
9  rs4435168 ENSG0229164;ENSG0225227;ENSG0211817
10 rs7035549
11 rs12707538 ENSG0186472

As you can see, the SNP with the identifier rs4435168 corresponds to 3
gene ids, rs927913 corresponds to 2 gene ids. As I'd like to perform a
join of several data.frames using the ensembl_gene_id later on, I'd
like to split columns with multiple gene identifiers into rows with
only one ensembl gene identifier each. So for the example of rs4435168
it should look like this (faked output):

snp   ensembl_gene_id
...
9  rs4435168 ENSG0229164
10rs4435168 ENSG0225227
11rs4435168 ENSG0211817
...

This is just a simple example. Finally there will be a lot of other
columns, which should be replicated like the snp column.

Does anyone know how to do this? I tried strsplit, which splits nicely
the multiple entries in column ensembl_gene_id. But how to go on?

I'd appreciate any kind of help very much!
Best regards from Munich,
Felix

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[R] mixdist package in R

[cindy Guo]

Hi, All,

I fitted a 3-component normal mixture model with the mixdist package in R.
How can I get the density of a new data after I fit the model? Is there any
function to do it?

Thanks,

Cindy

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Re: [R] moving text labels in plot

[cls59]

Luis Iván Ortiz Valencia wrote:
 
 Hi R users
 
 I need to specify some parameter input in plot code to move Y text label
 to
 left.
 
 plot(temp, develo_rate, xlab = expression(paste(Temperature (C^o,))),
 ylab = expression(paste(Development rate (d^-1,))),las=1,pch=19,
 xlim=c(0,32),ylim=c(0,0.03),xaxs = i, yaxs = i)
 
 Plot result is added.
 
 any help?
 
 Ivan
 
 


From the looks of your plot, the label is already to the left as far as it
can go. The problem is that the left margin of the plot is not big enough to
allow good separation between the axes label and the axes numbering.

You can adjust the plot margins by calling the par() function before
creating your plot. In this case you want to indicate that you are setting
the mar option and pass a vector of four numbers:

par( mar = c(bottom, left, top, right) )

Note that the defaults for mar are:

c( 5.1, 4.1, 4.1, 2.1 )


Hope that helps!

-Charlie

-
Charlie Sharpsteen
Undergraduate
Environmental Resources Engineering
Humboldt State University
-- 
View this message in context: 
http://www.nabble.com/moving-text-labels-in-plot-tp24668930p2467.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] labelling points plotted in a 2D plan

[Khaled OUANES]

Hello Tal!
Nothing showed up when I used those commands!
The plot still shows dots with no labels!
Thanks



On Sun, Jul 26, 2009 at 7:28 AM, Tal Galili tal.gal...@gmail.com wrote:

 Hi Khaled,
 Did my answer help ?




 On Sun, Jul 26, 2009 at 12:15 AM, Khaled OUANES koua...@gmail.com wrote:

 Thanks for the answer Tal!
 But I can't get it to work correctly! :(
 Please bear with me this is the first time I am using R! and I am in a
 rush
 to correct a paper
 in fact on the plane I am plotting a table
  fullpointed=read.table(fullpoints_backup.txt,h=F)
   plot(range(-2.5,0.95),range(0.00,1.00),type=n,axes=TRUE)
 and in this table there are 300 points
 I want to label the first 175 points with A and the others with S
 I couldn't figure how to configure correctly labels.to.plot -
 sample(c(A,B), 100, replace = T) and text(x, y , labels =
 labels.to.plot) ?

 for instance:
 0,48875 0,142857143  the point plotted will be labelled a
 0,409 0,142857143  the point plotted will be labelled a
 0,45611 0,25 labelled a
 0,49833 0,2  labelled a
  #the first 175

 0,61158 0,125labelled S
 0,5709 0,125labelled S
 0,53266 0,125labelled S

 # the remaining
 Regards




 On Sat, Jul 25, 2009 at 5:32 PM, Tal Galili tal.gal...@gmail.com wrote:

  Sure,
 
  Here is an example:
 
  # get some random data to play with
  x - runif(100)
  y - runif(100)
  labels.to.plot - sample(c(A,B), 100, replace = T)
 
  # set up the window, play them one by one to see what they do
  plot.window(ylim = c(0,1), xlim = c(0,1))
  plot.new()
  axis(1)
  axis(2)
  box()
 
  # plot the things you wished to plot, where you wanted them plotted
  text(x, y , labels = labels.to.plot)
 
 
 
  Cheers,
  Tal
 
 
 
 
 
 
 
 
 
 
  On Sat, Jul 25, 2009 at 7:20 PM, Khaled OUANES koua...@gmail.com
 wrote:
 
  hey
  thanks for the answer but I couldn't achieve it? would you explain a
 bit
  more?
  I have like 300 points to label!
  thanks
 
 
 
 
  --
  --
 
 
  My contact information:
  Tal Galili
  Phone number: 972-50-3373767
  FaceBook: Tal Galili
  My Blogs:
  http://www.r-statistics.com/
  http://www.talgalili.com
  http://www.biostatistics.co.il
 
 
 

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 --
 --


 My contact information:
 Tal Galili
 Phone number: 972-50-3373767
 FaceBook: Tal Galili
 My Blogs:
 http://www.r-statistics.com/
 http://www.talgalili.com
 http://www.biostatistics.co.il




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[R] Question about rpart decision trees (being used to predict customer churn)

[Robert Smith]

Hi,

I am using rpart decision trees to analyze customer churn. I am finding that
the decision trees created are not effective because they are not able to
recognize factors that influence churn. I have created an example situation
below. What do I need to do to for rpart to build a tree with the variable
experience? My guess is that this would happen if rpart used the loss matrix
while creating the tree.

 experience - as.factor(c(rep(good,90), rep(bad,10)))
 cancel - as.factor(c(rep(no,85), rep(yes,5), rep(no,5),
rep(yes,5)))
 table(experience, cancel)
  cancel
experience no yes
  bad   5   5
  good 85   5
 rpart(cancel ~ experience)
n= 100
node), split, n, loss, yval, (yprob)
  * denotes terminal node
1) root 100 10 no (0.900 0.100) *

I tried the following commands with no success.
rpart(cancel ~ experience, control=rpart.control(cp=.0001))
rpart(cancel ~ experience, parms=list(split='information'))
rpart(cancel ~ experience, parms=list(split='information'),
control=rpart.control(cp=.0001))
rpart(cancel ~ experience, parms=list(loss=matrix(c(0,1,1,0), nrow=2,
ncol=2)))

Thanks a lot for your help.

Best regards,
Robert

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Re: [R] running a .r script and saving the output to a file

[David Riebel]

Hello,

I am running R under Ubuntu 8.04.  I am trying to do numerous linear
fits to various subsets of my data set.  I am having trouble convincing
R to send the output from these fits to text files from within a script.
When I run my script, all the plots are created as postscript files in
the correct directory, and text files appear with the correct names for
all the summaries of the fits.  However, all of these files are blank!
If I copy/paste the sink commands from my script manually into the
command line, the files are created correctly, but if the same commands
are executed from within my script, no output is generated.

The output section of my script, which correctly creates the plots I
want, but not the text, is:

x_seq1_fit=lm(xagb$X5mag[x_seq1_filt]~x1per)
x_seq2_fit=lm(xagb$X5mag[x_seq2_filt]~x2per)

setwd(/home/driebel/sage/output/pl_fits/R/5mag/chem)
postscript(file=x_seq1_fit.ps)
plot(x1per,xagb$X5mag[x_seq1_filt],ylim=c(13,7))
abline(x_seq1_fit,col=red)
dev.off()
postscript(file=x_seq1_resid.ps)
plot(x1per,x_seq1_fit$res)
abline(h=0,col=red)
dev.off()

postscript(file=x_seq2_fit.ps)
plot(x2per,xagb$X5mag[x_seq2_filt],ylim=c(13,7))
abline(x_seq2_fit,col=red)
dev.off()
postscript(file=x_seq2_resid.ps)
plot(x2per,x_seq2_fit$res)
abline(h=0,col=red)
dev.off()

sink(file=x_seq2_fit.dat)
summary(x_seq2_fit)
sink()
sink(file=x_seq1_fit.dat)
summary(x_seq1_fit)
sink()


Is there something else one must do with sink from within a script?
Thanks for your help,

Dave

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[R] ROC curve using epicalc (after logistic regression)

[Clifford Long]

Dear R-help list,

I'm attempting to use the ROC routine from the epicalc package after
performing a logistic regression analysis.  My code is included after
the sessionInfo() result.  The datafile (GasketMelt1.csv) is attached.
 I updated both R and the epicalc packages and tried again before
sending this request.

sessionInfo result:

R version 2.9.1 (2009-06-26)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

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

other attached packages:
[1] caret_4.19  lattice_0.17-25 epicalc_2.9.1.2 survival_2.35-4
[5] foreign_0.8-36

loaded via a namespace (and not attached):
[1] grid_2.9.1  tools_2.9.1


Header information from package 'epicalc':
Package:epicalc
Version:2.9.1.2
Date:   2009-07-14


My code ...

#
#  Logistic Regression   (the model result is as expected)
#

dfile = 'GasketMelt1.csv'
gmelt.df = read.csv(dfile, header = TRUE, as.is = TRUE)
names(gmelt.df)

gmelt.df$p = gmelt.df$Pass / gmelt.df$Total

gmelt.glm = glm(p ~ Time + Temperature + Depth
+ Time*Temperature + Time*Depth + Temperature*Depth,
family = binomial(link = logit), data=gmelt.df, 
weight=Total)
summary(gmelt.glm)

#
#  ROC
#
library(epicalc)

lroc(gmelt.glm, graph = TRUE, line.col = red)


The error message:

 lroc(gmelt.glm, graph = TRUE, line.col = red)
Error in dimnames(x) - dn :
  length of 'dimnames' [2] not equal to array extent



Have I overlooked something?


Many thanks to anyone who might have a suggestion.

Cliff
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Re: [R] splitting multiple data in one column into multiple rows with one entry per column

[jim holtman]

Try this:

 x - read.table(textConnection(snp  ensembl_gene_id
+  rs8032583
+  rs1071600 ENSG0101605
+  rs13406898 ENSG0167165
+  rs7030479 ENSG0107249
+  rs1244414 ENSG0165629
+  rs1005636 ENSG0230681
+  rs927913 ENSG0151655;ENSG0227546
+  rs4832680
+  rs4435168 ENSG0229164;ENSG0225227;ENSG0211817
+  rs7035549
+  rs12707538 ENSG0186472),
header=TRUE, fill=TRUE)
 closeAllConnections()
 x.new - do.call(rbind, apply(x, 1, function(.row){
+ .ids - unlist(strsplit(.row[2], ';'))
+ # check for no data in second column; substitute a blank
+ if (length(.ids) == 0) return(cbind(.row[1], ))
+ else return(cbind(.row[1], .ids))
+ }))
 x.new
  .ids
snp  rs8032583  
snp  rs1071600  ENSG0101605
snp  rs13406898 ENSG0167165
snp  rs7030479  ENSG0107249
snp  rs1244414  ENSG0165629
snp  rs1005636  ENSG0230681
ensembl_gene_id1 rs927913   ENSG0151655
ensembl_gene_id2 rs927913   ENSG0227546
snp  rs4832680  
ensembl_gene_id1 rs4435168  ENSG0229164
ensembl_gene_id2 rs4435168  ENSG0225227
ensembl_gene_id3 rs4435168  ENSG0211817
snp  rs7035549  
snp  rs12707538 ENSG0186472



On Sun, Jul 26, 2009 at 3:26 PM, Felix
Müller-Sarnowskidrfl...@googlemail.com wrote:
 Dear R colleagues,

 I annotated a list of single nuclotide polymorphiosms (SNP) with the
 corresponding genes using biomaRt. The result is the following
 data.frame (pasted from R):

 snp                                 ensembl_gene_id
 1      rs8032583
 2      rs1071600                                 ENSG0101605
 3      rs13406898                                 ENSG0167165
 4      rs7030479                                 ENSG0107249
 5      rs1244414                                 ENSG0165629
 6      rs1005636                                 ENSG0230681
 7      rs927913                 ENSG0151655;ENSG0227546
 8      rs4832680
 9      rs4435168 ENSG0229164;ENSG0225227;ENSG0211817
 10     rs7035549
 11     rs12707538                                 ENSG0186472

 As you can see, the SNP with the identifier rs4435168 corresponds to 3
 gene ids, rs927913 corresponds to 2 gene ids. As I'd like to perform a
 join of several data.frames using the ensembl_gene_id later on, I'd
 like to split columns with multiple gene identifiers into rows with
 only one ensembl gene identifier each. So for the example of rs4435168
 it should look like this (faked output):

 snp                   ensembl_gene_id
 ...
 9      rs4435168 ENSG0229164
 10    rs4435168 ENSG0225227
 11    rs4435168 ENSG0211817
 ...

 This is just a simple example. Finally there will be a lot of other
 columns, which should be replicated like the snp column.

 Does anyone know how to do this? I tried strsplit, which splits nicely
 the multiple entries in column ensembl_gene_id. But how to go on?

 I'd appreciate any kind of help very much!
 Best regards from Munich,
 Felix

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] running a .r script and saving the output to a file

[Sarah Goslee]

Hi Dave,

I don't know about using sink(), but if you run your script in batch
mode all the output will be saved to a text file foryou.

R CMD BATCH infile outfile

(at the linux prompt, not from within R).

Sarah

On Sun, Jul 26, 2009 at 4:47 PM, David Riebeldrie...@pha.jhu.edu wrote:
 Hello,

 I am running R under Ubuntu 8.04.  I am trying to do numerous linear
 fits to various subsets of my data set.  I am having trouble convincing
 R to send the output from these fits to text files from within a script.
 When I run my script, all the plots are created as postscript files in
 the correct directory, and text files appear with the correct names for
 all the summaries of the fits.  However, all of these files are blank!
 If I copy/paste the sink commands from my script manually into the
 command line, the files are created correctly, but if the same commands
 are executed from within my script, no output is generated.

 The output section of my script, which correctly creates the plots I
 want, but not the text, is:

 x_seq1_fit=lm(xagb$X5mag[x_seq1_filt]~x1per)
 x_seq2_fit=lm(xagb$X5mag[x_seq2_filt]~x2per)

 setwd(/home/driebel/sage/output/pl_fits/R/5mag/chem)
 postscript(file=x_seq1_fit.ps)
 plot(x1per,xagb$X5mag[x_seq1_filt],ylim=c(13,7))
 abline(x_seq1_fit,col=red)
 dev.off()
 postscript(file=x_seq1_resid.ps)
 plot(x1per,x_seq1_fit$res)
 abline(h=0,col=red)
 dev.off()

 postscript(file=x_seq2_fit.ps)
 plot(x2per,xagb$X5mag[x_seq2_filt],ylim=c(13,7))
 abline(x_seq2_fit,col=red)
 dev.off()
 postscript(file=x_seq2_resid.ps)
 plot(x2per,x_seq2_fit$res)
 abline(h=0,col=red)
 dev.off()

 sink(file=x_seq2_fit.dat)
 summary(x_seq2_fit)
 sink()
 sink(file=x_seq1_fit.dat)
 summary(x_seq1_fit)
 sink()


 Is there something else one must do with sink from within a script?
 Thanks for your help,

 Dave




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

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[R] normal mixture model

[cindy Guo]

Hi, All,

I want to fit a normal mixture model. Which package in R is best for this? I
was using the package 'mixdist', but I need to group the data into groups
before fitting model, and different groupings seem to lead to different
results. What other package can I use which is stable? And are there
packages that can automatically determine the number of components?

Thank you,

Cindy

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

Re: [R] normal mixture model

[Christian Hennig]

You can use mclustBIC in package mclust (uses the BIC for deciding 
about the 
number of components and hierarchical clustering for initialisation).


Christian

On Sun, 26 Jul 2009, cindy Guo wrote:


Hi, All,

I want to fit a normal mixture model. Which package in R is best for this? I
was using the package 'mixdist', but I need to group the data into groups
before fitting model, and different groupings seem to lead to different
results. What other package can I use which is stable? And are there
packages that can automatically determine the number of components?

Thank you,

Cindy

[[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.



*** --- ***
Christian Hennig
University College London, Department of Statistical Science
Gower St., London WC1E 6BT, phone +44 207 679 1698
chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche

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

Re: [R] running a .r script and saving the output to a file

[jim holtman]

You have to explicitly 'print' the output.  At the command line there
is an implicit 'print'.  Try:

sink(file=x_seq2_fit.dat)
print(summary(x_seq2_fit))
sink()
sink(file=x_seq1_fit.dat)
print(summary(x_seq1_fit))
sink()

On Sun, Jul 26, 2009 at 4:47 PM, David Riebeldrie...@pha.jhu.edu wrote:
 Hello,

 I am running R under Ubuntu 8.04.  I am trying to do numerous linear
 fits to various subsets of my data set.  I am having trouble convincing
 R to send the output from these fits to text files from within a script.
 When I run my script, all the plots are created as postscript files in
 the correct directory, and text files appear with the correct names for
 all the summaries of the fits.  However, all of these files are blank!
 If I copy/paste the sink commands from my script manually into the
 command line, the files are created correctly, but if the same commands
 are executed from within my script, no output is generated.

 The output section of my script, which correctly creates the plots I
 want, but not the text, is:

 x_seq1_fit=lm(xagb$X5mag[x_seq1_filt]~x1per)
 x_seq2_fit=lm(xagb$X5mag[x_seq2_filt]~x2per)

 setwd(/home/driebel/sage/output/pl_fits/R/5mag/chem)
 postscript(file=x_seq1_fit.ps)
 plot(x1per,xagb$X5mag[x_seq1_filt],ylim=c(13,7))
 abline(x_seq1_fit,col=red)
 dev.off()
 postscript(file=x_seq1_resid.ps)
 plot(x1per,x_seq1_fit$res)
 abline(h=0,col=red)
 dev.off()

 postscript(file=x_seq2_fit.ps)
 plot(x2per,xagb$X5mag[x_seq2_filt],ylim=c(13,7))
 abline(x_seq2_fit,col=red)
 dev.off()
 postscript(file=x_seq2_resid.ps)
 plot(x2per,x_seq2_fit$res)
 abline(h=0,col=red)
 dev.off()

 sink(file=x_seq2_fit.dat)
 summary(x_seq2_fit)
 sink()
 sink(file=x_seq1_fit.dat)
 summary(x_seq1_fit)
 sink()


 Is there something else one must do with sink from within a script?
 Thanks for your help,

 Dave

 __
 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.




-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

__
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] normal mixture model

[cindy Guo]

Hi, Christian,

Thank you for the reply. I just tried. Does the function mclustBIC only give
the best model, or does it also do EM to get the cluster means and variances
according to the best model it picks? I didn't find it.  Is there a way to
automatically select the best number of components and do EM? Because I need
to do the normal mixture model in a loop (one EM at an iteration), so I want
it to do everything automatically.
Thanks,

Cindy

On Sun, Jul 26, 2009 at 3:46 PM, Christian Hennig chr...@stats.ucl.ac.ukwrote:

 You can use mclustBIC in package mclust (uses the BIC for deciding about
 the number of components and hierarchical clustering for initialisation).

 Christian


 On Sun, 26 Jul 2009, cindy Guo wrote:

   Hi, All,

 I want to fit a normal mixture model. Which package in R is best for this?
 I
 was using the package 'mixdist', but I need to group the data into groups
 before fitting model, and different groupings seem to lead to different
 results. What other package can I use which is stable? And are there
 packages that can automatically determine the number of components?

 Thank you,

 Cindy

[[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.htmlhttp://www.r-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


 *** --- ***
 Christian Hennig
 University College London, Department of Statistical Science
 Gower St., London WC1E 6BT, phone +44 207 679 1698
 chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche


[[alternative HTML version deleted]]

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

[R] Specify CRAN repository from command line

[Aaron Hicks]

Hi,

It feels like I should be able to do something like:

R CMD INSTALL lib='/usr/lib64/R/library' repos='http://proxy.url/cran' package

We have a bunch of servers (compute nodes in a Rocks cluster) in an isolated 
subnet, there is a basic pass-through proxy set up on the firewall (the head 
node) which just passes HTTP requests through to our nearest CRAN mirror.

when using install. packages it's easy to make R install from the repository 
with the repos='address' option, but I can't figure out how do this from the 
command line.

Is there a command line option for this? Currently I'm doing it using an R 
script, but that's causing issues because it's not 'visible' to the installer.

This would greatly streamline R installation with a standard package set.

Regards,

Aaron Hicks

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[R] how to use do.call together with cbind and get inside a function

[Sean Zhang]

Dear R-helpers:
I have a question related to using do.call to call cbind and get.

#the following works
vec1 - c(1,2)
vec2 - c(3,4)
ColNameVec - c('vec1','vec2')
mat - do.call(cbind,lapply(ColNameVec,get))
mat

#put code above into a function then it does not work
#before doing so, first remove vec1 and vec2 from global environment
rm(vec1,vec2)

test - function()
{
vec1 - c(1,2)
vec2 - c(3,4)
ColNameVec - c('vec1','vec2')
mat - do.call(cbind,lapply(ColNameVec,get))
return(mat)
}
test()

In my task, I have to run do.call(cbind,lapply(ColNameVec,get))
inside a function, can someone kindly help?

Many thanks in advance!

-Sean

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Re: [R] Specify CRAN repository from command line

[Dirk Eddelbuettel]

On 27 July 2009 at 14:55, Aaron Hicks wrote:
| It feels like I should be able to do something like:
| 
| R CMD INSTALL lib='/usr/lib64/R/library' repos='http://proxy.url/cran' package

Here's what I do using littler, you can substitute Rscript as well:

#!/usr/bin/env r
#
# a simple example to install one or more packages

if (is.null(argv) | length(argv)1) {

  cat(Usage: installr.r pkg1 [pkg2 pkg3 ...]\n)
  q()

}

## adjust as necessary, see help('download.packages')
repos - http://cran.us.r-project.org;

## this makes sense on Debian where no packages touch /usr/local
lib.loc - /usr/local/lib/R/site-library

install.packages(argv, lib.loc, repos, dependencies=TRUE)

That way, I just say 'install.r foo bar baz' and these packages, plus their
Depends:, will get installed.

[ That said, I actually don't do that anymore because we now have cran2deb so
I can just say 'apt-get install r-cran-foo r-cran-bar r-cran-baz' but that is
a special case for Debian and pretty recent as per the announcement a few
days ago. ]

| We have a bunch of servers (compute nodes in a Rocks cluster) in an isolated 
subnet, there is a basic pass-through proxy set up on the firewall (the head 
node) which just passes HTTP requests through to our nearest CRAN mirror.
| 
| when using install. packages it's easy to make R install from the repository 
with the repos='address' option, but I can't figure out how do this from the 
command line.
| 
| Is there a command line option for this? Currently I'm doing it using an R 
script, but that's causing issues because it's not 'visible' to the installer.
| 
| This would greatly streamline R installation with a standard package set.

The repos argument can otherwise be set in ~/.Rprofile or Rprofile.site.

Dirk

-- 
Three out of two people have difficulties with fractions.

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[R] Disable summary statistics in LaTeX tables using MEMISC package

[Shige Song]

Dear All,

The mtable function in memisc package is very useful in producing
publication quality tables directly from estimated models. There are cases
where only the estimated coefficients and standard errors are needed in the
table but not the summary staitstics such as N, value of the likelihood,
AIC, etc. The summary.stats=FALSE  option is supposed to do this, but
there seems to be some problems.

The easiest way to verify this is to run the anes48.R file, which is
distributed as part of the package, then change the last line from:

mtable(model1,model6,model7,summary.stats=c(Deviance,AIC,N))

to:

mtable(model1,model6,model7,summary.stats=FALSE)

I get the error message of Error in as.table.default(sumstats) : cannot
coerce into a table.

Any ideas? Thanks.

Best,
Shige

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[R] Superstring in text()

[Murray Pung]

I'd like to paste a superstring with a number in an object.

Thanks for any help.
Murray


mycor - cor(1:10,1:10)
plot(1:10,1:10)
text(8,2,paste(expression(R^2), = ,mycor))

[[alternative HTML version deleted]]

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Re: [R] how to use do.call together with cbind and get inside a function

[Gabor Grothendieck]

Use lapply(ColNameVec, get, environment())
so that it gets the objects from the current
environment. See: ?get ?environment

On Sun, Jul 26, 2009 at 11:16 PM, Sean Zhangseane...@gmail.com wrote:
 Dear R-helpers:
 I have a question related to using do.call to call cbind and get.

 #the following works
 vec1 - c(1,2)
 vec2 - c(3,4)
 ColNameVec - c('vec1','vec2')
 mat - do.call(cbind,lapply(ColNameVec,get))
 mat

 #put code above into a function then it does not work
 #before doing so, first remove vec1 and vec2 from global environment
 rm(vec1,vec2)

 test - function()
 {
 vec1 - c(1,2)
 vec2 - c(3,4)
 ColNameVec - c('vec1','vec2')
 mat - do.call(cbind,lapply(ColNameVec,get))
 return(mat)
 }
 test()

 In my task, I have to run do.call(cbind,lapply(ColNameVec,get))
 inside a function, can someone kindly help?

 Many thanks in advance!

 -Sean

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


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[R] Writing to a UDP server from R?

[Saptarshi Guha]

Hello,
I have used socketConnection to connect to a TCP server. I havent
figured out a way to do the same with a UDP server.
i.e I have a server listening on 9000, communicating via UDP. I would
like to , from R, send packets to this server,

This does not work
 u -  socketConnection('localhost',9000)
Error in socketConnection(localhost, 9000, blocking = F) :
  cannot open the connection
In addition: Warning message:
In socketConnection(localhost, 9000, blocking = F) :
  localhost:9000 cannot be opened


I have confirmed that something is indeed listening on the other side.
Any help would be appreciated.
Regards
Saptarshi

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Re: [R] Superstring in text()

[Andrej Kastrin]

mycor - cor(1:10,1:10)
plot(1:10,1:10)
text(8,2,bquote(R^2 == .(mycor)))

HTH, Andrej

Murray Pung wrote:

I'd like to paste a superstring with a number in an object.

Thanks for any help.
Murray


mycor - cor(1:10,1:10)
plot(1:10,1:10)
text(8,2,paste(expression(R^2), = ,mycor))

[[alternative HTML version deleted]]



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

Re: [R] Superstring in text()

[Andrej Kastrin]

Sorry,

there should be a caret symbol (^) between R and 2.

Murray Pung wrote:

I'd like to paste a superstring with a number in an object.

Thanks for any help.
Murray


mycor - cor(1:10,1:10)
plot(1:10,1:10)
text(8,2,paste(expression(R^2), = ,mycor))

[[alternative HTML version deleted]]



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

Re: [R] running a .r script and saving the output to a file

[Andrej Kastrin]

Check out: http://akastrin.wordpress.com/category/r/

David Riebel wrote:

Hello,

I am running R under Ubuntu 8.04.  I am trying to do numerous linear
fits to various subsets of my data set.  I am having trouble convincing
R to send the output from these fits to text files from within a script.
When I run my script, all the plots are created as postscript files in
the correct directory, and text files appear with the correct names for
all the summaries of the fits.  However, all of these files are blank!
If I copy/paste the sink commands from my script manually into the
command line, the files are created correctly, but if the same commands
are executed from within my script, no output is generated.

The output section of my script, which correctly creates the plots I
want, but not the text, is:

x_seq1_fit=lm(xagb$X5mag[x_seq1_filt]~x1per)
x_seq2_fit=lm(xagb$X5mag[x_seq2_filt]~x2per)

setwd(/home/driebel/sage/output/pl_fits/R/5mag/chem)
postscript(file=x_seq1_fit.ps)
plot(x1per,xagb$X5mag[x_seq1_filt],ylim=c(13,7))
abline(x_seq1_fit,col=red)
dev.off()
postscript(file=x_seq1_resid.ps)
plot(x1per,x_seq1_fit$res)
abline(h=0,col=red)
dev.off()

postscript(file=x_seq2_fit.ps)
plot(x2per,xagb$X5mag[x_seq2_filt],ylim=c(13,7))
abline(x_seq2_fit,col=red)
dev.off()
postscript(file=x_seq2_resid.ps)
plot(x2per,x_seq2_fit$res)
abline(h=0,col=red)
dev.off()

sink(file=x_seq2_fit.dat)
summary(x_seq2_fit)
sink()
sink(file=x_seq1_fit.dat)
summary(x_seq1_fit)
sink()


Is there something else one must do with sink from within a script?
Thanks for your help,

Dave



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