Re: [R] regular expression

2007-04-08 Thread Laurent Rhelp
Gabor Grothendieck a écrit :

On 4/7/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
  

Gabor Grothendieck a écrit :



I assume is to replace single quotes in double quoted strings with single 
quoted
strings containing double quotes in place of the single quotes

  

that's rignt



while leaving
double quoted strings not containing single quotes as they are (i.e. leave 
them
as double quoted).

  

No, at the end,  I have to replace all the double quotes () by simple
quote (')



In that case it simplifies to the following (same as before except
for definition of f):

library(gsubfn)

squote - ' # single quote. This is a double quote, single quote, double 
quote
dquote - '' # double quote  This is a single quote, double quote, single 
quote

s - [\...\ \...'...\]
f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']

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I have the following error message :

  gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
Erreur dans parse(file, n, text, prompt) : erreur de syntaxe dans 
c([\,replacement(...\

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Re: [R] regular expression

2007-04-08 Thread Gabor Grothendieck
I can't reproduce that error.  For me it works fine.  What version of
R and gsubfn are you using?  I switched to a French locale just to
be sure.  I tried pasting it into an R session and also sourcing
it but both worked fine for me.

 Sys.getlocale()
[1] 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

 Sys.putenv(LANGUAGE=FR);Sys.setlocale(LC_ALL,FR)
[1] 
LC_COLLATE=French_France.1252;LC_CTYPE=French_France.1252;LC_MONETARY=French_France.1252;LC_NUMERIC=C;LC_TIME=French_France.1252
 library(gsubfn)

 squote - ' # single quote. This is a double quote, single quote, double 
 quote
 dquote - '' # double quote  This is a single quote, double quote, single 
 quote

 s - [\...\ \...'...\]

 f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
 gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
[1] ['...' '...\...']
 R.version.string # XP
[1] R version 2.4.1 Patched (2006-12-30 r40331)
 packageDescription(gsubfn)$Version
[1] 0.3-3

On 4/8/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
 Gabor Grothendieck a écrit :

 On 4/7/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
 
 
 Gabor Grothendieck a écrit :
 
 
 
 I assume is to replace single quotes in double quoted strings with single 
 quoted
 strings containing double quotes in place of the single quotes
 
 
 
 that's rignt
 
 
 
 while leaving
 double quoted strings not containing single quotes as they are (i.e. leave 
 them
 as double quoted).
 
 
 
 No, at the end,  I have to replace all the double quotes () by simple
 quote (')
 
 
 
 In that case it simplifies to the following (same as before except
 for definition of f):
 
 library(gsubfn)
 
 squote - ' # single quote. This is a double quote, single quote, double 
 quote
 dquote - '' # double quote  This is a single quote, double quote, single 
 quote
 
 s - [\...\ \...'...\]
 f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
 gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
 
 __
 R-help@stat.math.ethz.ch 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.
 
 
 
 
 I have the following error message :

   gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
 Erreur dans parse(file, n, text, prompt) : erreur de syntaxe dans
 c([\,replacement(...\




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[R] Simulation of the Frailty of the Cox PH model

2007-04-08 Thread Mohammad Ehsanul Karim
Dear R-list users,

I am trying to do simulation of survival data to
enable it to run under frailty option. Below is the
function a that I am using. My questions are:

1. How do I modify it to get bigger (hopefully
significant) value of Variance of random effect?

2. What changes do I have to make in the function to
run it under correlated frailty model? (may be in
kinship package)

3. Is there any program to run frailty by Inverse
gaussian or stable family in R?

Thank you for your time.
Thanks in advance.

Mohammad Ehsanul Karim
wildscop at yahoo dot com
Institute of Statistical Research and Training
University of Dhaka



# ***
sim.data- function(g,m){
set.seed(123)
group - rep(1:m,rep(g,m))
Frailty - rep(rgamma(m,100,1),rep(g,m))
covariate - rbinom(g*m,1,.05)
stimes -
rweibull(g*m,1.1,1/(5*Frailty*exp((covariate)/.5)))
cens - 5 + 5*runif(25)
times - pmin(stimes, cens)
censored - as.numeric(cens  times)
data.mat -
cbind(group,covariate,times,censored,Frailty)
data.mat -
data.mat[rev(order(times)),1:length(data.mat[1,])]
data.fr - data.frame(data.mat)
return(data.fr)
}
# ***


# Example of 50 group each with 100 members
sim.fr-sim.data(50,100)


library(survival)
fit.c - coxph(Surv(times,censored) ~ covariate,data=
sim.fr)
# fit.c gives the Usual cox proportional hazards model
fit.gm.em - coxph(Surv(times,censored) ~ covariate +
frailty(group, dist='gamma', method='em'), data=
sim.fr)
# fit.gm.em gives the gamma frailty model by EM
algorithm

fit.c # result of Cox PH model
fit.gm.em # result of gamma frailty model

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Re: [R] regular expression

2007-04-08 Thread Laurent Rhelp
Gabor Grothendieck a écrit :

 I can't reproduce that error.  For me it works fine.  What version of
 R and gsubfn are you using?  I switched to a French locale just to
 be sure.  I tried pasting it into an R session and also sourcing
 it but both worked fine for me.

 Sys.getlocale()

 [1] 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


 Sys.putenv(LANGUAGE=FR);Sys.setlocale(LC_ALL,FR)

 [1] 
 LC_COLLATE=French_France.1252;LC_CTYPE=French_France.1252;LC_MONETARY=French_France.1252;LC_NUMERIC=C;LC_TIME=French_France.1252
  


 library(gsubfn)

 squote - ' # single quote. This is a double quote, single quote, 
 double quote
 dquote - '' # double quote  This is a single quote, double quote, 
 single quote

 s - [\...\ \...'...\]

 f - function(x)  chartr(paste(squote, dquote), paste(dquote, 
 squote), x)
 gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']

 [1] ['...' '...\...']

 R.version.string # XP

 [1] R version 2.4.1 Patched (2006-12-30 r40331)

 packageDescription(gsubfn)$Version

 [1] 0.3-3

 On 4/8/07, Laurent Rhelp [EMAIL PROTECTED] wrote:

 Gabor Grothendieck a écrit :

 On 4/7/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
 
 
 Gabor Grothendieck a écrit :
 
 
 
 I assume is to replace single quotes in double quoted strings with 
 single quoted
 strings containing double quotes in place of the single quotes
 
 
 
 that's rignt
 
 
 
 while leaving
 double quoted strings not containing single quotes as they are 
 (i.e. leave them
 as double quoted).
 
 
 
 No, at the end,  I have to replace all the double quotes () by simple
 quote (')
 
 
 
 In that case it simplifies to the following (same as before except
 for definition of f):
 
 library(gsubfn)
 
 squote - ' # single quote. This is a double quote, single quote, 
 double quote
 dquote - '' # double quote  This is a single quote, double quote, 
 single quote
 
 s - [\...\ \...'...\]
 f - function(x)  chartr(paste(squote, dquote), paste(dquote, 
 squote), x)
 gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
 
 __
 R-help@stat.math.ethz.ch 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.
 
 
 
 
 I have the following error message :

   gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
 Erreur dans parse(file, n, text, prompt) : erreur de syntaxe dans
 c([\,replacement(...\





I am sorry Gabor, I had the R 2.3.1 version. I downloaded the good 
versions (R and gsubfn). All works fine now.
Thank you very much for your nice solution !

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


Re: [R] regular expression

2007-04-08 Thread Laurent Rhelp
Gabor Grothendieck a écrit :

On 4/7/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
  

Gabor Grothendieck a écrit :



I assume is to replace single quotes in double quoted strings with single 
quoted
strings containing double quotes in place of the single quotes

  

that's rignt



while leaving
double quoted strings not containing single quotes as they are (i.e. leave 
them
as double quoted).

  

No, at the end,  I have to replace all the double quotes () by simple
quote (')



In that case it simplifies to the following (same as before except
for definition of f):

library(gsubfn)

squote - ' # single quote. This is a double quote, single quote, double 
quote
dquote - '' # double quote  This is a single quote, double quote, single 
quote

s - [\...\ \...'...\]
f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']

__
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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.


  


I discovered that it doesn't work when the pattern  is not on one 
line.
Here's an example of file read with readLines :

Lines - c(line one example,
line two example,
example \comment on one line it works \ end of the line,
example \ comment on two lines,
the end of the comment \ without ' inside,
example \ comment on one line with ' inside it works \ end of the line,
example \ comment on two lines with ',
inside it doesn't works \)

s - Lines
f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
gsubfn('[][^]*[]', f, s)

[1] line one 
example  
[2] line two 
example  
[3] example 'comment on one line it works ' end of the 
line   
[4] example \ comment on two 
lines   
[5] the end of the comment \ without ' 
inside
[6] example ' comment on one line with \ inside it works ' end of the 
line
[7] example \ comment on two lines with 
'
[8] inside it doesn't works \

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[R] Labelling boxplot with fivenumber summary

2007-04-08 Thread Daniel Siddle

Just wanted to say thanks very much.  I used Chuck's 2nd idea as I found it the 
easiest to understand as I'm still finding me feet with R.  Just for reference 
for anyone else, fn[1] and fn[5] actually pasted and gave the values at the 
whiskers (~1.5 IQR) so replaced them with a max and min function which returned 
the true max and min values and pasted at the correct heights.

Thanks once again for all the help.  Regards,

Daniel Siddle

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


Re: [R] regular expression

2007-04-08 Thread Gabor Grothendieck
Your example is a vector of strings not a string with mulitple lines.
To turn it into a string with mulitple lines use paste:

gsubfn('[][^]*[]', f, paste(Lines, collapse = \n))

If you want to break up the output to create a vector
of strings rather than a single multi-line string see
?strsplit

On 4/8/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
 Gabor Grothendieck a écrit :

 On 4/7/07, Laurent Rhelp [EMAIL PROTECTED] wrote:
 
 
 Gabor Grothendieck a écrit :
 
 
 
 I assume is to replace single quotes in double quoted strings with single 
 quoted
 strings containing double quotes in place of the single quotes
 
 
 
 that's rignt
 
 
 
 while leaving
 double quoted strings not containing single quotes as they are (i.e. leave 
 them
 as double quoted).
 
 
 
 No, at the end,  I have to replace all the double quotes () by simple
 quote (')
 
 
 
 In that case it simplifies to the following (same as before except
 for definition of f):
 
 library(gsubfn)
 
 squote - ' # single quote. This is a double quote, single quote, double 
 quote
 dquote - '' # double quote  This is a single quote, double quote, single 
 quote
 
 s - [\...\ \...'...\]
 f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
 gsubfn('[][^]*[]', f, s)  #   ['...' '...\...']
 
 __
 R-help@stat.math.ethz.ch 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.
 
 
 
 

 I discovered that it doesn't work when the pattern  is not on one
 line.
 Here's an example of file read with readLines :

 Lines - c(line one example,
 line two example,
 example \comment on one line it works \ end of the line,
 example \ comment on two lines,
 the end of the comment \ without ' inside,
 example \ comment on one line with ' inside it works \ end of the line,
 example \ comment on two lines with ',
 inside it doesn't works \)

 s - Lines
 f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)
 gsubfn('[][^]*[]', f, s)

 [1] line one
 example
 [2] line two
 example
 [3] example 'comment on one line it works ' end of the
 line
 [4] example \ comment on two
 lines
 [5] the end of the comment \ without '
 inside
 [6] example ' comment on one line with \ inside it works ' end of the
 line
 [7] example \ comment on two lines with
 '
 [8] inside it doesn't works \






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[R] data encapsulation with classes

2007-04-08 Thread Peter Williams
Hi All,

I'm new to R from a C and Octave/Matlab background.  I am trying to  
construct some classes in R to which I want to attach pieces of data.
First, is attr(obj, 'member name') - data the accepted way of doing  
this?

Second, having attached member data with attr is there anything wrong  
with accessing it by the asperand (@), i.e.

  t = 
  attr(t, 'test') - 4
  [EMAIL PROTECTED]
[1] 4

I can't find any documentation on using the asperand to access  
attributes like this.  I saw it in a piece of code as I was googling  
around trying to find some information on attaching data to a class  
in R.  Is this deprecated?  Is this poor form?  If anyone has some  
suggestions please let me know.

Thank you

__
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Re: [R] data encapsulation with classes

2007-04-08 Thread Roger Bivand
On Sun, 8 Apr 2007, Peter Williams wrote:

 Hi All,
 
 I'm new to R from a C and Octave/Matlab background.  I am trying to  
 construct some classes in R to which I want to attach pieces of data.
 First, is attr(obj, 'member name') - data the accepted way of doing  
 this?

No, it isn't. You seem to be trying to deduce new-style classes from a
representation used before R 2.4, but in any case it would not be
sensible. Please consult John M. Chambers. Programming with Data.
Springer, New York, 1998, and/or William N. Venables and Brian D. Ripley.
S Programming. Springer, New York, 2000, or for a shorter online resource:

http://www.stat.auckland.ac.nz/S-Workshop/Gentleman/Methods.pdf

 
 Second, having attached member data with attr is there anything wrong  
 with accessing it by the asperand (@), i.e.
 
   t = 
   attr(t, 'test') - 4
   [EMAIL PROTECTED]
 [1] 4
 
 I can't find any documentation on using the asperand to access  
 attributes like this.  I saw it in a piece of code as I was googling  
 around trying to find some information on attaching data to a class  
 in R.  Is this deprecated?  Is this poor form?  If anyone has some  
 suggestions please let me know.
 
 Thank you
 
 __
 R-help@stat.math.ethz.ch 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.
 

-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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Re: [R] data encapsulation with classes

2007-04-08 Thread hadley wickham
On 4/8/07, Roger Bivand [EMAIL PROTECTED] wrote:
 On Sun, 8 Apr 2007, Peter Williams wrote:

  Hi All,
 
  I'm new to R from a C and Octave/Matlab background.  I am trying to
  construct some classes in R to which I want to attach pieces of data.
  First, is attr(obj, 'member name') - data the accepted way of doing
  this?

 No, it isn't. You seem to be trying to deduce new-style classes from a
 representation used before R 2.4, but in any case it would not be
 sensible. Please consult John M. Chambers. Programming with Data.
 Springer, New York, 1998, and/or William N. Venables and Brian D. Ripley.
 S Programming. Springer, New York, 2000, or for a shorter online resource:

 http://www.stat.auckland.ac.nz/S-Workshop/Gentleman/Methods.pdf

Unfortunately, all of those references are at least 4 years out of
date when it comes to S4 methods.  Is there any comprehensive
reference of the current implementation of the S4 OO system apart from
the source code?

Hadley

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Re: [R] Labelling boxplot with fivenumber summary

2007-04-08 Thread Ranjan Maitra
You may wish to provide brief commented source code: that would be most helpful 
for people referring to it in the archives.

Many thanks and best wishes,
Ranjan

On Sun, 8 Apr 2007 11:43:47 + Daniel Siddle [EMAIL PROTECTED] wrote:

 
 Just wanted to say thanks very much.  I used Chuck's 2nd idea as I found it 
 the easiest to understand as I'm still finding me feet with R.  Just for 
 reference for anyone else, fn[1] and fn[5] actually pasted and gave the 
 values at the whiskers (~1.5 IQR) so replaced them with a max and min 
 function which returned the true max and min values and pasted at the correct 
 heights.
 
 Thanks once again for all the help.  Regards,
 
 Daniel Siddle
 
 __
 R-help@stat.math.ethz.ch 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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Re: [R] compile errors with rgl-0.70.570 on FreeBSD

2007-04-08 Thread Bernardo Rangel Tura
On Wed, 2007-04-04 at 21:12 +0200, Rainer Hurling wrote:
 Thank you for your work on rgl.
 
 Reading in another thread about your new (inofficial) version of rgl 
 (see below) I tried it with R version 2.5.0 alpha (2007-03-31 r40986) 
 under FreeBSD 7.0-CURRENT. Unfortunately I got the following error:
 
 ---
 #R CMD INSTALL rgl_0.70.570.tar.gz
 * Installing to library '/usr/local/lib/R/library'
 * Installing *source* package 'rgl' ...
 checking for libpng-config... yes
 configure: using libpng-config
 configure: using libpng dynamic linkage
 checking for X... libraries /usr/X11R6/lib, headers /usr/X11R6/include
 checking GL/gl.h usability... no
 checking GL/gl.h presence... no
 checking for GL/gl.h... no
 checking GL/glu.h usability... no
 checking GL/glu.h presence... no
 checking for GL/glu.h... no
 configure: error: missing required header GL/gl.h
 ERROR: configuration failed for package 'rgl'
 ** Removing '/usr/local/lib/R/library/rgl'
 ** Restoring previous '/usr/local/lib/R/library/rgl'
 ---

Rainer,

I have same problem in my Ubuntu.
I solve instaling OpenGl in system, in my case I install this debian
poackages:

freeglut3, freeglut3-dev, libgl1-mesa-dev, libglu1-mesa-dev, libice-dev,
libsm-dev, libx11-dev, libxau-dev, libxdmcp-dev, libxext-dev, libxt-dev,
mesa-common-dev, x11proto-core-dev, x11proto-input-dev, x11proto-kb-dev,
x11proto-xext-dev, xlibmesa-gl-dev and xtrans-dev

If you install similar packages in your system  I think your problem
will be solve.

-- 
Bernardo Rangel Tura, M.D, PhD
National Institute of Cardiolgy
Rio de Janeiro - Brasil

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Re: [R] compile errors with rgl-0.70.570 on FreeBSD

2007-04-08 Thread Rainer Hurling
Bernardo,

thank you for your tips. But before version 0.70.570 rgl was running 
fine on my system. All necessary GL files are installed.

Duncan Murdoch is working on a configure script to get rgl also work on 
Mac OSX. Some of his changes breaks things for FreeBSD. We tested it 
without posting on [EMAIL PROTECTED]

Duncan now tries to find a solution for all platforms.

Rainer Hurling


Bernardo Rangel Tura schrieb:
 On Wed, 2007-04-04 at 21:12 +0200, Rainer Hurling wrote:
 Thank you for your work on rgl.

 Reading in another thread about your new (inofficial) version of rgl 
 (see below) I tried it with R version 2.5.0 alpha (2007-03-31 r40986) 
 under FreeBSD 7.0-CURRENT. Unfortunately I got the following error:

 ---
 #R CMD INSTALL rgl_0.70.570.tar.gz
 * Installing to library '/usr/local/lib/R/library'
 * Installing *source* package 'rgl' ...
 checking for libpng-config... yes
 configure: using libpng-config
 configure: using libpng dynamic linkage
 checking for X... libraries /usr/X11R6/lib, headers /usr/X11R6/include
 checking GL/gl.h usability... no
 checking GL/gl.h presence... no
 checking for GL/gl.h... no
 checking GL/glu.h usability... no
 checking GL/glu.h presence... no
 checking for GL/glu.h... no
 configure: error: missing required header GL/gl.h
 ERROR: configuration failed for package 'rgl'
 ** Removing '/usr/local/lib/R/library/rgl'
 ** Restoring previous '/usr/local/lib/R/library/rgl'
 ---
 
 Rainer,
 
 I have same problem in my Ubuntu.
 I solve instaling OpenGl in system, in my case I install this debian
 poackages:
 
 freeglut3, freeglut3-dev, libgl1-mesa-dev, libglu1-mesa-dev, libice-dev,
 libsm-dev, libx11-dev, libxau-dev, libxdmcp-dev, libxext-dev, libxt-dev,
 mesa-common-dev, x11proto-core-dev, x11proto-input-dev, x11proto-kb-dev,
 x11proto-xext-dev, xlibmesa-gl-dev and xtrans-dev
 
 If you install similar packages in your system  I think your problem
 will be solve.


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[R] Confidence-Intervals.... help...

2007-04-08 Thread Jochen.F

Hi...

I have to use R to find out the 90% confidence-interval for the sensitivity
and specificity of the following diagnostic test:

A particular diagnostic test for multiple sclerosis was conducted on 20 MS
patients and 20 healthy subjects, 6 MS patients were classified as healthy
and 8 healthy subjects were classified as suffering from the MS.

Furthermore, I need to find the number of MS patients required for a
sensitivity of 1%...

Is there a simple R-command which can do that for me?

I am completely new to R...

Help please!

Jochen
-- 
View this message in context: 
http://www.nabble.com/Confidence-Intervals-help...-tf3544217.html#a9894014
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Confidence-Intervals.... help...

2007-04-08 Thread Sarah Goslee
Hm... sounds like a homework problem to me...

Maybe start by figuring out how to do it without R -
what's the approach, and how would you calculate it?
Then search R help for the possible key words you
came up with.

Sarah

On 4/8/07, Jochen.F [EMAIL PROTECTED] wrote:

 Hi...

 I have to use R to find out the 90% confidence-interval for the sensitivity
 and specificity of the following diagnostic test:

 A particular diagnostic test for multiple sclerosis was conducted on 20 MS
 patients and 20 healthy subjects, 6 MS patients were classified as healthy
 and 8 healthy subjects were classified as suffering from the MS.

 Furthermore, I need to find the number of MS patients required for a
 sensitivity of 1%...

 Is there a simple R-command which can do that for me?

 I am completely new to R...

 Help please!

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

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[R] Failure of mcsamp() but not mcmcsamp()

2007-04-08 Thread Michael Kubovy
Daer r-helpers,

Can anyone help with the following:

I get:

  mcmcsamp(txt.lmer3f)
Markov Chain Monte Carlo (MCMC) output:
Start = 1
End = 1
Thinning interval = 1


But:

  mcsamp(txt.lmer3f)
Error in as.bugs.array(sims, program = lmer, n.iter = n.iter,  
n.burnin = n.burnin,  :
error in parameter sigma.s in parameters.to.save


I had attached the arm package:
  library(arm)
Loading required package: R2WinBUGS

arm (Version 1.0-15, built: 2007-3-17)
Working directory is /Users/mk/Documents/projects/Mike/prediss/Exp-Text
Loading required package: foreign

  sessionInfo()
R version 2.4.1 (2006-12-18)
i386-apple-darwin8.8.1

locale:
C

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

other attached packages:
  foreign  armR2WinBUGS coda 
Hmisc  gmodels  effects lme4
 0.8-18 1.0-15  2.0-4 0.10-7  3.2-1  
2.13.2  1.0-9  0.9975-13
   Matrix  car   weavercodetools
digest   xtable latticeExtra  lattice
0.9975-11  1.2-1  1.0.1  0.0-3  0.2.3   
1.4-3  0.1-40.14-17
 gridBase MASS  JGR   iplots
JavaGDrJava
  0.4-3 7.2-33 1.4-15  1.0-5  0.3-6  
0.4-14

Thanks,
MK
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
 McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/

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Re: [R] Reasons to Use R

2007-04-08 Thread Johann Hibschman
On 4/6/07, Wilfred Zegwaard [EMAIL PROTECTED] wrote:

 I'm not a programmer, but I have the experience that R is good for
 processing large datasets, especially in combination with specialised
 statistics.

This I find a little surprising, but maybe it's just a sign that I'm
not experienced enough with R yet.

I can't use R for big datasets.  At all.  Big datasets take forever to
load with read.table, R frequently runs out of memory,  and nlm or
gnlm never seem to actually converge to answers.  By comparison, I can
point SAS and NLIN at this data without problem.  (Of course, SAS is
running on a pretty powerful dedicated machine with a big ram disk, so
that may be part of the problem.)

R's pass-by-value semantics also make it harder than it should be to
deal with where it's crucial that you not make a copy of the data
frame, for fear of running out of memory.  Pass-by-reference would
make implementing data transformations so much easier that I don't
really understand how pass-by-value became the standard.  (If there's
a trick to doing in-place transformations, I've not found it.)

Right now, I'm considering starting on a project involving some big
Monte Carlo integrations over the complicated posterior parameter
distributions of a nonlinear regression model, and I have the strong
feeling that R will just choke.

R's great for small projects, but as soon as you even a few hundred
megs of data, it seems to break down.

If I'm doing things wrong, please tell me.  :-)  SAS is a beast to work with.

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Re: [R] Reasons to Use R

2007-04-08 Thread Gabor Grothendieck
On 4/8/07, Johann Hibschman [EMAIL PROTECTED] wrote:
 R's pass-by-value semantics also make it harder than it should be to
 deal with where it's crucial that you not make a copy of the data
 frame, for fear of running out of memory.  Pass-by-reference would
 make implementing data transformations so much easier that I don't
 really understand how pass-by-value became the standard.  (If there's
 a trick to doing in-place transformations, I've not found it.)

Because R processes objects in memory I also would not rate it as
as strong as some other packages on very large data sets but you can
use databases which may make it less important in some cases and you
can get a certain amount of mileage out of R environments and as
64 bit computers become commonplace and memory sizes grow
larger and larger data sets will become easy to handle.

Regarding environments, also available are proto objects from the
proto package which are environments with slightly different semantics.
Even if you don't intend to use the proto package its got quite a bit
of documentation and supporting information that might be
helpful:

- home page:
  http://code.google.com/p/r-proto/
- overview (click on Wiki tab at home page) which includes article links
  that discuss OO and environments
- tutorial, reference card, reference manual, vignette (see Links box)

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[R] [Fwd: Re: regular expression]

2007-04-08 Thread Laurent Rhelp


---BeginMessage---

Uwe Ligges a écrit :

I guess your problem has been solved by last night's discussion with 
Gabor G.?


Uwe Ligges


Laurent Rhelp wrote:


Uwe Ligges a écrit :


Laurent Rhelp wrote:
 


Uwe Ligges a écrit :

 


Laurent Rhelp wrote:

   


Dear R-List,

I have a great many files in a directory and I would like to 
replace in every file the character  by the character ' and in 
the same time, I have to change ' by '' (i.e. the character ' 
twice and not the unique character ) when the character ' is 
embodied in .
 So, . becomes '.' and .'.. becomes 
'.''..'
Certainly, regular expression could help me but I am not able to 
use it.


How can I do that with R ?
  



In fact, you do not need to know anything about regular 
expressions in this case, since you are simply going to replace 
certain characters by others without any fuzzy restrictions:


x - \.'..\
cat(x, \n)
xn - gsub('', ', gsub(', '', x))
cat(xn, \n)


Uwe Ligges


   


Thank you very much

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Yes, You are right. So I wrote the code below (that I find a little 
awkward but it works).


##-

dirdata - getwd()
fichnames - list.files(path=paste(dirdata,\\initial\\,sep=))
  



see ?file.path to improve the above.


 


for( i in 1:length(fichnames)){
  



see ?seq to improve the above: seq(along = fichnames)
Or even better, just work on the names (see below).

 


filein - paste(dirdata,\\initial\\,fichnames[i],sep=)
  



again, file.path() is your friend

 


conin - file(filein)
open(conin)  


  nbrows - length( readLines(conin,n=-1) )
 


close(conin)
  



You can simply use readLines() with the filename which open the 
connection to a file itself. And I do not see why you want to read 
the file here. Since your code becomes really complicated now, let 
me suggest the following procedure (untested!):


dirdata - getwd()
fichnames - list.files(file.path(dirdata, initial))
for(i in fichnames){
temp - readLines(file.path(dirdata, initial, i))
temp - gsub('', ', gsub(', '', temp))
writeLines(temp, con = file.path(dirdata, result, i))
}

Uwe Ligges





 


fileout - paste(dirdata,\\result\\,fichnames[i],sep=)
conout - file(fileout,w)

conin - file(filein)
open(conin)


for( l in 1:nbrows )
{
  text - gsub('',',gsub(','',readLines(conin,n=1)))
  writeLines(con=conout,text=text)
}

close(conin)
close(conout)
}

##--
  



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Dear Uwe,

The code doesn't do what I want because I want to replace ' by '' 
only when the character ' is embodied in ..

So :
1.  becomes '
2. .'.. becomes '..''..'
3.  but '...'  has to stay  '...' and not ''..''

Did I miss something ?




   





Yes, Gabor gave me the end of the solution. Thank you Uwe and Gabor.
For the people who are interested in the solution I will recapitulate 
below the differents steps.


The objective was to replace in a lot of files single quotes in double 
quoted strings with single quoted strings containing double quotes in 
place of the single quotes. We have to allow for the fact that the 
pattern can be written on several lines in the files.


Let us list the steps to realize that :

 1. read the lines of every file in a vector of strings (readLines)
 2. transform this vector in a string with multiple lines 
(paste(,collapse=\n)

 3. use a regular expression on this string to do the replacement
 4. go back to a vector of strings (strsplit) to have again the initial 
file altered

 5. write the new file

##--
   
library(gsubfn)

squote - ' # single quote.
 # This is a double quote, single quote, double quote
dquote - '' # double quote
 #This is a single quote, double quote, single quote

f - function(x)  chartr(paste(squote, dquote), paste(dquote, squote), x)

dirdata - getwd()  # not necessary
fichnames - list.files(file.path(dirdata, \\initial),pattern=.PRC$) 
# to select only the files with .PRC extension for example

for(i in fichnames){
  Lines - readLines(file.path(dirdata, \\initial, i))
  temp - gsubfn('[][^]*[]', f, paste(Lines, collapse = \n))
  Lines - unlist(strsplit(temp,\n)) # strsplit returns a list not a 
character vector

  writeLines(Lines, con = file.path(dirdata, \\result, i))
}

Thank you very much again.



---End Message---

[R] How do I back transforme ordinary log-krigged prdiction values?

2007-04-08 Thread Zia Uddin Ahmed
I have a question to everybody.

After log10 transfprmation, I have done ordinary kriging in gstat in R? I
need to back trnasform the prediction values to orgiginal scale. How do I
do this in gstat in R?

Thanks

Zia
-- 
Zia Uddin Ahmed
915 Brad Field Hall
Department of Crop and Soil
Cornell University
Ithaca NY 14850
USA

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[R] Plot symbols dimensions

2007-04-08 Thread Cressoni, Massimo \(NIH/NHLBI\) [F]

I am writing some code to obtain publication-like plots (like the ones can be 
obtained with
SigmaPlot). I am not able to find the dimensions of a point. Parameters like 
cex make you able 
to make it bigger or smaller but I need to know the exact dimension (in 
relative coordinates :
if my plot is 640x480 a dot plotted with cex=1.5 will be...)

Massimo Cressoni

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Re: [R] Reasons to Use R

2007-04-08 Thread Wilfred Zegwaard
Dear Johann and Gabor,

It's what amounts to large datasets. There are hundreds of datasets R
can't handle, probably thousands or more. I noticed on my computer
(which is nothing more that an average PC) that R breaks down after 250
MB of memory. I also note that SPSS breaks down, Matlab, etc.

I'm not a SAS user, but I have worked in the past with SAS. It's very
good as a remember, but it's ten years ago. And it's a dollar machine
I've been told: you add dollars to SAS as you add dollars to a Porsche.
I haven't got it and for most statistical applications it isn't
necessary I've been told. R is sufficient for that. The datasets I use
are often not that big (the way I like it).
About three years ago I spoke to somebody who has worked with it and
said it's database system is excellent and statistical profound.
Someone with a PhD, so probably he is right.

Monte-Carlo simulations are computationally time-consuming, but probably
these can be done in R. I haven't seen any libaries for it (they might
be there). It has been done with S (the commercial counterpart of R), so
probably with R too. If you tie Monte Carlo simulaton with large
datasets you probably run into problems with a conventional R system.
What I've been told in those instances is buy a new computer / add
memory and buy a new processor... and don't smoke hashiesh.

That wasn't a good advice because the guy who told me smoked hashiesh
like hell and drank Pastis (blue liqor) like water. I kicked him out.
But that's another story.

Cheers,

Wilfred

(I drink wine and tailor made beer, and only on occasions. That's why.
His simulations were good I've been told.)

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[R] Could not fit correct values in discriminant analysis by bruto.

2007-04-08 Thread 川口 修治
Dear R-users,

I would like to use bruto function in mda package for flexible  
discriminant analysis.
Then, I tried, for example, following approach.

  x
 band1  band2band3
1   -1.206780  -1.448007  -1.084431
2   -0.294938  -0.113222  -0.95
3   -0.267303  -0.241567  -1.040979
4   -1.206780  -1.448007  -1.040979
5   -1.151518  -0.806286  -0.671630
6   -1.179146  -1.396670  -1.453775
7   -0.294938  -0.241567  -1.453775
8   -0.350200  -0.267239  -1.084431
9   -1.151518  -0.857623  -0.649901
10  1.362954  -1.396670  -2.235926
11 -0.239675   1.118883   1.457551
12 -0.294938  -1.268325  -0.497817
13 -0.294938  -0.729278  -0.106745
14 -1.123883  -0.703612  -0.150196
15  0.616905   1.144548   -0.150196
16 -0.267303  1.657930 1.044750
17  1.611637  1.041874  0.610225
18 -1.123883 -0.677941 0.262605
19 -0.239675 -0.626604-0.128473
20  2.274797   1.118883 1.805171

  y
[1] 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0

  fit - bruto(x,y)

But, obtained fit$fitted.values are enormously high (or low) .
Execution of bruto(x[,2:3], y) is done well (values are nearly 1 or 0).
Values of column 1 are wrong or appropriate option is needed?
I contacted the package maintainer, but the problem could not be solved.

Thanks

Shuji Kawaguchi

  R.version
platform   i386-apple-darwin8.8.1
arch   i386
os darwin8.8.1
system i386, darwin8.8.1
version.string R version 2.4.0 (2006-10-03)

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[R] Repeated Measures design using lme

2007-04-08 Thread Scott Norton
Hi, 
I have what I believe is a repeated-measures dataset that I'm trying to analyze 
using lme().  This is *not* homework, but an exercise in my trying to 
self-teach myself repeated-measure ANOVA for other *real* datasets that I have 
and that are extremely similar to the following design.

I'm fairly sure the dataset described below would work with lme() -- but it'd 
be great if anybody can confirm that after I describe the dataset below) 

The study involves measuring the effect of a drug on blood pressure. There were 
16 patients in all and 6 replicate measures per patient of their blood pressure 
on one week (one measure per day). Two weeks later, a drug was introduced to 8 
randomly selected patients in such a way that I had equal representation of the 
4 age groups among the two treatment groups. Then, another two weeks later, 6 
replicate measures per patient (per day) of blood pressure was retaken.  So 
each patient had 12 total measures whether they were in the treatment group or 
in the control group (6 reads (R1-R6) in the baseline-week and another 6 reads 
(R1-R6) in the post-treatment week).  

So, 
Background: 16 patients 
Response measure: Blood pressure 
Fixed Factor: 4 Age groups 
Fixed Factor: Drug vs. NoDrug 
Random factor: Day of the read (i.e. 6 replicate reads (R1-R6) at the baseline 
time, and 6 replicate reads (R1-R6) after the drug has had time to take effect) 
Random Factor: Subjects 1-16 

PatientAgeGroupBP(Blood Pressure)Read (replicate reads)
Pre/PostTreatmentWeekGroup
120-2983   R1   
 preTreat 
220-2981   R1   
 preControl 
320-2974   R1   
 preTreat 
420-2985   R1   
 preControl 
530-3982   R1   
 preTreat 
……………… 
320-2974   R2   
 preTreat 
……………… 
120-2983   R1   
 post   Treat 
220-2982   R1   
 post   Control 
320-2986   R1   
 post   Treat 
420-2984   R1   
 post   Control 
………………

I'm trying to do an analysis of variance to decide whether there is a 
measurable change in blood pressure between the Treat and Control groups.

Another issue is that some of the 16 patients didn't get all 6 replicate reads 
in their pre/post treatment weeks, so I need to include the na.omit function.

What I think I'm having the most trouble with is the repeated reads (R1 through 
R6) in the pre/post treatment weeks.  I'm fairly sure this is a random variable 
-- their order or identify (R1 in pre-treatment week has no relation to R1 in 
the post-treatment week, etc).  By placing Read as a random variable, am I 
covering myself there?
If I execute:

 summary(lme(BP ~ Group, random = ~ 1 | Patient, data = bloodpress, 
 na.action=na.omit))

I get a result, but I'm not sure it's correct -- do I need to tell the model 
about the Read factor (R1-R6 in pre/post weeks)?  

I'm really trying to set the right form of the lme() function call to decide 
1) if there is a statistical difference between the Treat/Control groups and,
2) if one takes into account AgeGroup, is there a statistical difference 
between Treat/Control Groups, and finally
3) if I don't see a statistical difference, can someone recommend an R function 
that might solve the supplemental question, given the noise in day-to-day 
blood pressure reads, and given that I wanted to have enough statistical power 
to observe a say, a 5% benefit in blood pressure, how many additional reads or 
patients I would need.

Basically, is lme() the proper function, and can someone offer any pointers on 
what my call to this function should look like to make the above to 
determinations?

Thanks!
-Scott

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[R] R:Maximum likelihood estimation using BHHH and BFGS

2007-04-08 Thread joey repice
Dear R users,

I am new to R. I would like to find *maximum likelihood estimators for psi
and alpha* based on the following *log likelihood function*, c is
consumption data comprising 148 entries:

fn-function(c,psi,alpha)

{

s1-sum(for(i in 1:n){(c[i]-(psi^(-1/alpha)*(lag(c[i],-1^2*

(lag(c[i],-1)^((-2)*(alpha+1))

)});

s2- sum(for(m in 1:n){log(lag(c[m],-1)^(((2)*alpha)+2))});

constant-(74*s1)-(1/2)*s2}



In particular I want to use the BHHH and the BFGS algorithms. I have found
the following packages that can be used for the task:  *mle{stats4},
optim{stats}, maxBHHH{micEcon}, maxBFGS{micEcon}*. I have documented some of
my attempts below ((a) package name (b) usage (c) my attempt and
corresponding error). In all humility I apologise for any bad coding, and
ask if anyone can *direct me in finding these estimators*.



Yours sincerely.





*(1a)  mle{stats4}

(b)  Usage:

 mle(minuslogl, start = formals(minuslogl), method = BFGS,

 fixed = list(), ...)

(For this I use the negative of the log-likelihood function,bn)*

*(c)  mle(start=list(psi=1,alpha=0),fn, method=BFGS,fixed=list(c=c))*

Error in optim(start, f, method = method, hessian = TRUE, ...) :

non-finite finite-difference value [1]



*(2a)  optim{stats}



(b)  optim(par, fn, gr = NULL,

  method = c(Nelder-Mead, BFGS, CG, L-BFGS-B, SANN),*

*

  lower = -Inf, upper = Inf,

  control = list(), hessian = FALSE, ...)



(c)   optim(c(c,1,0),fn, method=BFGS, control=list(c=c))*

Error: argument psi is missing, with no default



*(3a) maxBHHH{micEcon}

(b) Usage: maxBHHH(fn, grad = NULL, hess=NULL, theta, print.level
= 0, iterlim = 100,   ...)*

* *

*(c)  maxBHHH(fn(c,psi,alpha),theta=list(psi=1,alpha=0))*

 Error in psi^(-1/alpha) * (lag(c[i], -1)) :

time-series/vector length mismatch



(4a)  maxBFGS{micEcon}



(b)  Usage: maxBFGS(fn, grad = NULL, theta, print.level = 0, iterlim = 200,

 tol = 1e-06, ... )

*(ci)   maxBFGS(fn(c,psi,alpha),theta=list(psi=1,alpha=0))*

Error in psi^(-1/alpha) * (lag(c[i], -1)) :

time-series/vector length mismatch



*(cii) maxBFGS(fn,theta=list(psi=1,alpha=0))*

Error: argument psi is missing, with no default

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[R] Dealing with large nominal predictor in sem package

2007-04-08 Thread adschai
Hi,

I am using tsls function from sem package to estimate a model which includes 
large number of data. Among its predictors, it has a nominal data which has 
about 10 possible values. So I expand this parameter into 9-binary-value 
predictors with the coefficient of base value equals 0. I also have another 
continuous predictor. 

The problem is that, whenever I run the tsls, I will get 'System is 
computationally singular' error all the time. I'm wondering if there is anyway 
that I can overcome this problem? Please kindly suggest. Thank you so much in 
advance.

- adschai

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