date:20080214

I don't know if this you want but, you can try:

 table(data.frame(prev, iris$Species))

On 14/02/2008, Carla Rebelo [EMAIL PROTECTED] wrote:
 Hello!

  May you help me? I'm trying to do a contingency table using this

data(iris)
library(rpart)
modelo - rpart(Species ~., iris)
prev - predict(modelo, iris)

  Finally the contingency table
table(iris$Species, prev)
  But an error occurs:
  Error in table(iris$Species, prev) :
   all arguments must have the same length

  And I do not understand why, may you help me?

  Thank You!

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  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
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Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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[R] How to check cy5 and cy3 values were lowess normalized

2008-02-14 Thread Ng Stanley

Hi,

I have some microarray data,  cy5 and cy3 values are in log2. Is there a
way to check they have undergone lowess normalization ?

Thanks
Stanley

[[alternative HTML version deleted]]

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[R] Random termination of R with RODBC and PostgreSQL 8.2.6

2008-02-14 Thread Ulrich Leopold

Dear list,

I get random terminations of R when using RODBC with a remote PostgreSQL 8.2.6
database under Ubuntu Linux 7.10 32bit client.

The exact messages can be seen below. It seems that there is some memory
corruption going on. But it seems random to me and can neither be related to
addPK=T/F or fast=T/F. It can happend for all possible combinations. Besides I
do not really understand what these messages mean.

Any idea to what it is related exactly?

Thanks in advance,
Ulrich Leopold

Session details:

 library(RODBC)
 channel = odbcConnect(crtedb, uid = uleopold)
 channel
RODB Connection 1
Details:
  case=tolower
  DSN=crtedb
  DATABASE=valid
  SERVER=10.3.1.91
  PORT=5432
  SSLMODE=prefer
  UID=uleopold
  PWD=**
  ReadOnly=Yes
  FakeOidIndex=No
  ShowOidColumn=No
  RowVersioning=No
  ShowSystemTables=No
  ConnSettings=
  Fetch=100
  Socket=8192
  UnknownSizes=0
  MaxVarcharSize=254
  MaxLongVarcharSize=8190
  Debug=0
  CommLog=0
  Optimizer=1
  Ksqo=1
  UseDeclareFetch=0
  TextAsLongVarchar=1
  UnknownsAsLongVarchar=0
  BoolsAsChar=1
  Parse=0
  CancelAsFreeStmt=0
  ExtraSysTablePrefixes=dd_

  LFConversion=0
  UpdatableCursors=0
  DisallowPremature=0
  TrueIsMinus1=0
  BI=0
  ByteaAsLongVarBinary=0
  UseServerSidePrepare=0
  LowerCaseIdentifier=0

 sqlDrop(channel, clx_test)
character(0)
 sqlSave(channel, clx_test, varTypes=varspec, addPK=TRUE, fast=TRUE)
 sessionInfo()
R version 2.6.2 (2008-02-08)
i486-pc-linux-gnu

locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=en_US.UTF-8;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C

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

other attached packages:
[1] RODBC_1.2-3

loaded via a namespace (and not attached):
[1] rcompgen_0.1-17
 odbcClose(channel)
*** glibc detected *** /usr/lib/R/bin/exec/R: corrupted double-linked list:
0x082dd498 ***
=== Backtrace: =
/lib/tls/i686/cmov/libc.so.6[0xb7b8c3f9]
/lib/tls/i686/cmov/libc.so.6[0xb7b8db6e]
/lib/tls/i686/cmov/libc.so.6(cfree+0x90)[0xb7b91800]
/usr/lib/R/lib/libR.so(R_chk_free+0x21)[0xb7d687c1]
/usr/local/lib/R/site-library/RODBC/libs/RODBC.so[0xb742db8d]
/usr/local/lib/R/site-library/RODBC/libs/RODBC.so(RODBCClose+0x2a)[0xb742dd5a]
/usr/lib/R/lib/libR.so[0xb7d13ea2]
/usr/lib/R/lib/libR.so(Rf_eval+0x707)[0xb7d406a7]
/usr/lib/R/lib/libR.so[0xb7d43037]
/usr/lib/R/lib/libR.so(Rf_eval+0x444)[0xb7d403e4]
/usr/lib/R/lib/libR.so[0xb7d41967]
/usr/lib/R/lib/libR.so(Rf_eval+0x444)[0xb7d403e4]
/usr/lib/R/lib/libR.so(Rf_applyClosure+0x2a7)[0xb7d439b7]
/usr/lib/R/lib/libR.so(Rf_eval+0x2e2)[0xb7d40282]
/usr/lib/R/lib/libR.so(Rf_ReplIteration+0x19b)[0xb7d6632b]
/usr/lib/R/lib/libR.so(run_Rmainloop+0x107)[0xb7d666f7]
/usr/lib/R/lib/libR.so(Rf_mainloop+0x1c)[0xb7d6674c]
/usr/lib/R/bin/exec/R(main+0x46)[0x8048776]
/lib/tls/i686/cmov/libc.so.6(__libc_start_main+0xe0)[0xb7b3a050]
/usr/lib/R/bin/exec/R[0x8048691]
=== Memory map: 
08048000-08049000 r-xp  08:03 422045 /usr/lib/R/bin/exec/R
08049000-0804a000 rw-p  08:03 422045 /usr/lib/R/bin/exec/R
0804a000-09941000 rw-p 0804a000 00:00 0  [heap]
a770-a7721000 rw-p a770 00:00 0
a7721000-a780 ---p a7721000 00:00 0
a78ef000-a793e000 r-xp  08:03 259109 /usr/lib/libgcrypt.so.11.2.3
a793e000-a794 rw-p 0004e000 08:03 259109 /usr/lib/libgcrypt.so.11.2.3
a794-a7943000 r-xp  08:03 259082 /usr/lib/libgpg-error.so.0.3.0
a7943000-a7944000 rw-p 2000 08:03 259082 /usr/lib/libgpg-error.so.0.3.0
a7944000-a7953000 r-xp  08:03 266851 /usr/lib/libtasn1.so.3.0.9
a7953000-a7954000 rw-p e000 08:03 266851 /usr/lib/libtasn1.so.3.0.9
a7954000-a79be000 r-xp  08:03 730108 /usr/lib/libgnutls.so.13.3.0
a79be000-a79c4000 rw-p 0006a000 08:03 730108 /usr/lib/libgnutls.so.13.3.0
a79c4000-a79da000 r-xp  08:03 732700 /usr/lib/libsasl2.so.2.0.22
a79da000-a79db000 rw-p 00015000 08:03 732700 /usr/lib/libsasl2.so.2.0.22
a79db000-a79e6000 r-xp  08:03 732704 /usr/lib/liblber.so.2.0.130
a79e6000-a79e7000 rw-p b000 08:03 732704 /usr/lib/liblber.so.2.0.130
a79e7000-a79f6000 r-xp  08:03 712558
/lib/tls/i686/cmov/libresolv-2.6.1.so
a79f6000-a79f8000 rw-p f000 08:03 712558
/lib/tls/i686/cmov/libresolv-2.6.1.so
a79f8000-a79fa000 rw-p a79f8000 00:00 0
a79fa000-a7a01000 r-xp  08:03 259121 /usr/lib/libkrb5support.so.0.1
a7a01000-a7a02000 rw-p 6000 08:03 259121 /usr/lib/libkrb5support.so.0.1
a7a02000-a7a26000 r-xp  08:03 259108 /usr/lib/libk5crypto.so.3.1
a7a26000-a7a27000 rw-p 00024000 08:03 259108 /usr/lib/libk5crypto.so.3.1
a7a27000-a7a5b000 r-xp  08:03 732705 /usr/lib/libldap_r.so.2.0.130
a7a5b000-a7a5c000 rw-p 00034000 08:03 732705 /usr/lib/libldap_r.so.2.0.130
a7a5c000-a7a61000 r-xp  08:03

[R] Removing columns that are all NA from a matrix

2008-02-14 Thread Martin Waller

Hi,

I guess this might be a FAQ or something, and there's probably a nice 
simple way to do it, but I can't think of it:

Given a matrix, I want to remove columns that are _entirely_ filled with 
NAs (partial NAs are fine).

How please?

Thanks,

Martin

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Re: [R] Removing columns that are all NA from a matrix

2008-02-14 Thread Gabor Csardi

data - data[ , !apply(is.na(data), 2, all)]

(or something like that)

G.

On Thu, Feb 14, 2008 at 12:59:46PM +, Martin Waller wrote:
 Hi,
 
 I guess this might be a FAQ or something, and there's probably a nice 
 simple way to do it, but I can't think of it:
 
 Given a matrix, I want to remove columns that are _entirely_ filled with 
 NAs (partial NAs are fine).
 
 How please?
 
 Thanks,
 
 Martin
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Csardi Gabor [EMAIL PROTECTED]UNIL DGM

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Re: [R] contingency table

2008-02-14 Thread Prof Brian Ripley

Try looking at 'prev' (it is a matrix) and then reading ?predict.rpart to 
see why.  My guess is that you intended type=class.

On Thu, 14 Feb 2008, Carla Rebelo wrote:

 Hello!

 May you help me? I'm trying to do a contingency table using this

  data(iris)
  library(rpart)
  modelo - rpart(Species ~., iris)
  prev - predict(modelo, iris)

 Finally the contingency table
  table(iris$Species, prev)
 But an error occurs:
 Error in table(iris$Species, prev) :
  all arguments must have the same length

 And I do not understand why, may you help me?

 Thank You!

 __
 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.


-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] Removing columns that are all NA from a matrix

2008-02-14 Thread Chuck Cleland

On 2/14/2008 7:59 AM, Martin Waller wrote:
 Hi,
 
 I guess this might be a FAQ or something, and there's probably a nice 
 simple way to do it, but I can't think of it:
 
 Given a matrix, I want to remove columns that are _entirely_ filled with 
 NAs (partial NAs are fine).
 
 How please?

mymat[,which(colMeans(is.na(mymat))  1)]

 Thanks,
 
 Martin
 
 __
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-- 
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
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Re: [R] Removing columns that are all NA from a matrix

try this:

mat - matrix(rnorm(42), 6, 7)
mat[sample(42, 10)] - NA
mat[, c(3,5)] - NA

ind - colSums(is.na(mat)) != nrow(mat)
mat
mat[, ind]


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
 http://www.student.kuleuven.be/~m0390867/dimitris.htm


- Original Message - 
From: Martin Waller [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, February 14, 2008 1:59 PM
Subject: [R] Removing columns that are all NA from a matrix


 Hi,

 I guess this might be a FAQ or something, and there's probably a 
 nice
 simple way to do it, but I can't think of it:

 Given a matrix, I want to remove columns that are _entirely_ filled 
 with
 NAs (partial NAs are fine).

 How please?

 Thanks,

 Martin

 __
 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.
 


Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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Re: [R] Removing columns that are all NA from a matrix

2008-02-14 Thread Ted Harding

On 14-Feb-08 12:59:46, Martin Waller wrote:
 Hi,
 I guess this might be a FAQ or something, and there's
 probably a nice simple way to do it, but I can't think of it:
 
 Given a matrix, I want to remove columns that are _entirely_
 filled with NAs (partial NAs are fine).
 
 How please?
 
 Thanks,
 Martin

It probably isn't a FAQ (it's a somewhat special question)
but the following illustrates one way to do it:

  M-cbind(c(1,2,3,4),c(1,2,NA,4),c(NA,NA,NA,NA))
  M
## [,1] [,2] [,3]
##[1,]11   NA
##[2,]22   NA
##[3,]3   NA   NA
##[4,]44   NA
  apply(M,2,function(x){!all(is.na(x))})
##[1]  TRUE  TRUE FALSE
  M[,apply(M,2,function(x){!all(is.na(x))})]
## [,1] [,2]
##[1,]11
##[2,]22
##[3,]3   NA
##[4,]44

Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 14-Feb-08   Time: 13:14:20
-- XFMail --

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[R] Question: about loading a new package

2008-02-14 Thread Stefano Sofia

I am using R Version 2.3.1 (2006-06-01) in Linux with Ubuntu 7.4.

I installed a new package with 

R CMD INSTALL -l /usr/lib/R/library newpackage.tar.gz

Each time I run R I need to load this new package typing library(newpackage).
My question is: how can I automatically load this new package once for ever?
I read the manual but I have not been able to find the solution to that.

thank you
Stefano

[[alternative HTML version deleted]]

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[R] (RESOLVED) Re: Removing columns that are all NA from a matrix

2008-02-14 Thread Martin Waller

Thanks to all who responded - great stuff!

Martin

Martin Waller wrote:
 Hi,
 
 I guess this might be a FAQ or something, and there's probably a nice 
 simple way to do it, but I can't think of it:
 
 Given a matrix, I want to remove columns that are _entirely_ filled with 
 NAs (partial NAs are fine).
 
 How please?
 
 Thanks,
 
 Martin
 
 __
 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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Question: about loading a new package

2008-02-14 Thread Ted Harding

On 14-Feb-08 13:23:17, Stefano Sofia wrote:
 I am using R Version 2.3.1 (2006-06-01) in Linux with Ubuntu 7.4.
 I installed a new package with 
 
 R CMD INSTALL -l /usr/lib/R/library newpackage.tar.gz
 
 Each time I run R I need to load this new package typing
 library(newpackage).
 My question is: how can I automatically load this new package
 once for ever?
 I read the manual but I have not been able to find the solution to
 that.
 
 thank you
 Stefano

Since you are using Linux, your home directory (e.g. /home/stefano)
will contain a file .Rprofile (which you won't normally see unless
you run the Linux command ls -a).

Edit this file so that it includes the line

  library(newpackage)

(and, while you're at it, anything else that you would like
to be done every time you start R).

Hoping this helps,
Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 14-Feb-08   Time: 13:48:30
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Re: [R] Tinn-R not working well with latest R

2008-02-14 Thread Philippe Grosjean

Hello,

These are *warnings*. So, the functions should still work. However, 
there are changes made in R that probably need some lifting in the svIDE 
package. The whole SciViews bundle is currently reworked.

Unfortunately, I now work on Mac OS X, and it is a little bit more 
difficult for me to test this with Tinn-R. Could you give me more infos 
about the great features in Tinn-R that stop working in R 2.6.2, please?
Best,

Philippe Grosjean
..°}))
  ) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
  ) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
  ) ) ) ) )   Mons-Hainaut University, Belgium
( ( ( ( (
..

Farrel Buchinsky wrote:
 I recently installed R 2.6.2 and am getting errors on startup that 
 relate to svIDE being loaded by Tinn-R.
 
 
 Loading required package: tcltk 
 Loading Tcl/Tk interface ... done 
 Warning messages: 
 1: '\A' is an unrecognized escape in a character string  
 2: unrecognized escape removed from ;for Options\AutoIndent: 0=Off, 
 1=follow language scoping and 2=copy from previous line\n  
 3: In grep(paste([{]TclEval , topic, [}], sep = ), 
 tclvalue(.Tcl(dde services TclEval {})), : 
 argument 'useBytes = TRUE' will be ignored 
 Loading required package: svMisc 
 Loading required package: R2HTML 
  
 
 Any idea what is going on. 
 I use R 2.6.2 on windows xp
 
 I also started R without the profile that Tinn-R made.
 If I manualy enter library(svIDE) then I get.
   library(svIDE)
 Warning messages:
 1: '\A' is an unrecognized escape in a character string
 2: unrecognized escape removed from ;for Options\AutoIndent: 0=Off, 
 1=follow language scoping and 2=copy from previous line\n
 
 So the underlying problem may be svIDE
 see: http://tolstoy.newcastle.edu.au/R/e2/help/07/04/15738.html
 
 Apparently, because of this error, several great features in Tinn-R are 
 not working properly.
 Any solutions or workarounds?
 
 
 -- 
 Farrel

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[R] Analysis with spatstat and Kcross() requires to much memory

2008-02-14 Thread Rainer M Krug

Hi

I am running analysis with Kcross from the package spatstat and I am getting
a message that R can  not allocate enough memory for a vector of 900MB.
R seems to be running towards the 2GB limit per process.

The dataset is not to big (ca 3000 points) but the mask for the points is
extremely irregular (a buffer around roads which have been sampled) and I
can do trhe analysis if I use a rectangular mask covering the whole area
(which is obviously not correct).

My question is: is there a way of splitting the dataset or changing the mask
son that I can do the analysis, as I don't want to move to a 64bit computer
(if not necessary).

Thanks,

Rainer

My computrer:

SuSE 10, 32 bit, 1GB ram, 1GB swap.

R:
 version
   _
platform   i686-pc-linux-gnu
arch   i686
os linux-gnu
system i686, linux-gnu
status
major  2
minor  6.2
year   2008
month  02
day08
svn rev44383
language   R
version.string R version 2.6.2 (2008-02-08)


-- 
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT)

Plant Conservation Unit Department of Botany
University of Cape Town
Rondebosch 7701
South Africa

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[R] Kaplan Meier function

2008-02-14 Thread Eleni Christodoulou

Hi all,

I am trying to draw a Kaplan-Meier curve and I found online that Kaplan -
Meier estimates are computed with a function called km in the event package.
Is there an update for that because when I choose to download packages in
R,. there is no package called event, even though I have selected all the
repositories.

Thanks in advance,
Eleni

[[alternative HTML version deleted]]

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Re: [R] Kaplan Meier function

2008-02-14 Thread Kevin E. Thorpe

See the survfit() in the survival package which is one on the
recommended packages that should (I think) have come with R.

Eleni Christodoulou wrote:
 Hi all,
 
 I am trying to draw a Kaplan-Meier curve and I found online that Kaplan -
 Meier estimates are computed with a function called km in the event package.
 Is there an update for that because when I choose to download packages in
 R,. there is no package called event, even though I have selected all the
 repositories.
 
 Thanks in advance,
 Eleni
 


-- 
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED]  Tel: 416.864.5776  Fax: 416.864.6057

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Re: [R] Kaplan Meier function

2008-02-14 Thread Dieter Menne

Eleni Christodoulou elenichri at gmail.com writes:

 I am trying to draw a Kaplan-Meier curve and I found online that Kaplan -
 Meier estimates are computed with a function called km in the event package.
 Is there an update for that because when I choose to download packages in
 R,. there is no package called event, even though I have selected all the
 repositories.
 

The standard survival package is package survival. Try survfit, plot.survfit,
and summary(survfit). There are some fancy extension in package Design, but
better start with survival, Design-based KM have huge options list that can be
confusing at first.

Dieter

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Re: [R] Any mountain clustering method in R?

2008-02-14 Thread Ramon Diaz-Uriarte

Dear Josep Maria,

Maybe this is way too naive, but what about a hill-climbing algorithm
started from different places, and without (almost) any attempt to
avoid local maxima? I think that if you formulate your question like a
clustering problem, then you'll be using techniques that attempt to
solve problems that are not really of any concern for you original
question.

Regardless, I think you'll need to carefully think what constitute
legitimate peaks for you. Suppose the 1D data
2, 3, 4, 3, 4, 5, 4, 3, 2

is the first 4 a peak or not?

Best,

R.

On Thu, Feb 14, 2008 at 12:08 PM, Josep Maria Campanera Alsina
[EMAIL PROTECTED] wrote:
 Dear all,
  I wonder which R algorithm could perform a mountain clustering in an spatial
  grid, in other words, having the coordinates of a map (x,y ... or even more)
  and then the altitude/height at each point I would like to localise the
  peaks of that topological surface. Which R algorithm can perform this? I see
  it as a clustering problem where the peak should be at the center of each
  cluster.

  Thanks a lot,

  Josep Maria,

  -
  Josep Maria Campanera Alsina
  Investigador Juan de la Cierva
  Departament de Fisicoquímica
  Facultat de Farmàcia
  Avgda Joan XXIII, s/n
  08028 Barcelona · Catalonia · Spain
  Tel: +34 93 4035988
  Fax: +34 93 4035987
  [EMAIL PROTECTED]
  

 [[alternative HTML version deleted]]


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-- 
Ramon Diaz-Uriarte
Statistical Computing Team
Structural Biology and Biocomputing Programme
Spanish National Cancer Centre (CNIO)
http://ligarto.org/rdiaz

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Re: [R] Kaplan Meier function

check function survfit() in package survival.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
 http://www.student.kuleuven.be/~m0390867/dimitris.htm


- Original Message - 
From: Eleni Christodoulou [EMAIL PROTECTED]
To: r-help@r-project.org
Sent: Thursday, February 14, 2008 2:50 PM
Subject: [R] Kaplan Meier function


 Hi all,

 I am trying to draw a Kaplan-Meier curve and I found online that 
 Kaplan -
 Meier estimates are computed with a function called km in the event 
 package.
 Is there an update for that because when I choose to download 
 packages in
 R,. there is no package called event, even though I have selected 
 all the
 repositories.

 Thanks in advance,
 Eleni

 [[alternative HTML version deleted]]

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


Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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[R] Transposing by a group variable

2008-02-14 Thread Shubha Vishwanath Karanth

Hi R,

 

Can I transpose a data frame by a particular group variable?

 

For example:

d=data.frame(group=c(1,1,2,2,2),val=c(6,4,6,3,5))

And my output should be:

data.frame(group=c(1,2),v1=c(6,6),v2=c(4,3),v3=c(NA,5))

 

Many thanks,

Shubha

This e-mail may contain confidential and/or privileged i...{{dropped:13}}

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Re: [R] Kaplan Meier function

2008-02-14 Thread Eleni Christodoulou

Thank you all for the replies!

Eleni

On Thu, Feb 14, 2008 at 4:10 PM, Dimitris Rizopoulos 
[EMAIL PROTECTED] wrote:

 check function survfit() in package survival.

 Best,
 Dimitris

 
 Dimitris Rizopoulos
 Ph.D. Student
 Biostatistical Centre
 School of Public Health
 Catholic University of Leuven

 Address: Kapucijnenvoer 35, Leuven, Belgium
 Tel: +32/(0)16/336899
 Fax: +32/(0)16/337015
 Web: http://med.kuleuven.be/biostat/
 
 http://www.student.kuleuven.be/~m0390867/dimitris.htmhttp://www.student.kuleuven.be/%7Em0390867/dimitris.htm


 - Original Message -
 From: Eleni Christodoulou [EMAIL PROTECTED]
 To: r-help@r-project.org
 Sent: Thursday, February 14, 2008 2:50 PM
 Subject: [R] Kaplan Meier function


  Hi all,
 
  I am trying to draw a Kaplan-Meier curve and I found online that
  Kaplan -
  Meier estimates are computed with a function called km in the event
  package.
  Is there an update for that because when I choose to download
  packages in
  R,. there is no package called event, even though I have selected
  all the
  repositories.
 
  Thanks in advance,
  Eleni
 
  [[alternative HTML version deleted]]
 
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  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 


 Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm



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Re: [R] User defined split function in Rpart

2008-02-14 Thread Terry Therneau

  The question is about the direction vector in rpart.
  
  There are (at least) two preferred ways to lay out a tree, wrt the question 
of 
which obs are sent left and which right.

1. Send the smaller y values to the left.  In the final tree,  there will 
be 
a graphical ordering with smaller y's to the left and larger ones to the right. 
 
One has a left bad, right good orientation when traversing the tree.  I find 
that medical researchers often like this.
 
2. Send observations with x  cutpoint to the left.  Setting all elements 
of 
the direction vector to -1 will give this behavior.  

I happen to slightly prefer option 1, which of course means that it became 
the default behavior in rpart.  (For a categorical y with many levels, however, 
rpart orders on the percent of observations in category 1, which may not be 
particularly useful.)


Terry Therneau

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[R] data manipulation for plotting

2008-02-14 Thread [EMAIL PROTECTED]


   Hi,

   i'd like to plot some data that I have with the value on the x axis and freq
   on the y axis.

   So, I need to calculate the freq a value is seen within my data vector

   for example, say i have a vector of data

   data=c(1,1,1,4,5,5,6)

   I want

   values-c(1,4,5,6)

   freq-c(3,1,2,1)

   in order to enable me to plot this. Sorry, i'm new to R. What is standard
   procedure here for plotting the data vector?

   cheers

   Rich





   
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Re: [R] Kaplan Meier function

2008-02-14 Thread Matthias Gondan

Hi Eleni, hi list,

here is small sample program, the library is survival, it is included
in the standard R distribution.

 data(colon)
 s = survfit(Surv(time, status) ~ rx, data=colon)
 plot(s)
 plot(s, col=1:3)

By the way: Does anyone know a neat way to indicate the number of
patients under risk in
the curve? Like here:

http://www.drugs.com/pro/images/8c3bb783-1c3c-4d75-a502-2f7ee3632f41/inspra-image02.jpg

Best wishes,

Matthias

Eleni Christodoulou schrieb:
 Hi all,

 I am trying to draw a Kaplan-Meier curve and I found online that Kaplan -
 Meier estimates are computed with a function called km in the event package.
 Is there an update for that because when I choose to download packages in
 R,. there is no package called event, even though I have selected all the
 repositories.

 Thanks in advance,
 Eleni

   [[alternative HTML version deleted]]

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



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Re: [R] Transposing by a group variable

try this:

d - data.frame(group = c(1,1,2,2,2), val = c(6,4,6,3,5))

d$time - unlist(tapply(d$group, d$group,
function (x) seq(1, len = length(x
reshape(d, idvar = group, direction = wide)


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
 http://www.student.kuleuven.be/~m0390867/dimitris.htm


- Original Message - 
From: Shubha Vishwanath Karanth [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, February 14, 2008 3:22 PM
Subject: [R] Transposing by a group variable


 Hi R,



 Can I transpose a data frame by a particular group variable?



 For example:

 d=data.frame(group=c(1,1,2,2,2),val=c(6,4,6,3,5))

 And my output should be:

 data.frame(group=c(1,2),v1=c(6,6),v2=c(4,3),v3=c(NA,5))



 Many thanks,

 Shubha

 This e-mail may contain confidential and/or privileged 
 i...{{dropped:13}}

 __
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 http://www.R-project.org/posting-guide.html
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[R] Levene's test for homogeneity of variances (befor using ANOVA)

2008-02-14 Thread Kes Knave

Dear all

I have tried to find this function in R, but don't find it by searching in
the help function.

Anybody who knows if R has the function Levene's test for homogeneity of
variances?

Note: Im a R-begginer

Regards Kes

[[alternative HTML version deleted]]

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Re: [R] data manipulation for plotting

2008-02-14 Thread Gavin Simpson


On Thu, 2008-02-14 at 14:35 +, [EMAIL PROTECTED] wrote:
 Hi,
 
i'd like to plot some data that I have with the value on the x axis and 
 freq
on the y axis.
 
So, I need to calculate the freq a value is seen within my data vector
 
for example, say i have a vector of data
 
data=c(1,1,1,4,5,5,6)

Don't call your data 'data' ;-)

install.packages(fortunes)
library(fortunes)
fortune(dog)

 
I want
 
values-c(1,4,5,6)
 
freq-c(3,1,2,1)

use ?unique and ?table

my.dat - c(1,1,1,4,5,5,6)
values - unique(my.dat)
values
tab - table(dat)
tab
freq - as.numeric(tab)
freq

plot(tab) # for one way of plotting what you want

plot(freq ~ values) # for another

HTH

G

 
in order to enable me to plot this. Sorry, i'm new to R. What is standard
procedure here for plotting the data vector?
 
cheers
 
Rich
 
 
 
 
 

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Re: [R] Kaplan Meier function

2008-02-14 Thread Frank E Harrell Jr

Matthias Gondan wrote:
 Hi Eleni, hi list,
 
 here is small sample program, the library is survival, it is included
 in the standard R distribution.
 
 data(colon)
 s = survfit(Surv(time, status) ~ rx, data=colon)
 plot(s)
 plot(s, col=1:3)
 
 By the way: Does anyone know a neat way to indicate the number of
 patients under risk in
 the curve? Like here:

That's one of many options in the survplot function in the Design 
package, which uses the survival package.

Frank

 
 http://www.drugs.com/pro/images/8c3bb783-1c3c-4d75-a502-2f7ee3632f41/inspra-image02.jpg
 
 Best wishes,
 
 Matthias
 
 Eleni Christodoulou schrieb:
 Hi all,

 I am trying to draw a Kaplan-Meier curve and I found online that Kaplan -
 Meier estimates are computed with a function called km in the event package.
 Is there an update for that because when I choose to download packages in
 R,. there is no package called event, even though I have selected all the
 repositories.

 Thanks in advance,
 Eleni

  [[alternative HTML version deleted]]

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


 
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
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  Department of Biostatistics   Vanderbilt University

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Re: [R] Modelling disease spread

2008-02-14 Thread francogrex


Thanks Marcel,
In addition to your program and the reference to simecol, someone had
replied to my private email pointing out RLadyBug: An R package for
stochastic epidemic models which is on CRAN and which seems one of the most
relevant. I write it here as a reference for users doing a future search in
this archive.
Regards.


Bio7 wrote:
 
 The simecol package is maybe what you want.
  http://hhbio.wasser.tu-dresden.de/projects/simecol/
 http://hhbio.wasser.tu-dresden.de/projects/simecol/ 
 Another possibility is to use a program i've written.
 Here is a flash presentation maybe also interesting for you.
 
 http://www.uni-bielefeld.de/biologie/Oekosystembiologie/bio7app/flashtut/animaterplot.htm
 http://www.uni-bielefeld.de/biologie/Oekosystembiologie/bio7app/flashtut/animaterplot.htm
  
 

-- 
View this message in context: 
http://www.nabble.com/Modelling-disease-spread-tp15459834p15480798.html
Sent from the R help mailing list archive at Nabble.com.

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[R] RFC for package PopCon: a popularity contest for R and packages

2008-02-14 Thread Jeffrey Horner

(I posted this to the R-devel list yesterday, but I thought others on 
this list would be interested, so sorry for those who get it twice.)

Hello all,

I've developed a prototype package called PopCon (short for popularity 
contest), a package for tracking the popularity of R and its packages. 
I'd like this work to be similar in spirit to the Debian package 
popularity-contest: http://popcon.debian.org/.

Once Popcon is loaded, it captures two kinds of information from the 
user and stores it into a cache: the names of the libraries he/she 
loads, and the names of symbols requested from his/her code. Once the 
cache is full, the goal is to flush the data to a central server for 
storage, free for anyone to download and analyze. That's it. Pretty 
simple use and works behind the scenes. You can get the prototype here:

http://biostat.mc.vanderbilt.edu/twiki/pub/Main/JeffreyHorner/PopCon_0.1.tar.gz

And note that flushing of the cache is NOT TURNED ON and IT WON'T 
FORWARD ANY DATA ANYWHERE! It only gets deleted.

So, I envision all the software and data generated and stored to be 
licensed under a GPL and a Creative Commons license, or even public domain.

Thoughts? I'm looking for volunteers, because there are many issues to 
hash out. Here's a few of them:

1. Obviously storing IP addresses or any bit of personal information is 
out, but I'm interested in generating a permanent random key of some 
sort so that data from the same R installs can be tracked. I'm wondering 
if just md5 hashing the combination of R version, platform, and IP 
address would be appropriate and reproducible per R install. The debian 
package popularity-contest has the benefit of installing an '/etc' 
config file and generating the key once, while I'd like PopCon users to 
just call 'library(PopCon)' and do nothing else.

2. I'm willing to maintain the central server and work on the 
infrastructure, but help will definitely be needed. Also, if there's 
significant interested, maybe R core would be interested in this.

3. What exactly is PopCon tracking as far as symbol names go? It 
currently used an R_ObjectTable object attached to the search path to 
capture names, but is this the best way? see 
http://www.omegahat.org/RObjectTables/. It's also replacing 
base::getHook to trap library loads.

4. What else would be interesting to track? Some folks have suggested 
various bits of R.Version() output.

Here's what PopCon can currently do:

  library(PopCon)
  search()
  [1] .GlobalEnvpackage:PopCon.pcUDB
  [4] package:stats package:graphics  package:grDevices
  [7] package:utils package:datasets  package:methods
[10] Autoloads package:base

# Notice the above search entry .pcUDB. That's the R Object Table

  typeof(PopCon::getCache())
[1] character
  PopCon::getCache()
[1] .conflicts.OK search::

# Now the cache contains the name 'search', which I called above,
# and the double colon operator.

  library(cluster)
   any(PopCon::getCache()=='package:cluster')
[1] TRUE

# Package names are represented in the PopCon cache just like
# their name on the search path.


  PopCon::getCache()
   [1] .conflicts.OKsearch
   [3] ::   $.data.frame
   [5] $.default$.data.frame
   [7] $.defaultunique.integer
   [9] unique.numeric   $.data.frame
  [11] $.defaultunique.integer
  [13] unique.numeric   unique.character
  [15] unique.integer   unique.numeric
  [17] close.gzfile $.packageDescription2
  [19] $.default$.data.frame
  [21] $.defaultunique.integer
  [23] unique.numeric   unique.character
  [25] unique.integer   unique.numeric
  [27] close.gzfile $.packageDescription2
  [29] $.defaultunique.integer
  [31] unique.numeric   close.gzfile
  [33] names.simple.listnames.default
  [35] [.defaultas.character.simple.list
  [37] as.vector.simple.listas.vector.default
  [39] unique.character $.packageDescription2
  [41] $.default=.R_system_version
  [43] Ops.R_system_version =.package_version
  [45] Ops.package_version  =.numeric_version
  [47] =.package_version   Ops.package_version
  [49] =.numeric_version   unlist.R_system_version
  [51] unlist.package_version   unlist.numeric_version
  [53] unlist.default   unlist.package_version
  [55] unlist.numeric_version   unlist.default
  [57] as.list.R_system_version as.list.package_version
  [59] unique.integer   unique.numeric
  [61] as.list.R_system_version as.list.package_version
  [63] unique.integer   unique.numeric
  [65] as.list.package_version  unique.integer
  [67] unique.numeric   as.list.package_version
  [69] unique.integer   unique.numeric
  [71] =.default   $.packageDescription2
  [73] $.default.R_system_version
  [75]

[R] plot each column of a matrix or dataframe versus x in a single plot

2008-02-14 Thread tomaschwutz

How do a plot several columns of a matrix at once in a single plot
versus a single x-variable?

The default plot.matrix  or plot.dataframe commands plot each column
versus each other column in several sub-plots. I want to plot each
column versus a single other vector (x) as several lines or points in
one plot.

I can do it by hand: get the range of all variables (i.e. columns of
the matrix or dataframe), initiate an empty plot command, loop through
all the columns and issue a lines or points command.
I guess there is a predefined sophisticated function for that,
however, I was not able to find it.

Does someone know where I find such a function?

Best regards
tomaschwutz

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Re: [R] Levene's test for homogeneity of variances (befor using ANOVA)

2008-02-14 Thread Chuck Cleland

On 2/14/2008 9:47 AM, Kes Knave wrote:
 Dear all
 
 I have tried to find this function in R, but don't find it by searching in
 the help function.
 
 Anybody who knows if R has the function Levene's test for homogeneity of
 variances?
 
 Note: Im a R-begginer

   I wonder how you searched for a function.  For example, when I do the 
following:

RSiteSearch(levene, restrict=function)

   levene.test() functions in the car and lawstat packages are the first 
two hits.

 Regards Kes
 
   [[alternative HTML version deleted]]
 
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-- 
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
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Re: [R] Levene's test for homogeneity of variances (befor using ANOVA)

2008-02-14 Thread David Whiting

On Thu, Feb 14, 2008 at 03:47:35PM +0100, Kes Knave wrote:
 Dear all
 
 I have tried to find this function in R, but don't find it by searching in
 the help function.

Take a look at in R:

?RSiteSearch

Use this to search for it.

 
 Anybody who knows if R has the function Levene's test for homogeneity of
 variances?
 

Yes.


David




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

-- 
David Whiting, Ph.D.
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Institute of Health and Society, The Medical School, 
Newcastle University, Framlington Place, Newcastle upon Tyne, NE2 4HH. 
Tel: +44 191 222 7045;  Extn: 7375; Fax: +44 191 222 8211.
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Re: [R] Error in model.frame.default: invalid type (list) for variable ...

2008-02-14 Thread Klaus (Diego)

 I got it right now, It wasn't working because the matrix of predictors had
the same name as the dataset, so I modified the call
creating a dataset with the response and predictors and keeping the
data.frame using the same name

metodocentrod - data.frame(metodocentro)
resultadocentro - glm(Ycentro ~ metodocentro-1, family = binomial(link =
logit), data=metodocentrod)

Thanks for the help, because of your e-mail I could figure this out.

[[alternative HTML version deleted]]

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Re: [R] plot each column of a matrix or dataframe versus x in a singleplot

you could use the function matplot(), e.g.,

x - 1:20
vals - 2 + rep(1:5, each = 20) * rep(x, 5) + rnorm(100)
dim(vals) - c(20, 5)

matplot(x, vals, type = l, lty = 1)


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
 http://www.student.kuleuven.be/~m0390867/dimitris.htm


- Original Message - 
From: tomaschwutz [EMAIL PROTECTED]
To: r-help@r-project.org
Sent: Thursday, February 14, 2008 3:55 PM
Subject: [R] plot each column of a matrix or dataframe versus x in a 
singleplot


 How do a plot several columns of a matrix at once in a single plot
 versus a single x-variable?

 The default plot.matrix  or plot.dataframe commands plot each column
 versus each other column in several sub-plots. I want to plot each
 column versus a single other vector (x) as several lines or points 
 in
 one plot.

 I can do it by hand: get the range of all variables (i.e. columns of
 the matrix or dataframe), initiate an empty plot command, loop 
 through
 all the columns and issue a lines or points command.
 I guess there is a predefined sophisticated function for that,
 however, I was not able to find it.

 Does someone know where I find such a function?

 Best regards
 tomaschwutz

 __
 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.
 


Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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Re: [R] Transposing by a group variable

Try also:

reshape(cbind(d, time=unlist(sapply(table(d$group), seq))),
idvar=group, direction=wide)

On 14/02/2008, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote:
 Hi R,



  Can I transpose a data frame by a particular group variable?



  For example:

  d=data.frame(group=c(1,1,2,2,2),val=c(6,4,6,3,5))

  And my output should be:

  data.frame(group=c(1,2),v1=c(6,6),v2=c(4,3),v3=c(NA,5))



  Many thanks,

  Shubha

  This e-mail may contain confidential and/or privileged i...{{dropped:13}}

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



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] {lattice/grid} Error using packet 1 and traceback

2008-02-14 Thread Felix Andrews

You can tell Lattice to stop when an error occurs, like this:

lattice.options(panel.error=stop)
xyplot(1:10 ~ 1:10, panel=function(...) stop(foo))
# - Error in panel(x = 1:10, y = 1:10) : foo

That is a little more informative than the default panel.error,
because it tells you the conditionCall as well as the
conditionMessage.

But unfortunately, traceback() will still not work as you expect, it
will say something unhelpful about tryCatch() inside print.trellis().
As far as I know there is no way to recover the call stack from an
error once it has been caught by tryCatch. (I would be very happy to
be proved wrong...)

As a work-around, you could call the panel function directly. Probably
the best way to do that is to start with a lattice plot using the
default panel function, and then
trellis.focus()
do.call(my.panel.xyplot, trellis.panelArgs())

Felix


On Thu, Feb 14, 2008 at 9:14 PM, Wolfram Fischer [EMAIL PROTECTED] wrote:
 On some errors during the plot of a lattice/grid graphics, there is written
  a message like Error using packet 1: missing value where TRUE/FALSE needed
  into the concering panel and the next panel is plotted.

  Which option I could use to stop the execution to have a look
  at the error by a traceback?

  Regards - Wolfram

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




-- 
Felix Andrews / 安福立
PhD candidate
Integrated Catchment Assessment and Management Centre
The Fenner School of Environment and Society
The Australian National University (Building 48A), ACT 0200
Beijing Bag, Locked Bag 40, Kingston ACT 2604
http://www.neurofractal.org/felix/
3358 543D AAC6 22C2 D336 80D9 360B 72DD 3E4C F5D8

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Re: [R] plot each column of a matrix or dataframe versus x in a single plot

Hi Thomas,

Ckeck ?matplot. Here's a simple example:

# Data set
x=runif(100)# Axis
x1=rnorm(100,25,2) # Variable x_i
x2=rnorm(100,10,1)
 x3=rnorm(100,15,4)
 x4=rnorm(100,30,4)
DATA=cbind(x,x1,x2,x3,x4)

# Plot
matplot(x,DATA[,2:5],col=1:5,pch=1:5)


I hope this helps.


Jorge




On 2/14/08, tomaschwutz [EMAIL PROTECTED] wrote:

 How do a plot several columns of a matrix at once in a single plot
 versus a single x-variable?

 The default plot.matrix  or plot.dataframe commands plot each column
 versus each other column in several sub-plots. I want to plot each
 column versus a single other vector (x) as several lines or points in
 one plot.

 I can do it by hand: get the range of all variables (i.e. columns of
 the matrix or dataframe), initiate an empty plot command, loop through
 all the columns and issue a lines or points command.
 I guess there is a predefined sophisticated function for that,
 however, I was not able to find it.

 Does someone know where I find such a function?

 Best regards
 tomaschwutz

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

Re: [R] data manipulation for plotting

2008-02-14 Thread Abi Ghanem josephine

Hi rich,
 data - c(1,1,1,4,5,5,6)
 table(data)
data
1 4 5 6
3 1 2 1

hope it helps
josephine

[EMAIL PROTECTED] wrote:
Hi,

i'd like to plot some data that I have with the value on the x axis and 
 freq
on the y axis.

So, I need to calculate the freq a value is seen within my data vector

for example, say i have a vector of data

data=c(1,1,1,4,5,5,6)

I want

values-c(1,4,5,6)

freq-c(3,1,2,1)

in order to enable me to plot this. Sorry, i'm new to R. What is standard
procedure here for plotting the data vector?

cheers

Rich






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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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[R] write output in a custom format

2008-02-14 Thread baptiste Auguié

Hi,


I need to create a text file in the following format,

 1 100.0 0
  0 0
  1 1
  0 0
  1 1
 #
 1 100.0 0
  0 0
  0 1
  1 0
  1 1
...

where # is part of the format and not a R comment.

Each block (delimited by #) consists of a first line with three  
values, call it dose, and a list of (x,y) coordinates which are a  
matrix or data.frame,


 particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c 
 (0,1,0,1)))

 print(particle)



I'd like to establish a connection to a file and append to it a  
particle block in the format above, or even write the whole file at  
once.

Because different lines have a different number of elements, I  
couldn't get write.table to work in this case, and my attempts at sink 
(), dump(), writeLines(), writeChar() all turn into really dirty  
solutions. I have this feeling I'm overlooking a simple solution.

Any help welcome,


baptiste

_

Baptiste Auguié

Physics Department
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag
http://projects.ex.ac.uk/atto

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Re: [R] R equivalent of linux cut, paste, and grep ?

2008-02-14 Thread Greg Snow

Depending on what your final goal is, there may be a better approach,
but to do as you state below, here is one set of options:

1) use read.csv (or read.table) and use the colClasses argument to
specify keeping only column 5
2) use cbind to bind column 5 (now a data.frame) to another data.frame
(possibly read in using read.csv), this assumes that the row order
matches, merge is a possibility if they don't match.
3) use the subset command to select the rows, use the %in% operator to
compare col 5 to your list of desired values.

Hope this helps,

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Ng Stanley
 Sent: Thursday, February 14, 2008 1:03 AM
 To: r-help
 Subject: [R] R equivalent of linux cut, paste, and grep ?
 
 Hi,
 
 What functions should I use in order to perform the following steps:
 
 1) Load a particular column (e.g., column 5) from a csv file
 2) Append the column 5 to another csv file
 3) select those rows that match specific keywords in column 5
 
   [[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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Re: [R] plot each column of a matrix or dataframe versus x in a single plot

2008-02-14 Thread tomaschwutz

Thank you Dimitris, Jorge, and Mark (personal mail),

for the hint to matplot and the examples.
This was exactly what I was thinking of.

Thomas

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Re: [R] How to check cy5 and cy3 values were lowess normalized

2008-02-14 Thread Henrik Bengtsson

On Thu, Feb 14, 2008 at 4:57 AM, Ng Stanley [EMAIL PROTECTED] wrote:
Hi,

I have some microarray data, cy5 and cy3 values are in log2. Is there a
way to check they have undergone lowess normalization ?

Yes, go back ask the one who you got the data from.

Honestly, this is a serious reply, because why waste time second
guessing what data you are working with. I've seen people (including
me) spending way too much time to knowing exactly what data I am
working with. It is just a waste of time and in the end of the day it
should be possible to find out from the source.

To answer your question as is, if you know that the data is *either*
curve-fit (lowess is one of them) normalized *or* not at all, then a
log-ratio log-intensity scatter plot *might* show symptoms that are
very specific to curve-fit normalized data. These symptoms are
stronger the more curvature there was in the log-ratio log-intensity
plot for the beginning. The best way I illustrate this symptom is to
look at the symmetry of the normalized data down at the lower
intensities. See Panel B (curve-fit) on Slide 16 in

http://www.stat.berkeley.edu/hb/talks/BengtssonH_20080208-YetAnotherAffinePresentation.pdf

Going back to the (R,G) space on intensity scale will also reveal the
same symptoms, cf. Slide 11.

You'll find all the background in:

H. Bengtsson and O. Hössjer, Methodological study of affine
transformations of gene expression data with proposed robust
non-parametric multi-dimensional normalization method, BMC
Bioinformatics, 2006, 7:100.

Finally, you reach more bioinformaticians/statisticians in the field
if you send these questions to the http://www.bioconductor.org/
mailing list instead. Please continue the thread there if you have
follow-up questions.

Cheers

/Henrik

Thanks
Stanley

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Re: [R] write output in a custom format

Here is a start.  You basically have to interate through your data and
use 'cat' to write it out:

particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c(0,1,0,1)))
output - file(/tempxx.txt, w)
cat(particle$dose, \n, file=output, sep= )
for (i in 1:nrow(particle$pos)){
cat(particle$pos$x[i], particle$pos$y[i], \n, file=output, sep= )
}
cat(#\n, file=output, sep= )
close(output)

Here is what the file looks like:

1 100 0
0 0
1 1
0 0
1 1
#


On 2/14/08, baptiste Auguié [EMAIL PROTECTED] wrote:
 Hi,


 I need to create a text file in the following format,

  1 100.0 0
   0 0
   1 1
   0 0
   1 1
  #
  1 100.0 0
   0 0
   0 1
   1 0
   1 1
 ...

 where # is part of the format and not a R comment.

 Each block (delimited by #) consists of a first line with three
 values, call it dose, and a list of (x,y) coordinates which are a
 matrix or data.frame,


  particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c
  (0,1,0,1)))
 
  print(particle)



 I'd like to establish a connection to a file and append to it a
 particle block in the format above, or even write the whole file at
 once.

 Because different lines have a different number of elements, I
 couldn't get write.table to work in this case, and my attempts at sink
 (), dump(), writeLines(), writeChar() all turn into really dirty
 solutions. I have this feeling I'm overlooking a simple solution.

 Any help welcome,


 baptiste

 _

 Baptiste Auguié

 Physics Department
 University of Exeter
 Stocker Road,
 Exeter, Devon,
 EX4 4QL, UK

 Phone: +44 1392 264187

 http://newton.ex.ac.uk/research/emag
 http://projects.ex.ac.uk/atto

 __
 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 you are trying to solve?

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Re: [R] {lattice/grid} Error using packet 1 and traceback

2008-02-14 Thread Deepayan Sarkar

On 2/14/08, Felix Andrews [EMAIL PROTECTED] wrote:
 You can tell Lattice to stop when an error occurs, like this:

  lattice.options(panel.error=stop)
  xyplot(1:10 ~ 1:10, panel=function(...) stop(foo))
  # - Error in panel(x = 1:10, y = 1:10) : foo

  That is a little more informative than the default panel.error,
  because it tells you the conditionCall as well as the
  conditionMessage.

  But unfortunately, traceback() will still not work as you expect, it
  will say something unhelpful about tryCatch() inside print.trellis().
  As far as I know there is no way to recover the call stack from an
  error once it has been caught by tryCatch. (I would be very happy to
  be proved wrong...)

Right, and I've also thought that this was a bad thing. I'll see if I
can figure out a way to restore traceback functionality. If all else
fails, I'll just another option that disables the tryCatch mechanism.

-Deepayan

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Re: [R] Levene's test for homogeneity of variances (befor using ANOVA)

Hi Kes,

Try

 library(help=car)
 ?levene.test

or levene in http://www.rseek.org/ (first hit).

I hope this helps,

Jorge


On 2/14/08, Kes Knave [EMAIL PROTECTED] wrote:

 Dear all

 I have tried to find this function in R, but don't find it by searching in
 the help function.

 Anybody who knows if R has the function Levene's test for homogeneity of
 variances?

 Note: Im a R-begginer

 Regards Kes

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[R] plot matrix

2008-02-14 Thread Marek Bartkuhn

Dear R users,


I like to plot a matrix A which looks like this:


,1  ,2  ,3  ,4

1,  1   10  100 1000
2,  0.5 0.2 1.0 4.3
3,  0.1 0.2 0.3 0.5

.


where the 1st row is representing the X-axis values. The subsequent  
rows should be plotted on the y-axis. I would prefer to use the  
smoothScatter plotting function of the geneplotter package, but to  
begin with R I'd be happy to use any scatterplot with the given  
dataset provided in the matrix A.

So I tried:

plot(A[1,],A[2:3,])

or

plot(A[1,],A[2-3,])

both result in :

Error in xy.coords(x, y, xlabel, ylabel, log) :
   'x' and 'y' lengths differ

What do I have to do?

Marek

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[R] Cholmod error `matrix not positive definite'

2008-02-14 Thread Martijn Vandegehuchte

Dear R-users,

I'm new to R, so my apologies if this question doesn't make sense.

I've tried the following model in lmer, and it works perfectly:
model-lmer(aphids~densroot+zone+(1|zone/site), family=quasipoisson)

But if I try the exact same model with a different variable, totmas, the model 
looks as follows:
model-lmer(aphids~totmas+zone+(1|zone/site), family=quasipoisson)

Totmas is also a continuous variable just like densroot, but in this case I 
receive the following message:

CHOLMOD warning: Ãâ e
Error in objective(.par, ...) : 
  Cholmod error `matrix not positive definite' at 
file:../Supernodal/t_cholmod_super_numeric.c, line 613

Moreover, if I test yet another continuous variable vitality, to my surprise 
R just crashes completely.


This is a mystery to me, especially because the variables totmas or vitality 
don't give any problem when I build the exact same models in SAS with proc 
glimmix... 

Does someone have experience with this type of problem?

Thank you in advance,

Martijn.




-- 
Martijn Vandegehuchte 
Ghent University 
Department Biology 
Terrestrial Ecology Unit 
K.L.Ledeganckstraat 35 
B-9000 Ghent
telephone: +32 (0)9/264 50 84
e-mail: [EMAIL PROTECTED] 

website TEREC: www.ecology.ugent.be/terec

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[R] data frame question

2008-02-14 Thread joseph



Hi

I have a data frame df1 in which I would like to multiply col1
by 2.


The way I did it does not allow me to keep the old data
frame.


How can I do this and be able to create a new data frame
df2?


 df1= data.frame(col1= c(3, 5, NA, 1), col2= c(4, NA,6,
2))


 df1


  col1 col2


134


25   NA


3   NA6


412


 df1$col1=df1$col1*2


 df1


  col1 col2


164


2   10   NA


3   NA6


422





  

Be a better friend, newshound, and 


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[R] Problems with Rcmdr unter JGR (Windows XP)

2008-02-14 Thread Willi Nagl

I try  to start Rmcdr from JGR. The Rmcdr-Windows comes up correctly; but
the
 Menu-Bar in the Rcmdr-Window goes away, if I try to go into the
Rmcdr-Window.

 I hope, someone has a solution.

Regards, willi

-- 
0(049)753124283 privat
0(049)7531882641 büro

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Re: [R] Replacing a character string when finding substring match

Try this:

x[grep(vehicle, x, ignore=T)] - Vehicle

On 14/02/2008, Judith Flores [EMAIL PROTECTED] wrote:
 Dear R-experts,

   I need to replace the values of a vector(tx) with a
  word ('Vehicle') when the value of the vector contains
  the word 'vehicle'. Sometimes, the value could be 'MCT
  vehicle', or 'control-vehicle', etc.
   I tried gsub like this

  treatment-gsub('vehicle','Vehicle', tx,
  ignore.case=T)

But then I end up with values like 'MCT Vehicle' or
  'control-Vehicle'. I want the values to be one word:
  'Vehicle'.



  Thank you in advance for any help.






   
 
  Be a better friend, newshound, and

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



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] data manipulation for plotting

Here are a couple of ways depending on what you want.

data1=c(1,1,1,4,5,5,6)
plot(table(data1))
---
plotvec - as.vector(aa)
nns - names(aa)
plot(plotvec, xaxt=n)
mtext(nns, at=1:4, side=1)
---
--- [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:

 
Hi,
 
i'd like to plot some data that I have with the
 value on the x axis and freq
on the y axis.
 
So, I need to calculate the freq a value is seen
 within my data vector
 
for example, say i have a vector of data
 
data=c(1,1,1,4,5,5,6)
 
I want
 
values-c(1,4,5,6)
 
freq-c(3,1,2,1)
 
in order to enable me to plot this. Sorry, i'm
 new to R. What is standard
procedure here for plotting the data vector?
 
cheers
 
Rich
 
 
 
 
 

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 reproducible code.
 



[[elided trailing spam]]

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[R] Replacing a character string when finding substring match

2008-02-14 Thread Judith Flores

Dear R-experts,

  I need to replace the values of a vector(tx) with a
word ('Vehicle') when the value of the vector contains
the word 'vehicle'. Sometimes, the value could be 'MCT
vehicle', or 'control-vehicle', etc.
  I tried gsub like this 

treatment-gsub('vehicle','Vehicle', tx, 
ignore.case=T)

   But then I end up with values like 'MCT Vehicle' or
'control-Vehicle'. I want the values to be one word:
'Vehicle'.



Thank you in advance for any help.






  

Be a better friend, newshound, and

__
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Re: [R] apply on large arrays

2008-02-14 Thread Erich Neuwirth

  system.time({
+   tab2 - tab1 - with(pisa1, table(CNT,GENDER,ISCOF,ISCOM))
+ tab2[] - 0
+ tab2[which(tab1 == 1, arr.ind = TRUE)] - 1
+ tab3 - rowSums(tab2)
+ })
user  system elapsed
3.170.994.17
 
  system.time({
+   tab4 - rowSums(tab1 == 1)
+ })
user  system elapsed
1.020.181.20
 


And yes,
the results were identical.


[EMAIL PROTECTED] wrote:
 Was the answer the same as the one you were getting with the original
 code?
 
 How long did the original code take compared to these two versions?
 
 Cheers,
 Bill V. 
 
 
 Bill Venables
 CSIRO Laboratories
 PO Box 120, Cleveland, 4163
 AUSTRALIA
 Office Phone (email preferred): +61 7 3826 7251
 Fax (if absolutely necessary):  +61 7 3826 7304
 Mobile: +61 4 8819 4402
 Home Phone: +61 7 3286 7700
 mailto:[EMAIL PROTECTED]
 http://www.cmis.csiro.au/bill.venables/ 
 
 -Original Message-
 From: Erich Neuwirth [mailto:[EMAIL PROTECTED] 
 Sent: Thursday, 14 February 2008 5:08 PM
 To: Venables, Bill (CMIS, Cleveland)
 Subject: Re: [R] apply on large arrays
 
 Thanks, this version is definitely faster than the first one.
 system.time gives 0.13 instead of 0.79 seconds.
 
 
 
 [EMAIL PROTECTED] wrote:
 Hmm.  I think this could be faster still:

  tab1 - with(pisa1, table(CNT,GENDER,ISCOF,ISCOM))
  tab3 - rowSums(tab1 == 1)

 but check it...

 Bill Venables
 CSIRO Laboratories
 PO Box 120, Cleveland, 4163
 AUSTRALIA
 Office Phone (email preferred): +61 7 3826 7251
 Fax (if absolutely necessary):  +61 7 3826 7304
 Mobile: +61 4 8819 4402
 Home Phone: +61 7 3286 7700
 mailto:[EMAIL PROTECTED]
 http://www.cmis.csiro.au/bill.venables/ 

 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED]
 On Behalf Of Venables, Bill (CMIS, Cleveland)
 Sent: Thursday, 14 February 2008 10:30 AM
 To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
 Subject: Re: [R] apply on large arrays

 Your code is


  tab1 - with(pisa1, table(CNT,GENDER,ISCOF,ISCOM))
  tab2 - apply(tab1, 1:4, 
  function(x) ifelse(sum(x) == 1, 1, 0))
  tab3 - apply(tab2, 1, sum)

 As far as I can see, step 2, (the problematic one), merely replaces
 any
 entries in tab1 that are not equal to one by zeros.  I think this
 would
 do the same job a bit faster:

  tab2 - tab1 - with(pisa1, table(CNT,GENDER,ISCOF,ISCOM))
  tab2[] - 0
  tab2[which(tab1 == 1, arr.ind = TRUE)] - 1
  tab3 - rowSums(tab2)

 If you don't need to keep tab1, you would make things even better by
 removing it.

 Bill Venables.
  




 Bill Venables
 CSIRO Laboratories
 PO Box 120, Cleveland, 4163
 AUSTRALIA
 Office Phone (email preferred): +61 7 3826 7251
 Fax (if absolutely necessary):  +61 7 3826 7304
 Mobile: +61 4 8819 4402
 Home Phone: +61 7 3286 7700
 mailto:[EMAIL PROTECTED]
 http://www.cmis.csiro.au/bill.venables/ 

 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED]
 On Behalf Of Erich Neuwirth
 Sent: Thursday, 14 February 2008 9:52 AM
 To: r-help
 Subject: [R] apply on large arrays

 I have a big contingency table, approximately of size 60*2*500*500,
 and I need to count the number of cells containing a count of 1 for
 each
 of the factors values defining the first dimension.
 Here is my attempt:

 tab1-with(pisa1,table(CNT,GENDER,ISCOF,ISCOM))
 tab2-apply(tab1,1:4,function(x)ifelse(sum(x)==1,1,0))
 tab3-apply(tab2,1,sum)

 Computing tab2 is very slow.
 Is there a faster and/or more elegant way of doing this?
 

-- 
Erich Neuwirth, University of Vienna
Faculty of Computer Science
Computer Supported Didactics Working Group
Visit our SunSITE at http://sunsite.univie.ac.at
Phone: +43-1-4277-39464 Fax: +43-1-4277-39459

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Re: [R] Generalized nonlinear mixed model function?

2008-02-14 Thread Christian Ritz

Hi Philip,

your data are event times because you're monitoring the same trees in 
each plot over time, the event being death of a tree.

Therefore methods from survival analysis are more appropriate.
Start out having a look at the package survival, possibly considering 
a Cox model with adjustment for clusters or with strata (the function 
coxph()) or a parametric model using the function survreg().


Christian

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Re: [R] data manipulation for plotting

Try this:

values - c(1,1,1,4,5,5,6)

with(rle(values), plot(values, lengths))

But I think that you can use barplot:

barplot(table(values))

On 14/02/2008, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:

Hi,

i'd like to plot some data that I have with the value on the x axis and 
 freq
on the y axis.

So, I need to calculate the freq a value is seen within my data vector

for example, say i have a vector of data

data=c(1,1,1,4,5,5,6)

I want

values-c(1,4,5,6)

freq-c(3,1,2,1)

in order to enable me to plot this. Sorry, i'm new to R. What is standard
procedure here for plotting the data vector?

cheers

Rich






  __
  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.



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

__
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[R] Advice on analyzing a mixed effects survival model?

2008-02-14 Thread Kevin Crowston

I have an experiment I'm trying to analyze that's turning out to be  
more complicated than I anticipated, so I was hoping for some  
suggestions about how to handle it.


The lab experiment is a comparison between two search interfaces.  
After a little training, each subject performs 12 information search  
tasks, 6 with one interface and 6 with the other, and we measure time  
to complete and number of clicks.  The overall design is a latin  
square: with 12 subjects, the design has each task done before and  
after each other one the same number of times, and each is done 6  
times with one interface and 6 with the other. Maybe a table will make  
it clearer what is happening:


RunTasks
1-   1281 
117546 
10392
2-1127 
8411105 
9623
3-9102 
43761 
512118
4-326 
9510114 
87121
5-635 
2119810 
12417
6-5611 
382129 
11074
7-714 
12108911 
2536
8-81112 
51673 
42109
9-1049 
721312 
68511
10  -1158 
612312 
79410
11  -293 
106457 
111812
12  -4710 
191228 
31165


RunInterface
1-00111 
1000110
2-10101 
0101010
3-01010 
1010101
4-10101 
0101010
5-01010 
1010101
6-10101 
0101010
7-01010 
1010101
8-10101 
0011100
9-11100 
0101001
10  -000101 
010111
11  -110110 
101000
12  -010001 
110011



The resulting data look something like

subject run  seq  task   interface   time  clicks
1  1  1  120123  18
1  1  280 197  23
1  1  311 156  21

2  2   1   1 1 8710
.


I was planning originally to analyze the data with ANOVA: time (or  
probably log(time)) ~ task + subject + interface. Some tasks are  
harder than others, some subjects slower, but we control for those to  
see the effect of the interface. I did not plan to include an  
interaction

Re: [R] Transposing by a group variable

How do we know what value corresponds to what new
variable?  

library(reshape)
mm - melt(d, id=group)
cast(mm, group~value)

will give you something but not quite what you want.

--- Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:

 Hi R,
 
  
 
 Can I transpose a data frame by a particular group
 variable?
 
  
 
 For example:
 
 d=data.frame(group=c(1,1,2,2,2),val=c(6,4,6,3,5))
 
 And my output should be:
 

data.frame(group=c(1,2),v1=c(6,6),v2=c(4,3),v3=c(NA,5))
 
  
 
 Many thanks,
 
 Shubha
 
 This e-mail may contain confidential and/or
 privileged i...{{dropped:13}}
 
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 PLEASE do read the posting guide
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[R] Advice on analyzing a mixed effects survival model?

2008-02-14 Thread Kevin Crowston

Small correction: I meant to say that I had been reading:

J. C. Pinheiro and D. M. Bates (2000), “Mixed-Effects Models in S and  
S-Plus”, Springer.

Kevin Crowston

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[R] deleting certain observations in a data frame

2008-02-14 Thread Chang Liu


Hi, I'm wondering what the fastest way is to delete certain data points 
(observations) in a data frame. 
I have a vector of the indices/row.names I would like to delete. I have tried 
replacing list by list, but it always complains about different lengths, 
replacing list of length a with length b and so on.
 
Another way to think of it is that it's a generazation of na.rm I guess.
 
Thanks,
Karen
_


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Re: [R] deleting certain observations in a data frame

2008-02-14 Thread Peter Alspach

Karen

For indices, use the minus sign: yourData[-indicesToBeDeleted,]

For rownames, negate %in%:
yourData[!rownames(yourData)%in%namesToBeDeleted,] 

HTH

Peter Alspach

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Chang Liu
 Sent: Friday, 15 February 2008 9:50 a.m.
 To: r-help@r-project.org
 Subject: [R] deleting certain observations in a data frame
 
 
 Hi, I'm wondering what the fastest way is to delete certain 
 data points (observations) in a data frame. 
 I have a vector of the indices/row.names I would like to 
 delete. I have tried replacing list by list, but it always 
 complains about different lengths, replacing list of length 
 a with length b and so on.
  
 Another way to think of it is that it's a generazation of 
 na.rm I guess.
  
 Thanks,
 Karen
 _
 
 
   [[alternative HTML version deleted]]
 
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[R] Using Conditional AIC with lmer

2008-02-14 Thread Kyle Edwards

Hi all,

This was posted originally on r-sig-mixed-models, but I thought I  
would post here as well as it might be of more general interest.

With a colleague, I have been trying to implement the Conditional AIC  
described by Vaida and Blanchard 2005 Biometrika, Conditional Akaike  
information for mixed-effects models. This quantity is derived in a  
way analogous to the AIC, but is appropriate for scenarios where one  
is interested in the particular coefficient estimates for individual  
random effects. The formula for the asymptotic CAIC is given as

-2*log(likelihood of observed values, conditional on ML estimates of  
fixed effects and empirical Bayes estimates of random effects) + 2*K

where K = rho + 1, and rho = effective degrees of freedom = trace  
of the hat matrix mapping predicted values onto observed values.

After some thinking and some off-list advice, we have decided that  
appropriate code for CAIC is

CAIC - function(model) {

sigma - attr(VarCorr(model), 'sc')
observed - attr(model, 'y')
predicted - fitted(model)  
cond.loglik - sum(dnorm(observed, predicted, sigma, log=TRUE))

rho - hatTrace(model)
p - length(fixef(model))
N - nrow(attr(model, 'X'))
K.corr - N*(N-p-1)*(rho+1)/((N-p)*(N-p-2)) + N*(p+1)/((N-p)*(N-p-2))

CAIC - -2*cond.loglik + 2*K.corr

return(CAIC)

}

where K.corr is the finite-sample correction for K, for ML model fits.

I am posting this so that 1) This code can be of use to any other  
souls in the statistical wilderness trying to do model selection with  
mixed models, and 2) So that wiser minds can point out any errors in  
our approach.

Thanks,

Kyle

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Re: [R] Replacing columns in a data frame using a previous condition

It's, Jim. Thank you so much.

Jorge


On 2/14/08, jim holtman [EMAIL PROTECTED] wrote:

 Is this what you want to do?

  x - data.frame(a=1:10, b=1:10, c=1:10, d=1:10)
  z - cbind(c=11:20, d=11:20)
  z
   c  d
 [1,] 11 11
 [2,] 12 12
 [3,] 13 13
 [4,] 14 14
 [5,] 15 15
 [6,] 16 16
 [7,] 17 17
 [8,] 18 18
 [9,] 19 19
 [10,] 20 20
  x[,colnames(z)] - z[, colnames(z)]
  x
a  b  c  d
 1   1  1 11 11
 2   2  2 12 12
 3   3  3 13 13
 4   4  4 14 14
 5   5  5 15 15
 6   6  6 16 16
 7   7  7 17 17
 8   8  8 18 18
 9   9  9 19 19
 10 10 10 20 20
 


 On 2/14/08, Jorge Iván Vélez [EMAIL PROTECTED] wrote:
  Dear R-list,
 
  I'm working with a data frame which dimensions are
 
   dim(GERU)
  [1] 3468  318
 
  and looks like
 
   GERU[1:10,1:10]
ped ind par1 par2 sex sta rs7696470 rs7696470.1 rs1032896
 rs1032896.1
  1  USA5854   200   2   1 4   4
 1   1
  2  USA5854   312   1   1 4   4
 1   1
  3  USA5854   412   2   2 1   4
 1   3
  4  USA5854   512   1   2 4   2
 2   1
  5  USA5855   100   1   1 0   0
 0   0
  6  USA5855   200   2   2 1   0
 0   0
  7  USA5855   312   1   2 0   2
 0   0
  8  USA5855   412   1   1 2   0
 2   1
  9  USA5855   512   1   2 0   1
 0   0
  10 USA5856   100   1   13   3
 3   3
 
  What I would like to do is:
 
  1. Identify which column (from 6 to 318) has more than 4 categories (I
  solved that). In GERU would be rs7696470 and rs7696470.1.
  2. Using the columns in step 1, replace its entries equals to 2 for 3.
 For
  example, rs7696470 would be 4,4,1,4,0,1,0,3,0,3 and so on.
  3. Once replaced the entries, I need to rewrite the columns in GERU.
 
  Here is what I've done:
 
   # Function to identify columns with 3 or more categories
   tx=function(x) ifelse(dim(table(x))4,1,0)
 
   # Identifying the columns
   M4=apply(GUPN[,-c(1:6)],2,tx)
   names(which(MR==1))# Step 1
   [1] rs335322 rs335322.1   rs186750 rs186750.1
  rs1565901rs1565901.1  rs1565902
   [8] rs1565902.1  rs11131334   rs11131334.1 rs1948616
  rs1948616.1  rs4484334rs4484334.1
  [15] rs1497921rs1497921.1  rs1391320rs1391320.1
  rs1497913rs1497913.1  rs996208
  [22] rs996208.1
   # Step 2
   REPLACE=GUPN[,names(which(AR==1))]
   RES=apply(REPLACE,2,function(x) ifelse(x==2,3,x))
   RES[1:10,1:5]
rs335322 rs335322.1 rs186750 rs186750.1 rs1565901
  1 1  33  3 3
  2 1  13  3 3
  3 3  31  3 3
  4 1  33  3 3
  5 0  00  0 0
  6 0  00  0 0
  7 0  00  0 0
  8 0  00  0 0
  9 0  00  0 0
  101  33  3 1
 
  Now, the problem I have is replacing the columns in GERU by the columns
 in
  RES (step 3). At the end the dimension of the new data set should be
  3468x318. Any help would be greatly appreciated.
 
  Thanks you so much,
 
 
  Jorge
 
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Re: [R] Replacing columns in a data frame using a previous condition

Is this what you want to do?

 x - data.frame(a=1:10, b=1:10, c=1:10, d=1:10)
 z - cbind(c=11:20, d=11:20)
 z
   c  d
 [1,] 11 11
 [2,] 12 12
 [3,] 13 13
 [4,] 14 14
 [5,] 15 15
 [6,] 16 16
 [7,] 17 17
 [8,] 18 18
 [9,] 19 19
[10,] 20 20
 x[,colnames(z)] - z[, colnames(z)]
 x
a  b  c  d
1   1  1 11 11
2   2  2 12 12
3   3  3 13 13
4   4  4 14 14
5   5  5 15 15
6   6  6 16 16
7   7  7 17 17
8   8  8 18 18
9   9  9 19 19
10 10 10 20 20



On 2/14/08, Jorge Iván Vélez [EMAIL PROTECTED] wrote:
 Dear R-list,

 I'm working with a data frame which dimensions are

  dim(GERU)
 [1] 3468  318

 and looks like

  GERU[1:10,1:10]
   ped ind par1 par2 sex sta rs7696470 rs7696470.1 rs1032896 rs1032896.1
 1  USA5854   200   2   1 4   4 1   1
 2  USA5854   312   1   1 4   4 1   1
 3  USA5854   412   2   2 1   4 1   3
 4  USA5854   512   1   2 4   2 2   1
 5  USA5855   100   1   1 0   0 0   0
 6  USA5855   200   2   2 1   0 0   0
 7  USA5855   312   1   2 0   2 0   0
 8  USA5855   412   1   1 2   0 2   1
 9  USA5855   512   1   2 0   1 0   0
 10 USA5856   100   1   13   3 3   3

 What I would like to do is:

 1. Identify which column (from 6 to 318) has more than 4 categories (I
 solved that). In GERU would be rs7696470 and rs7696470.1.
 2. Using the columns in step 1, replace its entries equals to 2 for 3. For
 example, rs7696470 would be 4,4,1,4,0,1,0,3,0,3 and so on.
 3. Once replaced the entries, I need to rewrite the columns in GERU.

 Here is what I've done:

  # Function to identify columns with 3 or more categories
  tx=function(x) ifelse(dim(table(x))4,1,0)

  # Identifying the columns
  M4=apply(GUPN[,-c(1:6)],2,tx)
  names(which(MR==1))# Step 1
  [1] rs335322 rs335322.1   rs186750 rs186750.1
 rs1565901rs1565901.1  rs1565902
  [8] rs1565902.1  rs11131334   rs11131334.1 rs1948616
 rs1948616.1  rs4484334rs4484334.1
 [15] rs1497921rs1497921.1  rs1391320rs1391320.1
 rs1497913rs1497913.1  rs996208
 [22] rs996208.1
  # Step 2
  REPLACE=GUPN[,names(which(AR==1))]
  RES=apply(REPLACE,2,function(x) ifelse(x==2,3,x))
  RES[1:10,1:5]
   rs335322 rs335322.1 rs186750 rs186750.1 rs1565901
 1 1  33  3 3
 2 1  13  3 3
 3 3  31  3 3
 4 1  33  3 3
 5 0  00  0 0
 6 0  00  0 0
 7 0  00  0 0
 8 0  00  0 0
 9 0  00  0 0
 101  33  3 1

 Now, the problem I have is replacing the columns in GERU by the columns in
 RES (step 3). At the end the dimension of the new data set should be
 3468x318. Any help would be greatly appreciated.

 Thanks you so much,


 Jorge

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+1 513 646 9390

What is the problem you are trying to solve?

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Re: [R] Principal component analysis PCA

2008-02-14 Thread SNN


Thanks for the advice.

I tried to find the cov of my matrix using R and it ran out of memory. I am
not sure how to do double loop to create the covariace matrix?  Also is
doing prcomp( covariace matrix) the same as finding 
prcomp( original data ,matrix of snps)?

Thanks for your help,




Thomas Lumley wrote:
 
 On Wed, 13 Feb 2008, Wang, Zhaoming (NIH/NCI) [C] wrote:
 

 Try EIGENSTRAT http://www.nature.com/ng/journal/v38/n8/abs/ng1847.html
 
 The same approach as EIGENSTRAT is pretty straightforward in R.
 
 You need to create the covariance matrix of people (rather than of SNPs) 
 for the 0/1/2 genotype at each SNP and take the principal components of 
 that matrix.
 
 In this case the number of individuals is small enough that you should be 
 able to create the covariance matrix directly by matrix operations.  In 
 larger data sets where the entire data matrix doesn't fit in memory, you 
 need some sort of double loop.
 
   -thomas
 
 
 Zhaoming
 -Original Message-
 From: SNN [mailto:[EMAIL PROTECTED]
 Sent: Wednesday, February 13, 2008 9:14 PM
 To: r-help@r-project.org
 Subject: [R] Principal component analysis PCA


 Hi,

 I am trying to run PCA on a set of data with dimension 115*300,000. The
 columns represnt the snps and the row represent the individuals. so this
 is what i did.

 #load the data

 code-read.table(code.txt, sep='\t', header=F, nrows=30)

 # do PCA #

 pr-prcomp(code, retx=T, center=T)

 I am getting the following error message

 Error: cannot allocate vector of size 275.6 Mb

 I tried to increase the memory size :

 memory.size(4000)

 but it did not work, is there a solution for this ? or is there another
 software that can handle large data sets.

 Thanks


 --
 View this message in context:
 http://www.nabble.com/Principal-component-analysis-PCA-tp15472509p154725
 09.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 Thomas Lumley Assoc. Professor, Biostatistics
 [EMAIL PROTECTED] University of Washington, Seattle
 
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[R] Replacing columns in a data frame using a previous condition (SOLVED)

Dear R-list,

Thanks a lot for your help. Thanks to Jim, Dimitris and Phil. It's exactly
what I needed to do.


Jorge


On 2/14/08, Jorge Iván Vélez [EMAIL PROTECTED] wrote:
 Dear R-list,

 I'm working with a data frame which dimensions are

  dim(GERU)
 [1] 3468  318

 and looks like

  GERU[1:10,1:10]
   ped ind par1 par2 sex sta rs7696470 rs7696470.1 rs1032896
rs1032896.1
 1  USA5854   200   2   1 4   4 1
1
 2  USA5854   312   1   1 4   4 1
1
 3  USA5854   412   2   2 1   4 1
3
 4  USA5854   512   1   2 4   2 2
1
 5  USA5855   100   1   1 0   0 0
0
 6  USA5855   200   2   2 1   0 0
0
 7  USA5855   312   1   2 0   2 0
0
 8  USA5855   412   1   1 2   0 2
1
 9  USA5855   512   1   2 0   1 0
0
 10 USA5856   100   1   13   3 3
3

 What I would like to do is:

 1. Identify which column (from 6 to 318) has more than 4 categories (I
 solved that). In GERU would be rs7696470 and rs7696470.1.
 2. Using the columns in step 1, replace its entries equals to 2 for 3. For
 example, rs7696470 would be 4,4,1,4,0,1,0,3,0,3 and so on.
 3. Once replaced the entries, I need to rewrite the columns in GERU.

 Here is what I've done:

  # Function to identify columns with 3 or more categories
  tx=function(x) ifelse(dim(table(x))4,1,0)

  # Identifying the columns
  M4=apply(GUPN[,-c(1:6)],2,tx)
  names(which(MR==1))# Step 1
  [1] rs335322 rs335322.1   rs186750 rs186750.1
 rs1565901rs1565901.1  rs1565902
  [8] rs1565902.1  rs11131334   rs11131334.1 rs1948616
 rs1948616.1  rs4484334rs4484334.1
 [15] rs1497921rs1497921.1  rs1391320rs1391320.1
 rs1497913rs1497913.1  rs996208
 [22] rs996208.1
  # Step 2
  REPLACE=GUPN[,names(which(AR==1))]
  RES=apply(REPLACE,2,function(x) ifelse(x==2,3,x))
  RES[1:10,1:5]
   rs335322 rs335322.1 rs186750 rs186750.1 rs1565901
 1 1  33  3 3
 2 1  13  3 3
 3 3  31  3 3
 4 1  33  3 3
 5 0  00  0 0
 6 0  00  0 0
 7 0  00  0 0
 8 0  00  0 0
 9 0  00  0 0
 101  33  3 1

 Now, the problem I have is replacing the columns in GERU by the columns in
 RES (step 3). At the end the dimension of the new data set should be
 3468x318. Any help would be greatly appreciated.

 Thanks you so much,


 Jorge

[[alternative HTML version deleted]]

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Re: [R] Replacing columns in a data frame using a previous condition

try this:

GERU[6:318] - lapply(GERU[6:318], function (x) {
 if (length(unique(x[!is.na(x)])) = 5) x[x == 2] - 3
 x
})


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
  http://www.student.kuleuven.be/~m0390867/dimitris.htm


Quoting Jorge Iván Vélez [EMAIL PROTECTED]:

 Dear R-list,

 I'm working with a data frame which dimensions are

 dim(GERU)
 [1] 3468  318

 and looks like

 GERU[1:10,1:10]
ped ind par1 par2 sex sta rs7696470 rs7696470.1 rs1032896 rs1032896.1
 1  USA5854   200   2   1 4   4 1   1
 2  USA5854   312   1   1 4   4 1   1
 3  USA5854   412   2   2 1   4 1   3
 4  USA5854   512   1   2 4   2 2   1
 5  USA5855   100   1   1 0   0 0   0
 6  USA5855   200   2   2 1   0 0   0
 7  USA5855   312   1   2 0   2 0   0
 8  USA5855   412   1   1 2   0 2   1
 9  USA5855   512   1   2 0   1 0   0
 10 USA5856   100   1   13   3 3   3

 What I would like to do is:

 1. Identify which column (from 6 to 318) has more than 4 categories (I
 solved that). In GERU would be rs7696470 and rs7696470.1.
 2. Using the columns in step 1, replace its entries equals to 2 for 3. For
 example, rs7696470 would be 4,4,1,4,0,1,0,3,0,3 and so on.
 3. Once replaced the entries, I need to rewrite the columns in GERU.

 Here is what I've done:

 # Function to identify columns with 3 or more categories
 tx=function(x) ifelse(dim(table(x))4,1,0)

 # Identifying the columns
 M4=apply(GUPN[,-c(1:6)],2,tx)
 names(which(MR==1))# Step 1
  [1] rs335322 rs335322.1   rs186750 rs186750.1
 rs1565901rs1565901.1  rs1565902
  [8] rs1565902.1  rs11131334   rs11131334.1 rs1948616
 rs1948616.1  rs4484334rs4484334.1
 [15] rs1497921rs1497921.1  rs1391320rs1391320.1
 rs1497913rs1497913.1  rs996208
 [22] rs996208.1
 # Step 2
 REPLACE=GUPN[,names(which(AR==1))]
 RES=apply(REPLACE,2,function(x) ifelse(x==2,3,x))
 RES[1:10,1:5]
rs335322 rs335322.1 rs186750 rs186750.1 rs1565901
 1 1  33  3 3
 2 1  13  3 3
 3 3  31  3 3
 4 1  33  3 3
 5 0  00  0 0
 6 0  00  0 0
 7 0  00  0 0
 8 0  00  0 0
 9 0  00  0 0
 101  33  3 1

 Now, the problem I have is replacing the columns in GERU by the columns in
 RES (step 3). At the end the dimension of the new data set should be
 3468x318. Any help would be greatly appreciated.

 Thanks you so much,


 Jorge

   [[alternative HTML version deleted]]

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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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Re: [R] finding source for a function

2008-02-14 Thread Gabor Csardi

RSiteSearch is your friend. E.g.:

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/63365.html

and then click on 'Next in thread a couple of times

Gabor

On Thu, Feb 14, 2008 at 03:23:30PM -0600, Edna Bell wrote:
 Dear R Gurus:
 
 How do you get source for functions which say UseMethod when you
 type in their names, please?
 
 I've tried getAnywhere and getMethods...I thought that might produce them.
 
 Thanks in advance.
 
 Sincerely,
 Edna
 
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-- 
Csardi Gabor [EMAIL PROTECTED]UNIL DGM

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Re: [R] Principal component analysis PCA

2008-02-14 Thread Thomas Lumley

On Thu, 14 Feb 2008, SNN wrote:


 Thanks for the advice.

 I tried to find the cov of my matrix using R and it ran out of memory.

How did you do this? The covariance matrix is only 115x115, so it 
shouldn't run out of memory
   cov(t(code))
should work

If that doesn't work then
   tcrossprod(code)/30 - tcrossprod(rowMeans(code))
might.

 I am
 not sure how to do double loop to create the covariace matrix?  Also is
 doing prcomp( covariace matrix) the same as finding
 prcomp( original data ,matrix of snps)?

That's the point of the paper behind the EIGENSTRAT software, which is 
worth reading.  The eigenvalues are the same and the eigenvectors are 
related.  One way around gives the left singular vectors of the data 
matrix, the other gives the right singular vectors.


-thomas

Thomas Lumley   Assoc. Professor, Biostatistics
[EMAIL PROTECTED]   University of Washington, Seattle

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[R] finding source for a function

2008-02-14 Thread Edna Bell

Dear R Gurus:

How do you get source for functions which say UseMethod when you
type in their names, please?

I've tried getAnywhere and getMethods...I thought that might produce them.

Thanks in advance.

Sincerely,
Edna

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Re: [R] write output in a custom format

2008-02-14 Thread baptiste Auguié

Thanks for the input! It does work fine, however I'll have to do  
another loop to repeat this whole process quite a few times (10^3,  
10^4 particles maybe), so I was hoping for a solution without loop.  
Maybe I could reshape all the values into a big array, dump it to a  
file and replace some values using system(awk...). I just don't  
really know how to format the data, having different number of values  
for some lines. Would that be a sensible thing to do?

thanks,

baptiste




On 14 Feb 2008, at 16:49, jim holtman wrote:

 Here is a start.  You basically have to interate through your data and
 use 'cat' to write it out:

 particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c 
 (0,1,0,1)))
 output - file(/tempxx.txt, w)
 cat(particle$dose, \n, file=output, sep= )
 for (i in 1:nrow(particle$pos)){
 cat(particle$pos$x[i], particle$pos$y[i], \n, file=output,  
 sep= )
 }
 cat(#\n, file=output, sep= )
 close(output)

 Here is what the file looks like:

 1 100 0
 0 0
 1 1
 0 0
 1 1
 #


 On 2/14/08, baptiste Auguié [EMAIL PROTECTED] wrote:
 Hi,


 I need to create a text file in the following format,

 1 100.0 0
  0 0
  1 1
  0 0
  1 1
 #
 1 100.0 0
  0 0
  0 1
  1 0
  1 1
 ...

 where # is part of the format and not a R comment.

 Each block (delimited by #) consists of a first line with three
 values, call it dose, and a list of (x,y) coordinates which are a
 matrix or data.frame,


 particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c
 (0,1,0,1)))

 print(particle)



 I'd like to establish a connection to a file and append to it a
 particle block in the format above, or even write the whole file at
 once.

 Because different lines have a different number of elements, I
 couldn't get write.table to work in this case, and my attempts at  
 sink
 (), dump(), writeLines(), writeChar() all turn into really dirty
 solutions. I have this feeling I'm overlooking a simple solution.

 Any help welcome,


 baptiste

 _

 Baptiste Auguié

 Physics Department
 University of Exeter
 Stocker Road,
 Exeter, Devon,
 EX4 4QL, UK

 Phone: +44 1392 264187

 http://newton.ex.ac.uk/research/emag
 http://projects.ex.ac.uk/atto

 __
 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 you are trying to solve?

_

Baptiste Auguié

Physics Department
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag
http://projects.ex.ac.uk/atto

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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] Cholmod error `matrix not positive definite'

2008-02-14 Thread Douglas Bates

Could you tell us which version of the lme4 package you are using?
You can just send the output produced by

sessionInfo()

If you can make your data available so we can test it then please do
so.  If the data set is large you could send it to me in private email
and I will make it available on a web site.

I think that the development version of the lme4 package, available via

install.packages(lme4, repos = http://r-forge.r-project.org;)

should be resistant to that type of error but I am willing to be shown
otherwise.

On Thu, Feb 14, 2008 at 10:36 AM, Martijn Vandegehuchte
[EMAIL PROTECTED] wrote:
 Dear R-users,

  I'm new to R, so my apologies if this question doesn't make sense.

  I've tried the following model in lmer, and it works perfectly:
  model-lmer(aphids~densroot+zone+(1|zone/site), family=quasipoisson)

  But if I try the exact same model with a different variable, totmas, the 
 model looks as follows:
  model-lmer(aphids~totmas+zone+(1|zone/site), family=quasipoisson)

  Totmas is also a continuous variable just like densroot, but in this case I 
 receive the following message:

  CHOLMOD warning: ß†e
  Error in objective(.par, ...) :
   Cholmod error `matrix not positive definite' at 
 file:../Supernodal/t_cholmod_super_numeric.c, line 613

  Moreover, if I test yet another continuous variable vitality, to my 
 surprise R just crashes completely.


  This is a mystery to me, especially because the variables totmas or vitality 
 don't give any problem when I build the exact same models in SAS with proc 
 glimmix...

  Does someone have experience with this type of problem?

  Thank you in advance,

  Martijn.




  --
  Martijn Vandegehuchte
  Ghent University
  Department Biology
  Terrestrial Ecology Unit
  K.L.Ledeganckstraat 35
  B-9000 Ghent
  telephone: +32 (0)9/264 50 84
  e-mail: [EMAIL PROTECTED]

  website TEREC: www.ecology.ugent.be/terec

 [[alternative HTML version deleted]]


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[R] setting color ranges

2008-02-14 Thread samsr


Hi,

 I need to plot a matrix using image() such that negative values are easily
distinguishable from posittive values, while also maintaining a gradation in
color with magnitude. How can I set ranges for colors in order to achieve
this. Thanks.

Sam
-- 
View this message in context: 
http://www.nabble.com/setting-color-ranges-tp15489086p15489086.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] write output in a custom format

There is nothing wrong with a loop for handling this case.  Most of
your time is probably going to be spent writing out the files.  If you
don't want 'for' loops, you can use 'lapply', but I am not sure what
type of performance improvement you will see.  You are having to
make decisions on each particle on how to write it. You can also use
awk/perl as you indicated, but you would have to write the data out
for those programs.  You might take a test run and see.  I would guess
that by the time you format it for awk and then run awk, you could
have done the whole thing in R.  But it is your choice and there are
plenty of tools to choose from.

On 2/14/08, baptiste Auguié [EMAIL PROTECTED] wrote:
 Thanks for the input! It does work fine, however I'll have to do
 another loop to repeat this whole process quite a few times (10^3,
 10^4 particles maybe), so I was hoping for a solution without loop.
 Maybe I could reshape all the values into a big array, dump it to a
 file and replace some values using system(awk...). I just don't
 really know how to format the data, having different number of values
 for some lines. Would that be a sensible thing to do?

 thanks,

 baptiste




 On 14 Feb 2008, at 16:49, jim holtman wrote:

  Here is a start.  You basically have to interate through your data and
  use 'cat' to write it out:
 
  particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c
  (0,1,0,1)))
  output - file(/tempxx.txt, w)
  cat(particle$dose, \n, file=output, sep= )
  for (i in 1:nrow(particle$pos)){
  cat(particle$pos$x[i], particle$pos$y[i], \n, file=output,
  sep= )
  }
  cat(#\n, file=output, sep= )
  close(output)
 
  Here is what the file looks like:
 
  1 100 0
  0 0
  1 1
  0 0
  1 1
  #
 
 
  On 2/14/08, baptiste Auguié [EMAIL PROTECTED] wrote:
  Hi,
 
 
  I need to create a text file in the following format,
 
  1 100.0 0
   0 0
   1 1
   0 0
   1 1
  #
  1 100.0 0
   0 0
   0 1
   1 0
   1 1
  ...
 
  where # is part of the format and not a R comment.
 
  Each block (delimited by #) consists of a first line with three
  values, call it dose, and a list of (x,y) coordinates which are a
  matrix or data.frame,
 
 
  particle - list(dose=c(1,100.0,0),pos=data.frame(x=c(0,1,0,1),y=c
  (0,1,0,1)))
 
  print(particle)
 
 
 
  I'd like to establish a connection to a file and append to it a
  particle block in the format above, or even write the whole file at
  once.
 
  Because different lines have a different number of elements, I
  couldn't get write.table to work in this case, and my attempts at
  sink
  (), dump(), writeLines(), writeChar() all turn into really dirty
  solutions. I have this feeling I'm overlooking a simple solution.
 
  Any help welcome,
 
 
  baptiste
 
  _
 
  Baptiste Auguié
 
  Physics Department
  University of Exeter
  Stocker Road,
  Exeter, Devon,
  EX4 4QL, UK
 
  Phone: +44 1392 264187
 
  http://newton.ex.ac.uk/research/emag
  http://projects.ex.ac.uk/atto
 
  __
  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 you are trying to solve?

 _

 Baptiste Auguié

 Physics Department
 University of Exeter
 Stocker Road,
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What is the problem you are trying to solve?

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Re: [R] non-interactive connection to SQL Server

2008-02-14 Thread Moshe Olshansky

Yes, it does - thank you!
The only thing I forgot (and it took me a while to
find this out) was to separate the fields by
semicolon, i.e. the correct command is:

odbcDriverConnect(driver=SQL Server;
database=dataBaseName; wsid=myComputer;
server=dataBaseServer; uid=moshe; pwd=moshe)

--- Prof Brian Ripley [EMAIL PROTECTED] wrote:

 On Wed, 13 Feb 2008, Moshe Olshansky wrote:
 
  Hi everyone,
 
  I am afraid that I have already asked this
 question in
  the past (or at least I knew an answer to it) but
 I am
  unable to do it now.
  I have an SQL Server data base. I used the GUI
  interface of odbcDriverConnect to create a .dsn
 file
  for this data base and every time I want to
 connect I
  invoke odbcDriverConnect() which opens GUI from
 which
  I choose my .dsn file and the connection is
  established.
  Now I want to do this automatically (without GUI)
 and
  all my attempts fail. If I remember correctly the
 dsn
  must be a string containing all the connection
 details
  but it does not work for me.
 
 Clearly it does, as you did that via
 odbcDriverConnect.  Look at the 
 object it returns: it contains the string it used. 
 You can use that 
 directly.
 
 
  Could anybody help, please!
 
  Thank you!
  Moshe.
 
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 reproducible code.
 
 
 -- 
 Brian D. Ripley, 
 [EMAIL PROTECTED]
 Professor of Applied Statistics, 
 http://www.stats.ox.ac.uk/~ripley/
 University of Oxford, Tel:  +44 1865
 272861 (self)
 1 South Parks Road, +44 1865
 272866 (PA)
 Oxford OX1 3TG, UKFax:  +44 1865
 272595


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Re: [R] setting color ranges

2008-02-14 Thread hadley wickham

Hi Sam,

You might find it easier to use ggplot2 to do this.  See
http://had.co.nz/ggplot2/geom_tile.html for some examples.

Hadley

On Thu, Feb 14, 2008 at 2:59 PM, samsr [EMAIL PROTECTED] wrote:

  Hi,

   I need to plot a matrix using image() such that negative values are easily
  distinguishable from posittive values, while also maintaining a gradation in
  color with magnitude. How can I set ranges for colors in order to achieve
  this. Thanks.

  Sam
  --
  View this message in context: 
 http://www.nabble.com/setting-color-ranges-tp15489086p15489086.html
  Sent from the R help mailing list archive at Nabble.com.

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




-- 
http://had.co.nz/

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Re: [R] data frame question

Create the new data.frame and do the muliplying on it?

df2 - df1
df2[,1] - df2[,1]*2

--- joseph [EMAIL PROTECTED] wrote:

 
 
 Hi
 
 I have a data frame df1 in which I would like to
 multiply col1
 by 2.
 
 
 The way I did it does not allow me to keep the old
 data
 frame.
 
 
 How can I do this and be able to create a new data
 frame
 df2?
 
 
  df1= data.frame(col1= c(3, 5, NA, 1), col2= c(4,
 NA,6,
 2))
 
 
  df1
 
 
   col1 col2
 
 
 134
 
 
 25   NA
 
 
 3   NA6
 
 
 412
 
 
  df1$col1=df1$col1*2
 
 
  df1
 
 
   col1 col2
 
 
 164
 
 
 2   10   NA
 
 
 3   NA6
 
 
 422
 
 
 
 
 
  


 Be a better friend, newshound, and 
 
 
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Re: [R] plot each column of a matrix or dataframe versus x in a single plot

Perhaps matplot will do what you want?
?matplot
aa - matrix(1:25, nrow=5)
matplot(aa)

--- tomaschwutz [EMAIL PROTECTED] wrote:

 How do a plot several columns of a matrix at once in
 a single plot
 versus a single x-variable?
 
 The default plot.matrix  or plot.dataframe commands
 plot each column
 versus each other column in several sub-plots. I
 want to plot each
 column versus a single other vector (x) as several
 lines or points in
 one plot.
 
 I can do it by hand: get the range of all variables
 (i.e. columns of
 the matrix or dataframe), initiate an empty plot
 command, loop through
 all the columns and issue a lines or points command.
 I guess there is a predefined sophisticated function
 for that,
 however, I was not able to find it.
 
 Does someone know where I find such a function?
 
 Best regards
 tomaschwutz
 
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Re: [R] passing username and password to source()

2008-02-14 Thread [Ricardo Rodriguez] Your XEN ICT Team

Phil Spector wrote:
 Ricardo -
 The authentication can't be done through environmental variables --
 the only way is to send an Authorization header.   I believe the 
 environmental variables that Dieter is thinking of are the ones that
 are created on the server side based on the headers that the user
 agent sends.
   - Phil 

Hi Phil,

If I've well understood what you and the other colleagues have said is 
that whatever the method to write the authorization header I use, it is 
the only way of accessing programmatically my server.

I've installed Live HTTP headers, but I am still far from understanding 
all the messages in its window.

Please, could you point me in the right direction to learn what must I 
look for in Live HTTP headers log? Any example?

Thank you so much,

Ricardo

-- 
Ricardo Rodríguez
Your XEN ICT Team

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[R] forestplot() with large number of confidence intervals

2008-02-14 Thread array chip

Hi I am using forestplot() in rmeta package on a
dataset of 45 point estimates with corresponding
confidence intervals. The resulting plot was just too
large and plotted out of the graphic window that I can
not see whole picture. Is there anyway to fix this
problem?

Thanks


  

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Re: [R] setting color ranges

2008-02-14 Thread Achim Zeileis

On Thu, 14 Feb 2008, samsr wrote:


 Hi,

  I need to plot a matrix using image() such that negative values are easily
 distinguishable from posittive values, while also maintaining a gradation in
 color with magnitude. How can I set ranges for colors in order to achieve
 this. Thanks.

Look at the function diverge_hcl() in package vcd which provides
diverging palettes. You can find several examples in
  example(diverge_hcl, package = vcd)
  vignette(hcl-colors, package = vcd)

Further background information is available in this technical report by
Kurt Hornik, Paul Murrell and me
  http://epub.wu-wien.ac.at/dyn/openURL?id=oai:epub.wu-wien.ac.at:epub-wu-01_c87

Best,
Z

 Sam
 --
 View this message in context: 
 http://www.nabble.com/setting-color-ranges-tp15489086p15489086.html
 Sent from the R help mailing list archive at Nabble.com.

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

?write.table  and look at the row.names arguement
which is what they are called in this instance.

write.table(XX, file=XX.txt,quote=FALSE,sep=\t,
row.names=FALSE)


--- Roberto Olivares Hernandez [EMAIL PROTECTED] wrote:

 Hi,
 
 I used the write.table function to save data in txt
 file, and this is the output:
 
 
 V1  V2  V3  V4
 1   YAL005C  21  14  11
 2   YAL007C   2   1   4
 3   YAL012W   8  16   3
 4   YAL016W  24  23  23
 5   YAL019W   3   3   2
 6   YAL020C   2   4   2
 7   YAL021C   7   5   5
 8   YAL022C   3   1   2
 
 
 but I  need to remove the dimnames  (first column)
 
 I tried to use dimnames function to remove it and
 then save it, but still, the output is the same
 
 These are the command lines,
 
 XX #matrix
 dimnames(XX)-NULL
 write.table(XX,XX.txt,quote=FALSE,sep=\t)
 
 
 
 Thanks in advance
 Roberto
 
 
 
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Re: [R] data frame question

2008-02-14 Thread joseph

Thanks. I have another question:
In the following data frame df, I want to replace all values in col1 that are 
higher than 3 with NA.
df= data.frame(col1=c(1:5, NA),col2= c(2,NA,4:7))

- Original Message 
From: John Kane [EMAIL PROTECTED]
To: joseph [EMAIL PROTECTED]; r-help@r-project.org
Cc: r-help@r-project.org
Sent: Thursday, February 14, 2008 3:09:40 PM
Subject: Re: [R] data frame question


Create 
the 
new 
data.frame 
and 
do 
the 
muliplying 
on 
it?

df2 
- 
df1
df2[,1] 
- 
df2[,1]*2

--- 
joseph 
[EMAIL PROTECTED] 
wrote:

 
 
 
Hi
 
 
I 
have 
a 
data 
frame 
df1 
in 
which 
I 
would 
like 
to
 
multiply 
col1
 
by 
2.
 
 
 
The 
way 
I 
did 
it 
does 
not 
allow 
me 
to 
keep 
the 
old
 
data
 
frame.
 
 
 
How 
can 
I 
do 
this 
and 
be 
able 
to 
create 
a 
new 
data
 
frame
 
df2?
 
 
 
 
df1= 
data.frame(col1= 
c(3, 
5, 
NA, 
1), 
col2= 
c(4,
 
NA,6,
 
2))
 
 
 
 
df1
 
 
  
 
col1 
col2
 
 
 
1  
  
3  
  
4
 
 
 
2  
  
5  
 
NA
 
 
 
3  
 
NA  
  
6
 
 
 
4  
  
1  
  
2
 
 
 
 
df1$col1=df1$col1*2
 
 
 
 
df1
 
 
  
 
col1 
col2
 
 
 
1  
  
6  
  
4
 
 
 
2  
 
10  
 
NA
 
 
 
3  
 
NA  
  
6
 
 
 
4  
  
2  
  
2
 
 
 
 
 
  
  
  


 
Be 
a 
better 
friend, 
newshound, 
and 
 
 
 

[[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.
 



  
  
  
Connect 
with 
friends 
from 
any 
web 
browser 
- 
no 
download 
required. 
Try 
the 
new 

Canada 
Messenger 
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[R] Retrieving data frames from a for loop

2008-02-14 Thread Judith Flores

Dear R-helpers,

   I need to retrieve the data frames generated in a
for loop. What I have looks something like this:

where tab is a pre-existing data frame.

for (i in 1:20) {

 g-sample(rep(LETTERS[1:2],each=10))
 combination-data.frame(tab,g)

}

   I tried to name every single combination doing
this:

assign(paste('combination',i), combination)

  without success.

I need to retrieve every combination per separate.

Thank you once again for your help.


  

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[R] LMER

2008-02-14 Thread Daniel Malter

Hi,

I run the following models: 

1a. lmer(Y~X+(1|Subject),family=binomial(link=logit)) and
1b. lmer(Y~X+(1|Subject),family=binomial(link=logit),method=PQL)

Why does 1b produce results different from 1a? The reason why I am asking is
that the help states that PQL is the default of GLMMs

and

2. gamm(Y~X,family=binomial(link=logit),random=list(Subject=~1))

The interesting thing about the example below is, that gamm is also supposed
to fit by PQL. Interestingly, however, the GAMM fit yields about the
coefficient estimates of 1b. But the significance values of 1a. Any insight
would be greatly appreciated.


library(lme4)
library(mgcv)

Y=c(0,1,1,1,1,0,0,0,0,0,1,1,1,1,0,0,0,1,1,1,1)
X=c(1,2,3,4,3,1,0,0,2,3,3,2,4,3,2,1,1,3,4,2,3)
Subject=as.factor(c(1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7))
cbind(Y,X,Subject)

r1=lmer(Y~X+(1|Subject),family=binomial(link=logit))
summary(r1)

r2=lmer(Y~X+(1|Subject),family=binomial(link=logit),method=PQL)
summary(r2)

r3=gamm(Y~X,family=binomial(link=logit),random=list(Subject=~1))
summary(r3$gam)



-
cuncta stricte discussurus

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[R] Questions about EM algorithm

2008-02-14 Thread Hung-Hsuan Chen (Sean)

Dear all:
Assume I have 3 distributions, x1, x2, and x3.
x1 ~ normal(mu1, sd1)
x2 ~ normal(mu2, sd2)
x3 ~ normal(mu3, sd3)
y1 = x1 + x2
y2 = x1 + x3

Now that the data I can observed is only y1 and y2. It is
easy to estimate (mu1+m2), (mu1+mu3), (sd1^2+sd2^2) and
(sd1^2+sd3^2) by EM algorithm since
y1 ~ normal(mu1+mu2, sqrt(sd1^2+sd2^2)) and
y2 ~ normal(mu1+mu3, sqrt(sd1^2+sd3^2))

However, I want to estimate mu1, mu2, mu3, sd1, sd2, and sd3.
Is it possible to do so by EM algorithm (or any other estimation
methods like gibbs sampling or MLE) ?

Anyone can give me some sample code or suggestions? Thanks a
lot in advance.

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[R] Quantile Regression R squared

2008-02-14 Thread frank frank


Dear R,
I am currently trying to caculate the coefficient of determination for 
different quantile regression models. For example

fit-rq(Hrubra~SessileInvertebrates,tau=0.8, data=Q1) 
fit1-rq(Hrubra~SessileInvertebrates,tau=0.8, data=Q2)
etc 

Could someone please advise me how do you calculate the the sum of the weighted 
absolute deviations in the models for the formula

R = 1-(Sum(f)/Sum(r))

I have looked all through the help files but the only example I can find is in 
Stata.
Thanks in advance
Beth Strain


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Re: [R] Problems with Rcmdr unter JGR (Windows XP)

2008-02-14 Thread Michael Bibo

Willi Nagl willi.nagl at gmail.com writes:

 
 I try  to start Rmcdr from JGR. The Rmcdr-Windows comes up correctly; but
 the
  Menu-Bar in the Rcmdr-Window goes away, if I try to go into the
 Rmcdr-Window.
 
  I hope, someone has a solution.
 
 Regards, willi
 

My experience is that, under Windows,  you cannot run Rcmdr with either JGR or
(X)Emacs/ESS.  From John Fox's own 'An Introduction to ESS + XEmacs for Windows
Users of R': The Rcmdr package does not run reliably under XEmacs/ESS for
Windows.  I get the same instability (Rcmdr not responding and then R crashing)
with Rcmdr and JGR.

Under Linux, I have had Emacs/ESS and Rcmdr running together.  I don't know
about JGR, as I don't presently have it installed on a Linux box.

If you want to use Rcmdr under Windows, the combination of R-Gui, Tinn-R and
Rcmdr works well.

Michael Bibo
Research Officer
Queensland Health

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Re: [R] Retrieving data frames from a for loop