date:20070425

On 24/04/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
 On Tue, 24 Apr 2007, Hendrik Fuß wrote:

  Hi,
 
  I've been trying here to install Rmpi on an SGI IA-64 machine with 64
  processors, running SuSE Linux Enterprise Server 9, R 2.4.0 and
  lam-mpi 7.1.3. While I've read of similar problems on this list, I
  think I've got an entirely new set of error messages to contribute
  (see below). I'm not sure what the actual error is and what the @gprel
  relocation message is about. Any help greatly appreciated.

 I don't know for sure, but on many 64-bit OSes you cannot link code from
 static libraries into dynamic shared libraries, and that seems to be the
 case with ia64 Linux.  Almost certainly you need to re-compile LAM with
 -fPIC flags.

Yes, thanks a million, this solved the problem.

While Rmpi now works, there is another issue connected with this one:
I'm trying to use the papply package. However, when I do:

 library(papply)
 papply(list(1:10, 1:15, 1:20), sum)
1 slaves are spawned successfully. 0 failed.
master (rank 0, comm 1) of size 2 is running on: behemoth
slave1 (rank 1, comm 1) of size 2 is running on: behemoth
[1] Running serial version of papply\n

Papply only spawns one slave and then decides to run the serial
version instead. I'm not sure how to tell it to use all the 64
processors available.

Hendrik

-- 
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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Re: [R] regular expressions with grep() and negative indexing

2007-04-25 Thread Peter Dalgaard

Stephen Tucker wrote:
 Dear R-helpers,

 Does anyone know how to use regular expressions to return vector elements
 that don't contain a word? For instance, if I have a vector
   x - c(seal.0,seal.1-exclude)
 I'd like to get back the elements which do not contain the word exclude,
 using something like (I know this doesn't work) but:
   grep([^(exclude)],x)

 I can use 
   x[-grep(exclude,x)]
 for this case but then if I use this expression in a recursive function, it
 will not work for instances in which the vector contains no elements with
 that word. For instance, if I have
   x2 - c(dolphin.0,dolphin.1)
 then
   x2[-grep(exclude,x2)]
 will give me 'character(0)'

 I know I can accomplish this in several steps, for instance:
   myfunc - function(x) {
 iexclude - grep(exclude,x)
 if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
 # do stuff with x2 ...?
   }

 But this is embedded in a much larger function and I am trying to minimize
 intermediate variable assignment (perhaps a futile effort). But if anyone
 knows of an easy solution, I'd appreciate a tip.
   
It has come up a couple of times before, and yes, it is a bit of a pain.

Probably the quickest way out is

negIndex - function(i) 

   if(length(i))

   -i 

   else 

   TRUE

-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

__
R-help@stat.math.ethz.ch mailing list
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] regarding 3d Bar Plot


On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:

[EMAIL PROTECTED] wrote:


I have data in a two dimensional table. each row of the data adds
upto 100 ( hence they are percentages ).  it can be interpreted as
like this A - I are the matches and  P - X are the players. Thus
Player P scored 20% of the runs during this season in Match C, 60% in
Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
Matches and Z axis as the percentage(0 - 100%) Please help me in this
regards.
 
 snip


Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots with 2D
bases) written by a chap named Colin Goodall, from (at that
time) the University of Bristol and/or from Penn State.

I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code running
in R to see if it might solve guarav's problem.

The trouble is, all the guts of the procedure, *including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does not
exist in R.

Is there an equivalent subroutine in R that could be called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(

I looked around a bit for corresponding C or Fortran code but
obviously didn't know how to look properly.

I think that the Fortran code could be translated into raw R
and the call to segmtz changed to a call to segments() ---
but this would seem to be a lot of work.

Can anyone suggest a reasonably simple way of replacing the
call to segmtz in the Fortran?


I don't know how to do what you want, but I'd suggest working in R code 
rather than Fortran.  I did write a hist3d function for the djmrgl 
package (based on hist), mostly to show off the graphics, but haven't 
found it useful enough to port to rgl.  Here's a quick port, not good 
enough to use, but maybe it will give you a starting point.


Duncan Murdoch




hist3d -
function (x, y, xbreaks, ybreaks, freq= NULL, probability = !freq, 
include.lowest= TRUE,
  right= TRUE, 
  xlim = range(xbreaks), ylim = range(ybreaks), zlim = NULL,
  xlab = xname, ylab = yname, zlab,
  plot = TRUE, top = TRUE, nclass = NULL, ...)
{
if (!is.numeric(x) | !is.numeric(y))
stop(`x' and `y' must be numeric)
xname - deparse(substitute(x))
yname - deparse(substitute(y))
n - length(x - x[!is.na(x)])
use.xbr - !missing(xbreaks)
if(use.xbr) {
if(!missing(nclass))
warning(`nclass' not used when `xbreaks' specified)
}
else if(!is.null(nclass)  length(nclass) == 1)
xbreaks - nclass
use.xbr - use.xbr  (nB - length(xbreaks))  1
if(use.xbr)
xbreaks - sort(xbreaks)
else {  # construct vector of breaks
rx - range(x)
nnb -
if(missing(xbreaks)) 1 + log2(n)
else {  # breaks = `nclass'
if (is.na(xbreaks) | xbreaks  2)
stop(invalid number of xbreaks)
xbreaks
}
xbreaks - pretty (rx, n = nnb, min.n=1, eps.corr = 2)
}
nxB - length(xbreaks)
if(nxB = 1) ##-- Impossible !
stop(paste(hist3d: error, xbreaks=,format(xbreaks)))

storage.mode(x) - double
storage.mode(xbreaks) - double
use.ybr - !missing(ybreaks)
if(use.ybr) {
if(!missing(nclass))
warning(`nclass' not used when `ybreaks' specified)
}
else if(!is.null(nclass)  length(nclass) == 1)
ybreaks - nclass
use.ybr - use.ybr  (nB - length(ybreaks))  1
if(use.ybr)
ybreaks - sort(ybreaks)
else {  # construct vector of breaks
ry - range(y)
nnb -
if(missing(ybreaks)) 1 + log2(n)
else {  # breaks = `nclass'
if (is.na(ybreaks) | ybreaks  2)
stop(invalid number of ybreaks)
ybreaks
}
ybreaks - pretty (ry, n = nnb, min.n=1, eps.corr = 2)
}
nyB - length(ybreaks)
if(nyB = 1) ##-- Impossible !
stop(paste(hist3d: error, ybreaks=,format(ybreaks)))

storage.mode(y) - double
storage.mode(ybreaks) - double
counts - table(cut(x,xbreaks),cut(y,ybreaks))
if (sum(counts)  n)
stop(some data not counted; maybe breaks do not span range of data)
xh - diff(xbreaks)
if

Re: [R] regular expressions with grep() and negative indexing

2007-04-25 Thread jim holtman

Find the ones that match and then remove them from the full set with 'setdiff'.

 x - c(seal.0,seal.1-exclude)
 x.match - grep(exclude, x)  # find matches
 x.match
[1] 2
 setdiff(seq_along(x), x.match)  # exclude the matches
[1] 1



On 4/25/07, Peter Dalgaard [EMAIL PROTECTED] wrote:
 Stephen Tucker wrote:
  Dear R-helpers,
 
  Does anyone know how to use regular expressions to return vector elements
  that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
  I'd like to get back the elements which do not contain the word exclude,
  using something like (I know this doesn't work) but:
grep([^(exclude)],x)
 
  I can use
x[-grep(exclude,x)]
  for this case but then if I use this expression in a recursive function, it
  will not work for instances in which the vector contains no elements with
  that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
  then
x2[-grep(exclude,x2)]
  will give me 'character(0)'
 
  I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
  iexclude - grep(exclude,x)
  if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
  # do stuff with x2 ...?
}
 
  But this is embedded in a much larger function and I am trying to minimize
  intermediate variable assignment (perhaps a futile effort). But if anyone
  knows of an easy solution, I'd appreciate a tip.
 
 It has come up a couple of times before, and yes, it is a bit of a pain.

 Probably the quickest way out is

 negIndex - function(i)

   if(length(i))

   -i

   else

   TRUE

 --
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
  (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
 ~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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



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

What is the problem you are trying to solve?

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

Re: [R] regarding 3d Bar Plot

2007-04-25 Thread gyadav


Hi Duncan

I am restating the problem and thanks you for sending me such a good 
function histogram in 3d. Thanks for that but i think my problem has been 
misinterpreted. I just wanted a simple 3d bar Plot. Although I have not 
written anything for R but i will surely like to contribute to R and if i 
can assist someone in writing then it would be a great help to me.

Problem was :-)

I have data in a two dimensional table. each row of the data adds upto 100 

( hence they are percentages ). 
it can be interpreted as like this A - I are the matches and  P - X are 
the players. Thus Player P scored 20% of the runs during this season in 
Match C, 60% in Match D and remaining 20% in Match G. 

I want to plot 3-d bar plot, where X axis have players, Y axis have 
Matches and Z axis as the percentage(0 - 100%) 
Please help me in this regards. (Please note on my X and Y axes Numbers 
are not there instead alphabets)

A   B   C   D   E   F   G   H   I 
P   0   0   20  60  0   0   20  0   0 
Q   0   16.8674726.907631   11.646586   0 
12.449799   0.8032129   0   31.325301 
R   0   59.649123   10.526316   12.280702   0   0 
1.7543860   15.789474 
S   3.57909807  20.281556   33.404915   7.31329 0.584586 
5.9651631.1930327   0   27.678358 
T   0   0   0   0   0   0   0   0   0 
U   0   9.09090927.272727   18.181818   0 
36.363636   0   0   9.090909 
V   0   33.33   33.33   0   0   33.33 
0   0   0 
W   0   2.1881841.09409236.105033   0 
44.420131   5.2516411   0   10.940919 
X   0.05994234  51.550409   16.304315   6.9976680 
17.383277   0.5994234   0.4741439   6.630821 



Thanks in advance
-gaurav




Duncan Murdoch [EMAIL PROTECTED] 
25-04-07 04:42 PM

To
[EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject
Re: [R] regarding 3d Bar Plot






On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:
 [EMAIL PROTECTED] wrote:
 
 I have data in a two dimensional table. each row of the data adds
 upto 100 ( hence they are percentages ).  it can be interpreted as
 like this A - I are the matches and  P - X are the players. Thus
 Player P scored 20% of the runs during this season in Match C, 60% in
 Match D and remaining 20% in Match G.

 I want to plot 3-d bar plot, where X axis have players, Y axis have
 Matches and Z axis as the percentage(0 - 100%) Please help me in this
 regards.
 
  snip
 
Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots 
with 2D
bases) written by a chap named Colin Goodall, from (at 
that
time) the University of Bristol and/or from Penn State.
 
I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code 
running
in R to see if it might solve guarav's problem.
 
The trouble is, all the guts of the procedure, 
*including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does 
not
exist in R.
 
Is there an equivalent subroutine in R that could be 
called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(
 
I looked around a bit for corresponding C or Fortran code 
but
obviously didn't know how to look properly.
 
I think that the Fortran code could be translated into 
raw R
and the call to segmtz changed to a call to segments() 
---
but this would seem to be a lot of work.
 
Can anyone suggest a reasonably simple way of replacing 
the
call to segmtz in the Fortran?

I don't know how to do what you want, but I'd suggest working in R code 
rather than Fortran.  I did write a hist3d function for the djmrgl 
package (based on hist), mostly to show off the graphics, but haven't 
found it useful enough to port to rgl.  Here's a quick port, not good 
enough to use, but maybe it will give you a starting point.

Duncan Murdoch




hist3d -
function (x, y, xbreaks, ybreaks, freq= NULL, probability = !freq, 
include.lowest= TRUE,
  right= TRUE, 
  xlim = range(xbreaks), ylim = range(ybreaks), zlim = NULL,
  xlab = xname, ylab = yname, zlab,
  plot = TRUE, top

Re: [R] correlation table

Possible: yes - just calcuate correlation of as.numeric(your.factors)
Meaningful: no
It will depend on the coding for your factors, which may be absolutely 
arbitrally...

Petr

elyakhlifi mustapha napsal(a):
 hello,
 is it possible to create a correlation table between factors?
 
 
   
 ___
 
 
 
 
 
   [[alternative HTML version deleted]]
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 

-- 
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

__
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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] regarding 3d Bar Plot --- correction.

2007-04-25 Thread gyadav


Hi Rolf,

If it is possible then please share the code as i am not able to locate 
any pointers. Thanks in advance :-) cheers and chiao

regards
-gaurav




[EMAIL PROTECTED] 
24-04-07 07:27 PM

To
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
cc

Subject
Re: [R] regarding 3d Bar Plot --- correction.







I mis-spoke.  It seems I had two collections of functions in the same
directory.  One by Colin Goodall, and one by David Scott (I have no
record of where he is/was located).  It is the *latter* collection
that does all its work from within Fortran.

I'll have another look at what Colin Goodall actually wrote to see if
it could be useful to guarav.

 cheers,

  Rolf Turner
  [EMAIL PROTECTED]




DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}

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

Re: [R] Coercing data types for use in model.frame

2007-04-25 Thread Gabor Grothendieck

You could try to follow the code in the dyn package.  It intercepts
model.frame calls involving time series objects so that it can align
lagged objects:

e.g.
z - ts(seq(10)^2)
library(dyn)
dyn$lm(z ~ lag(z, -1))

It transforms the last line above to:

dyn(lm(dyn(z ~ lag(z, -1)))

and the inner dyn then produces a formula with class c(dyn, model.frame)
so that model.frame.dyn can intercept the call while the outer dyn adds
dyn to the class of the result so that anova.dyn, predict.dyn, etc. can be
used to intercept the result.

Thus for any lm-like function you just preface it with dyn$ as shown and
you get automatically alignment of time series or in your case you would
interception of the mChoice variables.

On 4/24/07, Frank E Harrell Jr [EMAIL PROTECTED] wrote:
 In the Hmisc package there is a new data class 'mChoice' for multiple
 choice variables.  There are format and as.numeric methods (the latter
 creates a matrix of dummy variables).  mChoice variables are not allowed
 by model.frame.  Is there a way to specify a conversion function that
 model.frame will use automatically?  I would use as.factor here.
 model.frame does not seem to use as.data.frame.foo for individual variables.

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

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


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

Re: [R] RE : for loops

I know that a looping solution was requested, but this is exactly what 
apply() should be used for...

Petr

justin bem napsal(a):
 You can see with this simple example. 
 
 matrix.t.test-function(mx){
 
 p-dim(mx)[2]  #number of column in the matrix
 n-dim(mx)[1]  #number of row
 
 n.tests- p*(p-1)/2 #Number of tests to be done
 
 tested.var -rep(,n.tests) #Keep rang of tested
 column
 r.t.stat-rep(0, n.tests)#contain t.stat
 r.p.val -rep(0, n.tests)#contain p.values
 
 ctst-1 #current test
 for (i in 1:(p-1)){
   for (j in (i+1):p){
 r.t.stat[ctst]-t.test(mx[,i],mx[,j])$statistic
 r.p.val [ctst] -t.test(mx[,i],mx[,j])$p.value
 tested.var [ctst]-paste(i,-,j)
 ctst-ctst+1
   }
 }
 
 result-data.frame(tested.var,r.t.stat,r.p.val)
 return(result)
 }
 
 matrix.t.test(matrix(rnorm(50),nr=10,nc=5))
 
 
 
 
 
 
 
 
 
 --- [EMAIL PROTECTED]
 [EMAIL PROTECTED] a écrit :
 
 Hello everybody
 I'm very new at using R so probably this is a very
 stupid question.
 I have a matrix of p columns and I have to
 calculate for each of them the two sample
 t-statistic and p-value and to save the results
 into two different vectors.
 I have divided my matrix into two submatrices:
 submatrix A containing the first n1 rows (p
 columns) and submatrix B containing the remaining
 n2 (total rows=n1+n2).
 How can I do this with for loop construction?
 Friendly regards
 Silvia



 --
 Passa a Infostrada. ADSL e Telefono senza limiti e
 senza canone Telecom
 http://click.libero.it/infostrada

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

 
 
 Justin BEM
 Elève Ingénieur Statisticien Economiste
 BP 294 Yaoundé.
 Tél (00237)9597295.
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 

-- 
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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

Re: [R] A very simple question

2007-04-25 Thread Gabor Grothendieck

If by objects you mean the packages you installed and assuming you
are on Windows then there are two batch files movedir.bat and copydir.bat
in the batchfiles distribution that will either move them from an older version
(they won't be available to the older version any more but its much faster and
they won't be taking up space twice) or copy them (they will now be in both
versions so you can use them from either but its slower and they will be taking
up space twice).  See the batchfiles home page:

   http://code.google.com/p/batchfiles/

which has a link to the download site and a link to the README which
you should be sure to read.

On 4/25/07, David Kaplan [EMAIL PROTECTED] wrote:
 Hi all,

 I just loaded R 2.50.  I want this version to bring the objects from the
 previous version.  How do I do that?

 Thanks

 David

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 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Help with saptial analysis (cluster)

2007-04-25 Thread ONKELINX, Thierry

Dear Fransico,

The distance matrix would be 102000 x 102000. So it would contain 1040400 
values. If you need one bit for each value, this would requier 9,7 GB. So the 
distance matrix won't fit in the RAM of your computer.

Cheers,

Thierry


ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology 
and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be 

Do not put your faith in what statistics say until you have carefully 
considered what they do not say.  ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of 
uncertainties, a surgery of suppositions. ~M.J.Moroney

 

 -Oorspronkelijk bericht-
 Van: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] Namens Francisco Pastor
 Verzonden: woensdag 25 april 2007 12:34
 Aan: r-help@stat.math.ethz.ch
 Onderwerp: [R] Help with saptial analysis (cluster)
 
 Hi R-users
 
 I'm a beginner with R and statistics, so I need some help to 
 start my data analysis. I've been reading some docs and 
 tutorials on R and cluster analysis.
 I've got a large dataset (102000 points) with values of 
 longitude, latitude and temperature and want to see if I can 
 find groups (clusters).
 
 Following some tutorials I can look for principal components 
 but get an error with calculation of distances:
 
  matriz.distancias-dist(comp.obs)
 Error in vector(double, length) : specified vector size is 
 too big (translated from spanish)
 
 So, my questions are: is the dataset too big? could you point 
 me to any docs explaining how to study spatially distributed 
 data (lon,lat,data)?
 
 Thanks in advance
 
 
 __
 _
 Francisco Pastor
 Meteorology department
 Fundación CEAM
 [EMAIL PROTECTED]
 http://www.gva.es/ceamet
 http://www.gva.es/ceam
 Paterna, Valencia, Spain
 __
 _
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Re: [R] prelim.norm() function not working

2007-04-25 Thread Inman, Brant A. M.D.


Thank you very much, that was indeed the problem. (And now that I read
more carefully the help page, it did in fact say that the input was a
data matrix and not a data frame.)

Brant
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Prof Brian Ripley
Sent: Wednesday, April 25, 2007 12:12 AM
To: Brant Inman
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] prelim.norm() function not working

Looks like you have a data frame where you need a matrix.  (The same 
issue occurs in most of Joe Schafer's packages, e.g. mix.)

Try as.matrix(usnews).

On Tue, 24 Apr 2007, Brant Inman wrote:

 R-experts:
 I am trying to reproduce some of Paul Allison's results in his little
 green book on missing data (Sage 2002).  The dataset for which I am
 having problems, usnews, can be found at:
 http://www.ats.ucla.edu/stat/books/md/default.htm.  I am working on a
 Windows machine with R 2.5 installed, all packages up-to-date.
 The problem has to do with the prelim.norm() function of the package
 norm.   Specifically, I need to use this pre-processing function to
 later use the EM algorithm and DA procedures in the norm package.  I
 am getting an error with the following code.
 --
 pre - prelim.norm(usnews)

 Error in as.double.default(list(csat = c(972L, 961L, NA, 881L, NA, NA,
:
(list) object cannot be coerced to 'double'

 -
 I have read the previous postings and I am wondering if the problem
 with prelim.norm is the size of the usnews dataset or the amount of
 missing data.

 

 dim(usnews)
 [1] 13027

 


 Does anyone have any ideas?  If not, are there alternatives to norm
 for implementing the MLE and EM methods of dealing with missing data?

 Thanks,

 Brant Inman
 Mayo Clinic

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-- 
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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[R] program avail. for simulating spatial patterns?

2007-04-25 Thread Wade Wall

Hi all,

I am wondering if there is a function available in R for simulating
spatial distribution of objects (plants in this case) in order to
simulate sampling of a population .  Specifically, I would like to be
able to change the spatial correlation of individuals.  I don't want
to reinvent the wheel if it already exists.

Thanks,

Wade

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[R] assigning two conditions to grep()

2007-04-25 Thread Abi Ghanem josephine

Hi,
i have a problem in assigning 2 conditions to grep()  ,
my data look like this:
DA 24 N7 Rad= 3.4 20 Sac= 0.93 Acc= 4.76
DA 24 N7 Rad= 3.4 14 Sac= 0.65 Acc= 3.33
DA 24 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DA 24 N7 Rad= 3.4 11 Sac= 0.51 Acc= 2.62
DG 23 N7 Rad= 3.4  8 Sac= 0.37 Acc= 1.91
DG 23 N7 Rad= 3.4  5 Sac= 0.23 Acc= 1.19
DG 23 N7 Rad= 3.4  0 Sac= 0.00 Acc= 0.00
DG 23 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  1 Sac= 0.04 Acc=  0.22
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
(it's a data.frame)

at first i wanted all the line begining with A 24:
data[grep(^24, data$V2)]
this works
and than i wanted to exctract all the lines with G23 N7,
neither the column 23 and the column N7 are unique
so i tried this
 data[grep(^23*N7, data),]
but doesn't work
not either
x[(grep(^N7, as.character(x$V3))) (grep(^23, x$V2)),]
he just returns everything.

thank u for any help,
josephine

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Re: [R] regarding 3d Bar Plot

On 4/25/2007 7:56 AM, [EMAIL PROTECTED] wrote:
 Hi Duncan
 
 I am restating the problem and thanks you for sending me such a good 
 function histogram in 3d. Thanks for that but i think my problem has been 
 misinterpreted. I just wanted a simple 3d bar Plot. Although I have not 
 written anything for R but i will surely like to contribute to R and if i 
 can assist someone in writing then it would be a great help to me.
 
 Problem was :-)
 
 I have data in a two dimensional table. each row of the data adds upto 100 
 
 ( hence they are percentages ). 
 it can be interpreted as like this A - I are the matches and  P - X are 
 the players. Thus Player P scored 20% of the runs during this season in 
 Match C, 60% in Match D and remaining 20% in Match G. 
 
 I want to plot 3-d bar plot, where X axis have players, Y axis have 
 Matches and Z axis as the percentage(0 - 100%) 
 Please help me in this regards. (Please note on my X and Y axes Numbers 
 are not there instead alphabets)

The plot.histogram function I sent does most of what you want.  The 
hist3d function calculates the matrix of counts that it plots, and 
plot.histogram plots the resulting bar chart.

Duncan Murdoch


 
 A   B   C   D   E   F   G   H   I 
 P   0   0   20  60  0   0   20  0   0 
 Q   0   16.8674726.907631   11.646586   0 
 12.449799   0.8032129   0   31.325301 
 R   0   59.649123   10.526316   12.280702   0   0 
 1.7543860   15.789474 
 S   3.57909807  20.281556   33.404915   7.31329 0.584586 
 5.9651631.1930327   0   27.678358 
 T   0   0   0   0   0   0   0   0   0 
 U   0   9.09090927.272727   18.181818   0 
 36.363636   0   0   9.090909 
 V   0   33.33   33.33   0   0   33.33 
 0   0   0 
 W   0   2.1881841.09409236.105033   0 
 44.420131   5.2516411   0   10.940919 
 X   0.05994234  51.550409   16.304315   6.9976680 
 17.383277   0.5994234   0.4741439   6.630821 
 
 
 
 Thanks in advance
 -gaurav
 
 
 
 
 Duncan Murdoch [EMAIL PROTECTED] 
 25-04-07 04:42 PM
 
 To
 [EMAIL PROTECTED]
 cc
 [EMAIL PROTECTED], r-help@stat.math.ethz.ch
 Subject
 Re: [R] regarding 3d Bar Plot
 
 
 
 
 
 
 On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:
 [EMAIL PROTECTED] wrote:

 I have data in a two dimensional table. each row of the data adds
 upto 100 ( hence they are percentages ).  it can be interpreted as
 like this A - I are the matches and  P - X are the players. Thus
 Player P scored 20% of the runs during this season in Match C, 60% in
 Match D and remaining 20% in Match G.

 I want to plot 3-d bar plot, where X axis have players, Y axis have
 Matches and Z axis as the percentage(0 - 100%) Please help me in this
 regards.
  snip

Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots 
 with 2D
bases) written by a chap named Colin Goodall, from (at 
 that
time) the University of Bristol and/or from Penn State.

I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code 
 running
in R to see if it might solve guarav's problem.

The trouble is, all the guts of the procedure, 
 *including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does 
 not
exist in R.

Is there an equivalent subroutine in R that could be 
 called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(

I looked around a bit for corresponding C or Fortran code 
 but
obviously didn't know how to look properly.

I think that the Fortran code could be translated into 
 raw R
and the call to segmtz changed to a call to segments() 
 ---
but this would seem to be a lot of work.

Can anyone suggest a reasonably simple way of replacing 
 the
call to segmtz in the Fortran?
 
 I don't know how to do what you want, but I'd suggest working in R code 
 rather than Fortran.  I did write a hist3d function for the djmrgl 
 package (based on hist), mostly to show off the graphics, but haven't 
 found it useful enough to port to rgl.  Here's a quick port, not good 
 enough to use, but maybe it will give you a

Re: [R] program avail. for simulating spatial patterns?

2007-04-25 Thread ONKELINX, Thierry

Have a look at packages in the spatial taskview (spatstat, splancs).

Cheers,

Thierry



ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be 

Do not put your faith in what statistics say until you have carefully
considered what they do not say.  ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney

 

 -Oorspronkelijk bericht-
 Van: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] Namens Wade Wall
 Verzonden: woensdag 25 april 2007 15:03
 Aan: r-help@stat.math.ethz.ch
 Onderwerp: [R] program avail. for simulating spatial patterns?
 
 Hi all,
 
 I am wondering if there is a function available in R for 
 simulating spatial distribution of objects (plants in this 
 case) in order to simulate sampling of a population .  
 Specifically, I would like to be able to change the spatial 
 correlation of individuals.  I don't want to reinvent the 
 wheel if it already exists.
 
 Thanks,
 
 Wade
 
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 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
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Re: [R] regarding 3d Bar Plot

2007-04-25 Thread hadley wickham

 I am restating the problem and thanks you for sending me such a good
 function histogram in 3d. Thanks for that but i think my problem has been
 misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
 written anything for R but i will surely like to contribute to R and if i
 can assist someone in writing then it would be a great help to me.

 Problem was :-)

 I have data in a two dimensional table. each row of the data adds upto 100

 ( hence they are percentages ).
 it can be interpreted as like this A - I are the matches and  P - X are
 the players. Thus Player P scored 20% of the runs during this season in
 Match C, 60% in Match D and remaining 20% in Match G.

 I want to plot 3-d bar plot, where X axis have players, Y axis have
 Matches and Z axis as the percentage(0 - 100%)
 Please help me in this regards. (Please note on my X and Y axes Numbers
 are not there instead alphabets)

I suggest that you don't use a 3d bar chart.  3d bar charts are
generally hard to interpret for two reasons: large bars will obscure
small bars behind them, and it is very difficult to judge the true
length of the bars. I suggest you try creating a series of 2d bar
charts instead - you are far more likely to be able to interpret them
easily.

For this data, you might also want to look into fluctuation diagrams.

Hadley

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[R] R CMD CHECK and require() / library() methods

2007-04-25 Thread Crispin Miller

Hi,
 
I have a piece of code that decides at runtime whether to load a data
package (and which package to load).
This is then done with a call to:
 
library(x)
 
(where x is a character variable containing the package name).
 
This causes R CMD check to throw out a warning:
'library' or 'required' calls not declared from:
x
 
Does anyone have any suggestions as to a fix or workaround for this?
 
Crispin
 



This email is confidential and intended solely for the use o...{{dropped}}

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[R] How to identify and exclude the outliers with R?

2007-04-25 Thread Shao

Hello, everyone,

I want to ask a simple question.
If I have a set  of data,and I want to identify how many outliers there are
in the data.Which packages and functions can I use?

Thanks.

Shao chunxuan.

[[alternative HTML version deleted]]

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[R] creating random numbers

2007-04-25 Thread raymond chiruka

 l want to create a  column of 1 and 2 randomly what command should l use 
  eg
  treatment  strata
  
  1  1
  2  0
  1  1
  2  1
  2  0  
  2  1
  2  0
  1  0
  these should be created randomly
  
  secondly if l have something like 
  for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l add an if statement 
in  side the c maybe to compare the 2 exponential  numbers to create  another  
variable

   
-


[[alternative HTML version deleted]]

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Re: [R] regarding 3d Bar Plot

On 4/25/2007 9:26 AM, hadley wickham wrote:
 I am restating the problem and thanks you for sending me such a good
 function histogram in 3d. Thanks for that but i think my problem has been
 misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
 written anything for R but i will surely like to contribute to R and if i
 can assist someone in writing then it would be a great help to me.

 Problem was :-)

 I have data in a two dimensional table. each row of the data adds upto 100

 ( hence they are percentages ).
 it can be interpreted as like this A - I are the matches and  P - X are
 the players. Thus Player P scored 20% of the runs during this season in
 Match C, 60% in Match D and remaining 20% in Match G.

 I want to plot 3-d bar plot, where X axis have players, Y axis have
 Matches and Z axis as the percentage(0 - 100%)
 Please help me in this regards. (Please note on my X and Y axes Numbers
 are not there instead alphabets)
 
 I suggest that you don't use a 3d bar chart.  3d bar charts are
 generally hard to interpret for two reasons: large bars will obscure
 small bars behind them, and it is very difficult to judge the true
 length of the bars. I suggest you try creating a series of 2d bar
 charts instead - you are far more likely to be able to interpret them
 easily.

I agree with this.

Duncan Murdoch

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

Re: [R] NA and NaN randomForest

2007-04-25 Thread Liaw, Andy

Hi Clayton,

If you use the formula interface, then it should do what you want:

R library(randomForest)
randomForest 4.5-18 
Type rfNews() to see new features/changes/bug fixes.
R iris1 - iris[-(1:5),]
R iris2 - iris[1:5,]
R iris2[1, 3] - NA
R iris2[3, 1] - NA
R iris.rf - randomForest(Species ~ ., iris1)
R predict(iris.rf, iris2[-5])
[1] NA   setosa NA   setosa setosa
Levels: setosa versicolor virginica

The problem, of course, is that the formula interface is not suitable
for data with large number of variables.  I'll look into doing the same
thing in the default method.

Andy


From: [EMAIL PROTECTED]
 
 Dear R-help,
 
 This is about randomForest's handling of NA and NaNs in test set data.
 Currently, if the test set data contains an NA or NaN then 
 predict.randomForest will skip that row in the output.
 
 I would like to change that behavior to outputting an NA.
 
 Can this be done with flags to randomForest?
 If not can some sort of wrapper be built to put the NAs back in?
 
 thanks,
 
 Clayton
 _
 
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Re: [R] creating random numbers

2007-04-25 Thread Doran, Harold

sample(1:2, 10, replace=TRUE) 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of raymond chiruka
 Sent: Wednesday, April 25, 2007 9:45 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] creating random numbers

  l want to create a  column of 1 and 2 randomly what 
 command should l use
   eg
   treatment  strata

   1  1
   2  0
   1  1
   2  1
   2  0  
   2  1
   2  0
   1  0
   these should be created randomly

   secondly if l have something like
   for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l 
 add an if statement in  side the c maybe to compare the 2 
 exponential  numbers to create  another  variable

 -

   [[alternative HTML version deleted]]

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

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

Re: [R] Random Number Generator of Park and Miller

2007-04-25 Thread David Forrest

On Tue, 24 Apr 2007, gracezhang wrote:


 Hi,

 I failed to search for R package providing random number generator of Park
 and Miller.
 Anyone know any R package supporting this kind of function?

rng.lcg-function(x,p1=16807,p2=0,N=2147483647){(x*p1+p2)%%N}

Dave
-- 
  Dr. David Forrest
  [EMAIL PROTECTED](804)684-7900w
  [EMAIL PROTECTED] (804)642-0662h
http://maplepark.com/~drf5n/

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[R] print format - fixed number of digits

Hi,

is it possible to format real (double) numbers in a data frame to an 
exact format?
I need something like format(..., digits=5) but this is a suggestion 
only and I need exactly 5 digits. Example where there are more than 5 
digits printed follows.

I can do it via
cbind(as.numeric(sprintf(%.5f,column_1)), 
as.numeric(sprintf(%.5f,column_2)),...)
but it is really annoying and there must be an easier solution.

Petr

  x - as.data.frame(matrix(rnorm(6),nrow=3)/100)

  x
 V1   V2
1  0.002640759 -0.002335782
2 -0.003960130  0.010373135
3 -0.007079349 -0.005792717

  format(x,digits=5,scientific=FALSE)
   V1 V2
1  0.0026408 -0.0023358
2 -0.0039601  0.0103731
3 -0.0070793 -0.0057927
 


-- 
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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[R] Split column of concatenated data

2007-04-25 Thread Street N.R.

Hi

I have a column of concatenated information stored in an RG object in the limma 
package and I need to split this information and then paste the first two 
pieces of data in each case back into two columns of the RG object.

This is how I am currently doing this


gene.info.split-strsplit(RG$genes$Name,,,fixed=TRUE)

for (h in 1: length(gene.info.split)){
RG$genes$ID[h]-gene.info.split[h][[1]][1]
RG$genes$Name[h]-gene.info.split[h][[1]][2]
}


However, this is very slow and presumably 'messy'. The problem is that there 
are an inconsistent number of comma separated entries in the original Name 
column so I cannot do

gene.info.split-as.data.frame(strsplit(RG$genes$Name,,,fixed=TRUE))

because I get the error message 

Error in data.frame(c(OligoCy3, SP Control poplar 48pin, A24, no length 
information : 
arguments imply differing number of rows: 4, 6, 5, 1

I also can't figure out how to usefully put the [list] data into a matrix (my 
ignorance I am sure).

Ideally I would be able to put each comma separated item into a column and then 
simply paste the first and second columns over the RG$genes$Name and 
RG$genes$ID columns respectively (and do away with the for loop).

Some cases in the original RG$genes$Name has only one piece of information (ie 
no commas) so I would need a way to fill any blanks with an NA value

If anyone can help me, it would be much appreciated

Nat Street



---
Nathaniel Street
 University of Southampton
 Plants and Environment Lab
  School of Biological Sciences
   Basset Crescent East
   Southampton
 SO16 7PX
   tel: +44 (0) 2380 594268
  fax: +44 (0) 2380 594269
[EMAIL PROTECTED]

    http://www.populus.biol.soton.ac.uk/~nat 
http://del.icio.us/n.r.street

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[R] barchart producing incorrect number of barcharts when columns renamed

2007-04-25 Thread Colm G. Connolly

Hi everybody,

I'm having problems with the barchart command in the lattice package.

I'm creating barcharts from matrices with with anything from 20 to 71  
columns. When I leave the column names alone, that is they are set in  
the read.table command (and inherited by subsequent commands) the  
correct number barcharts is created by the print(barchart(...))  
command. However, when I reset the column names by means of a scan  
command, the number of barcharts drawn by the same command is  
incorrect: it is always too few. The scan commands produce lists the  
same length as the number of columns for which I want barcharts.

In total I've got four pages with barcharts. The numbers in the table  
below indicate the number of barcharts per page. The numbers without  
() are the number of barcharts that I expect (and get when I don't  
reset the column titles). The numbers in () are the numbers of  
barcharts I get when I reset the column titles.

Not ClippedClipped
Errs   20 (18) 71 (46)
Stops 32 (24) 46 (36)

The following is the code used to create the barcharts with sample  
text output below it.

library('lattice')
rm(list=ls())

textFontSize=6;
pdf('../data/cocaineBarcharts.pdf', paper='a4')
fontsize=trellis.par.get(fontsize);
fontsize$text=textFontSize;
trellis.par.set(fontsize, fontsize);

resultsDirs=c(../data/Group.results.noclipping, ../data/ 
Group.results.clipped);
#resultsDirs=c(../data/Group.results.clipped);

for (resultsDir in resultsDirs) {
   cat(resultsDir, \n)
   if (any(grep(clipped, resultsDir)))
 {
   clipping=(Clipped);
 }
   else
 {
   clipping=(NOT Clipped);
 }

   roi.errs=read.table(paste(resultsDir,  
allGroupsROI.acrossGroupWithinEvent.errs, sep=/), header=T, sep=);
   roi.errs.names=names(roi.errs);
 # ctrl
   roi.errs.ctrl-roi.errs[roi.errs[,Group]==ctrl, 4:length 
(roi.errs)]
   roi.errs.ctrl.subjects=roi.errs[roi.errs[,Group]==ctrl, 2]
 # short
   roi.errs.short-roi.errs[roi.errs[,Group]==short, 4:length 
(roi.errs)]
   roi.errs.short.subjects=roi.errs[roi.errs[,Group]==short, 2]
 # long
   roi.errs.long-roi.errs[roi.errs[,Group]==long, 4:length 
(roi.errs)]
   roi.errs.long.subjects=roi.errs[roi.errs[,Group]==long, 2]


   roi.stops=read.table(paste(resultsDir,  
allGroupsROI.acrossGroupWithinEvent.stops, sep=/), header=T,  
sep=);
   roi.stops.names=names(roi.stops);
 # ctrl
   roi.stops.ctrl-roi.stops[roi.stops[,Group]==ctrl, 4:length 
(roi.stops)]
   roi.stops.ctrl.subjects=roi.stops[roi.stops[,Group]==ctrl, 2]
 # short
   roi.stops.short-roi.stops[roi.stops[,Group]==short, 4:length 
(roi.stops)]
   roi.stops.short.subjects=roi.stops[roi.stops[,Group]==short, 2]
 # long
   roi.stops.long-roi.stops[roi.stops[,Group]==long, 4:length 
(roi.stops)]
   roi.stops.long.subjects=roi.stops[roi.stops[,Group]==long, 2]


#matrixToPlot=as.matrix(roi.errs.ctrl[1:5,])
#yylim=c(floor(min(matrixToPlot)), ceiling(max(matrixToPlot)))
#barplot(matrixToPlot, col=c(2:6), beside=T, ylim=yylim,  
names.arg=colnames(roi.errs.ctrl),
#border=c(2:6), legend.text=roi.errs$Subject[1:5])

   roi.errs.ctrl.matrix=as.matrix(roi.errs.ctrl)
   roi.errs.short.matrix=as.matrix(roi.errs.short)
   roi.errs.long.matrix=as.matrix(roi.errs.long)

   roi.stops.ctrl.matrix=as.matrix(roi.stops.ctrl)
   roi.stops.short.matrix=as.matrix(roi.stops.short)
   roi.stops.long.matrix=as.matrix(roi.stops.long)

#
### errors
#

#  pdf(paste(resultsDir, 'errorsByGroup.pdf', sep=/), paper='a4')
#  fontsize=trellis.par.get(fontsize);
#  fontsize$text=textFontSize;
#  trellis.par.set(fontsize, fontsize);

   roi.errs.ctrl.means=colMeans(roi.errs.ctrl.matrix)
   roi.errs.short.means=colMeans(roi.errs.short.matrix)
   roi.errs.long.means=colMeans(roi.errs.long.matrix)
   yylim=c(floor(min(roi.errs[, 4:length(roi.errs)])), ceiling(max 
(roi.errs[, 4:length(roi.errs)])))

   errs.Means=rbind(roi.errs.ctrl.means, roi.errs.short.means,  
roi.errs.long.means)
   rownames(errs.Means)=c('control', 'short', 'long')
   cat(errs.Means dimensions before col name change , dim 
(errs.Means), \n);
   colnames(errs.Means) = scan(paste(resultsDir,  
clusterLocations.errs.csv, sep=/), sep=,, what=character)
   cat(errs.Means dimensions after col name change , dim 
(errs.Means), \n);

   print(barchart(errs.Means, groups=rownames(errs.Means), xlab='Mean  
Intensity',
  main=paste(Mean Cluster Intensity for Errors,  
clipping),
  ylab='Group', col=rainbow(3), border=rainbow(3)))
#  dev.off()

Re: [R] simulate values

2007-04-25 Thread Soare Marcian-Alin

Hello,

Yes I tried arima.sim and everything works fine. Thanks for the help!

Alin Soare

2007/4/25, Leeds, Mark (IED) [EMAIL PROTECTED]:

 he's just being a wise guy bcause I'm sure you meant to have an
 epsilon_t on the end. And he
 Surely knows that also. Did you
 Check out ?arima.sim.


 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Uwe Ligges
 Sent: Wednesday, April 25, 2007 3:15 AM
 To: Soare Marcian-Alin
 Cc: R-help@stat.math.ethz.ch
 Subject: Re: [R] simulate values



 Soare Marcian-Alin wrote:
  Hello,
 
  I want to simulate 100 values of the ARMA Process with this function:
 
  x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 *
  x[i-2] +
  0.3 * x[i-3]

 There is no kind of noise in your model, hence no need to do any
 simulation so far ...

 Uwe Ligges


 
  which possibilities do I have?
 
  Alin Soare
 
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[R] levelplot and unequal cell sizes

2007-04-25 Thread Waichler, Scott R

I am using levelplot() from lattice with grids that have unequal cell
sizes.  This means that the boundary between two cells is not always
half-way between nodes, as levelplot() assumes.  The result is that some
cell sizes are rendered incorrectly, which can be painfully obvious if
using relatively large cells.  Is there any work-around?  I am using the
conditioning capability of lattice and therefore image() would not be a
good way to go.

Thanks, 
Scott Waichler
Pacific Northwest National Laboratory
scott.waichler _at_ pnl.gov

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Re: [R] assigning two conditions to grep()

2007-04-25 Thread Abi Ghanem josephine

janek0 wrote:

Dnia 25-04-2007, śro o godzinie 15:04 +0200, Abi Ghanem josephine
napisał(a):

use data[grep(^24|N7, data$V2)]
and see ?regexp
  


thanks for replying
but actually my problem is that the column containing 23 is in data$V2 
and the one containing N7 is in data$V3,
so the line doen't work i just have all the line containing G23 N7  and 
G23 O6
and i want to separate the two.

josephine

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Re: [R] R CMD CHECK and require() / library() methods

2007-04-25 Thread Prof Brian Ripley

On Wed, 25 Apr 2007, Crispin Miller wrote:

 Hi,

 I have a piece of code that decides at runtime whether to load a data
 package (and which package to load).
 This is then done with a call to:

 library(x)

 (where x is a character variable containing the package name).

 This causes R CMD check to throw out a warning:
 'library' or 'required' calls not declared from:
 x

Which version of R is this?  All I can find say 'require'.


 Does anyone have any suggestions as to a fix or workaround for this?

That call should be

library(x, character.only=TRUE)

and that will in R 2.5.0 stop the warning AFAIK.

-- 
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] How to solve difficult equations?

2007-04-25 Thread Robert A LaBudde

At 03:15 AM 4/25/2007, francogrex wrote:

This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points not of
opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function in R that can
solve this? thanks.

Actually, if you're solving fn(a)==0, then some trivial algebra leads 
to a linear equation with a=0.001754.

Why are you trying to solve this numerically? Is it a single instance 
of a larger, more general problem?


Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: [EMAIL PROTECTED]
Least Cost Formulations, Ltd.URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239Fax: 757-467-2947

Vere scire est per causas scire

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Re: [R] creating random numbers

?sample
sample(c(1,2),10,replace=TRUE)

P.

raymond chiruka napsal(a):
  l want to create a  column of 1 and 2 randomly what command should l use 
   eg
   treatment  strata
   
   1  1
   2  0
   1  1
   2  1
   2  0  
   2  1
   2  0
   1  0
   these should be created randomly
   
   secondly if l have something like 
   for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l add an if 
 statement in  side the c maybe to compare the 2 exponential  numbers to 
 create  another  variable
 

 -
 
 
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-- 
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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Re: [R] R CMD CHECK and require() / library() methods

2007-04-25 Thread Crispin Miller

Many thanks! 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Prof 
 Brian Ripley
 Sent: 25 April 2007 16:03
 To: Crispin Miller
 Cc: R-help@stat.math.ethz.ch
 Subject: Re: [R] R CMD CHECK and require() / library() methods

 On Wed, 25 Apr 2007, Crispin Miller wrote:

  Hi,

  I have a piece of code that decides at runtime whether to 
 load a data 
  package (and which package to load).
  This is then done with a call to:

  library(x)

  (where x is a character variable containing the package name).

  This causes R CMD check to throw out a warning:
  'library' or 'required' calls not declared from:
  x

 Which version of R is this?  All I can find say 'require'.

My mistake - it was a typo it says: 'require'

  Does anyone have any suggestions as to a fix or workaround for this?

 That call should be

 library(x, character.only=TRUE)

 and that will in R 2.5.0 stop the warning AFAIK.

It works - much appreciated...

Crispin

This email is confidential and intended solely for the use o...{{dropped}}

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Re: [R] regular expressions with grep() and negative indexing

2007-04-25 Thread Peter Dalgaard

Peter Dalgaard wrote:
 Stephen Tucker wrote:
   
 Dear R-helpers,

 Does anyone know how to use regular expressions to return vector elements
 that don't contain a word? For instance, if I have a vector
   x - c(seal.0,seal.1-exclude)
 I'd like to get back the elements which do not contain the word exclude,
 using something like (I know this doesn't work) but:
   grep([^(exclude)],x)

 I can use 
   x[-grep(exclude,x)]
 for this case but then if I use this expression in a recursive function, it
 will not work for instances in which the vector contains no elements with
 that word. For instance, if I have
   x2 - c(dolphin.0,dolphin.1)
 then
   x2[-grep(exclude,x2)]
 will give me 'character(0)'

 I know I can accomplish this in several steps, for instance:
   myfunc - function(x) {
 iexclude - grep(exclude,x)
 if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
 # do stuff with x2 ...?
   }

 But this is embedded in a much larger function and I am trying to minimize
 intermediate variable assignment (perhaps a futile effort). But if anyone
 knows of an easy solution, I'd appreciate a tip.
   
 
 It has come up a couple of times before, and yes, it is a bit of a pain.

 Probably the quickest way out is

 negIndex - function(i) 

if(length(i))

-i 

else 

TRUE

   
... which of course needs braces if typed on the command line

negIndex - function(i) 
{
   if(length(i))
   -i 
   else 
   TRUE
}

And I should probably also have said that it works like this:

 x2 - c(dolphin.0,dolphin.1)
 x2[-grep(exclude,x2)]
character(0)
 x2[negIndex(grep(exclude,x2))]
[1] dolphin.0 dolphin.1



-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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Re: [R] assigning two conditions to grep()

2007-04-25 Thread chris

On 25/04/07, Abi Ghanem josephine [EMAIL PROTECTED] wrote:

 Hi,
 i have a problem in assigning 2 conditions to grep()  ,
 my data look like this:
 DA 24 N7 Rad= 3.4 20 Sac= 0.93 Acc= 4.76
 DA 24 N7 Rad= 3.4 14 Sac= 0.65 Acc= 3.33
 DA 24 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
 DA 24 N7 Rad= 3.4 11 Sac= 0.51 Acc= 2.62
 DG 23 N7 Rad= 3.4  8 Sac= 0.37 Acc= 1.91
 DG 23 N7 Rad= 3.4  5 Sac= 0.23 Acc= 1.19
 DG 23 N7 Rad= 3.4  0 Sac= 0.00 Acc= 0.00
 DG 23 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
 DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
 DG 23 O6 Rad= 3.3  1 Sac= 0.04 Acc=  0.22
 DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
 DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
 (it's a data.frame)

 at first i wanted all the line begining with A 24:
 data[grep(^24, data$V2)]
 this works
 and than i wanted to exctract all the lines with G23 N7,
 neither the column 23 and the column N7 are unique
 so i tried this
 data[grep(^23*N7, data),]
 but doesn't work



how about
data[ intersect( grep(^24, data$V2), grep(N7,data$V3) ) , ]
?

C.

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[R] Getting Confused

2007-04-25 Thread Steiner, Julien

Hello,

 

I'm getting confused with my experience of R installing.

 

I had R installed on January without any trouble. (I just had to install
gcc4.1.1) 

 

Now I'd like to install a packages which requires tcl/tk. So basically I
need to reconfigure and re install R right after having installed
tcl/tk.

 

So I installed tcl/tk I run the process to install R but I receive this
error : 

 

checking for dummy main to link with Fortran libraries...

none

checking for Fortran name-mangling scheme... configure:

error: cannot compile a simple Fortran program See `config.log' for more
details.

 

 

I checked in the config.log and the fact is that there's no fortran
compiler installed. But don't gcc already have a fortran compiler in
it? 

 

 

If somebody could help I would be thankful and especially if somebody
has a clue why it worked without any error before and now yes.

 

 

 

Thanks a lot

 

 

 

Julien Steiner


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[R] Box Ljung Statistics

2007-04-25 Thread gyadav


Hi All R Experts,

I met with below mentioned statistics in paper Stock Index Volatility 
Forecasting with High Frequency Data
by Eugenie Hol, Siem Jan Koopman 
http://ideas.repec.org/p/dgr/uvatin/20020068.html

I would like to ask that what is Box-Ljung portmantacau statistic based 
on N squared autocorrelation ?
Is it same as Box-Ljung Statistics of stats package ?
Further, please tell me how to compute it ?

I have a return series of an Index.
Please help me in this i am not able to get the statistics what is given 
in the paper for S  P 100:-)

   Sayonara With Smile  With Warm Regards :-)

  G a u r a v   Y a d a v
  Assistant Manager,
  Economic Research  Surveillance Department,
  Clearing Corporation Of India Limited.

  Address: 5th, 6th, 7th Floor, Trade Wing 'C',  Kamala City, S.B. Marg, 
Mumbai - 400 013
  Telephone(Office): - +91 022 6663 9398 ,  Mobile(Personal) (0)9821286118
  Email(Office) :- [EMAIL PROTECTED] ,  Email(Personal) :- 
[EMAIL PROTECTED]



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[R] Unsubscription Confirmation

2007-04-25 Thread Egoldsystem

Thank you for subscribing. You have now unsubscribed and no more messages will 
be sent.

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Re: [R] NA and NaN randomForest

2007-04-25 Thread clayton . springer

Hi Andy,

It worked for classification, but not regression. For example:

 iris1 - iris[-(1:5),]
 iris2 - iris[(1:5),]
 iris2[1,3] - NA
 iris2[3,1] - NA
 iris_sum - sum (iris$Sepal.Length + iris$Sepa.Width + iris$Petal.Length 
+ iris$Petal.Width)
 iris_sum1 -  iris_sum[-(1:5)]
 iris_sum2 -  iris_sum[(1:5)]
 iris_sum.rf - randomForest (iris_sum1 ~ ., iris1[,c(1:4)])
   predict (iris_sum.rf, iris2[-5])
   predict (iris_sum.rf, iris2[-5])
[1]  9.556591  9.589573 10.104155

# Just to be clear I was hoping for behavior like the linear model has:

 iris_sum.lm - lm (iris_sum1 ~ ., iris1[,c(1:4)])
  predict (iris_sum.lm, iris2[-5])
   12345 
  NA  9.5   NA  9.4 10.2 

In the event that this is not available in the regression part of 
randomForest, is a work around possible?



thanks,

Clayton




Liaw, Andy [EMAIL PROTECTED] 
04/25/2007 09:59 AM

To
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
cc

Subject
RE: [R] NA and NaN randomForest






Hi Clayton,

If you use the formula interface, then it should do what you want:

R library(randomForest)
randomForest 4.5-18 
Type rfNews() to see new features/changes/bug fixes.
R iris1 - iris[-(1:5),]
R iris2 - iris[1:5,]
R iris2[1, 3] - NA
R iris2[3, 1] - NA
R iris.rf - randomForest(Species ~ ., iris1)
R predict(iris.rf, iris2[-5])
[1] NA   setosa NA   setosa setosa
Levels: setosa versicolor virginica

The problem, of course, is that the formula interface is not suitable
for data with large number of variables.  I'll look into doing the same
thing in the default method.

Andy


From: [EMAIL PROTECTED]
 
 Dear R-help,
 
 This is about randomForest's handling of NA and NaNs in test set data.
 Currently, if the test set data contains an NA or NaN then 
 predict.randomForest will skip that row in the output.
 
 I would like to change that behavior to outputting an NA.
 
 Can this be done with flags to randomForest?
 If not can some sort of wrapper be built to put the NAs back in?
 
 thanks,
 
 Clayton
 _
 
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Re: [R] regular expressions with grep() and negative indexing

2007-04-25 Thread Tony Plate

I use regexpr() instead of grep() in cases like this, e.g.:

x2[regexpr(exclude,x2)==-1]

(regexpr returns a vector of the same length as character vector given 
it, so there's no problem with it returning a zero length vector)

-- Tony Plate

Peter Dalgaard wrote:
 Stephen Tucker wrote:
 Dear R-helpers,

 Does anyone know how to use regular expressions to return vector elements
 that don't contain a word? For instance, if I have a vector
   x - c(seal.0,seal.1-exclude)
 I'd like to get back the elements which do not contain the word exclude,
 using something like (I know this doesn't work) but:
   grep([^(exclude)],x)

 I can use 
   x[-grep(exclude,x)]
 for this case but then if I use this expression in a recursive function, it
 will not work for instances in which the vector contains no elements with
 that word. For instance, if I have
   x2 - c(dolphin.0,dolphin.1)
 then
   x2[-grep(exclude,x2)]
 will give me 'character(0)'

 I know I can accomplish this in several steps, for instance:
   myfunc - function(x) {
 iexclude - grep(exclude,x)
 if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
 # do stuff with x2 ...?
   }

 But this is embedded in a much larger function and I am trying to minimize
 intermediate variable assignment (perhaps a futile effort). But if anyone
 knows of an easy solution, I'd appreciate a tip.
   
 It has come up a couple of times before, and yes, it is a bit of a pain.
 
 Probably the quickest way out is
 
 negIndex - function(i) 
 
if(length(i))
 
-i 
 
else 
 
TRUE


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[R] R News, volume 7, issue 1 is now available

2007-04-25 Thread Torsten Hothorn


Hi

The October 2006 issue of R News is now available on CRAN under the
Documentation/Newsletter link.

Torsten
(on behalf of the R News Editorial Board)

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Re: [R] How to identify and exclude the outliers with R?

2007-04-25 Thread Horace Tso

It depends on the nature of your data set. There is a package simply called 
'outliers', which has the Grubbs/Dixon/Cochran tests. There is also the 
Bonferroni outlier test in 'car' package. I'm sure there are more in the 
hundreds of packages on CRAN.

HTH

Horace

 Shao [EMAIL PROTECTED] 4/25/2007 6:27:37 AM 
Hello, everyone,

I want to ask a simple question.
If I have a set  of data,and I want to identify how many outliers there are
in the data.Which packages and functions can I use?

Thanks.

Shao chunxuan.

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Re: [R] Getting Confused [Broadcast]

2007-04-25 Thread Liaw, Andy

If you are serious in getting useful help, please do try to follow
suggestions in the Posting Guide.  You have not told us anything about
your OS, the versions of R you tried to install, and exactly what you
typed to build/install them.

Many Linux distro by default do not install the Fortran part of GCC, so
don't be surprised if that's the case for you (if you are trying to do
this on some version of Linux).

Andy

From: Steiner, Julien
 
 Hello,
 
  
 
 I'm getting confused with my experience of R installing.
 
  
 
 I had R installed on January without any trouble. (I just had 
 to install
 gcc4.1.1) 
 
  
 
 Now I'd like to install a packages which requires tcl/tk. So 
 basically I
 need to reconfigure and re install R right after having installed
 tcl/tk.
 
  
 
 So I installed tcl/tk I run the process to install R but I 
 receive this
 error : 
 
  
 
 checking for dummy main to link with Fortran libraries...
 
 none
 
 checking for Fortran name-mangling scheme... configure:
 
 error: cannot compile a simple Fortran program See 
 `config.log' for more
 details.
 
  
 
  
 
 I checked in the config.log and the fact is that there's no fortran
 compiler installed. But don't gcc already have a fortran compiler in
 it? 
 
  
 
  
 
 If somebody could help I would be thankful and especially if somebody
 has a clue why it worked without any error before and now yes.
 
  
 
  
 
  
 
 Thanks a lot
 
  
 
  
 
  
 
 Julien Steiner
 
 
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Re: [R] R News, volume 7, issue 1 is now available

2007-04-25 Thread Earl F. Glynn

Torsten Hothorn [EMAIL PROTECTED] wrote in message 
news:[EMAIL PROTECTED]

 The October 2006 issue of R News is now available on CRAN under the
 Documentation/Newsletter link.

Direct links are useful:

R News
http://cran.r-project.org/doc/Rnews/

April 2007 Issue:
http://cran.r-project.org/doc/Rnews/Rnews_2007-1.pdf

efg

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Re: [R] Problem installing Rmpi with lam on SGI SLES9

2007-04-25 Thread Martin Morgan

Hendrik,

Are you starting the lam daemons before starting R?

% lamboot

You might need to specify a 'hosts' argument to lamboot. The default
way Rmpi calls lamboot is with no arguments, and this might simply
create a single lam daemon.

I don't usually use papply, but glancing at it's code suggests that it
does require(Rmpi) and then decides based on the result of
mpi.comm.size what to do. So to debug, load Rmpi and try
mpi.comm.size. As a work-around, I think it should be possible to

 library(Rmpi)
 mpi.spawn.Rslaves(nslaves=64) # maybe a little bold, initially!

before making your first papply call.

Hope that helps

Martin

Hendrik Fuß [EMAIL PROTECTED] writes:

 On 24/04/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
 On Tue, 24 Apr 2007, Hendrik Fuß wrote:

  Hi,
 
  I've been trying here to install Rmpi on an SGI IA-64 machine with 64
  processors, running SuSE Linux Enterprise Server 9, R 2.4.0 and
  lam-mpi 7.1.3. While I've read of similar problems on this list, I
  think I've got an entirely new set of error messages to contribute
  (see below). I'm not sure what the actual error is and what the @gprel
  relocation message is about. Any help greatly appreciated.

 I don't know for sure, but on many 64-bit OSes you cannot link code from
 static libraries into dynamic shared libraries, and that seems to be the
 case with ia64 Linux.  Almost certainly you need to re-compile LAM with
 -fPIC flags.

 Yes, thanks a million, this solved the problem.

 While Rmpi now works, there is another issue connected with this one:
 I'm trying to use the papply package. However, when I do:

 library(papply)
 papply(list(1:10, 1:15, 1:20), sum)
 1 slaves are spawned successfully. 0 failed.
 master (rank 0, comm 1) of size 2 is running on: behemoth
 slave1 (rank 1, comm 1) of size 2 is running on: behemoth
 [1] Running serial version of papply\n

 Papply only spawns one slave and then decides to run the serial
 version instead. I'm not sure how to tell it to use all the 64
 processors available.

 Hendrik

 -- 
 Hendrik Fuß
 PhD student
 Systems Biology Research Group

 University of Ulster, School of Biomedical Sciences
 Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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-- 
Martin Morgan
Bioconductor / Computational Biology
http://bioconductor.org

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Re: [R] levelplot and unequal cell sizes

2007-04-25 Thread hadley wickham

On 4/25/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
 I am using levelplot() from lattice with grids that have unequal cell
 sizes.  This means that the boundary between two cells is not always
 half-way between nodes, as levelplot() assumes.  The result is that some
 cell sizes are rendered incorrectly, which can be painfully obvious if
 using relatively large cells.  Is there any work-around?  I am using the
 conditioning capability of lattice and therefore image() would not be a
 good way to go.

You might be able to use the tile plot in ggplot, which allows you to
specify the size of each tile (it assumes they're all the same size by
default).  Have a look at ?ggtile, or if you provide more info about
your data, I could provide a worked example.

Regards,

Hadley

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Re: [R] R News, volume 7, issue 1 is now available

2007-04-25 Thread Romain Francois

Earl F. Glynn wrote:
 Torsten Hothorn [EMAIL PROTECTED] wrote in message 
 news:[EMAIL PROTECTED]
   
 The October 2006 issue of R News is now available on CRAN under the
 Documentation/Newsletter link.
 

 Direct links are useful:

 R News
 http://cran.r-project.org/doc/Rnews/

 April 2007 Issue:
 http://cran.r-project.org/doc/Rnews/Rnews_2007-1.pdf

 efg
   
... And maybe the TOC:

Editorial

Viewing Binary Files with the hexView Package
FlexMix: An R Package for Finite Mixture Modelling
Using R to Perform the AMMI Analysis on Agriculture Variety Trials
Inferences for Ratios of Normal Means
Working with Unknown Values
A New Package for Fitting Random Effect Models
Augmenting R with Unix Tools
POT: Modelling Peaks Over a Threshold
Backtests

Review of John Verzani’s Book Using R for Introductory Statistics
DSC 2007
New Journal: Annals of Applied Statistics
Forthcoming Events: useR! 2007
Changes in R 2.5.0
Changes on CRAN
R Foundation News
R News Referees 2006


-- 
Mango Solutions
data analysis that delivers

Tel:  +44(0) 1249 467 467
Fax:  +44(0) 1249 467 468
Mob:  +44(0) 7813 526 123

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Re: [R] Coercing data types for use in model.frame

2007-04-25 Thread Frank E Harrell Jr

Prof Brian Ripley wrote:
 Moved to R-devel 
 
 What is the 'data class'?  In particular what is its underlying type? 
 And where in model.frame[.default] are you trying to use it (in the 
 formula, data, in ..., etc).
 
 This is an example of where some reproducible code and the error 
 messages would be very helpful indeed.
 
 Brian

Brian,

Sorry - this was one of those too late in the day errors.  The problem 
was in a function called just before model.frame.  model.frame seems to 
work fine with an object of class c('mChoice', 'labelled').  It keeps 
mChoice variables as mChoice.  After model.frame is finished I'll change 
such variables to factors or matrices.

Frank


 
 On Tue, 24 Apr 2007, Frank E Harrell Jr wrote:
 
 In the Hmisc package there is a new data class 'mChoice' for multiple
 choice variables.  There are format and as.numeric methods (the latter
 creates a matrix of dummy variables).  mChoice variables are not allowed
 by model.frame.  Is there a way to specify a conversion function that
 model.frame will use automatically?  I would use as.factor here.
 model.frame does not seem to use as.data.frame.foo for individual 
 variables.

 Thanks
 Frank

 


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

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Re: [R] Box Ljung Statistics

2007-04-25 Thread Rogerio Porto

Gaurav,

 I met with below mentioned statistics in paper Stock Index Volatility 
 Forecasting with High Frequency Data
 by Eugenie Hol, Siem Jan Koopman 
 http://ideas.repec.org/p/dgr/uvatin/20020068.html
 
 I would like to ask that what is Box-Ljung portmantacau statistic based 
 on N squared autocorrelation ?
 Is it same as Box-Ljung Statistics of stats package ?

Yes, it seems the same. But note that the paper computes the
statistic for the raw data and also for the squared data.

 Further, please tell me how to compute it ?

If you mean R, use the Box.test() function. If you mean theory,
see any good book on time series like Brockwell and Davis'
Introduction to Time Series and Forecasting.

 I have a return series of an Index.
 Please help me in this i am not able to get the statistics what is given 
 in the paper for S  P 100:-)

I can't help you here since I don't have the 5-minute data used in
the paper.

Rogerio

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[R] Scripting graph generation

2007-04-25 Thread Mike Huber

Hi,

I'm looking to automate the generation of some graphs in R.  I can't seem to
figure out how to script R, and change the output device of hist() or plot()
to create a .gif or .png file.  This seems like something that is probably
really simple, and I've just overlooked the call do do it.  Can anyone point
me in the right direction, or maybe send a sample script?

thanks,
--Mike Huber

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[R] GLS terminology question not related to R

2007-04-25 Thread Leeds, Mark \(IED\)

This is a terminology question not related to R. The literature often
says that OLS is inefficient relative to GLS if the residuals in
the system are correlated ( and the RHS sides of each are not identical
). Does this mean that OLS overestimates residual and coefficient
variances , underestimates them or just gets them wrong and the
direction is not known ? Thanks.


This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}

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[R] [R-pkgs] new package adegenet

2007-04-25 Thread Thibaut Jombart

The new package *adegenet* (linked to the ade4 package for multivariate 
analysis) has been released on CRAN.

Its main focus is on molecular marker data handling for multivariate 
analysis.

Adegenet offers data import/export functions (from GENETIX, Genepop, 
Fstat and to the packages genetics and hierfstat) as well as several 
data handling tools. It aims at facilitating the access to multivariate 
methods as well as to usual population genetics methods (HWE tests, 
G-statistic tests, ...).

Suggestions, questions and contributions are welcome !

More information is available on the adegenet website: 
http://pbil.univ-lyon1.fr/software/adegenet/. 
http://pbil.univ-lyon1.fr/software/adegenet/

Regards,

Thibaut Jombart.
-- 
##
Thibaut JOMBART
CNRS UMR 5558 - Laboratoire de Biométrie et Biologie Evolutive
Universite Lyon 1
43 bd du 11 novembre 1918
69622 Villeurbanne Cedex
Tél. : 04.72.43.29.35
Fax : 04.72.43.13.88
[EMAIL PROTECTED]
http://biomserv.univ-lyon1.fr/sitelabo/pageperso.php?id_personne=178

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Re: [R] GLS terminology question not related to R

2007-04-25 Thread Thomas Lumley

On Wed, 25 Apr 2007, Leeds, Mark (IED) wrote:

 This is a terminology question not related to R. The literature often
 says that OLS is inefficient relative to GLS if the residuals in
 the system are correlated ( and the RHS sides of each are not identical
 ). Does this mean that OLS overestimates residual and coefficient
 variances , underestimates them or just gets them wrong and the
 direction is not known ? Thanks.

It does not mean either.

It means that the true variance of the OLS estimates is greater than the 
true variance of the GLS estimates.

A separate issue is whether the estimated variance of an OLS estimator is 
greater or less than the true variance of the OLS estimator.  This can go 
either way.

-thomas

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Re: [R] levelplot and unequal cell sizes

2007-04-25 Thread Deepayan Sarkar

On 4/25/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
 I am using levelplot() from lattice with grids that have unequal cell
 sizes.  This means that the boundary between two cells is not always
 half-way between nodes, as levelplot() assumes.

levelplot() is not supposed to make any such assumptions. Can you
provide a reproducible example please?

-Deepayan

 The result is that some
 cell sizes are rendered incorrectly, which can be painfully obvious if
 using relatively large cells.  Is there any work-around?  I am using the
 conditioning capability of lattice and therefore image() would not be a
 good way to go.

 Thanks,
 Scott Waichler
 Pacific Northwest National Laboratory
 scott.waichler _at_ pnl.gov

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[R] help

2007-04-25 Thread Natalie O'Toole

Hi all,

I have 2 questions:

1)How do I calculate the mean on an imported txt file? I've imported the 
file below and that's what it looks like imported. How do I then calcuate 
the mean, median, or mode on the column LeafArea using the desktop R 
package?

Any help would be greatly appreciated!!

Thanks,

Nat

   LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness
1 13 0.12   0.12   0.21  0.00
2 13 0.17   0.17   0.36  0.00
3 13 0.21   0.05   0.47  0.16
4 13 0.11   0.14   0.23  0.21
5 23 0.03   0.27   0.16  0.60
6 23 0.08   0.20   0.15  0.75
7 23 0.22   0.05   0.24  1.09
8 23 0.20   0.10   0.26  1.30
9 23 0.18   0.10   0.33  1.33
1023 0.14   0.07   0.19  1.36
1123 0.16   0.13   0.22  1.38
1223 0.18   0.06   0.25  1.39
1323 0.05   0.00   0.07  1.40
1423 0.11   0.01   0.21  1.41
1523 0.22   0.04   0.31  1.41
1623 0.09   0.10   0.13  1.44
1723 0.09   0.10   0.13  1.44
1823 0.13   0.08   0.19  1.46
1923 0.15   0.13   0.22  1.47
2023 0.15   0.03   0.22  1.47
2123 0.21   0.01   0.31  1.48
2213 0.21   0.14   0.32  1.50
2323 0.10   0.00   0.15  1.50
2413 0.26   0.60   0.40  1.53
2523 0.12   0.18   0.20  1.54
2623 0.20   0.15   0.31  1.55
2713 0.19   0.16   0.31  1.60
2813 0.13   0.00   0.21  1.62
2913 0.13   0.01   0.21  1.62
3013 0.37   0.27   0.60  1.62
3123 0.11   0.09   0.18  1.64
3223 0.14   0.00   0.23  1.64
3323 0.15   0.08   0.21  1.64
3423 0.20   0.10   0.33  1.65
3523 0.15   0.01  -0.25  1.67
3623 0.17   0.06   0.29  1.67
3723 0.13   0.08   0.22  1.69
3813 0.16   0.31   0.27  1.70
3913 0.21   0.01   0.40  1.70
4013 0.14   0.07   0.29  1.71
4123 0.14   0.00   0.24  1.71
4223 0.21   0.14   0.35  1.71
4323 0.11   0.09   0.19  1.73
4423 0.15   0.01   0.26  1.73
4523 0.19   0.11   0.33  1.74
4610 0.28   0.27   0.50  1.79
4713 0.10   0.01   0.18  1.80
4823 0.05   0.00   0.09  1.80
4913 0.12   0.11   0.22  1.83
5013 0.20   0.05   0.37  1.85
5123 0.14   0.14   0.26  1.86
5213 0.15   0.07   0.28  1.87
5313 0.15   0.01   0.28  1.87
5423 0.12   0.08   0.23  1.92
5523 0.15   0.00   0.29  1.93
5613 0.17   0.00   0.34  2.00
5713 0.21   0.02   0.42  2.00
5823 0.13   0.08   0.26  2.00
5913 0.16   0.06   0.32  2.05
6013 0.14   0.14   0.29  2.07
6123 0.12   0.08   0.25  2.08
6213 0.17   0.06   0.36  2.12
6313 0.13   0.08   0.28  2.13
6413 0.20   0.10   0.43  2.15
6513 0.26   0.08   0.56  2.15
6613 0.20   0.10   0.44  2.20
6713 0.19   0.11   0.42  2.21
6813 0.08   0.00   0.18  2.25
6913 0.12   0.00   0.27  2.25
7013 0.12   0.08   0.27

Re: [R] Scripting graph generation

2007-04-25 Thread Rajarshi Guha

On Wed, 2007-04-25 at 13:21 -0400, Mike Huber wrote:
 Hi,
 
 I'm looking to automate the generation of some graphs in R.  I can't seem to
 figure out how to script R, and change the output device of hist() or plot()
 to create a .gif or .png file.  This seems like something that is probably
 really simple, and I've just overlooked the call do do it.  Can anyone point
 me in the right direction, or maybe send a sample script?

To plot multiple PNG files

m - matrix(runif(100*10), nrow=10)
png(filename = 'plot%03d.png')
for (i in 1:nrow(m)) {
  plot(m[i,])
}
dev.off()

To plot multiple graphs as individual pages of a PDF file

m - matrix(runif(100*10), nrow=10)
pdf(file = 'plot.pdf')
for (i in 1:nrow(m)) {
  plot(m[i,])
}
dev.off()

HTH,

---
Rajarshi Guha [EMAIL PROTECTED]
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
In matrimony, to hesitate is sometimes to be saved.
-- Butler

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[R] Самый доступный и недорогой метод похудения

2007-04-25 Thread Francesco Barnett

«ÇÎËÎÒÀß ÑÅÐÜÃÀ» - ÈÇÁÀÂËÅÍÈÅ ÎÒ ËÈØÍÈÕ ÊÈËÎÃÐÀÌÌÎÂ (4 ÊÃ Â ÌÅÑ.)!

ÌÈËËÈÎÍÛ ËÞÄÅÉ ÈÇÁÀÂÈËÈÑÜ ÎÒ ËÈØÍÈÕ ÊÈËÎÃÐÀÌÌÎÂ ÁËÀÃÎÄÀÐß ÇÎËÎÒÎÉ_ÑÅÐÜÃÅ
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[R] dmnorm not meant for 1024-dimensional data?

2007-04-25 Thread Daniel Elliott

Hello,

I have some data generated by a simple mixture of Gaussians (more like
K-means) and (as a test) am using dmnorm to calculate the probability
of each data point coming from each Gaussian.  However, I get only
zero probabilities.

This code works in low dimensions (tried 2 and 3 already).  I have run
into many implementations that do not work in high dimension, but I
thought that I was safe with dmnorm because it has an option to
compute the log of the probability.

So, is dmnorm not intended to be used with data of such high dimensionality?

Thank you,

dan elliott

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Re: [R] negative number to positive number

2007-04-25 Thread J . delasHeras

Quoting H. Paul Benton [EMAIL PROTECTED]:

 Hello all,

 I know this is a pretty easy question but I can't find it in S poetry or
 R help.

 How can I make a negative number positive. Such as
 -5 to be +5
 I tried +(-5), but that didn't work.

 So no, I don't mean taking a -5^2 just to get a positive number.
 This is in a function so it's not just -5 it's x. :)

 Thanks,

 Paul

how about just multiplying it by -1??? :-)

-5*(-1)

Jose

-- 
Dr. Jose I. de las Heras  Email: [EMAIL PROTECTED]
The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374
Institute for Cell  Molecular BiologyFax:   +44 (0)131 6507360
Swann Building, Mayfield Road
University of Edinburgh
Edinburgh EH9 3JR
UK

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[R] Help on 'find.BIB' function

2007-04-25 Thread Jason Parcon

Hello everyone,
   
  I am trying to use the 'find.BIB' function to construct a balanced incomplete 
block design.  When I ran the example given in the help file 
(find.BIB(10,30,4)), I obtained the following error message:
   
  Error in optBlock(~., withinData = factor(1:trt), blocksize = rep(k, b)) : 
object .Random.seed not found

  I investigated what the optBlock function is doing but the manual / help 
files did not give me any information regarding the above error.
   
  I hope somebody can help me regarding this matter.
   
  Best regards,
   
  Jason Parcon
   

   
-


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Re: [R] barchart producing incorrect number of barcharts when columns renamed

2007-04-25 Thread Deepayan Sarkar

On 4/25/07, Colm G. Connolly [EMAIL PROTECTED] wrote:
 Hi everybody,

 I'm having problems with the barchart command in the lattice package.

 [... ... ...]

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You seem to have missed this footer that appears in every r-help
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long, shot.

-Deepayan

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Re: [R] Help on 'find.BIB' function

2007-04-25 Thread Chuck Cleland

Jason Parcon wrote:
 Hello everyone,

   I am trying to use the 'find.BIB' function to construct a balanced 
 incomplete block design.  When I ran the example given in the help file 
 (find.BIB(10,30,4)), I obtained the following error message:

   Error in optBlock(~., withinData = factor(1:trt), blocksize = rep(k, b)) : 
 object .Random.seed not found
 
   I investigated what the optBlock function is doing but the manual / help 
 files did not give me any information regarding the above error.

   I hope somebody can help me regarding this matter.

  The following seems to work for me:

 library(crossdes)
Loading required package: AlgDesign
Loading required package: gtools
Loading required package: MASS

 set.seed(671969)

 find.BIB(10,30,4)
  [,1] [,2] [,3] [,4]
 [1,]457   10
 [2,]123   10
 [3,]156   10
 [4,]289   10
 [5,]3567
 [6,]349   10
 [7,]1589
 [8,]1679
 [9,]1247
[10,]268   10
[11,]2357
[12,]1679
[13,]267   10
[14,]1239
[15,]2568
[16,]2459
[17,]3468
[18,]158   10
[19,]2478
[20,]369   10
[21,]1246
[22,]378   10
[23,]2359
[24,]145   10
[25,]4689
[26,]479   10
[27,]1378
[28,]3456
[29,]5789
[30,]1348

  I get the same error you report if I don't do the set.seed() step.

 sessionInfo()
R version 2.4.1 Patched (2007-03-31 r41127)
i386-pc-mingw32

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

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

other attached packages:
 crossdes  MASSgtools AlgDesign
  1.0-7  7.2-33   2.3.1   1.0-7

   Best regards,

   Jason Parcon

 

 -
 
 
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-- 
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NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
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tel: (732) 512-0171 (M, W, F)
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[R] help

2007-04-25 Thread Natalie O'Toole

Hi,

Would anyone know how to calculate the modal value of LeafArea?

Thank-you very much!!

Nat

__

Hi all,

I have 2 questions:

1)How do I calculate the mean on an imported txt file? I've imported the 
file below and that's what it looks like imported. How do I then calcuate 
the mean, median, or mode on the column LeafArea using the desktop R 
package?

Any help would be greatly appreciated!!

Thanks,

Nat

   LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness
1 13 0.12   0.12   0.21  0.00
2 13 0.17   0.17   0.36  0.00
3 13 0.21   0.05   0.47  0.16
4 13 0.11   0.14   0.23  0.21
5 23 0.03   0.27   0.16  0.60
6 23 0.08   0.20   0.15  0.75
7 23 0.22   0.05   0.24  1.09
8 23 0.20   0.10   0.26  1.30
9 23 0.18   0.10   0.33  1.33
1023 0.14   0.07   0.19  1.36
1123 0.16   0.13   0.22  1.38
1223 0.18   0.06   0.25  1.39
1323 0.05   0.00   0.07  1.40
1423 0.11   0.01   0.21  1.41
1523 0.22   0.04   0.31  1.41
1623 0.09   0.10   0.13  1.44
1723 0.09   0.10   0.13  1.44
1823 0.13   0.08   0.19  1.46
1923 0.15   0.13   0.22  1.47
2023 0.15   0.03   0.22  1.47
2123 0.21   0.01   0.31  1.48
2213 0.21   0.14   0.32  1.50
2323 0.10   0.00   0.15  1.50
2413 0.26   0.60   0.40  1.53
2523 0.12   0.18   0.20  1.54
2623 0.20   0.15   0.31  1.55
2713 0.19   0.16   0.31  1.60
2813 0.13   0.00   0.21  1.62
2913 0.13   0.01   0.21  1.62
3013 0.37   0.27   0.60  1.62
3123 0.11   0.09   0.18  1.64
3223 0.14   0.00   0.23  1.64
3323 0.15   0.08   0.21  1.64
3423 0.20   0.10   0.33  1.65
3523 0.15   0.01  -0.25  1.67
3623 0.17   0.06   0.29  1.67
3723 0.13   0.08   0.22  1.69
3813 0.16   0.31   0.27  1.70
3913 0.21   0.01   0.40  1.70
4013 0.14   0.07   0.29  1.71
4123 0.14   0.00   0.24  1.71
4223 0.21   0.14   0.35  1.71
4323 0.11   0.09   0.19  1.73
4423 0.15   0.01   0.26  1.73
4523 0.19   0.11   0.33  1.74
4610 0.28   0.27   0.50  1.79
4713 0.10   0.01   0.18  1.80
4823 0.05   0.00   0.09  1.80
4913 0.12   0.11   0.22  1.83
5013 0.20   0.05   0.37  1.85
5123 0.14   0.14   0.26  1.86
5213 0.15   0.07   0.28  1.87
5313 0.15   0.01   0.28  1.87
5423 0.12   0.08   0.23  1.92
5523 0.15   0.00   0.29  1.93
5613 0.17   0.00   0.34  2.00
5713 0.21   0.02   0.42  2.00
5823 0.13   0.08   0.26  2.00
5913 0.16   0.06   0.32  2.05
6013 0.14   0.14   0.29  2.07
6123 0.12   0.08   0.25  2.08
6213 0.17   0.06   0.36  2.12
6313 0.13   0.08   0.28  2.13
6413 0.20   0.10   0.43  2.15
6513 0.26   0.08   0.56  2.15
6613 0.20   0.10   0.44  2.20
6713 0.19   0.11   0.42  2.21
6813 0.08   0.00   0.18  2.25
691

Re: [R] negative number to positive number

2007-04-25 Thread Clint Bowman

?abs

Clint BowmanINTERNET:   [EMAIL PROTECTED]
Air Dispersion Modeler  INTERNET:   [EMAIL PROTECTED]
Air Quality Program VOICE:  (360) 407-6815
Department of Ecology   FAX:(360) 407-7534

USPS:   PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Wed, 25 Apr 2007 [EMAIL PROTECTED] wrote:

 Quoting H. Paul Benton [EMAIL PROTECTED]:

  Hello all,
 
  I know this is a pretty easy question but I can't find it in S poetry or
  R help.
 
  How can I make a negative number positive. Such as
  -5 to be +5
  I tried +(-5), but that didn't work.
 
  So no, I don't mean taking a -5^2 just to get a positive number.
  This is in a function so it's not just -5 it's x. :)
 
  Thanks,
 
  Paul

 how about just multiplying it by -1??? :-)

 -5*(-1)

 Jose

 --
 Dr. Jose I. de las Heras  Email: [EMAIL PROTECTED]
 The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374
 Institute for Cell  Molecular BiologyFax:   +44 (0)131 6507360
 Swann Building, Mayfield Road
 University of Edinburgh
 Edinburgh EH9 3JR
 UK

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Re: [R] Problem installing Rmpi with lam on SGI SLES9

On 25/04/07, Martin Morgan [EMAIL PROTECTED] wrote:
 Hendrik Fuß [EMAIL PROTECTED] writes:
  I'm trying to use the papply package. However, when I do:
 
  library(papply)
  papply(list(1:10, 1:15, 1:20), sum)
  1 slaves are spawned successfully. 0 failed.
  master (rank 0, comm 1) of size 2 is running on: behemoth
  slave1 (rank 1, comm 1) of size 2 is running on: behemoth
  [1] Running serial version of papply\n
 
  Papply only spawns one slave and then decides to run the serial
  version instead. I'm not sure how to tell it to use all the 64
  processors available.

 Hendrik,

 Are you starting the lam daemons before starting R?

 % lamboot

 You might need to specify a 'hosts' argument to lamboot. The default
 way Rmpi calls lamboot is with no arguments, and this might simply
 create a single lam daemon.

Thanks, that was a pointer in the right direction.

The solution is to edit the file /etc/lam/lam-bhost.def and specify
the number of cpus (see man bhost)

cheers
Hendrik

-- 
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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Re: [R] GLS terminology question not related to R

2007-04-25 Thread Peter Dalgaard

Thomas Lumley wrote:
 On Wed, 25 Apr 2007, Leeds, Mark (IED) wrote:

   
 This is a terminology question not related to R. The literature often
 says that OLS is inefficient relative to GLS if the residuals in
 the system are correlated ( and the RHS sides of each are not identical
 ). Does this mean that OLS overestimates residual and coefficient
 variances , underestimates them or just gets them wrong and the
 direction is not known ? Thanks.
 

 It does not mean either.

 It means that the true variance of the OLS estimates is greater than the 
 true variance of the GLS estimates.
   
Yes, and to complicate things further that is not necessarily true if 
many parameters go into determining the variances and covariances 
necessary for GLS. (Cue recent discussion comparing T^2 and F tests).
 A separate issue is whether the estimated variance of an OLS estimator is 
 greater or less than the true variance of the OLS estimator.  This can go 
 either way.



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Re: [R] Fwd: dmnorm not meant for 1024-dimensional data?

2007-04-25 Thread Adelchi Azzalini

On Wed, Apr 25, 2007 at 01:25:09PM -0500, Daniel Elliott wrote:
 Hello.  Thank you for your mnormt package.  Below is an email I sent to
 R-help regarding the function dmnorm.
 
 Any help you can provide would be greatly appreciated.
 

Hello.

You have not provided information on what specifically you 
have tried as for parameters and evaluation points, so I have
chosen a particularly simple case, namely the 1024-dimensional
density with independent standard marginals, using this code

d- 1024
x - 1
const- 0.5*log(2*pi)
log.pdf -0
for (i in 1:d)   log.pdf - log.pdf - x^2/2 -const
#
S - diag(d)
mu- rep(0,d)
X - rep(x,d)
log.pdf2 - dmnorm(X,mu,S, log=TRUE)
cat(log.pdf, log.pdf2, abs(log.pdf-log.pdf2),\n)

and the outcome was this one:
  -1453 -1453 2.137e-11 
which seems quite decent to me. Obviously, you must not take exp()
of this log-density, as otherwise a 0 is indeed produced, but 
the reason is not in the package, rather in the possibility of
representing that number using standard floating-point
numerical devices.

best regards,

Adelchi Azzalini


 Thank you.
 
 - dan elliott
 
 -- Forwarded message --
 From: Daniel Elliott [EMAIL PROTECTED]
 Date: Apr 25, 2007 1:21 PM
 Subject: dmnorm not meant for 1024-dimensional data?
 To: r-help@stat.math.ethz.ch
 
 Hello,
 
 I have some data generated by a simple mixture of Gaussians (more like
 K-means) and (as a test) am using dmnorm to calculate the probability
 of each data point coming from each Gaussian.  However, I get only
 zero probabilities.
 
 This code works in low dimensions (tried 2 and 3 already).  I have run
 into many implementations that do not work in high dimension, but I
 thought that I was safe with dmnorm because it has an option to
 compute the log of the probability.
 
 So, is dmnorm not intended to be used with data of such high dimensionality?
 
 Thank you,
 
 dan elliott

-- 
Adelchi Azzalini  [EMAIL PROTECTED]
Dipart.Scienze Statistiche, Università di Padova, Italia
tel. +39 049 8274147,  http://azzalini.stat.unipd.it/

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[R] applying rbind to list elements

Hi,

I have a list of n data.frames (or matrices) which I would like to
convert to a single data.frame using rbind:

   x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )

Is there a simple way to do this?

thanks
Hendrik

-- 
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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Re: [R] applying rbind to list elements

On 4/25/2007 4:09 PM, Hendrik Fuß wrote:
 Hi,
 
 I have a list of n data.frames (or matrices) which I would like to
 convert to a single data.frame using rbind:
 
x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )
 
 Is there a simple way to do this?

do.call(rbind, l).

Duncan Murdoch

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Re: [R] applying rbind to list elements