[R] GAM Package: preplot.gam taking a **long** time

2006-08-14 Thread Paul Check
Hi: I have a large data set that I'm testing and I'm finding that 
preplot.gam is taking a very long amount of time to compute (like, more 
than 20 minutes). My machine is 32-bit, Debian unstable, 4GB memory, 
dual Xeon 3GHz. While the data set is very large, the gam() procedure is 
able to compute the model without any trouble, in a minute or so. Is 
there any reason why preplot.gam would be so slow? I am not using 
newdata in the preplot.gam function, so I am assuming that the memory 
is not a problem. Or could it be?

Is it normal for preplot.gam to take so long on large data sets? I have 
not had this experience with S-Plus, on a lower quality machine, same data.

Thanks, Paul

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[R] [R-pkgs] Random Survival Forest 1.0.0 is now available.

2006-08-14 Thread K. B. Udaya
Dear useRs,

Release 1.0.0 of the new R package 'randomSurvivalForest' is now available
on CRAN and its mirrors.  The package implements Ishwaran and Kogalur's
Random Survival Forests algorithm for right censored survival data.  The
algorithm is closely patterned after Breiman's random forests, but suitably
modified for the survival setting.  Some key features are:

o An ensemble cumulative hazard is constructed
from binary recursive survival trees grown
under different splitting rules.

o Mortality estimates interpretable in terms
of total number of deaths are provided.

o An out-of-bag estimate of Harrell's concordance
index is provided for assessing prediction.

Detailed information on usage can be found in the manual.

Thank you.

Hemant Ishwaran,  Udaya B. Kogalur

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Re: [R] Gower Similarity Coefficient

2006-08-14 Thread Chuck Cleland
Timothy Rye wrote:
 I'm interested in clustering my data using the Gower Similarity Coefficient,
 and I was wondering if R is capable of using that metric
 
 Timothy Rye

RSiteSearch(Gower) points to a number of relevant messages in the
archives.  It also reveals gdist() in the mvpart package, vegdist() in
the vegan package, and dist.binary() in the ade4 package, which may do
what you want.

   [[alternative HTML version deleted]]
 
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-- 
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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Re: [R] split a y-axis to show data on different scales

2006-08-14 Thread Jim Lemon
Rashmi Mathur wrote:
 Hello,
 
 How do I split a y-axis to plot data on different scales?
 
 Eg:
 
 x - 1:10
 y - c(-0.01,0.79,0.74,0.55,-0.67,0.32,-0.47,-0.05,723,759)
 plot(x,y)
 
 I'd like to show these data on the same plot, but the way it's written, all
 contrast in the first 8 data points is lost.  Can R split a y-axis for me?
 
Hi Rashmi,

Although Hadley's answer is relevant (displaying vastly different ranges 
of data can be dangerous) you might find that gap.plot in the plotrix 
package will do the dirty deed.

Jim

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[R] posting

2006-08-14 Thread adlai burman
[EMAIL PROTECTED]

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[R] naive help with setting node attributes

2006-08-14 Thread adlai burman
I have been trying for a LONG time to figure out how to do what I 
imagine is a fairly simple graph rendering issue. Can anyone help me 
figure out how to take two LARGE graphs G1 and G2 (not random) and set 
their node/edge attributes separately and uniquely such that when they 
are joined, G1 and G2 will retain their unique attributes under 
rendering?

Thanks,
AJ

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Re: [R] naive help with setting node attributes

2006-08-14 Thread Gabor Csardi
AJ,

hmmm, which package are you using? This might help answering the question,
which is quite hard to decypher in this form. At least for me.

Gabor

On Mon, Aug 14, 2006 at 02:25:17AM -0800, adlai burman wrote:
 I have been trying for a LONG time to figure out how to do what I 
 imagine is a fairly simple graph rendering issue. Can anyone help me 
 figure out how to take two LARGE graphs G1 and G2 (not random) and set 
 their node/edge attributes separately and uniquely such that when they 
 are joined, G1 and G2 will retain their unique attributes under 
 rendering?
 
 Thanks,
 AJ
 
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-- 
Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK

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[R] column to row

2006-08-14 Thread yohannes alazar
Dear mailing list
I have a data in two columns and how can i convert it to one row . thank you
in advance

inpute

1 2
3 4
5 6
7 8
9 1


out put

1 2 3 4 5 6 7 8 9 1

[[alternative HTML version deleted]]

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Re: [R] column to row

2006-08-14 Thread Paul Smith
On 8/14/06, yohannes alazar [EMAIL PROTECTED] wrote:
 I have a data in two columns and how can i convert it to one row . thank you
 in advance

 inpute

 1 2
 3 4
 5 6
 7 8
 9 1


 out put

 1 2 3 4 5 6 7 8 9 1

An example follows:

 input - matrix(1:10,5,2)
 input
 [,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5   10
 output - c(input[,1],input[,2])
 output
 [1]  1  2  3  4  5  6  7  8  9 10


Paul

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[R] Calculating trace of products

2006-08-14 Thread Søren Højsgaard
Dear all, 
I need to calculate tr(A B), tr(A B A B) and similar quantities **fast** where 
the matrices A, B are symmetrical. I've searched for built-in functions for 
that purpose, but without luck. Can anyone help?
Thanks in advance
Søren

[[alternative HTML version deleted]]

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Re: [R] column to row

2006-08-14 Thread Dimitris Rizopoulos
probably something like:

mat - matrix(c(1:9, 1), 5, byrow = TRUE)
c(t(mat))


I hope it helps.

Best,
Dimitris


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

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


- Original Message - 
From: yohannes alazar [EMAIL PROTECTED]
To: r-help r-help@stat.math.ethz.ch
Sent: Monday, August 14, 2006 1:47 PM
Subject: [R] column to row


 Dear mailing list
 I have a data in two columns and how can i convert it to one row . 
 thank you
 in advance

 inpute

 1 2
 3 4
 5 6
 7 8
 9 1


 out put

 1 2 3 4 5 6 7 8 9 1

 [[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.
 


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

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Re: [R] column to row

2006-08-14 Thread Neuro LeSuperHéros
input - matrix(1:10,5,2)
input
 [,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5   10
as.vector(input)
[1]  1  2  3  4  5  6  7  8  9 10

Neuro


From: yohannes alazar [EMAIL PROTECTED]
To: r-help r-help@stat.math.ethz.ch
Subject: [R] column to row
Date: Mon, 14 Aug 2006 12:47:25 +0100

Dear mailing list
I have a data in two columns and how can i convert it to one row . thank 
you
in advance

inpute

1 2
3 4
5 6
7 8
9 1


out put

1 2 3 4 5 6 7 8 9 1

   [[alternative HTML version deleted]]

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Re: [R] Calculating trace of products

2006-08-14 Thread roger koenker
I would suspect that something simple like

sum(diag(crossprod(A,B)))

would be quite competitive...

url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820


On Aug 14, 2006, at 6:58 AM, Søren Højsgaard wrote:

 Dear all,
 I need to calculate tr(A B), tr(A B A B) and similar quantities  
 **fast** where the matrices A, B are symmetrical. I've searched for  
 built-in functions for that purpose, but without luck. Can anyone  
 help?
 Thanks in advance
 Søren

   [[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
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Re: [R] Calculating trace of products

2006-08-14 Thread Dimitris Rizopoulos
check the following:

A - matrix(rnorm(100*100), 100, 100); A - A + t(A)
B - matrix(rnorm(100*100), 100, 100); B - B + t(B)

sum(diag(A %*% B))
sum(A * B)

system.time(for(i in 1:1) out - sum(diag(A %*% B)))
system.time(for(i in 1:1) out - sum(A * B))


I hope it helps.

Best,
Dimitris


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

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


- Original Message - 
From: Søren Højsgaard [EMAIL PROTECTED]
To: R-help list r-help@stat.math.ethz.ch
Sent: Monday, August 14, 2006 1:58 PM
Subject: [R] Calculating trace of products


Dear all,
I need to calculate tr(A B), tr(A B A B) and similar quantities 
**fast** where the matrices A, B are symmetrical. I've searched for 
built-in functions for that purpose, but without luck. Can anyone 
help?
Thanks in advance
Søren

[[alternative HTML version deleted]]







 __
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 http://www.R-project.org/posting-guide.html
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Re: [R] Gower Similarity Coefficient

2006-08-14 Thread Martin Maechler
 Chuck == Chuck Cleland [EMAIL PROTECTED]
 on Mon, 14 Aug 2006 04:53:55 -0400 writes:

Chuck Timothy Rye wrote:
 I'm interested in clustering my data using the Gower
 Similarity Coefficient, and I was wondering if R is
 capable of using that metric
 
 Timothy Rye

Chuck RSiteSearch(Gower) points to a number of relevant
Chuck messages in the archives.  It also reveals gdist() in
Chuck the mvpart package, vegdist() in the vegan package,
Chuck and dist.binary() in the ade4 package, which may do
Chuck what you want.

The daisy() function from the package cluster
(which is *recommended* hence part of every complete R installation)
is also based on Gower's (dis)similarity coefficient.

So you don't need to install a new package
[But I need to add the word 'Gower' to a better place on
 daisy()'s help page ...]

Martin Maechler, ETH Zurich.

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Re: [R] Gower Similarity Coefficient

2006-08-14 Thread Martin Maechler
 Martin == Martin Maechler [EMAIL PROTECTED]
 on Mon, 14 Aug 2006 14:30:39 +0200 writes:

 Chuck == Chuck Cleland [EMAIL PROTECTED]
 on Mon, 14 Aug 2006 04:53:55 -0400 writes:

Chuck Timothy Rye wrote:
 I'm interested in clustering my data using the Gower
 Similarity Coefficient, and I was wondering if R is
 capable of using that metric
 
 Timothy Rye

Chuck RSiteSearch(Gower) points to a number of relevant
Chuck messages in the archives.  It also reveals gdist() in
Chuck the mvpart package, vegdist() in the vegan package,
Chuck and dist.binary() in the ade4 package, which may do
Chuck what you want.

Martin The daisy() function from the package cluster
Martin (which is *recommended* hence part of every complete
Martin R installation) is also based on Gower's
Martin (dis)similarity coefficient.

Martin So you don't need to install a new package [But I
Martin need to add the word 'Gower' to a better place on
Martin daisy()'s help page ...]

Hmm, actually, it's already there.

And if you use help.search() smartly 
[I use lib.loc = R system library in order to not search in
 the more than 1000 CRAN and bioconductor packages we have installed],

 help.search(Gower, agrep = FALSE, lib.loc = tail(.libPaths(), 1))

I get exactly the correct match

   Help files with alias or concept or title matching $,1rx(BGower$,1ry(B 
using
   regular expression matching:

   daisy(cluster)  Dissimilarity Matrix Calculation

   Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.


Martin Maechler

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[R] mtext uses the typographical descender to align text

2006-08-14 Thread Andreas Svensson


Hello

One sometimes (quite often really ) marvel at the choice of defaults in 
R's graphic engine.
For some obscure reason, mtext uses the typographical  descender (bottom 
of letters) to align text. That is: gG will end up slightly higher 
that GG. Depending on the font, Q might end up higher than O.

(for explanation of baseline  descender see: 
http://www.paratype.com/help/term/terms.asp?code=88)

Example:

y1 - rnorm(30)
y2 - rnorm(30)
group - as.factor(rep(1:6, each=10))
y - c(y1,y2)
testdata - data.frame(group, y)
plot (y ~group, axes=F, xlab=why does R align the BOTTOM of the 
letters??? )
box()
mtext(expression(italic(Normal 1))   ,1 ,  line=1, at=1)
mtext(expression(italic(Higher 1 ))  ,1 ,  line=1, at=2)
mtext(expression(italic(Normal 3 ))  ,1 ,  line=1, at=3)
mtext(expression(italic(Higher 2 ))  ,1 ,  line=1, at=4)
mtext(expression(italic(Normal 3 ))  ,1 ,  line=1, at=5)
mtext(expression(italic(Higher 3 ))  ,1 ,  line=1, at=6)


As the word Higher includes a descending letter (g), this factor name 
ends up higher.
I must say I have never encountered a software that uses the 
typographical descender instead of the baseline to align text .

Does anyone know how to make R use the baseline instead? Perhaps using  
adj or padj? Or do I have to do something silly as adding
g, col=white
to each mtext-line to trick R into aligning the names.

Cheers
Andreas

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[R] Lattice barchart with different fill pattern

2006-08-14 Thread Rafael Duarte
Dear list,
I am new to lattice plots.
I want to make a barchart with 10 and more levels.
I need to use a grey scale for printing purposes.

The problem is that with 10 or more levels in factors it is very 
difficult to distinguish each level in the plot and legend, since the 
greys are very similar (some levels have value of zero and don't appear).

Here is an example of my problem:

df - data.frame(year = rep(1996:2005,10),
spe = c(rep(aa,10), rep(bb,10), rep(cc,10), rep(dd,10), 
rep(ee,10), rep(ff,10), rep(gg,10), 
rep(hh,10),rep(ii,10),rep(jj,10)),
value = sample( c(0:10),100 , replace=TRUE)
  )

require(lattice)
barchart( value ~ factor(year), groups = spe, data=df, stack = TRUE, 
main=, xlab=,
  auto.key = list(points = FALSE, rectangles = TRUE, space = 
right ),
  par.settings = list(superpose.polygon = list(col = 
gray.colors(10) ), scales = list(cex=0.8, rot=c(90,0,0)))
)

I was thinking that by changing the fill pattern, the different levels 
could be better distinguished.

I have read the lattice help and searched in the mailing list archives, 
but did no find any solution for lattice barchart.

Is there a way to change the fill pattern of the bar levels or any other 
approach to help in the identification of the different levels in the 
barchart?

Many thanks.

Rafael Duarte

OS: windows XP

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Re: [R] lasso for variable selection

2006-08-14 Thread Liaw, Andy
For importance it's probably best to stick with absolute values of
coefficients, instead of value of the penalty parameter for which the
coefficients changed to non-zero.

Friedman skipped a lot of details on his rule ensemble in that talk, due to
time constraint.  In his implementation he was using his own algorithm,
PathSeeker, for which paper and software are available on his web page.
PathSeeker is a different generalization of LASSO than LAR.

HTH,
Andy 

From: zubin
 
 Attended JSM last week and Friedman mentioned the use of 
 LASSO for variable selection (he uses it for rules 
 ensembles).  I am an econometrician and not familiar with, i 
 started running the examples in 
 R this week and you get to the plots section of the LARS package.   
 Plots of beta/max(beta)  vs standardized coefficients.  How 
 does one interpret them?  u see plots of each variable 
 converging to zero at different times - its pretty cool - but 
 can i use this for variable importance?
 
 for variable selection - i have a group of correlated 
 variables that we need to determine importance in predicting 
 change of a Y variable.
 
 -zubin
 
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Re: [R] Auto-save possible in R?

2006-08-14 Thread Liaw, Andy
You could try something like:

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/0334.html

replacing the time-stamp with save.image() or save().  Not sure how it works
in BATCH though...

HTH,
Andy 

From: John Morrow
 
 Hello fellow R'ers, I have a simple calculation with a very 
 large data set being generated (34.9 million values) on a 
 somewhat unreliable XP box that will likely take ~ 74hrs.  I 
 wanted to know if there is a way to have my script 
 automatically save.image() throughout the calculation in 
 case of a crash.  This could be on the basis of output 
 generated or time elapsed.  I checked the archive, and only 
 got a hint of it from:
 https://stat.ethz.ch/pipermail/r-help/1997-May/001611.html
 
  
 
 Any quick suggestions would be greatly appreciated,
 
  
 
 John Morrow
 
  
 
 
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Re: [R] rpvm/snow packages on a cluster with dual-processor machi nes

2006-08-14 Thread Liaw, Andy
That's what I've tried before, on three dual-Xeon boxes, so I know it worked
(as documented a that time).

Andy 

From: Paul Y. Peng
 
 Luke Tierney just reminded me that makeCluster() can take a 
 number greater than the number of machines in a cluster. It 
 seems to be a solution to this problem. But I haven't tested it yet.
 
 Paul.
 
 
 Ryan Austin wrote:
  Hi,
  
  Adding a node twice gives a duplicate node error.
  However, adding the parameter sp=2000 to your pvm hostfile should 
  enable dual processors.
  
  Ryan
  
  Liaw, Andy wrote:
  
  Caveat: I've only played with this a couple of years ago... 
 
  I believe you can just add each host _twice_ (or as many 
 times as the 
  number of CPUs at that host) to get both CPUs to work.
 
  Andy
 
  From: Paul Y. Peng
   
 
  Hi,
 
  does anybody know how to use the dual processors in the 
 machines of 
  a cluster? I am using R with rpvm and snow packages. I 
 usually start 
  pvm daemon and add host machines first, and then run R to 
 start my 
  computing work. But I find that only one processor in 
 each machine 
  is used in this way and the other one always stays idle. Is there 
  any simple way to tell pvm to use the two processors at the same 
  time? In other words, I would like to see two copies of R 
 running on 
  each machine's two processors when using pvm. Any hints/help are 
  greatly appreciated.
 
  Paul.
 
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[R] left-justified fixed-width format

2006-08-14 Thread roger bos
I need to save data in fixed-width format without headers and reading the
help archive leads me to believe that sprintf is pretty much the only way to
do this.  My question is, is there anyway to change the output so the text
in each column is left justified instead of right justified?  My code sample
is below where comb is the data frame.  TIA, Roger


out - sprintf(%6s %22s %8s %15s %7s %5s, comb$tic,
substr(as.character(comb$conm),
1, 22), comb$cusip, comb$type, comb$exchange, comb$currency)
write.table(out, file=c:/pit/portia_test.txt, row.names=FALSE,
col.names=FALSE, quote=FALSE)

[[alternative HTML version deleted]]

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[R] lme() F-values disagree with aov()

2006-08-14 Thread William Simpson
I have used lme() on data from a between-within subjects experiment. The correct
ANOVA table is known because this is a textbook example (Experimental Design by
Roger Kirk Chapter 12: Split-Plot Factorial Design). The lme() F-values differ 
from
the known results. Please help me understand why.

d-read.table(kirkspf2.dat,header=TRUE)
for(j in 1:4) d[,j] - factor(d[,j])  ### Make vars into type factor

##lme() results
library(nlme)
fit-lme(y~a*b*c,random=~1|s, data=d)
anova(fit)

##correct anova table
##subjects are nested within a; a between, b  c within
fit2-aov(y ~ a*b*c + Error(s/(c*b)), data=d)
summary(fit2)

I suspect I need a different random=... statement in lme().
Thanks very much for any help
Bill

The data file is attached -- kirkspf2.dat
Here it is again:

s a c b y
1 1 1 1 3
1 1 1 2 7
1 1 2 1 4
1 1 2 2 7
2 1 1 1 6
2 1 1 2 8
2 1 2 1 5
2 1 2 2 8
3 1 1 1 3
3 1 1 2 7
3 1 2 1 4
3 1 2 2 9
4 1 1 1 3
4 1 1 2 6
4 1 2 1 3
4 1 2 2 8
5 2 1 1 1
5 2 1 2 5
5 2 2 1 2
5 2 2 2 10
6 2 1 1 2
6 2 1 2 6
6 2 2 1 3
6 2 2 2 10
7 2 1 1 2
7 2 1 2 5
7 2 2 1 4
7 2 2 2 9
8 2 1 1 2
8 2 1 2 6
8 2 2 1 3
8 2 2 2 11
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Re: [R] mtext uses the typographical descender to align text

2006-08-14 Thread Peter Dalgaard
Andreas Svensson [EMAIL PROTECTED] writes:

 
 
 Hello
 
 One sometimes (quite often really ) marvel at the choice of defaults in 
 R's graphic engine.
 For some obscure reason, mtext uses the typographical  descender (bottom 
 of letters) to align text. That is: gG will end up slightly higher 
 that GG. Depending on the font, Q might end up higher than O.
 
 (for explanation of baseline  descender see: 
 http://www.paratype.com/help/term/terms.asp?code=88)
snip
 mtext(expression(italic(Normal 1))   ,1 ,  line=1, at=1)
snap
 As the word Higher includes a descending letter (g), this factor name 
 ends up higher.
 I must say I have never encountered a software that uses the 
 typographical descender instead of the baseline to align text .
 
 Does anyone know how to make R use the baseline instead? Perhaps using  
 adj or padj? Or do I have to do something silly as adding
 g, col=white
 to each mtext-line to trick R into aligning the names.

Notice that it only happens when plotting math expressions. Using 

 mtext(Normal 1   ,1 ,  line=1, at=1,font=3)
 mtext(Higher 1   ,1 ,  line=1, at=2, font=3)
etc.

does not exhibit the same effect. 

I suppose the idea is that math expressions can grow arbitrarily tall
or deep, and that the design decision (wise or not) has been to align
on the bounding box of the whole enchilada. For a workaround, if you
do need to do it with expressions, notice the phantom() construction.

-- 
   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] Lattice barchart with different fill pattern

2006-08-14 Thread Deepayan Sarkar
On 8/14/06, Rafael Duarte [EMAIL PROTECTED] wrote:
 Dear list,
 I am new to lattice plots.
 I want to make a barchart with 10 and more levels.
 I need to use a grey scale for printing purposes.

 The problem is that with 10 or more levels in factors it is very
 difficult to distinguish each level in the plot and legend, since the
 greys are very similar (some levels have value of zero and don't appear).

 Here is an example of my problem:

 df - data.frame(year = rep(1996:2005,10),
 spe = c(rep(aa,10), rep(bb,10), rep(cc,10), rep(dd,10),
 rep(ee,10), rep(ff,10), rep(gg,10),
 rep(hh,10),rep(ii,10),rep(jj,10)),
 value = sample( c(0:10),100 , replace=TRUE)
  )

Nothing to do with your problem, but 'rep' has an argument 'each'
which would shorten the call above.

 require(lattice)
 barchart( value ~ factor(year), groups = spe, data=df, stack = TRUE,
 main=, xlab=,
  auto.key = list(points = FALSE, rectangles = TRUE, space =
 right ),
  par.settings = list(superpose.polygon = list(col =
 gray.colors(10) ), scales = list(cex=0.8, rot=c(90,0,0)))
 )

 I was thinking that by changing the fill pattern, the different levels
 could be better distinguished.

 I have read the lattice help and searched in the mailing list archives,
 but did no find any solution for lattice barchart.

 Is there a way to change the fill pattern of the bar levels or any other
 approach to help in the identification of the different levels in the
 barchart?

Unfortunately, no. lattice depends on the grid package for rendering,
and last time I checked, grid didn't have support for fill patterns.

Deepayan

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Re: [R] left-justified fixed-width format

2006-08-14 Thread Prof Brian Ripley
On Mon, 14 Aug 2006, roger bos wrote:

 I need to save data in fixed-width format without headers and reading the
 help archive leads me to believe that sprintf is pretty much the only way to
 do this.  My question is, is there anyway to change the output so the text
 in each column is left justified instead of right justified?  My code sample
 is below where comb is the data frame.  TIA, Roger

format can do this, easily.  And for sprintf, use the '-' flag documented 
on the help page(!):

 sprintf(%22s, foo)
[1]foo
 sprintf(%-22s, foo)
[1] foo   

-- 
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] lme() F-values disagree with aov()

2006-08-14 Thread i.m.s.white
Your lme statement is OK. To get the usual split-plot anova, your aov
statement should be

fit2 - aov(y ~ a*b*c + Error(s), data = d)

This gives the same F-values as lme.


On Mon, Aug 14, 2006 at 10:27:19AM -0400, William Simpson wrote:
 I have used lme() on data from a between-within subjects experiment. The 
 correct
 ANOVA table is known because this is a textbook example (Experimental Design 
 by
 Roger Kirk Chapter 12: Split-Plot Factorial Design). The lme() F-values 
 differ from
 the known results. Please help me understand why.
 
 d-read.table(kirkspf2.dat,header=TRUE)
 for(j in 1:4) d[,j] - factor(d[,j])  ### Make vars into type factor
 
 ##lme() results
 library(nlme)
 fit-lme(y~a*b*c,random=~1|s, data=d)
 anova(fit)
 
 ##correct anova table
 ##subjects are nested within a; a between, b  c within
 fit2-aov(y ~ a*b*c + Error(s/(c*b)), data=d)
 summary(fit2)
 
 I suspect I need a different random=... statement in lme().
 Thanks very much for any help
 Bill
 
 The data file is attached -- kirkspf2.dat
 Here it is again:
 
 s a c b y
 1 1 1 1 3
 1 1 1 2 7
 1 1 2 1 4
 1 1 2 2 7
 2 1 1 1 6
 2 1 1 2 8
 2 1 2 1 5
 2 1 2 2 8
 3 1 1 1 3
 3 1 1 2 7
 3 1 2 1 4
 3 1 2 2 9
 4 1 1 1 3
 4 1 1 2 6
 4 1 2 1 3
 4 1 2 2 8
 5 2 1 1 1
 5 2 1 2 5
 5 2 2 1 2
 5 2 2 2 10
 6 2 1 1 2
 6 2 1 2 6
 6 2 2 1 3
 6 2 2 2 10
 7 2 1 1 2
 7 2 1 2 5
 7 2 2 1 4
 7 2 2 2 9
 8 2 1 1 2
 8 2 1 2 6
 8 2 2 1 3
 8 2 2 2 11

 __
 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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-- 

*I.White   *
*University of Edinburgh   *
*Ashworth Laboratories, West Mains Road*
*Edinburgh EH9 3JT *
*Fax: 0131 650 6564   Tel: 0131 650 5490   *
*E-mail: [EMAIL PROTECTED]  *

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Re: [R] lme() F-values disagree with aov()

2006-08-14 Thread William Simpson
 Your lme statement is OK. To get the usual split-plot anova, your aov
 statement should be

 fit2 - aov(y ~ a*b*c + Error(s), data = d)

No, this gives wrong F-values. By wrong I mean it does not agree with the
published table.

Table 12.10-2, page 559:
 Number of obs =  32 R-squared =  0.9920
 Root MSE  = .559017 Adj R-squared =  0.9589

Source |  Partial SSdf   MS   F Prob  F
---+
 Model | 233.62525   9.345  29.90 0.0002
   |
 a |   3.125 1   3.125   2.00 0.2070
   s|a |   9.375 6  1.5625
---+
 b |  162.00 1  162.00 199.38 0.
   a*b |   6.125 1   6.125   7.54 0.0335
 b*s|a |   4.875 6   .8125
---+
 c |   24.50 1   24.50  61.89 0.0002
   a*c |  10.125 1  10.125  25.58 0.0023
 c*s|a |   2.375 6  .39583
---+
   b*c |8.00 18.00  25.60 0.0023
 a*b*c |   3.125 1   3.125  10.00 0.0195
   |
  Residual |   1.875 6   .3125
---+
 Total |  235.5031  7.59677419

Bill

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Re: [R] How to order or sort a data.frame

2006-08-14 Thread John Kane

--- Dimitrios Rizopoulos
[EMAIL PROTECTED] wrote:

 try the following:
 
 mdf - data.frame(us.state, count, year, month)
 mdf[order(mdf$year, mdf$month), ]
 


Thansk to Dimitris and Dieter. This has helped since
seems to have shown me a way around the problem. It
just means that I have to sort the data earlier.  

What my example did not show clearly is that when I
subset the data the variable 'month' which is numeric
in the original dataframe becomes a factor.  I was
wondering if there was a way to sort the factor so
that I would get a numeric sort.  So far I have not
been able to see how to coerce the factor month into
a numeric when using 'order

Thanks for helpl.

Hi Hadley, 
I have not had time to check out the reshape but
thanks.




 Ph.D. Student
 Biostatistical Centre
 School of Public Health
 Catholic University of Leuven
 
 Address: Kapucijnenvoer 35, Leuven, Belgium
 Tel: +32/(0)16/336899
 Fax: +32/(0)16/337015
 Web: http://med.kuleuven.be/biostat/
  

http://www.student.kuleuven.be/~m0390867/dimitris.htm
 
 
 Quoting John Kane [EMAIL PROTECTED]:
 
  I have a dataframe where I would like to order
 first
  by  variable, year, and then within that variable
 by
  month.
 
  So far the only way that I have seen to do this is
 to
  order by year and then subset year and sort by
 month
  and then do an rbind to get things back together.
 
  Is this the right approach?
 
  Example:
 
  us.state -rep(California, 23)
 
  count -
 c(774,283,774,283,508,283,774,283,602,283,
 
  774,508,0,602,330,283,283,283,602,301,126, NA,301)
 
  year - c(2002,  2003, 2001, 2002, 2001, 2002,
 2001,
  2002, 2002, 2003,
2002, 2002, 2001,  2002, 2001, 2002,
 2001,
  2002, 2001, 2002,
2001, 2001, 2002)
 
  month - c( 1, 1, 10, 10, 11, 11, 12, 12,
 
  2, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9,
 9,
  9)
 
 
 
 
  df - data.frame (cbind(us.state,count, year,
 month))
 
  # ordering as a factor works here
 
  df1 - df[order(df$year),]
 
  df1
 
 
 
  df2 - subset(df1, year==2001)
 
 
 
  # ordering as a factor works but not a good
  appearance.
 
 
  df3 - df2[order(as.numeric(df2$month)),]
 
  df3
 
 
 
  This works but  month is ordered as  a factor
 and I
  would prefer to coerce it into a numeric for
  presentation purposes but
   df3 - df2[order(as.numeric(df2$month)),] does
 not
  seem to work,  nor has a couple of other things
 I've
  tried.
 
  Any suggestions gratefully received.
 
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  PLEASE do read the posting guide
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 Disclaimer:
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[R] help with glmmPQL

2006-08-14 Thread Matthew A. Krutky
Hello Folks- 
 
Is there a way to create confidence bands with 'glmmPQL' ??? 
 
I am trying to decide a way to best plot the model which we created with the
glmmPQL function in R.   I would like 
to plot my actual averaged data points within 95 % confidence intervals from
the model. Plotting
the model is easy, but determining confidence bands is not. 
 
Here is my model:
 
ratiomodel-glmmPQL(ratio~as.factor(joint)*time, random = ~ 1 | subject,
family = Gamma(link =
identity),alldata3)
 
If I use 'lm' and pass the command 'int = c ' 'to create this model I can
easily find and plot this type of confidence band for 'ratio~time'.  But I
need to take into account 'as.factor(joint)', and in fact I can
produce confidence bands with 'glm' by passing in 'se.fit = TRUE', but the
problem is I need to
make subject a random variable, and take into account my ratio with the
Gamma distribution.  
 
Is there a way to create confidence bands with 'glmmPQL' ??? '
as.factor(joint)' has 3 levels, so I 
would like to produce this linear model with three levels and confidence
bands for comparison of
the levels of 'joint'.  
 
Any Help at all with my problem would be greatly appreciated!!

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[R] solving non-linear system of equations

2006-08-14 Thread HAKAN DEMIRTAS
Didn't get any useful response to the following question. Trying again.

I can't seem to get computationally stable estimates for the following
system:

Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination
of the first three powers of a standard normal variable.) Assuming that
E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious
algebraic calculations:

1) b^2+6bd+2c^2+15d^2=1
2) 2c(b^2+24bd+105d^2+2)=E(Y^3)
3) 24[bd+c^2(1+b^2+28bd)+d^2(12+48bd+141c^2+225d^2)]=E(Y^4)-3

Obviously, a=-c. Suppose that distributional form of Y is given so we know
E(Y^3) and E(Y^4). In other words, we have access to the third and fourth
raw moments. How do we solve for these four coefficients? I reduced the
number of unknowns/equations to two, and subsequently used a grid
approach. It works well when I am close to the center of the support, but
fails at the tails. Any ideas?

Hakan Demirtas

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Re: [R] lme() F-values disagree with aov()

2006-08-14 Thread Peter Dalgaard
William Simpson [EMAIL PROTECTED] writes:

  Your lme statement is OK. To get the usual split-plot anova, your aov
  statement should be
 
  fit2 - aov(y ~ a*b*c + Error(s), data = d)
 
 No, this gives wrong F-values. By wrong I mean it does not agree with the
 published table.

Well, it's the model that is equivalent to your lme() model

Thing is that you want to add random effects of s:b and s:c, which are
crossed factors, so somewhat tricky to code with lme() (this sort of
thing is easier in lmer() from the lme4 packages). 

The generic way to handle this in lme() is via something like

   random=list(s=pdBlocked(list(
 pdIdent(~1),
 pdIdent(~b-1),
 pdIdent(~c-1

You probably won't get the degrees of freedom right, though. 

 
 Table 12.10-2, page 559:
  Number of obs =  32 R-squared =  0.9920
  Root MSE  = .559017 Adj R-squared =  0.9589
 
 Source |  Partial SSdf   MS   F Prob  F
 ---+
  Model | 233.62525   9.345  29.90 0.0002
|
  a |   3.125 1   3.125   2.00 0.2070
s|a |   9.375 6  1.5625
 ---+
  b |  162.00 1  162.00 199.38 0.
a*b |   6.125 1   6.125   7.54 0.0335
  b*s|a |   4.875 6   .8125
 ---+
  c |   24.50 1   24.50  61.89 0.0002
a*c |  10.125 1  10.125  25.58 0.0023
  c*s|a |   2.375 6  .39583
 ---+
b*c |8.00 18.00  25.60 0.0023
  a*b*c |   3.125 1   3.125  10.00 0.0195
|
   Residual |   1.875 6   .3125
 ---+
  Total |  235.5031  7.59677419
 
 Bill
 
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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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-- 
   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] lme() F-values disagree with aov()

2006-08-14 Thread William Simpson
Thanks very much Peter!
  Your lme statement is OK. To get the usual split-plot anova, your aov
  statement should be
 
  fit2 - aov(y ~ a*b*c + Error(s), data = d)

 No, this gives wrong F-values. By wrong I mean it does not agree with the
 published table.

 Well, it's the model that is equivalent to your lme() model
Yes, I thought my lme() model was wrong but couldn't figure out how to do it 
properly.

 Thing is that you want to add random effects of s:b and s:c, which are
 crossed factors, so somewhat tricky to code with lme() (this sort of
 thing is easier in lmer() from the lme4 packages).
I am happy to do it that way if you show me how...

 The generic way to handle this in lme() is via something like

random=list(s=pdBlocked(list(
  pdIdent(~1),
  pdIdent(~b-1),
  pdIdent(~c-1

I see. Thanks very much Peter!!

Bill

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Re: [R] mtext uses the typographical descender to align text

2006-08-14 Thread Andreas Svensson
Hi

The reason I use expression is to get italics. Perhaps this is possible 
some other way?

/A

Peter Dalgaard wrote:

Andreas Svensson [EMAIL PROTECTED] writes:

  



Hello

One sometimes (quite often really ) marvel at the choice of defaults in 
R's graphic engine.
For some obscure reason, mtext uses the typographical  descender (bottom 
of letters) to align text. That is: gG will end up slightly higher 
that GG. Depending on the font, Q might end up higher than O.

(for explanation of baseline  descender see: 
http://www.paratype.com/help/term/terms.asp?code=88)


snip
  

mtext(expression(italic(Normal 1))   ,1 ,  line=1, at=1)


snap
  

As the word Higher includes a descending letter (g), this factor name 
ends up higher.
I must say I have never encountered a software that uses the 
typographical descender instead of the baseline to align text .

Does anyone know how to make R use the baseline instead? Perhaps using  
adj or padj? Or do I have to do something silly as adding
g, col=white
to each mtext-line to trick R into aligning the names.



Notice that it only happens when plotting math expressions. Using 

 mtext(Normal 1   ,1 ,  line=1, at=1,font=3)
 mtext(Higher 1   ,1 ,  line=1, at=2, font=3)
etc.

does not exhibit the same effect. 

I suppose the idea is that math expressions can grow arbitrarily tall
or deep, and that the design decision (wise or not) has been to align
on the bounding box of the whole enchilada. For a workaround, if you
do need to do it with expressions, notice the phantom() construction.

  


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[R] Question on .Options$max.print

2006-08-14 Thread Hans-Joerg Bibiko
Hi,


I have a tiny question concerning .Options$max.print


I have to set up this value to a greater value than 1 because I  
want to concatenate my output of a function to one single string (for  
connivence).

I did this via .Options$max.print - 64000 or options(max.print=64000)

Then I call out - paste(out, blabla) several times, but nchar(out)  
is never larger than 1.

I read in the help about '.Options' that this is not yet used in base  
R. Could this be my problem?


Thanks for any hint

Hans

BTW I believe, there is a typo within the help page about '.Options'

...
The ‘factory-fresh’ default settings of some of these options are
...
max.print 1000
...


This should be
...
max.print 1
...

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Re: [R] mtext uses the typographical descender to align text

2006-08-14 Thread Uwe Ligges


Andreas Svensson wrote:
 Hi
 
 The reason I use expression is to get italics. Perhaps this is possible 
 some other way?


1. Yes, see ?mtext:

plot(1:10)
mtext(Hello World, 3, font=3)


2. expression() has to use bounding boxes for several reasons, in 
particular fractions spring to mind at once. Solutions as Peter suggested:
   plot(1:10)
   mtext(expression(Hello World * phantom(g)), 3)

Uwe Ligges



 /A
 
 Peter Dalgaard wrote:
 
 Andreas Svensson [EMAIL PROTECTED] writes:

  

 

 Hello

 One sometimes (quite often really ) marvel at the choice of defaults in 
 R's graphic engine.
 For some obscure reason, mtext uses the typographical  descender (bottom 
 of letters) to align text. That is: gG will end up slightly higher 
 that GG. Depending on the font, Q might end up higher than O.

 (for explanation of baseline  descender see: 
 http://www.paratype.com/help/term/terms.asp?code=88)


 snip
  

 mtext(expression(italic(Normal 1))   ,1 ,  line=1, at=1)


 snap
  

 As the word Higher includes a descending letter (g), this factor name 
 ends up higher.
 I must say I have never encountered a software that uses the 
 typographical descender instead of the baseline to align text .

 Does anyone know how to make R use the baseline instead? Perhaps using  
 adj or padj? Or do I have to do something silly as adding
 g, col=white
 to each mtext-line to trick R into aligning the names.


 Notice that it only happens when plotting math expressions. Using 

 mtext(Normal 1   ,1 ,  line=1, at=1,font=3)
 mtext(Higher 1   ,1 ,  line=1, at=2, font=3)
 etc.

 does not exhibit the same effect. 

 I suppose the idea is that math expressions can grow arbitrarily tall
 or deep, and that the design decision (wise or not) has been to align
 on the bounding box of the whole enchilada. For a workaround, if you
 do need to do it with expressions, notice the phantom() construction.

  

 
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Re: [R] split a y-axis to show data on different scales

2006-08-14 Thread Spencer Graves
  The pro's and con's of using scale breaks were discussed by 
Cleveland (1985) The Elements of Graphing Data (Wadsworth, pp. 85-91, 
149).  I don't know what Cleveland said about this is the second edition 
of this book, but I believe there are times when scale breaks are 
appropriate, but the display should make this nonstandard transition 
very clear;  otherwise, it can be seriously misleading.

  Hope this helps.
  Spencer Graves

Jim Lemon wrote:
 Rashmi Mathur wrote:
 Hello,

 How do I split a y-axis to plot data on different scales?

 Eg:

 x - 1:10
 y - c(-0.01,0.79,0.74,0.55,-0.67,0.32,-0.47,-0.05,723,759)
 plot(x,y)

 I'd like to show these data on the same plot, but the way it's written, all
 contrast in the first 8 data points is lost.  Can R split a y-axis for me?

 Hi Rashmi,
 
 Although Hadley's answer is relevant (displaying vastly different ranges 
 of data can be dangerous) you might find that gap.plot in the plotrix 
 package will do the dirty deed.
 
 Jim
 
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Re: [R] Question on .Options$max.print

2006-08-14 Thread Martin Maechler
 HansJB == Hans-Joerg Bibiko [EMAIL PROTECTED]
 on Mon, 14 Aug 2006 17:56:30 +0200 writes:

HansJB Hi, I have a tiny question concerning
HansJB .Options$max.print


HansJB I have to set up this value to a greater value than
HansJB 1 because I want to concatenate my output of a
HansJB function to one single string (for connivence).

HansJB I did this via .Options$max.print - 64000 or
HansJB options(max.print=64000)

HansJB Then I call out - paste(out, blabla) several times,
HansJB but nchar(out) is never larger than 1.

HansJB I read in the help about '.Options' that this is not
HansJB yet used in base R. Could this be my problem?

well, it at least makes clear that your assumption that 
this option would influence your paste() must be wrong.

I don't think there's any option influencing paste()
and I hope there won't every be any.
Options typically should only influence ``output formatting''
but not the result of a ``computational'' (i.e. non-printing/plotting)
function.

HansJB Thanks for any hint

The posting guide -- and every footer of all R-help posting asks
for a small reproducible example of R code.
So please do provide one
[and, BTW, keep this thread on R-help; do not reply privately..]

HansJB Hans

HansJB BTW I believe, there is a typo within the help page
HansJB about '.Options'

HansJB ...  The ?factory-fresh? default settings of some
HansJB of these options are ...  max.print 1000 ...


HansJB This should be ...  max.print 1 ...

Yes, thank you.

Martin Maechler, ETH Zurich

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[R] Attempt to access unmapped memory

2006-08-14 Thread Alexandre Aguiar
Hi,

I am usiing R 2.3.1 under Linux kernel 2.6.11 with libreadline/libhistory 5.1. 
Bothe R and readline were compiled without quircks with gcc 3.3.3 and g77.

Every time I try to edit command line by using del key the following error 
happens:

8--
 ipacks - instaleld.packages()
 *** caught segfault ***
address (nil), cause 'memory not mapped'

Possible actions:
1: abort (with core dump)
2: normal R exit
3: exit R without saving workspace
4: exit R saving workspace
Selection: Segmentation fault
8--

As I have readline 4.3 installed too, I adjusted the symlinks to point to 5.1. 
No problem with bash. This setup works fine in another machine with same OS 
and versions.

Any clues?

Thanks in advance.


-- 


Alexandre Aguiar
Independent consultant for medical research
SPS Consultoria
Voice: +55-11-9320-2046
Fax: +55-11-5549-8760

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Re: [R] How to reply to a thread if receiving R-help mails in digest form

2006-08-14 Thread Michael Dewey
At 14:21 13/08/2006, Dirk Enzmann wrote:
I receive R-help messages in digest form that makes it difficult to 
answer a post. I noticed that my answer is not added to the thread 
(instead, a new thread is started) although I use the same subject 
line (starting with Re: ) as the original post. Is there a 
solution (I prefer the digest to separate messages for several 
reasons and don't want to change my email reader)?

Dirk, I asked some while ago for people to tell me how they read 
digests in their news reader so I could put together a list of 
helpful hints. Unfortunately I got nowhere. In the email client I use 
(Eudora) there is an option to receive digests as attachments (as Ted 
Harding has also outlined in another email. It is rather hidden so I 
suspect if you can do it in Thunderbird it will be similarly hidden. 
Try looking for such an option and let me know if you find it.


The way I answer post up to now is:
1) I press the reply button of my email program (Mozilla / 
Thunderbird, Windows)
2) I delete all contents of the digest except for the post 
(including name and mail address of the posting person) I want to 
answer so that the original question will be included (cited) in my answer.
3) I add the email address to the individual sender to cc: to the 
automatically generated address of the R-help list.
4) I replace the automatically generated subject line (for example 
Re: R-help Digest, Vol 42, Issue 13 by Re:  followed by a copy 
of the original subject line of the post.
5) I write my answer and send the mail to the mailing list.

It's not that this is tedious - the problem is that the thread is 
broken. Is there a better way even if I want to keep receiving 
messages in digest form? The posting guide is silent about this.

Dirk

*
Dr. Dirk Enzmann
Institute of Criminal Sciences
Dept. of Criminology
Edmund-Siemers-Allee 1
D-20146 Hamburg
Germany

phone: +49-(0)40-42838.7498 (office)
+49-(0)40-42838.4591 (Billon)
fax:   +49-(0)40-42838.2344
email: [EMAIL PROTECTED]
www: 
http://www2.jura.uni-hamburg.de/instkrim/kriminologie/Mitarbeiter/Enzmann/Enzmann.html



Michael Dewey
http://www.aghmed.fsnet.co.uk

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Re: [R] Vector Join

2006-08-14 Thread Michael Dewey
At 13:23 13/08/2006, Michael Zatorsky wrote:
Hi,

I'm working on producing a simple cumulative frequency
distribution.

Thanks to the help of the good people on this list I
now have four vectors that I'd like to join/relate
into a table. e.g.


v1 - myHistogram$breaks # classes
v2 - myHistogram$counts # freqs
v3 - cumsum(v2) # cumulative freq
v4 - ((v3 / length(myData)) * 100)  # cumulative %

data.frame(v1 = myHistogram$breaks, v2 = myHistogram$counts, and so on ...)



What is the recommend approach to turning these into a
single table with four columns?  ie effectively doing
a relational join on row id?

The goal is to ultimately have the data with one row
per class in a format I can write out to a text file
as:

   v1v2v3v4
   v1v2v3v4
   etc...

Any advice will be appreciated.

Regards
Michael.







Coming soon: Celebrity Survivor - 11 celebrities, 25 days, unlimited drama

Michael Dewey
[EMAIL PROTECTED]
http://www.aghmed.fsnet.co.uk/home.html

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Re: [R] solving non-linear system of equations

2006-08-14 Thread Spencer Graves
  Are you saying you did not receive my reply stamped 8/10/2006 2:12 
AM in my Sent folder, or that my reply was not useful?  In case the 
former is correct, my comments were as follows:

Have you tried writing a function to compute SS = sum of squares 
deviations between the the left and right hand sides of your three 
equations, then using 'optim'?  See also Venables and Ripley (2002) 
Modern Applied Statistics with S, 4th ed. (Springer).

p.s.  I don't see how it's obvious that 'a=-c'.

  If you think these comments are not useful, I would appreciate the 
courtesy of a reply.  I have solved many superficially similar problems 
in this way, and I would like to know why this would not work for your 
case.

   hope this helps.
   Spencer Graves   

HAKAN DEMIRTAS wrote:
 Didn't get any useful response to the following question. Trying again.
 
 I can't seem to get computationally stable estimates for the following
 system:
 
 Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination
 of the first three powers of a standard normal variable.) Assuming that
 E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious
 algebraic calculations:
 
 1) b^2+6bd+2c^2+15d^2=1
 2) 2c(b^2+24bd+105d^2+2)=E(Y^3)
 3) 24[bd+c^2(1+b^2+28bd)+d^2(12+48bd+141c^2+225d^2)]=E(Y^4)-3
 
 Obviously, a=-c. Suppose that distributional form of Y is given so we know
 E(Y^3) and E(Y^4). In other words, we have access to the third and fourth
 raw moments. How do we solve for these four coefficients? I reduced the
 number of unknowns/equations to two, and subsequently used a grid
 approach. It works well when I am close to the center of the support, but
 fails at the tails. Any ideas?
 
 Hakan Demirtas
 
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[R] missing data treatment in MCMC pack

2006-08-14 Thread Mariagiulia Matteucci
Hello, 
does anyone know something about missing data in the MCMC pack? 
thank you,
Mariagiulia



Mariagiulia Matteucci
Dipartimento di Scienze Statistiche “Paolo Fortunati”
Università di Bologna
Via Belle Arti, 41 
40126 Bologna (ITALY)
e-mail: [EMAIL PROTECTED]
TEL: +39 051 264182
FAX: +39 051 232153

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[R] Making R script to run in a console

2006-08-14 Thread Ronaldo Reis-Jr.
Hi,

is possible to make a R script to run under a console without open the R 
environment?

Something like this example.R

#!/usr/bin/R

function(name=Put here your name) {
print(name)
}

In a console I make
./example.R name=Ronaldo Reis Júnior
then program print my name.

It is possible?

Thanks
Ronaldo
-- 
A jury consists of twelve persons chosen to decide who has the better lawyer.
-- Robert Frost
--
 Prof. Ronaldo Reis Júnior
|  .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
| : :'  : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia
| `. `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
|   `- Fone: (38) 3229-8190 | [EMAIL PROTECTED]
| ICQ#: 5692561 | LinuxUser#: 205366

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Re: [R] solving non-linear system of equations

2006-08-14 Thread Peter Dalgaard
Spencer Graves [EMAIL PROTECTED] writes:

 p.s.  I don't see how it's obvious that 'a=-c'.

  Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination
  of the first three powers of a standard normal variable.) Assuming that
  E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious
  algebraic calculations:

Take means on both sides: EY = a+0+c+0 = 0


-- 
   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] How to order or sort a data.frame

2006-08-14 Thread Dimitrios Rizopoulos
Quoting John Kane [EMAIL PROTECTED]:


 --- Dimitrios Rizopoulos
 [EMAIL PROTECTED] wrote:

 try the following:

 mdf - data.frame(us.state, count, year, month)
 mdf[order(mdf$year, mdf$month), ]



 Thansk to Dimitris and Dieter. This has helped since
 seems to have shown me a way around the problem. It
 just means that I have to sort the data earlier.

 What my example did not show clearly is that when I
 subset the data the variable 'month' which is numeric
 in the original dataframe becomes a factor.  I was
 wondering if there was a way to sort the factor so
 that I would get a numeric sort.  So far I have not
 been able to see how to coerce the factor month into
 a numeric when using 'order

for converting factors to numeric check R FAQ 7.10.

Best,
Dimitris




 Thanks for helpl.

 Hi Hadley,
 I have not had time to check out the reshape but
 thanks.




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

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


 http://www.student.kuleuven.be/~m0390867/dimitris.htm


 Quoting John Kane [EMAIL PROTECTED]:

  I have a dataframe where I would like to order
 first
  by  variable, year, and then within that variable
 by
  month.
 
  So far the only way that I have seen to do this is
 to
  order by year and then subset year and sort by
 month
  and then do an rbind to get things back together.
 
  Is this the right approach?
 
  Example:
 
  us.state -rep(California, 23)
 
  count -
 c(774,283,774,283,508,283,774,283,602,283,
 
  774,508,0,602,330,283,283,283,602,301,126, NA,301)
 
  year - c(2002,  2003, 2001, 2002, 2001, 2002,
 2001,
  2002, 2002, 2003,
2002, 2002, 2001,  2002, 2001, 2002,
 2001,
  2002, 2001, 2002,
2001, 2001, 2002)
 
  month - c( 1, 1, 10, 10, 11, 11, 12, 12,
 
  2, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9,
 9,
  9)
 
 
 
 
  df - data.frame (cbind(us.state,count, year,
 month))
 
  # ordering as a factor works here
 
  df1 - df[order(df$year),]
 
  df1
 
 
 
  df2 - subset(df1, year==2001)
 
 
 
  # ordering as a factor works but not a good
  appearance.
 
 
  df3 - df2[order(as.numeric(df2$month)),]
 
  df3
 
 
 
  This works but  month is ordered as  a factor
 and I
  would prefer to coerce it into a numeric for
  presentation purposes but
   df3 - df2[order(as.numeric(df2$month)),] does
 not
  seem to work,  nor has a couple of other things
 I've
  tried.
 
  Any suggestions gratefully received.
 
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  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 reproducible code.
 
 



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



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[R] ARMA(1,1) for panel data

2006-08-14 Thread Tom Boonen
Dear List,

I am new to TS-Modeling in R. I would like to fit an ARMA(1,1) model
for a balanced panel, running Y on a full set of unit and year dummies
using an arma(1,1) for the disturbance:

y_it=unit.dummies+yeardummies+e_it

where: e_it=d*e_it-1+u_it+q*u_it-1

How can I fit this model in R? arma() does not seem to take covariates
(or I don't understand how to specify the function so that it would).
Thank you very much.

Best, Tom

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[R] CircStats help

2006-08-14 Thread Isabelle Deguise
Hello,

I am quite new to R and although I have read some material  
(apparently not enough) I'm getting very frustrated. I'm trying to do  
some simple circular statistics in R (CircStats, primarily circ.mean)  
and I'm confused about the results. What is a negative radian or  
degree mean in R? I thought it would be a coterminal angle, but that  
doesn't make sense with my data points. I'm sorry if this is trivial  
to most of you, but I really appreciate any help.

I have tried to obtain the book this package was based on  
(Jammalamadaka and SenGupta 2001 Topics in Circular Statistics) but  
it is not available at any of my university libraries.

Here is an example. The values of x are directions in degrees of  
one amphibian over 6 days and I'd just like to simply calculate the  
mean direction:

  x - c(5.009684,37.814266, 295.722970, 0.00,  
326.463366,242.678840)
  radx - rad(x)
  circ.mean(radx)
[1] -0.4283351
  deg(-0.4283351)
[1] -24.54179

I would just like this number to be converted to the appropriate  
angle between 0-360 degrees. Can someone help me?

Sincerely,

Isabelle Deguise
MSc candidate





[[alternative HTML version deleted]]

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Re: [R] CircStats help

2006-08-14 Thread Ben Bolker
Isabelle Deguise ideguise at interchange.ubc.ca writes:

 
 Hello,
 
   
   x - c(5.009684,37.814266, 295.722970, 0.00,  
 326.463366,242.678840)
   radx - rad(x)
   circ.mean(radx)
 [1] -0.4283351
   deg(-0.4283351)
 [1] -24.54179
 
 I would just like this number to be converted to the appropriate  
 angle between 0-360 degrees. Can someone help me?
 

  Just add 360 to the result ... when in doubt, you can also look at what
the function is actually doing by typing the name of the
function by itself, e.g.

 circ.mean

In this case it shows that the function is just adding
up the sines and cosines of the individual angles and
then taking the arc-tangent.  With this size problem
you can work through the details and understand how
it all works ...

  Ben Bolker

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Re: [R] Attempt to access unmapped memory

2006-08-14 Thread Prof Brian Ripley
On Mon, 14 Aug 2006, Alexandre Aguiar wrote:

 Hi,
 
 I am usiing R 2.3.1 under Linux kernel 2.6.11 with 
 libreadline/libhistory 5.1.

That is known to be buggy.  Either patch it or downdate to 5.0.

 Bothe R and readline were compiled without quircks with gcc 3.3.3 and g77.
 
 Every time I try to edit command line by using del key the following error 
 happens:
 
 8--
  ipacks - instaleld.packages()
  *** caught segfault ***
 address (nil), cause 'memory not mapped'
 
 Possible actions:
 1: abort (with core dump)
 2: normal R exit
 3: exit R without saving workspace
 4: exit R saving workspace
 Selection: Segmentation fault
 8--
 
 As I have readline 4.3 installed too, I adjusted the symlinks to point to 
 5.1. 
 No problem with bash. This setup works fine in another machine with same OS 
 and versions.
 
 Any clues?
 
 Thanks in advance.
 
 
 

-- 
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] Fast way to load multiple files

2006-08-14 Thread Peter Eiger
Hi,

Instead of having to program a loop to load several workspaces in a directory, 
it would be nice to store the filenames in a list filelist and then to apply 
load to this list
lapply( filelist, load)
Unfortunately, although it seems that R is loading the files, the contained 
objects are not available in the workspace afterwards.
Any hints what I'm doing wrong or how to circumvent the problem?
Peter
--

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Re: [R] How to order or sort a data.frame

2006-08-14 Thread John Kane

--- Dimitrios Rizopoulos
[EMAIL PROTECTED] wrote:

 Quoting John Kane [EMAIL PROTECTED]:
 
 
  --- Dimitrios Rizopoulos
  [EMAIL PROTECTED] wrote:
 
  try the following:
 
  mdf - data.frame(us.state, count, year, month)
  mdf[order(mdf$year, mdf$month), ]
 
 
 
  Thansk to Dimitris and Dieter. This has helped
 since
  seems to have shown me a way around the problem.
 It
  just means that I have to sort the data earlier.
 
  What my example did not show clearly is that when
 I
  subset the data the variable 'month' which is
 numeric
  in the original dataframe becomes a factor.  I was
  wondering if there was a way to sort the factor so
  that I would get a numeric sort.  So far I have
 not
  been able to see how to coerce the factor month
 into
  a numeric when using 'order
 
 for converting factors to numeric check R FAQ 7.10.
 
 Best,
 Dimitris

Argh!  Of course. I should never trust my memory!

Thanks.
 

 
 
 
  Thanks for helpl.
 
  Hi Hadley,
  I have not had time to check out the reshape but
  thanks.
 
 
 
 
  Ph.D. Student
  Biostatistical Centre
  School of Public Health
  Catholic University of Leuven
 
  Address: Kapucijnenvoer 35, Leuven, Belgium
  Tel: +32/(0)16/336899
  Fax: +32/(0)16/337015
  Web: http://med.kuleuven.be/biostat/
 
 
 

http://www.student.kuleuven.be/~m0390867/dimitris.htm
 
 
  Quoting John Kane [EMAIL PROTECTED]:
 
   I have a dataframe where I would like to order
  first
   by  variable, year, and then within that
 variable
  by
   month.
  
   So far the only way that I have seen to do this
 is
  to
   order by year and then subset year and sort by
  month
   and then do an rbind to get things back
 together.
  
   Is this the right approach?
  
   Example:
  
   us.state -rep(California, 23)
  
   count -
  c(774,283,774,283,508,283,774,283,602,283,
  
   774,508,0,602,330,283,283,283,602,301,126,
 NA,301)
  
   year - c(2002,  2003, 2001, 2002, 2001, 2002,
  2001,
   2002, 2002, 2003,
 2002, 2002, 2001,  2002, 2001, 2002,
  2001,
   2002, 2001, 2002,
 2001, 2001, 2002)
  
   month - c( 1, 1, 10, 10, 11, 11, 12, 12,
  
   2, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8,
 9,
  9,
   9)
  
  
  
  
   df - data.frame (cbind(us.state,count, year,
  month))
  
   # ordering as a factor works here
  
   df1 - df[order(df$year),]
  
   df1
  
  
  
   df2 - subset(df1, year==2001)
  
  
  
   # ordering as a factor works but not a good
   appearance.
  
  
   df3 - df2[order(as.numeric(df2$month)),]
  
   df3
  
  
  
   This works but  month is ordered as  a factor
  and I
   would prefer to coerce it into a numeric for
   presentation purposes but
df3 - df2[order(as.numeric(df2$month)),] does
  not
   seem to work,  nor has a couple of other things
  I've
   tried.
  
   Any suggestions gratefully received.
  
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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.
  
  
 
 
 
  Disclaimer:
  http://www.kuleuven.be/cwis/email_disclaimer.htm
 
 
 
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 reproducible code.
 
 
 
 
 
 Disclaimer:
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Re: [R] Fast way to load multiple files

2006-08-14 Thread Berton Gunter
A reproducible example here would help (please see posting guide). A guess:
is your filelist a list of (quoted) character strings? Correct pathnames to
the files with correct separators for your OS?

-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
The business of the statistician is to catalyze the scientific learning
process.  - George E. P. Box
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Peter Eiger
 Sent: Monday, August 14, 2006 1:00 PM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Fast way to load multiple files
 
 Hi,
 
 Instead of having to program a loop to load several 
 workspaces in a directory, it would be nice to store the 
 filenames in a list filelist and then to apply load to this list
 lapply( filelist, load)
 Unfortunately, although it seems that R is loading the files, 
 the contained objects are not available in the workspace afterwards.
 Any hints what I'm doing wrong or how to circumvent the problem?
 Peter
 --
 
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 http://www.R-project.org/posting-guide.html
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Re: [R] Making R script to run in a console

2006-08-14 Thread Marc Schwartz (via MN)
On Mon, 2006-08-14 at 14:12 -0300, Ronaldo Reis-Jr. wrote:
 Hi,
 
 is possible to make a R script to run under a console without open the R 
 environment?
 
 Something like this example.R
 
 #!/usr/bin/R
 
 function(name=Put here your name) {
 print(name)
 }
 
 In a console I make
 ./example.R name=Ronaldo Reis Júnior
 then program print my name.
 
 It is possible?
 
 Thanks
 Ronaldo

Ronaldo,

You might want to review these web pages:

http://wiki.r-project.org/rwiki/doku.php?id=developers:rinterp

http://kavaro.fi/mediawiki/index.php/Using_R_from_the_shell


HTH,

Marc Schwartz

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Re: [R] Fast way to load multiple files

2006-08-14 Thread Prof Brian Ripley
On Mon, 14 Aug 2006, Berton Gunter wrote:

 A reproducible example here would help (please see posting guide). A guess:
 is your filelist a list of (quoted) character strings? Correct pathnames to
 the files with correct separators for your OS?

I think the issue is (from the help page)

Usage:

 load(file, envir = parent.frame())
^^
Arguments:

file: a (readable binary) connection or a character string giving
  the name of the file to load.

   envir: the environment where the data should be loaded.

and so they were not loaded into .GlobalEnv. Try

lapply(filelist, load, envir=.GlobalEnv)

which works for me.

  -Original Message-
  From: [EMAIL PROTECTED] 
  [mailto:[EMAIL PROTECTED] On Behalf Of Peter Eiger
  Sent: Monday, August 14, 2006 1:00 PM
  To: r-help@stat.math.ethz.ch
  Subject: [R] Fast way to load multiple files
  
  Hi,
  
  Instead of having to program a loop to load several 
  workspaces in a directory, it would be nice to store the 
  filenames in a list filelist and then to apply load to this list
  lapply( filelist, load)
  Unfortunately, although it seems that R is loading the files, 
  the contained objects are not available in the workspace afterwards.
  Any hints what I'm doing wrong or how to circumvent the problem?

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


Re: [R] Making R script to run in a console

2006-08-14 Thread Romain Francois
Le 14.08.2006 19:12, Ronaldo Reis-Jr. a écrit :
 Hi,

 is possible to make a R script to run under a console without open the R 
 environment?

 Something like this example.R

 #!/usr/bin/R

 function(name=Put here your name) {
 print(name)
 }

 In a console I make
 ./example.R name=Ronaldo Reis Júnior
 then program print my name.

 It is possible?

 Thanks
 Ronaldo
   
Hi,

take a look at that wiki page : 
http://wiki.r-project.org/rwiki/doku.php?id=developers:rinterp

Cheers,

Romain

-- 
+---+
| Romain FRANCOIS   |
| R Graph Gallery : http://addictedtor.free.fr/graphiques   |
`---+

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Re: [R] Call for Beta Testers: R+ (read R plus) for Solaris and Linux:

2006-08-14 Thread Bernardo Rangel tura
At 12:21 PM 8/8/2006, [EMAIL PROTECTED] wrote:

We look forward to hearing your comments and inputs on R+ ... please
feel free to suggest a final name for our commercially supported R.

I don´t understanding this mail.

Is possible exist a commercial version of R?

If R source is licensed for GPL as free software 
other people can make a commercial version?


[]s
Tura

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Re: [R] Making R script to run in a console

2006-08-14 Thread Alexandre Aguiar
Em Seg 14 Ago 2006 17:58, Marc Schwartz (via MN) escreveu:
 http://kavaro.fi/mediawiki/index.php/Using_R_from_the_shell

I made a small change to the wrapper example by implementing dynamic 
allocation of memory and sizing the command line buffer to 32768 that is the 
real maximum size (at least under bash/sh :-).

#include stdio.h /* file i/o functions */
#include string.h /* string functions */
#include stdlib.h /* malloc and exit codes */

/* 
 * R_HOME has to be defined. Usually it is /usr/lib/R
 * todo: copy all command line params to R
*/
int main(int argc, char **argv)
{
 /* should be enough for all */
char *cmd, l;
FILE *out;

/* without this if a segmentation fault will happen 
if no parameters are passed */
if(argc == 1)
return(EXIT_FAILURE);

/* dynamically allocates memory for command line buffer */
cmd=(char *)malloc(32768);
/* assemble command line */
strcpy(cmd,/usr/lib/R/bin/exec/R --vanilla --slave  );
strcat(cmd,argv[1]);
 
/* do the real job */
out=popen(cmd,r);
l = fgetc(out);
printf(%c,l);
while(l != EOF) 
{
l = fgetc(out);
if(l !=EOF)
printf(%c,l);
}
/* cleanup: close files and free memory */
pclose(out);
free((void *)cmd);
return(EXIT_SUCCESS);
}


-- 


Alexandre Aguiar
Independent consultant for medical research
SPS Consultoria
Voice: +55-11-9320-2046
Fax: +55-11-5549-8760

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[R] Help with workaround for: Function '`[`' is not in the derivatives table

2006-08-14 Thread Earl F. Glynn
# This works fine:
 a - 1

 b - 2

 c - 3



 E - expression(a * exp(b*X) + c)



 X - c(0.5, 1.0, 2.0)



 eval(E)

[1]  5.718282 10.389056 57.598150



 D(E, b)

a * (exp(b * X) * X)

 eval(D(E, b))

[1]   1.359141   7.389056 109.196300



# But if (a,b,c) are replaced with (A[1], A[2], A[3]), how can I get a 
derivative using D?



 A - c(1, 2, 3)

 E - expression(A[1] * exp(A[2]*X) + A[3])

 X - c(0.5, 1.0, 2.0)

 eval(E)

[1]  5.718282 10.389056 57.598150



# Why doesn't this work?  Any workarounds?

 D(E, A[2])

Error in D(E, A[2]) : Function '`[`' is not in the derivatives table



If I want to have a long vector of coefficients, A, (perhaps dozens) how can 
I use D to compute partial derivatives?



Thanks for any help with this.



efg



Earl F. Glynn

Scientific Programmer

Stowers Institute for Medical Research

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[R] Alexandre MENICACCI/Daix/RED/GroupeFournier est absent(e).

2006-08-14 Thread a . menicacci




Je serai absent(e) du  14/08/2006 au 21/08/2006.

Je répondrai à votre message dès mon retour.

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Re: [R] Call for Beta Testers: R+ (read R plus) for Solaris and L inux:

2006-08-14 Thread Liaw, Andy
Note that it's commercially supported, not just commercial.  Also,
commercial does not mean closed source.  Open source (and even GPL) and
commercial need not be mutually exclusive.  Remember that the free in
free software is as in freedom rather than as in beer (as far as GPL is
concern, at least).
 
Having commercial support for R may not be a bad thing:  It may well help
adaptation in the corporate world.  (Linux probably would not have made
inroads into large corps if not for companies providing commercial support
for it.)  It would be great if those who profited from R can contribute back
to the project in some way.
 
Just my $0.02...
 
Andy

  _  

From: [EMAIL PROTECTED] on behalf of Bernardo Rangel tura
Sent: Mon 8/14/2006 5:25 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] Call for Beta Testers: R+ (read R plus) for Solaris and
Linux: [Broadcast]



At 12:21 PM 8/8/2006, [EMAIL PROTECTED] wrote: 

We look forward to hearing your comments and inputs on R+ ... please 
feel free to suggest a final name for our commercially supported R. 

I don´t understanding this mail. 

Is possible exist a commercial version of R? 

If R source is licensed for GPL as free software 
other people can make a commercial version? 


[]s 
Tura 

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

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Re: [R] Help with workaround for: Function '`[`' is not in thederivatives table

2006-08-14 Thread Berton Gunter
I think this is the sort of problem which is most elegantly handled by
computing on the language. Here is an INelegant solution: 

  A - c(1, 2, 3)

 for(i in 1:3)assign(paste('A',i,sep=''),A[i])

  E - expression(A1 * exp(A2*X) + A3) ## could also use substitute() here,
I think
## instead of explicitly assigning the coefficients

 X - c(0.5, 1.0, 2.0)

  eval(E)
[1]  5.718282 10.389056 57.598150

  D(E, A2)
A1 * (exp(A2 * X) * X)


Bert Gunter
Genentech
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Earl F. Glynn
 Sent: Monday, August 14, 2006 3:44 PM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Help with workaround for: Function '`[`' is not 
 in thederivatives table
 
 # This works fine:
  a - 1
 
  b - 2
 
  c - 3
 
 
 
  E - expression(a * exp(b*X) + c)
 
 
 
  X - c(0.5, 1.0, 2.0)
 
 
 
  eval(E)
 
 [1]  5.718282 10.389056 57.598150
 
 
 
  D(E, b)
 
 a * (exp(b * X) * X)
 
  eval(D(E, b))
 
 [1]   1.359141   7.389056 109.196300
 
 
 
 # But if (a,b,c) are replaced with (A[1], A[2], A[3]), how 
 can I get a 
 derivative using D?
 
 
 
  A - c(1, 2, 3)
 
  E - expression(A[1] * exp(A[2]*X) + A[3])
 
  X - c(0.5, 1.0, 2.0)
 
  eval(E)
 
 [1]  5.718282 10.389056 57.598150
 
 
 
 # Why doesn't this work?  Any workarounds?
 
  D(E, A[2])
 
 Error in D(E, A[2]) : Function '`[`' is not in the derivatives table
 
 
 
 If I want to have a long vector of coefficients, A, (perhaps 
 dozens) how can 
 I use D to compute partial derivatives?
 
 
 
 Thanks for any help with this.
 
 
 
 efg
 
 
 
 Earl F. Glynn
 
 Scientific Programmer
 
 Stowers Institute for Medical Research
 
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[R] merge 2 data frame based on more than 2 variables

2006-08-14 Thread Wensui Liu
Dear Lister,

I understand merge() can be used to join 2 data frames based on 1 variable.
But how about merge based on more than 2 variables?

Thank you so much!

-- 
WenSui Liu
(http://spaces.msn.com/statcompute/blog)
Senior Decision Support Analyst
Health Policy and Clinical Effectiveness
Cincinnati Children Hospital Medical Center

[[alternative HTML version deleted]]

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Re: [R] merge 2 data frame based on more than 2 variables

2006-08-14 Thread Simon Blomberg
Wensui Liu wrote:
 Dear Lister,

 I understand merge() can be used to join 2 data frames based on 1 variable.
 But how about merge based on more than 2 variables?

 Thank you so much!

   
Just specify the 2 (or more) variable names in a column vector for by)

merge(dat1, dat2, by= c(VarA, VarB))

assuming both data frames have columns VarA and VarB.

-- 
Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
Centre for Resource and Environmental Studies
The Australian National University
Canberra ACT 0200
Australia
T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au
F: +61 2 6125 0757
CRICOS Provider # 00120C

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Re: [R] merge 2 data frame based on more than 2 variables

2006-08-14 Thread Wensui Liu
what if the names are different in 2 data frames?


On 8/14/06, Simon Blomberg [EMAIL PROTECTED] wrote:
 Wensui Liu wrote:
  Dear Lister,
 
  I understand merge() can be used to join 2 data frames based on 1 variable.
  But how about merge based on more than 2 variables?
 
  Thank you so much!
 
 
 Just specify the 2 (or more) variable names in a column vector for by)

 merge(dat1, dat2, by= c(VarA, VarB))

 assuming both data frames have columns VarA and VarB.

 --
 Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
 Centre for Resource and Environmental Studies
 The Australian National University
 Canberra ACT 0200
 Australia
 T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au
 F: +61 2 6125 0757
 CRICOS Provider # 00120C




-- 
WenSui Liu
(http://spaces.msn.com/statcompute/blog)
Senior Decision Support Analyst
Health Policy and Clinical Effectiveness
Cincinnati Children Hospital Medical Center

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[R] help: cannot allocate vector of length 828310236

2006-08-14 Thread T Mu
Hi all,

I was trying a probit regression using polr() and got this message,

Error in model.matrix.default(Terms, m, contrasts) :
cannot allocate vector of length 828310236

The data is about 20M (a few days ago I asked a question about large file,
thank you for responses, then I use MS Access to select those columns I
would use).

R is 2.3.1, Windows XP, 512M Ram.

I am going to read some help on memory use in R, but hope anybody can give
me some quick hints.

Is it because iphysical memory runs out, or some other things could be wrong
with data or polr()?
Does R use virtual memory? If so, what options can I set?
If not, can R deal with really huge data (except adding RAM according to
data size)? If this is the case, it is too bad that I have to tell my boss
to go back to SAS. Now it is not a speed issue yet.

Thank you.

[[alternative HTML version deleted]]

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Re: [R] merge 2 data frame based on more than 2 variables

2006-08-14 Thread Simon Blomberg
Then instead of by, use by.x and by.y to specifiy the variable names 
separately for both data frames. See ?merge, especially the examples.

Wensui Liu wrote:
 what if the names are different in 2 data frames?


 On 8/14/06, Simon Blomberg [EMAIL PROTECTED] wrote:
   
 Wensui Liu wrote:
 
 Dear Lister,

 I understand merge() can be used to join 2 data frames based on 1 variable.
 But how about merge based on more than 2 variables?

 Thank you so much!


   
 Just specify the 2 (or more) variable names in a column vector for by)

 merge(dat1, dat2, by= c(VarA, VarB))

 assuming both data frames have columns VarA and VarB.

 --
 Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
 Centre for Resource and Environmental Studies
 The Australian National University
 Canberra ACT 0200
 Australia
 T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au
 F: +61 2 6125 0757
 CRICOS Provider # 00120C


 


   


-- 
Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat.
Centre for Resource and Environmental Studies
The Australian National University
Canberra ACT 0200
Australia
T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au
F: +61 2 6125 0757
CRICOS Provider # 00120C

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