Re: [R] isolate elements in vector that match one of many possible values

2008-09-09 Thread Adam D. I. Kramer

Check out ?match, ?"%in%"


x <- c(1,2,3,4)
y <- c(1,2,4)
match(y,x)

[1] 1 2 4




--Adam

On Mon, 8 Sep 2008, Andrew Barr wrote:


Hi all,

I want to get the index numbers of all elements of a vector which match any
of a long series of possible values.  Say x <- c(1,2,3,4) and I want to know
which values are equal to 1, 2 or 4.  I could do

which(x == 1 | x==2 | x==4)
[1] 1 2 4

This gets really ugly though, when the list of values of interest is really
long.  Is there a nicer way to do this?  Something akin to the MySQL
construction in(), as in

#MySQL script example
Select * from table where parameter in(x,y,z);

Thanks!

--
W. Andrew Barr
Biological Anthropology
University of Texas at Austin

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Re: [R] make methods work in lapply - remove lapply's environment

2008-09-09 Thread Prof Brian Ripley
This is a side-effect of lapply being in the base namespace and not 
evaluating its arguments, as explained on its help page which also points 
out that using a wrapper is sometimes needed.  It also points out that 
code has been written that relies on the current behaviour.


On Mon, 8 Sep 2008, Tim Hesterberg wrote:


I've defined my own version of summary.default,
that gives a better summary for highly skewed vectors.

If I call
 summary(x)
the method is used.

If I call
 summary(data.frame(x))
the method is not used.

I've traced this to lapply; this uses the new method:
 lapply(list(x), function(x) summary(x))
and this does not:
 lapply(list(x), summary)

If I make a copy of lapply, WITHOUT the environment,
then the method is used.

lapply <- function (X, FUN, ...) {
   FUN <- match.fun(FUN)
   if (!is.vector(X) || is.object(X))
   X <- as.list(X)
   .Internal(lapply(X, FUN))
}

I'm curious to hear reactions to this.
There is a March 2006 thread
   object size vs. file size
in which Duncan Murdoch wrote:

Functions in R consist of 3 parts: the formals, the body, and the
environment. You can't remove any part, but you can change it.

That is exactly what I want to do, remove the environment, so that
when I define a better version of some function that the better
version is used.

Here's a function to automate the process:
copyFunction <- function(Name){
 # Copy a function, without its environment.
 # Name should be quoted
 # Return the copy
 file <- tempfile()
 on.exit(unlink(file))
 dput(get(Name), file = file)
 f <- source(file)$value
 f
}
lapply <- copyFunction("lapply")

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Brian D. Ripley,  [EMAIL PROTECTED]
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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] naive variance in GEE

2008-09-09 Thread Prof Brian Ripley

On Mon, 8 Sep 2008, Qiong Yang wrote:

The standard error from logistic regression is slightly different from the 
naive SE from GEE under independence working correlation structure.


Shouldn't they be identical? Anyone has insight about this?


They are computed quantities from iterations with different stopping 
criteria.  The coefficients are not 'identical' either.


Your example is incorrect (the first line) and not reproducible (no seed 
is set, no library gee), so we don't know what you saw.  But with


set.seed(1)
a <- rbinom(1000, 1, 0.2)
b <- rbinom(1000, 2, 0.1)
c <- rbinom(1000, 10, 0.5)
library(gee)
summary(gee(a ~ b, id=c, family="binomial", corstr="independence"))$coef
summary(glm(a ~ b, family="binomial"))$coef

the differences I see are negligible.  I suggest you talk to your 
supervisor about some courses on numerical methods.




Thanks,
Qiong

a<-rbinom(1000,1)
b<-rbinom(1000,2,0.1)
c<-rbinom(1000,10,0.5)
summary(gee(a~b, id=c,family="binomial",corstr="independence"))$coef
summary(glm(a~b,family="binomial"))

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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] S.O.S "try" doesnot work in boot?

2008-09-09 Thread ctu

First thanks for Jinsong's suggestions
I would like to do a bootstrap in a nonlinear model. But it fails to  
converge in most of time. (it did converge if I just use nls without  
boot). Thus, I use "try" function to resolve my problem. This  
following code is from Jinsong's suggestion.


h1a.nls<-nls(density~nmf(time, alpha, delta, psi, tau, gamma),data=h1a,
start=c(alpha=0.3, delta=0.08869, psi=1.26523, tau=3.93919,
   gamma=-1.41927))


h1a.data<-data.frame(h1a,res=resid(h1a.nls),fitted=fitted(h1a.nls))
h1a.fun<-function(data,i){
 d<-data
 d$density<-d$fitted+d$res[i]
 try(update(h1a.nls,data=d),silent=T)
 if(!inherits(h1a.nls,"try-error")) h1a.coef<-coef(h1a.nls)
 else h1a.coef<-NA
 h1a.coef
 }
h1a.boot<-boot(h1a.data, statistic = h1a.fun, R=1000)

h1a.boot


ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = h1a.data, statistic = h1a.fun, R = 1000)
Bootstrap Statistics :
   original  biasstd. error
t1*  0.27892590   0   0
t2*  0.08869433   0   0
t3*  1.26523275   0   0
t4*  3.93919567   0   0
t5* -1.41926966   0   0
all of the values of each column in h1a.boot$t are the same.
Is anyone know to how I can solve this problem?
Appreciate in advance

Chunhao

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Re: [R] correct lme syntax for this problem?

2008-09-09 Thread ONKELINX, Thierry
Dear Matthew,

First of all I'm forwarding this to R-SIG-Mixed, which is a more
appropriate list for your question.
Using a mixed effect with only 5 levels is a borderline situation.
Douglas Bates recommends at least 6 levels in order to get a more or
less reliable estimate. So I would consider the populations as fixed
effects. Do you have repeated measurements of individuals within your
populations? If you do you could use those as random effects.

Your anova tests whether the variances of the random slope on SPI is
zero. I think you might want this:

mod1 <- lm(height ~ SPI * population + covariate1 + covariate2)
mod2 <- lm(height ~ SPI + population + covariate1 + covariate2)
anova(mod1, mod2)

HTH,

Thierry



ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research 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 

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Matthew Keller
Verzonden: dinsdag 9 september 2008 1:10
Aan: R Help
Onderwerp: [R] correct lme syntax for this problem?

Hello all,

I am about to send off a manuscript and, although I am fairly
confident I have used the lme function correctly, I want to be 100%
sure. Could some kind soul out there put my mind at ease?

I am simply interested in whether a predictor (SPI) is related to
height. However, there are five different populations, and each may
differ in mean level of height as well as the relationship between SPI
and height. Thus, I also want to a) account for mean level differences
in height and b) check whether the relationship between height and SPI
is different between the groups. I hope this is sufficient
information.

height, SPI, covariate1, and covariate2 are numeric. population is a
factor with 5 levels. Here are the steps I took:

summary(mod1 <- lme(height ~ SPI + covariate1 + covariate2, random = ~
SPI | population))

summary(mod2 <- lme(height ~ SPI + covariate1 + covariate2, random = ~
1 | population))

anova(mod1,mod2) #this checks whether there is evidence for IQ & SPI
being related differently between the 5 populations.

Is this correct? THANKS!

Matt


-- 
Matthew C Keller
Asst. Professor of Psychology
University of Colorado at Boulder
www.matthewckeller.com

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Re: [R] S.O.S "try" doesnot work in boot?

2008-09-09 Thread Prof Brian Ripley
Returning NA (of the correct length, not length 1) will not help you, as 
all the derived statistics from the bootstrap runs will be NA.


But here you never looked at the result of try.

On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


First thanks for Jinsong's suggestions
I would like to do a bootstrap in a nonlinear model. But it fails to converge 
in most of time. (it did converge if I just use nls without boot). Thus, I 
use "try" function to resolve my problem. This following code is from 
Jinsong's suggestion.


h1a.nls<-nls(density~nmf(time, alpha, delta, psi, tau, gamma),data=h1a,
  start=c(alpha=0.3, delta=0.08869, psi=1.26523, tau=3.93919, 
gamma=-1.41927))


h1a.data<-data.frame(h1a,res=resid(h1a.nls),fitted=fitted(h1a.nls))
h1a.fun<-function(data,i){
d<-data
d$density<-d$fitted+d$res[i]
try(update(h1a.nls,data=d),silent=T)
if(!inherits(h1a.nls,"try-error")) h1a.coef<-coef(h1a.nls)


h1a.nls is the original fit, not the result of try().


else h1a.coef<-NA
h1a.coef
}
h1a.boot<-boot(h1a.data, statistic = h1a.fun, R=1000)

h1a.boot


ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = h1a.data, statistic = h1a.fun, R = 1000)
Bootstrap Statistics :
 original  biasstd. error
t1*  0.27892590   0   0
t2*  0.08869433   0   0
t3*  1.26523275   0   0
t4*  3.93919567   0   0
t5* -1.41926966   0   0
all of the values of each column in h1a.boot$t are the same.
Is anyone know to how I can solve this problem?
Appreciate in advance

Chunhao

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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] How do I compute interactions with anova.mlm ?

2008-09-09 Thread Schadwinkel, Stefan
Hi,

I wish to compute multivariate test statistics for a within-subjects repeated 
measures design with anova.mlm. 

This works great if I only have two factors, but I don't know how to compute 
interactions with more than two factors. 
I suspect, I have to create a new "grouping" factor and then test with this 
factor to get these interactions (as it is hinted in R News 2007/2), 
but I don't really know how to use this approach. 

Here is my current code:

Two Factors: fac1, fac2

mlmfit <- lm(mydata~1)
mlmfit0 <- update(mlmfit, ~0)

% test fac1, works, produces same output as SAS
anova(mlmfit, mlmfit0, M = ~ fac1 + fac2, X = ~ fac2, idata = idata, test = 
"Wilks")

% test fac1*fac2 interaction, also works, also the same output as SAS
anova(mlmfit, mlmfit0, X = ~ fac1 + fac2, idata = idata, test = "Wilks")



Three Factors: fac1, fac2, fac3 

mlmfit <- lm(mydata~1)
mlmfit0 <- update(mlmfit, ~0)

% test fac1, works, same as SAS
anova(mlmfit, mlmfit0, M = ~ fac1 + fac2 + fac3, X = ~ fac2 + fac3, idata = 
idata, test = "Wilks")



Now, I try to compute the interactions the same way, but this doesn't work:

% fac1*fac2
anova(mlmfit, mlmfit0, M = ~ fac1 + fac2 + fac3, X = ~ fac3, idata = idata, 
test = "Wilks") 

% fac1*fac2*fac3
anova(mlmfit, mlmfit0, X = ~ fac1 + fac2 + fac3, idata = idata, test = "Wilks")


Both of these above differ quite much from the SAS output and I suspect, my 
understanding of X and M is somewhat flawed. 

I would be very happy, if someone could tell me how to compute the two 
interactions above and an interaction of N factors in general.

I would also be interested in computing linear contrasts using the T matrix and 
anova.mlm.

Thank you very much,

Stefan 

 

--
Stefan Schadwinkel, Dipl.-Inf.
Neurologische Klinik
Sektion Biomagnetismus
Universität Heidelberg
Im Neuenheimer Feld 400
69120 Heidelberg

Telefon:  06221 - 56 5196
Email:[EMAIL PROTECTED] 

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[R] exporting tapply objects to csv-files

2008-09-09 Thread Kunzler, Andreas
Dear Everyone,

I try to create a cvs-file with different results form the table function.

Imagine a data-frame with two vectors a and b where b is of the class factor. 

I use the tapply function to count a for the different values of b.

tapply(a,b,table)

and I use the table function to have a look of the frequencies as a total

table(a)

I would like to put both results together in one txt or csv file that I can 
import to e.g. Excel.

The export file should have a layout like

1,2,3,4,5,6,7 (possible values of a)
3,6,7,8,8,8,1 (Counts of a total)
1,2,3,4,5,3,0 (Counts of a where b==A)
2,4,4,4,3,5,1 (Counts of a where b==B)

I tried to change the class of the table result to a matrix but I could not 
find a way to use the results of tapply. I use tapply because b has 15 
different values.

Thanx

Andreas Kunzler

Bundeszahnärztekammer (BZÄK)
Chausseestraße 13
10115 Berlin

Tel.: 030 40005-113
Fax:  030 40005-119

E-Mail: [EMAIL PROTECTED] 

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Re: [R] Read from url requiring authentication?

2008-09-09 Thread Damien
René Sachse wrote:

> Damien schrieb:
>
> > I'm looking into opening an url on a server which requires
> >authentication.
>
> Under a Windows Operating System you could try to start R with the
> --internet2 option. This worked in my case.

Thanks René it did the trick for me too!

Best Regards,
Damien

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Re: [R] Read from url requiring authentication?

2008-09-09 Thread Damien
On 8 Sep, 20:15, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> On Mon, 8 Sep 2008, Damien wrote:
> > Hi all,
>
> > I'm looking into opening an url on a server which requires
> >authentication.
>
> > After failing to find some kind of connection structure to fill in I
> > turned to explicitly stating the credentials in the url itself (e.g.
> > "http://username:[EMAIL PROTECTED]").
> > Sadly this didn't do the trick either and both source() and url()
> > failed trying to resolve the username ()
>
> > Is there anything I missed in the documentation/internet/groups?
> > If not could I maybe add to the existing R functions as it doesn't
> > seem too far of a stretch to allow the username and password in the
> > url string fed to the web server?
>
> Look at the RCurl package: it is more like download.file than url, though,
> and you could perhaps wse the "wget" method of download.file.

Thank you for the quick reply,

it seems that the argument "--internet2" did solve my immediate
problem
but I'll have a look at RCurl too.

Best Regards,
Damien

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Re: [R] Compiling date

2008-09-09 Thread David Scott

On Mon, 8 Sep 2008, Megh Dal wrote:


Hi,

I have following kind of dataset (all are dates) in my Excel sheet.

09/08/08
09/05/08
09/04/08
09/02/08
09/01/08
29/08/2008
28/08/2008
27/08/2008
26/08/2008
25/08/2008
22/08/2008
21/08/2008
20/08/2008
18/08/2008
14/08/2008
13/08/2008
08/12/08
08/11/08
08/08/08
08/07/08

However I want to use R to compile those data to make all dates in same 
format. Can anyone please tell me any automated way for doing that?




Well you have to read them in as character first. Then use sub to make the 
two digit years into four digits. The following could probably be improved 
by a regular expression whiz, but works:



strngs <- c("06/05/08","23/11/2008")
sub("([0-9][0-9]/[0-9][0-9]/)([0-9][0-9]$)","\\120\\2",strngs)

[1] "06/05/2008" "23/11/2008"


David Scott



_
David Scott Department of Statistics, Tamaki Campus
The University of Auckland, PB 92019
Auckland 1142,NEW ZEALAND
Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000
Email:  [EMAIL PROTECTED]

Graduate Officer, Department of Statistics
Director of Consulting, Department of Statistics

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Re: [R] Memory allocation problem (during kmeans)

2008-09-09 Thread Peter Dalgaard
rami batal skrev:
> Dear all,
>
> I am trying to apply kmeans clusterring on a data file (size is about 300
> Mb)
>
> I read this file using
>
> x=read.table('file path' , sep=" ")
>
> then i do kmeans(x,25)
>
> but the process stops after two minutes with an error :
>
> Error: cannot allocate vector of size 907.3 Mb
>
> when i read the archive i notice that the best solution is to use a 64bit
> OS.
>
> "Error messages beginning cannot allocate vector of size indicate a failure
> to obtain memory, either because the size exceeded the address-space limit
> for a process or, more likely, because the system was unable to provide the
> memory. Note that on a 32-bit OS there may well be enough free memory
> available, but not a large enough contiguous block of address space into
> which to map it. "
>
> the problem that I have two machines with two OS (32bit and 64bit) and when
> i used the 64bit OS the same error remains.
>
> Thank you if you have any suggestions to me and excuse me because i am a
> newbie.
>
> Here the default information for the 64bit os:
>
>   
>> sessionInfo()
>> 
> R version 2.7.1 (2008-06-23)
> x86_64-redhat-linux-gnu
>
>   
>> gc()
>> 
>  used (Mb) gc trigger (Mb) max used (Mb)
> Ncells 137955  7.4 35 18.7   35 18.7
> Vcells 141455  1.1 786432  6.0   601347  4.6
>
> I tried also to start R using the options to control the available memory
> and the result still the same. or maybe i don't assign the correct values.
>
>   
It might be a good idea first to work out what the actual memory
requirements are. 64 bits does not help if you are running out of RAM
(+swap).

-- 
   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 do I compute interactions with anova.mlm ?

2008-09-09 Thread Peter Dalgaard
Schadwinkel, Stefan skrev:
> Hi,
>
> I wish to compute multivariate test statistics for a within-subjects repeated 
> measures design with anova.mlm. 
>
> This works great if I only have two factors, but I don't know how to compute 
> interactions with more than two factors. 
> I suspect, I have to create a new "grouping" factor and then test with this 
> factor to get these interactions (as it is hinted in R News 2007/2), 
> but I don't really know how to use this approach. 
>
> Here is my current code:
>
> Two Factors: fac1, fac2
>
> mlmfit <- lm(mydata~1)
> mlmfit0 <- update(mlmfit, ~0)
>
> % test fac1, works, produces same output as SAS
> anova(mlmfit, mlmfit0, M = ~ fac1 + fac2, X = ~ fac2, idata = idata, test = 
> "Wilks")
>
> % test fac1*fac2 interaction, also works, also the same output as SAS
> anova(mlmfit, mlmfit0, X = ~ fac1 + fac2, idata = idata, test = "Wilks")
>
>
>
> Three Factors: fac1, fac2, fac3 
>
> mlmfit <- lm(mydata~1)
> mlmfit0 <- update(mlmfit, ~0)
>
> % test fac1, works, same as SAS
> anova(mlmfit, mlmfit0, M = ~ fac1 + fac2 + fac3, X = ~ fac2 + fac3, idata = 
> idata, test = "Wilks")
>
>
>
> Now, I try to compute the interactions the same way, but this doesn't work:
>
> % fac1*fac2
> anova(mlmfit, mlmfit0, M = ~ fac1 + fac2 + fac3, X = ~ fac3, idata = idata, 
> test = "Wilks") 
>
> % fac1*fac2*fac3
> anova(mlmfit, mlmfit0, X = ~ fac1 + fac2 + fac3, idata = idata, test = 
> "Wilks")
>
>
> Both of these above differ quite much from the SAS output and I suspect, my 
> understanding of X and M is somewhat flawed. 
>
> I would be very happy, if someone could tell me how to compute the two 
> interactions above and an interaction of N factors in general.
>
>   
You need to ensure that the difference between the X and M models is the
relevant interaction, so something like

M=~fac1*fac2*fac3
X=~fac1*fac2*fac3 - fac1:fac2:fac3

should test for fac1:fac2:fac3

If the within-subject design is fac1*fac2*fac3 with one observation per
cell (NB!), then you can omit M. X can also be written as
~fac1*fac2+fac2*fac3+fac1*fac3 or ~(fac1+fac2+fac3)^2.

For the next step, use, e.g.,

M=~fac1*fac2+fac2*fac3+fac1*fac3
X=~fac2*fac3+fac1*fac3

to test significance of fac1:fac2 (notice that the main effects are
still in X becaus of the meaning of the "*" operator in R).


> I would also be interested in computing linear contrasts using the T matrix 
> and anova.mlm.
>
> Thank you very much,
>
> Stefan 
>
>  
>
> --
> Stefan Schadwinkel, Dipl.-Inf.
> Neurologische Klinik
> Sektion Biomagnetismus
> Universität Heidelberg
> Im Neuenheimer Feld 400
> 69120 Heidelberg
>
> Telefon:  06221 - 56 5196
> Email:[EMAIL PROTECTED] 
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>   


-- 
   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] match problem by rownames

2008-09-09 Thread Charilaos Skiadas

As suggested in ?"[.data.frame", try:

dat[match('a1', rownames(dat)),]


Haris Skiadas
Department of Mathematics and Computer Science
Hanover College

On Sep 9, 2008, at 2:41 AM, Xianming Wei wrote:


Hi all,

While dat['a1',] and dat['a10',] produce the same results in the
following example, I'd like dat['a1',] to return NAs.

dat <- data.frame(x1 = paste(letters[1:5],10, sep=''), x2=rnorm(5))
rownames(dat) <- dat$x1
dat['a1',]
dat['a10',]


sessionInfo()

R version 2.7.2 (2008-08-25)
i386-pc-mingw32

locale:
LC_COLLATE=English_Australia.1252;LC_CTYPE=English_Australia. 
1252;LC_MON
ETARY=English_Australia.1252;LC_NUMERIC=C;LC_TIME=English_Australia. 
1252


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


other attached packages:
[1] lattice_0.17-13

loaded via a namespace (and not attached):
[1] grid_2.7.2




Regards,
Xianming



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Re: [R] how to split a data framed with sequences

2008-09-09 Thread jim holtman
Is this what you want:

> my.df <- data.frame(a = c(1:5, 1:10, 1:20), b = runif(35))
> split(my.df, c(0, cumsum(diff(my.df$a) < 0)))
$`0`
  a b
1 1 0.2655087
2 2 0.3721239
3 3 0.5728534
4 4 0.9082078
5 5 0.2016819

$`1`
a  b
6   1 0.89838968
7   2 0.94467527
8   3 0.66079779
9   4 0.62911404
10  5 0.06178627
11  6 0.20597457
12  7 0.17655675
13  8 0.68702285
14  9 0.38410372
15 10 0.76984142

$`2`
a  b
16  1 0.49769924
17  2 0.71761851
18  3 0.99190609
19  4 0.38003518
20  5 0.77744522
21  6 0.93470523
22  7 0.21214252
23  8 0.65167377
24  9 0.1210
25 10 0.26722067
26 11 0.38611409
27 12 0.01339033
28 13 0.38238796
29 14 0.86969085
30 15 0.34034900
31 16 0.48208012
32 17 0.59956583
33 18 0.49354131
34 19 0.18621760
35 20 0.82737332

>


On Tue, Sep 9, 2008 at 5:38 AM, David Carslaw
<[EMAIL PROTECTED]> wrote:
>
> Hi all,
>
> Given a data frame:
>
> my.df <- data.frame(a = c(1:5, 1:10, 1:20), b = runif(35))
>
> I want to split it by "a" such that I end up with a list containing 3
> components i.e. the first containing a = 1 to 5, the second a = 1 to 10 etc.
> In other words, sets of sequences of a.
>
> I can't seem to find the right form using the split function - can you help?
>
> Much appreciated.
>
> David
>
>
>
> -
> Institute for Transport Studies
> University of Leeds
> --
> View this message in context: 
> http://www.nabble.com/how-to-split-a-data-framed-with-sequences-tp19388964p19388964.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

What is the problem that you are trying to solve?

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[R] how to split a data framed with sequences

2008-09-09 Thread David Carslaw

Hi all,

Given a data frame:

my.df <- data.frame(a = c(1:5, 1:10, 1:20), b = runif(35))

I want to split it by "a" such that I end up with a list containing 3
components i.e. the first containing a = 1 to 5, the second a = 1 to 10 etc.
In other words, sets of sequences of a.

I can't seem to find the right form using the split function - can you help? 

Much appreciated.

David



-
Institute for Transport Studies
University of Leeds
-- 
View this message in context: 
http://www.nabble.com/how-to-split-a-data-framed-with-sequences-tp19388964p19388964.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] match problem by rownames

2008-09-09 Thread Dimitris Rizopoulos
try this:

dat <- data.frame(x1 = paste(letters[1:5],10, sep=''), x2=rnorm(5))
row.names(dat) <- dat$x1

dat['a1' %in% row.names(dat), ]
dat['a10'  %in% row.names(dat), ]


I hope it helps.

Best,
Dimitris


> Hi all,
>
> While dat['a1',] and dat['a10',] produce the same results in the
> following example, I'd like dat['a1',] to return NAs.
>
> dat <- data.frame(x1 = paste(letters[1:5],10, sep=''), x2=rnorm(5))
> rownames(dat) <- dat$x1
> dat['a1',]
> dat['a10',]
>
>> sessionInfo()
> R version 2.7.2 (2008-08-25)
> i386-pc-mingw32
>
> locale:
> LC_COLLATE=English_Australia.1252;LC_CTYPE=English_Australia.1252;LC_MON
> ETARY=English_Australia.1252;LC_NUMERIC=C;LC_TIME=English_Australia.1252
>
> attached base packages:
> [1] stats graphics  grDevices utils datasets  methods   base
>
>
> other attached packages:
> [1] lattice_0.17-13
>
> loaded via a namespace (and not attached):
> [1] grid_2.7.2
>>
>
> Regards,
> Xianming
>
>
>
> DISCLAIMER:\ For details of our e-mail disclaimer, pleas...{{dropped:15}}
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043399
Fax: +31/(0)10/7044657

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[R] plotting group means

2008-09-09 Thread Erich Studerus
Hi all,

 

I want to plot the grouped means of some variables. The dependent variables
and the grouping factor are stored in different columns. I want to draw a
simple line-plot of means, in which the x-axis represents the variables and
y-axis represents the means. The means of the groups should be connected by
lines. So far, the only function that I could find comes closest to what I'm
looking for, is the error.bars.by-function in the psych-package. To know,
what I'm looking for, just type:

 

library(psych)
x <- matrix(rnorm(500),ncol=20)
y <- sample(4,25 ,replace=TRUE)
x <- x+y
error.bars.by(x,y,ci=0)

 

Now, I want to put a legend for the grouping factor of this graph. I also
would like to manipulate the linetypes and colors of the lines. I've read
the documentation, but it was not clear to me, how to do this. Are there
other plotting functions in R, which can do the same?

 

Erich

 



Erich Studerus
Lic. Phil. Klinische Psychologie
Psychiatric University Hospital Zurich
Division of Clinical Research
Lenggstr. 31
CH-8008 Zurich
Switzerland
Mail: [EMAIL PROTECTED]
Office: +41 44 384 26 66
Mobile: +41 76 563 31 54


 


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Re: [R] Question about multiple regression

2008-09-09 Thread Gustaf Rydevik
On Mon, Sep 8, 2008 at 7:47 PM, Dimitri Liakhovitski <[EMAIL PROTECTED]> wrote:
> Thank you everyone for your responses. I'll answer several questions.
>
> 1. >  Disclaimer: I have **NO IDEA** of the details of what you want
> to do or why
>> -- but I am willing to bet that there are better ways of doing it than  1.8
>> mm multiple refressions that take 270 secs each!! (which I find difficult to
>> believe in itself -- are you sure you are doing things right? Something
>> sounds very fishy here: R's regression code is typically very fast).
> I probably should not bore everyone, but just to explain where the
> large number is coming from. I have an experimental design with 7
> factors. Each factor has between 3 and 5 levels. Once you cross them
> all, you end up with 18,000 cells. For each cell, I want to generate a
> sample of N=100. For each sample I have to analyze the data using 3
> different statistical methods of analysis (the goal of the
> Monte-Carlo) is to compare those methods. One of the methods requires
> running of up to ~32,000 simple multiple regressions - yes just for
> one sample and it's not a mistake. I test-ran one such analysis for a
> sample with N=800 and 15 predictors and it took 270 seconds. R was
> actually very fast - it ran each of the individual regressions in
> about 0.008 seconds. Still I need something faster.
>
> 2. Sorry - what was the formula sum(lm.fit(x,y))$residuals^2) for? For
> example, using it on my data, I got a value of 36,644...
>
> 3. I know that for similarly challenging situations people did used
> Fortran compilers. So, anyone heard of a free Fortran library or an
> efficient piece of code?
>
> Thank you!
> Dimitri
>


Have you considered the fact that 32000 regressions simply takes a lot of time?
I don't really have anything to go by, but it sounds unlikely that you
will be able to cut computing time by more than, say, ten times to 27
second. That would still leave you with 4 months of running a
computer.

Perhaps an alternative approach would be to get access to stronger
(super)computers, either at a university, or buying access. A quick
googling turns up http://www.clusterondemand.com/ for example.

Anyhow, good luck with your project! I'm sure the R list would be very
interested to hear of how you solved your problem.

Regards,

Gustaf


-- 
Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE
skype:gustaf_rydevik

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Re: [R] Compiling date

2008-09-09 Thread Henrique Dallazuanna
Try this:

strptime(x, ifelse(nchar(x) == 8, '%d/%m/%y', '%d/%m/%Y'))

On Tue, Sep 9, 2008 at 3:48 AM, Megh Dal <[EMAIL PROTECTED]> wrote:

> Hi,
>
> I have following kind of dataset (all are dates) in my Excel sheet.
>
> 09/08/08
> 09/05/08
> 09/04/08
> 09/02/08
> 09/01/08
> 29/08/2008
> 28/08/2008
> 27/08/2008
> 26/08/2008
> 25/08/2008
> 22/08/2008
> 21/08/2008
> 20/08/2008
> 18/08/2008
> 14/08/2008
> 13/08/2008
> 08/12/08
> 08/11/08
> 08/08/08
> 08/07/08
>
> However I want to use R to compile those data to make all dates in same
> format. Can anyone please tell me any automated way for doing that?
>
> __
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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.
>



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

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Re: [R] Compiling date

2008-09-09 Thread Dr Eberhard Lisse
Why not Format -> Cell in Excell?

el

on 9/9/08 1:03 PM Henrique Dallazuanna said the following:
> Try this:
> 
> strptime(x, ifelse(nchar(x) == 8, '%d/%m/%y', '%d/%m/%Y'))
> 
> On Tue, Sep 9, 2008 at 3:48 AM, Megh Dal <[EMAIL PROTECTED]> wrote:
> 
>> Hi,
>>
>> I have following kind of dataset (all are dates) in my Excel sheet.
>>
>> 09/08/08
>> 09/05/08
>> 09/04/08
>> 09/02/08
>> 09/01/08
>> 29/08/2008
>> 28/08/2008
>> 27/08/2008
>> 26/08/2008
>> 25/08/2008
>> 22/08/2008
>> 21/08/2008
>> 20/08/2008
>> 18/08/2008
>> 14/08/2008
>> 13/08/2008
>> 08/12/08
>> 08/11/08
>> 08/08/08
>> 08/07/08
>>
>> However I want to use R to compile those data to make all dates in same
>> format. Can anyone please tell me any automated way for doing that?

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Re: [R] exporting tapply objects to csv-files

2008-09-09 Thread Henrique Dallazuanna
Try creating a new object:

tb <- rbind(table(a), do.call(rbind.data.frame, tapply(a, b, table)))
names(tb) <- unique(a)

then write to csv by write.table.

On Tue, Sep 9, 2008 at 5:48 AM, Kunzler, Andreas <[EMAIL PROTECTED]> wrote:

> Dear Everyone,
>
> I try to create a cvs-file with different results form the table function.
>
> Imagine a data-frame with two vectors a and b where b is of the class
> factor.
>
> I use the tapply function to count a for the different values of b.
>
> tapply(a,b,table)
>
> and I use the table function to have a look of the frequencies as a total
>
> table(a)
>
> I would like to put both results together in one txt or csv file that I can
> import to e.g. Excel.
>
> The export file should have a layout like
>
> 1,2,3,4,5,6,7 (possible values of a)
> 3,6,7,8,8,8,1 (Counts of a total)
> 1,2,3,4,5,3,0 (Counts of a where b==A)
> 2,4,4,4,3,5,1 (Counts of a where b==B)
>
> I tried to change the class of the table result to a matrix but I could not
> find a way to use the results of tapply. I use tapply because b has 15
> different values.
>
> Thanx
>
> Andreas Kunzler
> 
> Bundeszahnärztekammer (BZÄK)
> Chausseestraße 13
> 10115 Berlin
>
> Tel.: 030 40005-113
> Fax:  030 40005-119
>
> E-Mail: [EMAIL PROTECTED]
>
> __
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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.
>



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

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Re: [R] plotting group means

2008-09-09 Thread Chuck Cleland
On 9/9/2008 6:49 AM, Erich Studerus wrote:
> Hi all,
> 
>  
> 
> I want to plot the grouped means of some variables. The dependent variables
> and the grouping factor are stored in different columns. I want to draw a
> simple line-plot of means, in which the x-axis represents the variables and
> y-axis represents the means. The means of the groups should be connected by
> lines. So far, the only function that I could find comes closest to what I'm
> looking for, is the error.bars.by-function in the psych-package. To know,
> what I'm looking for, just type:
> 
>  
> 
> library(psych)
> x <- matrix(rnorm(500),ncol=20)
> y <- sample(4,25 ,replace=TRUE)
> x <- x+y
> error.bars.by(x,y,ci=0)
> 
>  
> 
> Now, I want to put a legend for the grouping factor of this graph. I also
> would like to manipulate the linetypes and colors of the lines. I've read
> the documentation, but it was not clear to me, how to do this. Are there
> other plotting functions in R, which can do the same?

  Here is an approach which uses xyplot() in the lattice package and
shows how to control line types and colors:

mydf <- data.frame(x=rep(paste("Group", 1:4, sep=""), 6),
   v=rep(paste("Variable", 1:6, sep=""), each=4),
   y=runif(24))

library(lattice)

xyplot(y ~ v, groups = x, data = mydf, type="b",
  xlab="Dependent Variables", ylab="Mean",
  auto.key=list(lines=TRUE, points=TRUE, space="right"),
  par.settings = list(superpose.symbol =
   list(pch=c(16,8,1,5),
col=c("black","red","green","blue"),
lty=c(1,2,3,4)),
  superpose.line =
   list(col=c("black","red","green","blue"),
lty=c(1,2,3,4

> Erich
> 
>  
> 
> 
> 
> Erich Studerus
> Lic. Phil. Klinische Psychologie
> Psychiatric University Hospital Zurich
> Division of Clinical Research
> Lenggstr. 31
> CH-8008 Zurich
> Switzerland
> Mail: [EMAIL PROTECTED]
> Office: +41 44 384 26 66
> Mobile: +41 76 563 31 54
> 
> 
>  
> 
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code. 

-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
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Re: [R] plotting group means

2008-09-09 Thread ONKELINX, Thierry
Dear Erich,

Have a look at ggplot2

library(ggplot2)
dataset <- expand.grid(x = 1:20, y = factor(LETTERS[1:4]), value = 1:10)
dataset$value <- rnorm(nrow(dataset), sd = 0.5) + as.numeric(dataset$y)
plotdata <- aggregate(dataset$value, list(x = dataset$x, y = dataset$y),
mean)
plotdata <- merge(plotdata, aggregate(dataset$value, list(x = dataset$x,
y = dataset$y), sd))
plotdata$min <- plotdata$x.x - plotdata$x.y
plotdata$max <- plotdata$x.x + plotdata$x.y
ggplot(plotdata, aes(x = x, y = x.x, colour = y, min = min, max = max))
+ geom_pointrange() + geom_line() + geom_point()

HTH,

Thierry 




ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research 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 

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Erich Studerus
Verzonden: dinsdag 9 september 2008 12:49
Aan: r-help@r-project.org
Onderwerp: [R] plotting group means

Hi all,



I want to plot the grouped means of some variables. The dependent
variables
and the grouping factor are stored in different columns. I want to draw
a
simple line-plot of means, in which the x-axis represents the variables
and
y-axis represents the means. The means of the groups should be connected
by
lines. So far, the only function that I could find comes closest to what
I'm
looking for, is the error.bars.by-function in the psych-package. To
know,
what I'm looking for, just type:



library(psych)
x <- matrix(rnorm(500),ncol=20)
y <- sample(4,25 ,replace=TRUE)
x <- x+y
error.bars.by(x,y,ci=0)



Now, I want to put a legend for the grouping factor of this graph. I
also
would like to manipulate the linetypes and colors of the lines. I've
read
the documentation, but it was not clear to me, how to do this. Are there
other plotting functions in R, which can do the same?



Erich





Erich Studerus
Lic. Phil. Klinische Psychologie
Psychiatric University Hospital Zurich
Division of Clinical Research
Lenggstr. 31
CH-8008 Zurich
Switzerland
Mail: [EMAIL PROTECTED]
Office: +41 44 384 26 66
Mobile: +41 76 563 31 54





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Re: [R] write dataframes

2008-09-09 Thread Williams, Robin
Hi,
Just a thought.
You wrote:
ob1<-object1$ORF
ob2<-object2$ORF
and then use cbind like,
HG<-cbind(on1,ob2)
but there is an error. Is there any other function I can use? 

  If you copied and pasted this from R, then your problem is 
Hg <- cbind(on1,ob2)
  You mean 
Hg <- cbind(ob1,ob2) 
  So perhaps just a typo.
HTH,
Robin Williams 
Met Office summer intern - Health Forecasting 
[EMAIL PROTECTED] 
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Roberto 
Olivares-Hernández
Sent: Tuesday, September 09, 2008 12:47 PM
To: r-help@r-project.org
Subject: [R] write dataframes

Hi,

After manipulate my data I have ended up with 5 different data frames with 
different number of observations but the same number of variables (columns)

An example, if I write str(object1), I see this,

data.frame':   47 obs. of  3 variables:
 $ ORF: Factor w/ 245 levels "YAL038W","YAL054C",..: 10 19 38 39 44 
45 50 51 59 60 ...
 $ mRNA   : num  0.891 1.148 1.202 1.479 1.445 ...
 $ Protein: num  1.230 1.288 1.175 0.724 0.851 ..

str(object2)
'data.frame':   21 obs. of  3 variables:
 $ ORF: Factor w/ 245 levels "YAL038W","YAL054C",..: 11 25 40 55 66 
78 104 119 141 153 ...
 $ mRNA   : num  0.794 0.741 0.676 1.047 0.912 ...
 $ Protein: num  0.427 0.363 0.468 0.501 0.661 ...

using the  column $ORF from each object , how can I  compose/write the results 
in a file that contains columns with different length ?

I have tried to generate objects like
ob1<-object1$ORF
ob2<-object2$ORF
and then use cbind like,
HG<-cbind(on1,ob2)
but there is an error. Is there any other function I can use?

Thanks for the help

Roberto

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[R] write dataframes

2008-09-09 Thread Roberto Olivares-Hernández

Hi,

After manipulate my data I have ended up with 5 different data frames  
with different number of observations but the same

number of variables (columns)

An example, if I write str(object1), I see this,

data.frame':   47 obs. of  3 variables:
$ ORF: Factor w/ 245 levels "YAL038W","YAL054C",..: 10 19 38 39 44 
45 50 51 59 60 ...

$ mRNA   : num  0.891 1.148 1.202 1.479 1.445 ...
$ Protein: num  1.230 1.288 1.175 0.724 0.851 ..

str(object2)
'data.frame':   21 obs. of  3 variables:
$ ORF: Factor w/ 245 levels "YAL038W","YAL054C",..: 11 25 40 55 66 
78 104 119 141 153 ...

$ mRNA   : num  0.794 0.741 0.676 1.047 0.912 ...
$ Protein: num  0.427 0.363 0.468 0.501 0.661 ...

using the  column $ORF from each object , how can I  compose/write the 
results in a file that contains columns with different length ?


I have tried to generate objects like
ob1<-object1$ORF
ob2<-object2$ORF
and then use cbind like,
HG<-cbind(on1,ob2)
but there is an error. Is there any other function I can use?

Thanks for the help

Roberto

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Re: [R] Compiling date

2008-09-09 Thread stephen sefick
this is day month year?
look at chron or maybe the easiest is to use excel to change the format

On Tue, Sep 9, 2008 at 7:12 AM, Dr Eberhard Lisse <[EMAIL PROTECTED]> wrote:
> Why not Format -> Cell in Excell?
>
> el
>
> on 9/9/08 1:03 PM Henrique Dallazuanna said the following:
>> Try this:
>>
>> strptime(x, ifelse(nchar(x) == 8, '%d/%m/%y', '%d/%m/%Y'))
>>
>> On Tue, Sep 9, 2008 at 3:48 AM, Megh Dal <[EMAIL PROTECTED]> wrote:
>>
>>> Hi,
>>>
>>> I have following kind of dataset (all are dates) in my Excel sheet.
>>>
>>> 09/08/08
>>> 09/05/08
>>> 09/04/08
>>> 09/02/08
>>> 09/01/08
>>> 29/08/2008
>>> 28/08/2008
>>> 27/08/2008
>>> 26/08/2008
>>> 25/08/2008
>>> 22/08/2008
>>> 21/08/2008
>>> 20/08/2008
>>> 18/08/2008
>>> 14/08/2008
>>> 13/08/2008
>>> 08/12/08
>>> 08/11/08
>>> 08/08/08
>>> 08/07/08
>>>
>>> However I want to use R to compile those data to make all dates in same
>>> format. Can anyone please tell me any automated way for doing that?
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] PCA and % variance explained

2008-09-09 Thread Mark Difford

Hi Paul,

>> how do you view how much each component contributes to variance in the
>> dataset ...

It helps to read the help:

?princomp
?prcomp

Contributions to inertia or variance are reported as standard deviations
(princomp.obj$sdev). So square these values to get the variance accounted
for by each component. For the rest, see loadings sub Value in the help
file.

Regards, Mark.


pgseye wrote:
> 
> After doing a PCA using princomp, how do you view how much each component
> contributes to variance in the dataset. I'm still quite new to the theory
> of PCA - I have a little idea about eigenvectors and eigenvalues (these
> determine the variance explained?). Are the eigenvalues related to
> loadings in R?
> 
> Thanks,
> 
> Paul
> 

-- 
View this message in context: 
http://www.nabble.com/PCA-and---variance-explained-tp19388970p19391466.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] plotting group means

2008-09-09 Thread Jim Lemon

Hi Erich,
Have a look at brkdn.plot in the plotrix package.

Jim

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Re: [R] plotting group means

2008-09-09 Thread hadley wickham
On Tue, Sep 9, 2008 at 6:56 AM, ONKELINX, Thierry
<[EMAIL PROTECTED]> wrote:
> Dear Erich,
>
> Have a look at ggplot2
>
> library(ggplot2)
> dataset <- expand.grid(x = 1:20, y = factor(LETTERS[1:4]), value = 1:10)
> dataset$value <- rnorm(nrow(dataset), sd = 0.5) + as.numeric(dataset$y)

Or with stat_summary:

qplot(x, value, data=dataset, colour=y, group = y) +
stat_summary(geom="line", fun="mean",size=2)


-- 
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Re: [R] R_USER - in which file should I include it?

2008-09-09 Thread Eduardo M. A. M.Mendes
Hello

Many thanks.  It works just fine.

How about the packages issue?  That is, same thing for the installation
path.

Cheers

Ed


-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] 
Sent: Monday, September 08, 2008 10:01 PM
To: Eduardo M. A. M.Mendes
Cc: r-help@r-project.org
Subject: Re: [R] R_USER - in which file should I include it?

Try adding this at the end of your etc/Rprofile.site file.  That file
should already be there so you don't have to create it,
just edit it.

cat("Hello from Rprofile.site\n")
setwd("C:/Users/eduardo/Documents")

You may need to edit it as Administrator.   You should
see the Hello message in which case you will know that the
Rprofile.site file is being run.

That should work unless Tinn-R runs R in such a way as to
ignore Rprofile.site.

On Mon, Sep 8, 2008 at 8:11 PM, Eduardo M. A. M.Mendes
<[EMAIL PROTECTED]> wrote:
> Hello
>
> I am not sure whether R starts from the same dir.  For instance:
>
> a) if I double-click on R-2.7.2 icon and then issue the command getwd(),
the
> result is:
>
>> getwd()
> [1] "C:/Users/eduardo/Documents"
>
> b) If R starts from within Tinn-R, the result is:
>
> getwd()
> [1] "C:/Program Files/R/R-2/bin"
>
> I want that no matter which calling R method I am using if I issue the
> command getwd() (first command) the result is:
>
> "C:/Users/eduardo/Documents/R"
>
>
> Moreover all new packages go to "C:/Users/eduardo/Documents/R/win-library"
>
>
> Thanks
>
> Ed
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Monday, September 08, 2008 8:57 PM
> To: Eduardo M. A. M.Mendes
> Cc: r-help@r-project.org
> Subject: Re: [R] R_USER - in which file should I include it?
>
> Could you explain more clearly what you mean by "the same"?
> Do you mean that each time you click on R 2.7.2 icon on your
> desktop that running this from the R console:
>
> getwd()
>
> is the same directory on each startup?  Isn't that already the case?
> I don't think you need to set any environment variables at all.  If
> you don't set
> any environment variables then what specifically is happening that
> you don't want to happen?
>
> On Mon, Sep 8, 2008 at 7:10 PM, Eduardo M. A. M.Mendes
> <[EMAIL PROTECTED]> wrote:
>> Hello
>>
>> I am a newbie.  I had my R upgraded from 2.7.1 to 2.7.2 and in doing so I
>> decided to install all 2.7 versions under c:\program files\R\2.7 from now
>> on (2.7.1 is located under .\2.7.1)
>>
>> Although I don't like the idea (I am running Vista), I have edited
>> etc\Renviron.site to contain:
>>
>>
>> R_USER=c:/Users/eduardo/Documents/R
>> R_LIBS_USER=c:/Users/eduardo/Documents/R/win-library/2.7
>>
>> As far as R starting always from the same location, that is,
>> c:/Users/eduardo/Documents/R, etc\Renviron.site didn't help.  So I wonder
>> whether someone from the list could help me to:
>>
>> a) force R to start always from the same location
>> b) force R to install all new packages in the same location
>>
>>
>> Many thanks
>>
>> Ed
>>
>> PS. Before sending this email, I read windows FAQ and browsed the
archives
>> (too many posts in the subject!).
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Hardwarefor R cpu 64 vs 32, dual vs quad

2008-09-09 Thread Nic Larson
Need to buy fast computer for running R on. Today we use 2,8 MHz intel D cpu
and the calculations takes around 15 days. Is it possible to get the same
calculations down to minutes/hours by only changing the hardware?
Should I go for an really fast dual 32 bit cpu and run R over linux or xp or
go for an quad core / 64 bit cpu?
Is it effective to run R on 64 bit (and problem free
(running/installing))???
Have around 2000-3000 euro to spend
Thanx for any tip

[[alternative HTML version deleted]]

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Re: [R] plotting group means

2008-09-09 Thread Erich Studerus
Thanks for all the suggestions, but it seems, that all these functions need
a rearrangement of my data, since in my case, the dependent variables are in
different columns. The error.bars.by-function seems to be the only plotting
function, that does not need a rearrangement. Are there other functions,
which can do that or is there an easy way to rearrange the columns into one?

Thanks

Erich


-Ursprüngliche Nachricht-
Von: hadley wickham [mailto:[EMAIL PROTECTED] 
Gesendet: Dienstag, 9. September 2008 15:02
An: ONKELINX, Thierry
Cc: Erich Studerus; r-help@r-project.org
Betreff: Re: [R] plotting group means

On Tue, Sep 9, 2008 at 6:56 AM, ONKELINX, Thierry
<[EMAIL PROTECTED]> wrote:
> Dear Erich,
>
> Have a look at ggplot2
>
> library(ggplot2)
> dataset <- expand.grid(x = 1:20, y = factor(LETTERS[1:4]), value = 1:10)
> dataset$value <- rnorm(nrow(dataset), sd = 0.5) + as.numeric(dataset$y)

Or with stat_summary:

qplot(x, value, data=dataset, colour=y, group = y) +
stat_summary(geom="line", fun="mean",size=2)


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

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


Re: [R] R_USER - in which file should I include it?

2008-09-09 Thread Gabor Grothendieck
You might look at ?.libPaths
(note the dot) and play around with adding a .libPaths command
to your Rprofile.site and again you may need Administrator rights
when editing it.  If that does not help then you can try clarifying
the problem.   In particular what "the same" refers to and what
is happening now and what you want to happen.

On Tue, Sep 9, 2008 at 9:14 AM, Eduardo M. A. M.Mendes
<[EMAIL PROTECTED]> wrote:
> Hello
>
> Many thanks.  It works just fine.
>
> How about the packages issue?  That is, same thing for the installation
> path.
>
> Cheers
>
> Ed
>
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Monday, September 08, 2008 10:01 PM
> To: Eduardo M. A. M.Mendes
> Cc: r-help@r-project.org
> Subject: Re: [R] R_USER - in which file should I include it?
>
> Try adding this at the end of your etc/Rprofile.site file.  That file
> should already be there so you don't have to create it,
> just edit it.
>
> cat("Hello from Rprofile.site\n")
> setwd("C:/Users/eduardo/Documents")
>
> You may need to edit it as Administrator.   You should
> see the Hello message in which case you will know that the
> Rprofile.site file is being run.
>
> That should work unless Tinn-R runs R in such a way as to
> ignore Rprofile.site.
>
> On Mon, Sep 8, 2008 at 8:11 PM, Eduardo M. A. M.Mendes
> <[EMAIL PROTECTED]> wrote:
>> Hello
>>
>> I am not sure whether R starts from the same dir.  For instance:
>>
>> a) if I double-click on R-2.7.2 icon and then issue the command getwd(),
> the
>> result is:
>>
>>> getwd()
>> [1] "C:/Users/eduardo/Documents"
>>
>> b) If R starts from within Tinn-R, the result is:
>>
>> getwd()
>> [1] "C:/Program Files/R/R-2/bin"
>>
>> I want that no matter which calling R method I am using if I issue the
>> command getwd() (first command) the result is:
>>
>> "C:/Users/eduardo/Documents/R"
>>
>>
>> Moreover all new packages go to "C:/Users/eduardo/Documents/R/win-library"
>>
>>
>> Thanks
>>
>> Ed
>>
>> -Original Message-
>> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
>> Sent: Monday, September 08, 2008 8:57 PM
>> To: Eduardo M. A. M.Mendes
>> Cc: r-help@r-project.org
>> Subject: Re: [R] R_USER - in which file should I include it?
>>
>> Could you explain more clearly what you mean by "the same"?
>> Do you mean that each time you click on R 2.7.2 icon on your
>> desktop that running this from the R console:
>>
>> getwd()
>>
>> is the same directory on each startup?  Isn't that already the case?
>> I don't think you need to set any environment variables at all.  If
>> you don't set
>> any environment variables then what specifically is happening that
>> you don't want to happen?
>>
>> On Mon, Sep 8, 2008 at 7:10 PM, Eduardo M. A. M.Mendes
>> <[EMAIL PROTECTED]> wrote:
>>> Hello
>>>
>>> I am a newbie.  I had my R upgraded from 2.7.1 to 2.7.2 and in doing so I
>>> decided to install all 2.7 versions under c:\program files\R\2.7 from now
>>> on (2.7.1 is located under .\2.7.1)
>>>
>>> Although I don't like the idea (I am running Vista), I have edited
>>> etc\Renviron.site to contain:
>>>
>>>
>>> R_USER=c:/Users/eduardo/Documents/R
>>> R_LIBS_USER=c:/Users/eduardo/Documents/R/win-library/2.7
>>>
>>> As far as R starting always from the same location, that is,
>>> c:/Users/eduardo/Documents/R, etc\Renviron.site didn't help.  So I wonder
>>> whether someone from the list could help me to:
>>>
>>> a) force R to start always from the same location
>>> b) force R to install all new packages in the same location
>>>
>>>
>>> Many thanks
>>>
>>> Ed
>>>
>>> PS. Before sending this email, I read windows FAQ and browsed the
> archives
>>> (too many posts in the subject!).
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>
>

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


[R] PCA and % variance explained

2008-09-09 Thread pgseye

After doing a PCA using princomp, how do you view how much each component
contributes to variance in the dataset. I'm still quite new to the theory of
PCA - I have a little idea about eigenvectors and eigenvalues (these
determine the variance explained?). Are the eigenvalues related to loadings
in R?

Thanks,

Paul
-- 
View this message in context: 
http://www.nabble.com/PCA-and---variance-explained-tp19388970p19388970.html
Sent from the R help mailing list archive at Nabble.com.

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


Re: [R] plotting group means

2008-09-09 Thread hadley wickham
On Tue, Sep 9, 2008 at 8:38 AM, Erich Studerus
<[EMAIL PROTECTED]> wrote:
> Thanks for all the suggestions, but it seems, that all these functions need
> a rearrangement of my data, since in my case, the dependent variables are in
> different columns. The error.bars.by-function seems to be the only plotting
> function, that does not need a rearrangement. Are there other functions,
> which can do that or is there an easy way to rearrange the columns into one?

Try:

library(reshape)
melt(x)

Hadley

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

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Re: [R] Vorticity and Divergence

2008-09-09 Thread Ravi Varadhan
Both vorticity and divergence are defined in terms of partial derivatives.
You can compute these derivatives using the `grad' function in "numDeriv"
package.

U <- function(X) { your U function}
V <- function(X) { your V function} 
# where X = c(x,y)

library(numDeriv)

grU <- function(X) grad(X, func=U)
grV <- function(X) grad(X, func=V)

# For a 2-dimensional vector field

vortivcity <- function(X) grV(X)[2] - grU(X)[1]
divergence <- function(X) grU(X)[1] + grV(X)[2]


# Here is an example:

U <- function(X) X[1]^2 + X[1] * X[2]
V <- function(X) X[2]^2 - X[1] * X[2]

> vorticity(c(2,1))
[1] -5
> divergence(c(2,1))
[1] 5
>

Does this help?

Ravi.

---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: [EMAIL PROTECTED]

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Igor Oliveira
Sent: Monday, September 08, 2008 11:37 AM
To: r-help@r-project.org
Subject: [R] Vorticity and Divergence

Hi all,

I have some wind data (U and V components) and I would like to compute
Vorticity and Divergence of these fields. Is there any R function that can
easily do that?

Thanks in advance for any help

Igor Oliveira
CSAG, Dept. Environmental & Geographical Science, University of Cape Town,
Private Bag X3, Rondebosch 7701. Tel.: +27 (0)21 650 5774
South Africa Fax: +27 (0)21 650 5773
  http:///www.csag.uct.ac.za/~igor

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


[R] Question

2008-09-09 Thread Veronique.Pinard
Hi,
 
I'm trying to verify the assumption of homogeneity of variance of residuals in 
an ANOVA with levene.test. I don't know how to define the groups. I have 3 
factors : A, B and C(AxB).
 
What do I have to change or to add in the command to set that I'm working with 
the residuals and to set the groups?
 
library(car)
attach(anova.sns2)
levene.test(residuals ~ ???)


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Re: [R] R_USER - in which file should I include it?

2008-09-09 Thread Eduardo M. A. M.Mendes
Many thanks. I shall look at it. In case I run into trouble again, I'll try
to clarify the "the same".

Ed
 

-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] 
Sent: Tuesday, September 09, 2008 10:46 AM
To: Eduardo M. A. M.Mendes
Cc: r-help@r-project.org
Subject: Re: [R] R_USER - in which file should I include it?

You might look at ?.libPaths
(note the dot) and play around with adding a .libPaths command
to your Rprofile.site and again you may need Administrator rights
when editing it.  If that does not help then you can try clarifying
the problem.   In particular what "the same" refers to and what
is happening now and what you want to happen.

On Tue, Sep 9, 2008 at 9:14 AM, Eduardo M. A. M.Mendes
<[EMAIL PROTECTED]> wrote:
> Hello
>
> Many thanks.  It works just fine.
>
> How about the packages issue?  That is, same thing for the installation
> path.
>
> Cheers
>
> Ed
>
>
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Monday, September 08, 2008 10:01 PM
> To: Eduardo M. A. M.Mendes
> Cc: r-help@r-project.org
> Subject: Re: [R] R_USER - in which file should I include it?
>
> Try adding this at the end of your etc/Rprofile.site file.  That file
> should already be there so you don't have to create it,
> just edit it.
>
> cat("Hello from Rprofile.site\n")
> setwd("C:/Users/eduardo/Documents")
>
> You may need to edit it as Administrator.   You should
> see the Hello message in which case you will know that the
> Rprofile.site file is being run.
>
> That should work unless Tinn-R runs R in such a way as to
> ignore Rprofile.site.
>
> On Mon, Sep 8, 2008 at 8:11 PM, Eduardo M. A. M.Mendes
> <[EMAIL PROTECTED]> wrote:
>> Hello
>>
>> I am not sure whether R starts from the same dir.  For instance:
>>
>> a) if I double-click on R-2.7.2 icon and then issue the command getwd(),
> the
>> result is:
>>
>>> getwd()
>> [1] "C:/Users/eduardo/Documents"
>>
>> b) If R starts from within Tinn-R, the result is:
>>
>> getwd()
>> [1] "C:/Program Files/R/R-2/bin"
>>
>> I want that no matter which calling R method I am using if I issue the
>> command getwd() (first command) the result is:
>>
>> "C:/Users/eduardo/Documents/R"
>>
>>
>> Moreover all new packages go to
"C:/Users/eduardo/Documents/R/win-library"
>>
>>
>> Thanks
>>
>> Ed
>>
>> -Original Message-
>> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
>> Sent: Monday, September 08, 2008 8:57 PM
>> To: Eduardo M. A. M.Mendes
>> Cc: r-help@r-project.org
>> Subject: Re: [R] R_USER - in which file should I include it?
>>
>> Could you explain more clearly what you mean by "the same"?
>> Do you mean that each time you click on R 2.7.2 icon on your
>> desktop that running this from the R console:
>>
>> getwd()
>>
>> is the same directory on each startup?  Isn't that already the case?
>> I don't think you need to set any environment variables at all.  If
>> you don't set
>> any environment variables then what specifically is happening that
>> you don't want to happen?
>>
>> On Mon, Sep 8, 2008 at 7:10 PM, Eduardo M. A. M.Mendes
>> <[EMAIL PROTECTED]> wrote:
>>> Hello
>>>
>>> I am a newbie.  I had my R upgraded from 2.7.1 to 2.7.2 and in doing so
I
>>> decided to install all 2.7 versions under c:\program files\R\2.7 from
now
>>> on (2.7.1 is located under .\2.7.1)
>>>
>>> Although I don't like the idea (I am running Vista), I have edited
>>> etc\Renviron.site to contain:
>>>
>>>
>>> R_USER=c:/Users/eduardo/Documents/R
>>> R_LIBS_USER=c:/Users/eduardo/Documents/R/win-library/2.7
>>>
>>> As far as R starting always from the same location, that is,
>>> c:/Users/eduardo/Documents/R, etc\Renviron.site didn't help.  So I
wonder
>>> whether someone from the list could help me to:
>>>
>>> a) force R to start always from the same location
>>> b) force R to install all new packages in the same location
>>>
>>>
>>> Many thanks
>>>
>>> Ed
>>>
>>> PS. Before sending this email, I read windows FAQ and browsed the
> archives
>>> (too many posts in the subject!).
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>
>

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Re: [R] Gumbell distribution - minimum case

2008-09-09 Thread Aaron Mackey
If you mean you want an EVD with a fat left tail (instead of a fat
right tail), then can;t you just multiply all the values by -1 to
"reverse" the distribution?  A new location parameter could then shift
the distribution wherever you want along the number line ...

-Aaron

On Mon, Sep 8, 2008 at 5:22 PM, Richard Gwozdz <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I would like to sample from a Gumbell (minimum) distribution.  I have
> installed package {evd} but the Gumbell functions there appear to refer to
> the maximum case.  Unfortunately, setting the scale parameter negative does
> not appear to work.
>
> Is there a separate package for the Gumbell minimum?
>
>
> --
> _
> Rich Gwozdz
> Fire and Mountain Ecology Lab
> College of Forest Resources
> University of Washington
> cell: 206-769-6808 office: 206-543-9138
> [EMAIL PROTECTED]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] How does predict.lm work?

2008-09-09 Thread Williams, Robin
Hi, 
  Please could someone explain how this element of predict.lm works?
>From the help file 
`
newdata   
An optional data frame in which to look for variables with which to
predict. If omitted, the fitted values are used.
' 
  Does this dataframe (newdata) need to have the same variable names as
was used in the original data frame used to fit the model? Or will R
just look across consecutive columns of newdata, and apply them to the
call as appropriate?
  For example, if I have fitted a model with four variables
(x1,x2,x3,x4) in my original dataframe, and then have a second dataframe
which I want to supply to the newdata argument in predict.lm with
variable names (x5, x6, x7, x8), do I need to change the variable names
in my newdata dataframe to match those of the original dataframe? Or
will R treat x5 as x1, x6 as x2, etc, when using predict.lm? 
  I would like to know so that I can design the structure of some
somewhat larger dataframes in a manner which will make using predict.lm
straight forward and quick.
Hope this makes sense.
Many thanks for any help. 
   Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED] 
 

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Re: [R] How does predict.lm work?

2008-09-09 Thread Gabor Grothendieck
Just try it:

> BOD # built in data frame
  Time demand
118.3
22   10.3
33   19.0
44   16.0
55   15.6
67   19.8
> BOD.lm <- lm(demand ~ Time, BOD)
> predict(BOD.lm, list(Time = 10))
   1
25.73571
> predict(BOD.lm, list(10))
Error in eval(expr, envir, enclos) : object "Time" not found


On Tue, Sep 9, 2008 at 10:59 AM, Williams, Robin
<[EMAIL PROTECTED]> wrote:
> Hi,
>  Please could someone explain how this element of predict.lm works?
> >From the help file
> `
> newdata
> An optional data frame in which to look for variables with which to
> predict. If omitted, the fitted values are used.
> '
>  Does this dataframe (newdata) need to have the same variable names as
> was used in the original data frame used to fit the model? Or will R
> just look across consecutive columns of newdata, and apply them to the
> call as appropriate?
>  For example, if I have fitted a model with four variables
> (x1,x2,x3,x4) in my original dataframe, and then have a second dataframe
> which I want to supply to the newdata argument in predict.lm with
> variable names (x5, x6, x7, x8), do I need to change the variable names
> in my newdata dataframe to match those of the original dataframe? Or
> will R treat x5 as x1, x6 as x2, etc, when using predict.lm?
>  I would like to know so that I can design the structure of some
> somewhat larger dataframes in a manner which will make using predict.lm
> straight forward and quick.
> Hope this makes sense.
> Many thanks for any help.
>   Robin Williams
> Met Office summer intern - Health Forecasting
> [EMAIL PROTECTED]
>
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] How does predict.lm work?

2008-09-09 Thread Marc Schwartz
on 09/09/2008 09:59 AM Williams, Robin wrote:
> Hi, 
>   Please could someone explain how this element of predict.lm works?
>>From the help file 
> `
> newdata 
> An optional data frame in which to look for variables with which to
> predict. If omitted, the fitted values are used.
> ' 
>   Does this dataframe (newdata) need to have the same variable names as
> was used in the original data frame used to fit the model? 

Yes. Also, see the Note in ?predict.lm:

"Variables are first looked for in newdata and then searched for in the
usual way (which will include the environment of the formula used in the
fit). A warning will be given if the variables found are not of the same
length as those in newdata if it was supplied."


It also says "Variables", not columns.

> Or will R
> just look across consecutive columns of newdata, and apply them to the
> call as appropriate?

No.

>   For example, if I have fitted a model with four variables
> (x1,x2,x3,x4) in my original dataframe, and then have a second dataframe
> which I want to supply to the newdata argument in predict.lm with
> variable names (x5, x6, x7, x8), do I need to change the variable names
> in my newdata dataframe to match those of the original dataframe? 

Yes.

> Or
> will R treat x5 as x1, x6 as x2, etc, when using predict.lm? 
>   I would like to know so that I can design the structure of some
> somewhat larger dataframes in a manner which will make using predict.lm
> straight forward and quick.
> Hope this makes sense.
> Many thanks for any help. 

HTH,

Marc Schwartz

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[R] printing all rows

2008-09-09 Thread ANJAN PURKAYASTHA
Hi,
my data table has 38939 rows. R prints  the first 1 columns and then
prints an error message:[ reached getOption("max.print") -- omitted 27821
rows ]].
is it possible to set the maxprint parameter so that R prints all the rows?

tia,
anjan

-- 
=
anjan purkayastha, phd
bioinformatics analyst
whitehead institute for biomedical research
nine cambridge center
cambridge, ma 02142

purkayas [at] wi [dot] mit [dot] edu
703.740.6939

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Re: [R] Hardwarefor R cpu 64 vs 32, dual vs quad

2008-09-09 Thread Prof Brian Ripley

On Tue, 9 Sep 2008, Nic Larson wrote:


Need to buy fast computer for running R on. Today we use 2,8 MHz intel D cpu
and the calculations takes around 15 days. Is it possible to get the same
calculations down to minutes/hours by only changing the hardware?


No: you would need to arrange to parallelize the computations.  I'd be 
surprised if you got a computer within your budget that was 3x faster on a 
single CPU than your current one, and R will only use (unaided) one CPU 
for most tasks (the exception being some matrix algebra).



Should I go for an really fast dual 32 bit cpu and run R over linux or xp or
go for an quad core / 64 bit cpu?
Is it effective to run R on 64 bit (and problem free
(running/installing))???


All answered in the R-admin manual, so please RTFM.


Have around 2000-3000 euro to spend
Thanx for any tip

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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] puzzle about contrasts

2008-09-09 Thread Kenneth Knoblauch

Hi,

I'm trying to redefine the contrasts for a linear model.
With a 2 level factor, x, with levels A and B, a two level
factor outputs A and B - A from an lm fit, say
lm(y ~ x). I would like to set the contrasts so that
the coefficients output are -0.5 (A + B) and B - A,
but I can't get the sign correct for the first coefficient
(Intercept).

Here is a toy example,

set.seed(12161952)
y <- rnorm(10)
x <- factor(rep(letters[1:2], each = 5))
##  so  A and B =
tapply(y, x, mean)

a  b
-0.719  0.8323837

## and with treatment contrasts
coef(lm(y ~ x))  ## A and B - A

(Intercept)  xb
 -0.719   1.5522724

Then, I try to redefine the contrasts

### would like contrasts: -0.5 (A + B) and B - A
D1 <- matrix( c(-0.5, -0.5,
-1, 1),
2, 2, byrow = TRUE)
C1 <- solve(D1)
Cnt <- C1[, -1]
contrasts(x) <- Cnt
coef(lm(y ~ x))

(Intercept)  x1
 0.05624745  1.55227241

but note that the desired value is
-0.5 * sum(tapply(y, x, mean))

[1] -0.05624745

I note that the first column of C1 is -1's not +1's
and that working by hand, if I tamper with the model matrix

mm <- model.matrix(y ~ x)
mm[, 1] <- -1

mm
   (Intercept)   x1
1   -1 -0.5
2   -1 -0.5
3   -1 -0.5
4   -1 -0.5
5   -1 -0.5
6   -1  0.5
7   -1  0.5
8   -1  0.5
9   -1  0.5
10  -1  0.5
attr(,"assign")
[1] 0 1
attr(,"contrasts")
attr(,"contrasts")$x
  [,1]
a -0.5
b  0.5

solve(t(mm) %*% mm) %*% t(mm) %*% y  ##Yes, I know. Use QR
   [,1]
(Intercept) -0.05624745
x1   1.55227241

gives the correct sign.

So, I guess my question reduces to how one would set the
contrasts for the model.matrix to be correct
for this to work out correctly?

Thank you.

Ken


--
Ken Knoblauch
Inserm U846
Institut Cellule Souche et Cerveau
Département Neurosciences Intégratives
18 avenue du Doyen Lépine
69500 Bron
France
tel: +33 (0)4 72 91 34 77
fax: +33 (0)4 72 91 34 61
portable: +33 (0)6 84 10 64 10
http://www.sbri.fr

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Re: [R] exporting tapply objects to csv-files

2008-09-09 Thread hadley wickham
On Tue, Sep 9, 2008 at 3:48 AM, Kunzler, Andreas <[EMAIL PROTECTED]> wrote:
> Dear Everyone,
>
> I try to create a cvs-file with different results form the table function.
>
> Imagine a data-frame with two vectors a and b where b is of the class factor.
>
> I use the tapply function to count a for the different values of b.
>
> tapply(a,b,table)
>
> and I use the table function to have a look of the frequencies as a total
>
> table(a)
>
> I would like to put both results together in one txt or csv file that I can 
> import to e.g. Excel.
>
> The export file should have a layout like
>
> 1,2,3,4,5,6,7 (possible values of a)
> 3,6,7,8,8,8,1 (Counts of a total)
> 1,2,3,4,5,3,0 (Counts of a where b==A)
> 2,4,4,4,3,5,1 (Counts of a where b==B)
>
> I tried to change the class of the table result to a matrix but I could not 
> find a way to use the results of tapply. I use tapply because b has 15 
> different values.

An alternative would be to use reshape (http://had.co.nz/reshape):

mydf <- data.frame( a = sample(7, 100, rep = T), b =
sample(letters[1:15], 100, rep = T))

library(reshape)
mydf$value <- 1
cast(mydf, b ~ a, sum, margins="row.major", fill = 0)

Regards,

Hadley

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

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Re: [R] PCA and % variance explained

2008-09-09 Thread ngottlieb
I did PCA stuff years there is a thing that is called a scree score 
Which will give an indication of the number of PCA's and the variance
explained.

Might want to web search on scree score and PCA.



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of pgseye
Sent: Tuesday, September 09, 2008 5:39 AM
To: r-help@r-project.org
Subject: [R] PCA and % variance explained


After doing a PCA using princomp, how do you view how much each
component contributes to variance in the dataset. I'm still quite new to
the theory of PCA - I have a little idea about eigenvectors and
eigenvalues (these determine the variance explained?). Are the
eigenvalues related to loadings in R?

Thanks,

Paul
--
View this message in context:
http://www.nabble.com/PCA-and---variance-explained-tp19388970p19388970.h
tml
Sent from the R help mailing list archive at Nabble.com.

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This information is being sent at the recipient's reques...{{dropped:16}}

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[R] Linear Modeling the best alternative

2008-09-09 Thread stephen sefick
I have a data set of mean velocity, discharge, and mean depth.  I need
to find out which model best fits them out of log linear, linear, some
other kind of model...  Using excel I have found that linear is not
that bad and log10(discharge) vs. the other two variables (I am trying
to predict velocity and depth from discharge) is not that bad either.
How do I test and see which one of these models is better...  better
R-squared...  I know this is a stats question and not particularly an
R question, but I will use R for the models vetting process.
any ideas would be greatly appreciated,
-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] puzzle about contrasts

2008-09-09 Thread Prof Brian Ripley

-0.5*(A+B) is not a contrast, which is the seat of your puzzlement.

All you can get from y ~ x is an intercept (a column of ones) and a single 
'contrast' column for 'x'.


If you use y ~ 0+x you can get two columns for 'x', but R does not give 
you an option of what columns in the case: see the source of contrasts(). 
So you would need to replace contrasts(), which I think will be hard as 
model.matrix.default will look in the 'stats' namespace.  It would 
probably be easier to create the model matrix yourself.


On Tue, 9 Sep 2008, Kenneth Knoblauch wrote:


Hi,

I'm trying to redefine the contrasts for a linear model.
With a 2 level factor, x, with levels A and B, a two level
factor outputs A and B - A from an lm fit, say
lm(y ~ x). I would like to set the contrasts so that
the coefficients output are -0.5 (A + B) and B - A,
but I can't get the sign correct for the first coefficient
(Intercept).

Here is a toy example,

set.seed(12161952)
y <- rnorm(10)
x <- factor(rep(letters[1:2], each = 5))
##  so  A and B =
tapply(y, x, mean)

  a  b
-0.719  0.8323837

## and with treatment contrasts
coef(lm(y ~ x))  ## A and B - A

(Intercept)  xb
-0.719   1.5522724

Then, I try to redefine the contrasts

### would like contrasts: -0.5 (A + B) and B - A
D1 <- matrix( c(-0.5, -0.5,
-1, 1),
2, 2, byrow = TRUE)
C1 <- solve(D1)
Cnt <- C1[, -1]
contrasts(x) <- Cnt
coef(lm(y ~ x))

(Intercept)  x1
0.05624745  1.55227241

but note that the desired value is
-0.5 * sum(tapply(y, x, mean))

[1] -0.05624745

I note that the first column of C1 is -1's not +1's
and that working by hand, if I tamper with the model matrix

mm <- model.matrix(y ~ x)
mm[, 1] <- -1

mm
 (Intercept)   x1
1   -1 -0.5
2   -1 -0.5
3   -1 -0.5
4   -1 -0.5
5   -1 -0.5
6   -1  0.5
7   -1  0.5
8   -1  0.5
9   -1  0.5
10  -1  0.5
attr(,"assign")
[1] 0 1
attr(,"contrasts")
attr(,"contrasts")$x
[,1]
a -0.5
b  0.5

solve(t(mm) %*% mm) %*% t(mm) %*% y  ##Yes, I know. Use QR
 [,1]
(Intercept) -0.05624745
x1   1.55227241

gives the correct sign.

So, I guess my question reduces to how one would set the
contrasts for the model.matrix to be correct
for this to work out correctly?

Thank you.

Ken


--
Ken Knoblauch
Inserm U846
Institut Cellule Souche et Cerveau
Département Neurosciences Intégratives
18 avenue du Doyen Lépine
69500 Bron
France
tel: +33 (0)4 72 91 34 77
fax: +33 (0)4 72 91 34 61
portable: +33 (0)6 84 10 64 10
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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)
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[R] passing graph image data from remote Rserve

2008-09-09 Thread Patil, Prasad
Hello,

 

I am using Rserve to create a dedicated computational back-engine. I
generate and pass an array of data to a java application on a separate
server. I was wondering if the same is possible for an image. I believe
that Rserve supports passing certain R objects and JRclient can cast
these objects into their Java counterparts. If I generate a barplot in R
(remotely), can I pass the graph image back to the Java application for
display? Currently, I am reduced to saving the graph as a .pdf locally,
passing the .pdf's filepath to the Java application and allowing the
application access to the file, which is not an ideal structure.

 

Thanks for your help,

 

Prasad


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[R] Modality Test

2008-09-09 Thread Amin W. Mugera

Dear Readers:

I have two issues in nonparametric statistical analysis that i need
help:

First, does R have a package that can implement the multimodality test,
e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
an earlier thread (sometime in 2003) where someone was trying to write
a code for the Silverman test of multimodality. Is there any other
tests that can enable me to know how many modes are in a distribution?

Second, i would like to test whether two distributions are equal. Does R
have a  package than can implement the Li (1996) test of the equality
of two distributions? Is there any other test i can use rather than the
Li test?

Thank you in advance for your help.

Amin Mugera
Graduate Student
AgEcon Dept. Kansas State University

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Re: [R] passing graph image data from remote Rserve

2008-09-09 Thread Patil, Prasad
I believe I have found my solution, so please disregard. Thanks


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Re: [R] Linear Modeling the best alternative

2008-09-09 Thread Ben Bolker
stephen sefick  gmail.com> writes:

> 
> I have a data set of mean velocity, discharge, and mean depth.  I need
> to find out which model best fits them out of log linear, linear, some
> other kind of model...  Using excel I have found that linear is not
> that bad and log10(discharge) vs. the other two variables (I am trying
> to predict velocity and depth from discharge) is not that bad either.
> How do I test and see which one of these models is better...  better
> R-squared...  I know this is a stats question and not particularly an
> R question, but I will use R for the models vetting process.
> any ideas would be greatly appreciated,


  AIC is not bad, but see
http://www.unc.edu/courses/2006spring/ecol/145/001/docs/lectures/lecture18.htm
for computing AIC to compare models where some have transformed
response variables ...

  Ben Bolker

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Re: [R] puzzle about contrasts

2008-09-09 Thread Peter Dalgaard
Prof Brian Ripley skrev:
> -0.5*(A+B) is not a contrast, which is the seat of your puzzlement.
>
> All you can get from y ~ x is an intercept (a column of ones) and a
> single 'contrast' column for 'x'.
>
> If you use y ~ 0+x you can get two columns for 'x', but R does not
> give you an option of what columns in the case: see the source of
> contrasts(). So you would need to replace contrasts(), which I think
> will be hard as model.matrix.default will look in the 'stats'
> namespace.  It would probably be easier to create the model matrix
> yourself.
>
Or accept the default and do the parameter transformations yourself.

l <- lm(y~x)
T <- rbind(
c(-1,-.5),
c(0,1))

c2 <- T%*%coef(l)
V2 <- T%*%vcov(l) %*% t(T)

cbind(coef=c(c2), s.e.=sqrt(diag(V2)))


> On Tue, 9 Sep 2008, Kenneth Knoblauch wrote:
>
>> Hi,
>>
>> I'm trying to redefine the contrasts for a linear model.
>> With a 2 level factor, x, with levels A and B, a two level
>> factor outputs A and B - A from an lm fit, say
>> lm(y ~ x). I would like to set the contrasts so that
>> the coefficients output are -0.5 (A + B) and B - A,
>> but I can't get the sign correct for the first coefficient
>> (Intercept).
>>
>> Here is a toy example,
>>
>> set.seed(12161952)
>> y <- rnorm(10)
>> x <- factor(rep(letters[1:2], each = 5))
>> ##  so  A and B =
>> tapply(y, x, mean)
>>
>>   a  b
>> -0.719  0.8323837
>>
>> ## and with treatment contrasts
>> coef(lm(y ~ x))  ## A and B - A
>>
>> (Intercept)  xb
>> -0.719   1.5522724
>>
>> Then, I try to redefine the contrasts
>>
>> ### would like contrasts: -0.5 (A + B) and B - A
>> D1 <- matrix( c(-0.5, -0.5,
>> -1, 1),
>> 2, 2, byrow = TRUE)
>> C1 <- solve(D1)
>> Cnt <- C1[, -1]
>> contrasts(x) <- Cnt
>> coef(lm(y ~ x))
>>
>> (Intercept)  x1
>> 0.05624745  1.55227241
>>
>> but note that the desired value is
>> -0.5 * sum(tapply(y, x, mean))
>>
>> [1] -0.05624745
>>
>> I note that the first column of C1 is -1's not +1's
>> and that working by hand, if I tamper with the model matrix
>>
>> mm <- model.matrix(y ~ x)
>> mm[, 1] <- -1
>>
>> mm
>>  (Intercept)   x1
>> 1   -1 -0.5
>> 2   -1 -0.5
>> 3   -1 -0.5
>> 4   -1 -0.5
>> 5   -1 -0.5
>> 6   -1  0.5
>> 7   -1  0.5
>> 8   -1  0.5
>> 9   -1  0.5
>> 10  -1  0.5
>> attr(,"assign")
>> [1] 0 1
>> attr(,"contrasts")
>> attr(,"contrasts")$x
>> [,1]
>> a -0.5
>> b  0.5
>>
>> solve(t(mm) %*% mm) %*% t(mm) %*% y  ##Yes, I know. Use QR
>>  [,1]
>> (Intercept) -0.05624745
>> x1   1.55227241
>>
>> gives the correct sign.
>>
>> So, I guess my question reduces to how one would set the
>> contrasts for the model.matrix to be correct
>> for this to work out correctly?
>>
>> Thank you.
>>
>> Ken
>>
>>
>> -- 
>> Ken Knoblauch
>> Inserm U846
>> Institut Cellule Souche et Cerveau
>> Département Neurosciences Intégratives
>> 18 avenue du Doyen Lépine
>> 69500 Bron
>> France
>> tel: +33 (0)4 72 91 34 77
>> fax: +33 (0)4 72 91 34 61
>> portable: +33 (0)6 84 10 64 10
>> http://www.sbri.fr
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> 
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>   


-- 
   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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[R] Help with 'spectrum'

2008-09-09 Thread rkevinburton
For the command 'spectrum' I read:

The spectrum here is defined with scaling 1/frequency(x), following S-PLUS. 
This makes the spectral density a density over the range (-frequency(x)/2, 
+frequency(x)/2], whereas a more common scaling is 2π and range (-0.5, 0.5] 
(e.g., Bloomfield) or 1 and range (-π, π]. 


Forgive my ignorance but I am having a hard time interpreting this. Does this 
mean that in the spectrum output every element of the $spec array is scaled by 
1/frequency(x)? I am having a hard time determing what is meant by 
'frequency'.Say I define a time series for a year with samples for every day. I 
input a 'frequency' of 365 (which in my mind is the period). On the output of 
'spectrum' would this mean that every element of the $spec array is scaled by 
1/365? There is a corresponding frequency array on the output from 'spectrum'. 
If the frequency is 365 and an element in the frequency array output from 
'spectrum' is .1 am I to assume that the period is 36.5 and a corresponding sin 
wave would be sin(2 * pi * 36.5/365)?

Thank you in advance for helping me clear up some confusion.

Kevin

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Re: [R] puzzle about contrasts

2008-09-09 Thread Peter Dalgaard
Peter Dalgaard skrev:
> Prof Brian Ripley skrev:
>   
>> -0.5*(A+B) is not a contrast, which is the seat of your puzzlement.
>>
>> All you can get from y ~ x is an intercept (a column of ones) and a
>> single 'contrast' column for 'x'.
>>
>> If you use y ~ 0+x you can get two columns for 'x', but R does not
>> give you an option of what columns in the case: see the source of
>> contrasts(). So you would need to replace contrasts(), which I think
>> will be hard as model.matrix.default will look in the 'stats'
>> namespace.  It would probably be easier to create the model matrix
>> yourself.
>>
>> 
> Or accept the default and do the parameter transformations yourself.
>
> l <- lm(y~x)
> T <- rbind(
> c(-1,-.5),
> c(0,1))
>
> c2 <- T%*%coef(l)
> V2 <- T%*%vcov(l) %*% t(T)
>
> cbind(coef=c(c2), s.e.=sqrt(diag(V2)))
>   

I forgot: Also have a look at estimable() from the gmodels packages.

-- 
   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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[R] : writeMat

2008-09-09 Thread erola pairo
I write a .mat file using the writeMat() command, but when i try to load it
in Matlab it says that  file may be corrupt. I did it a month ago and it
worked. It exists any option that I can change for making the file readable
to Matlab?

> A  <-  c(1:10)
> dim(A) <- c(2,5)
> library(R.matlab)
> writeMat('A.mat', A=A)

And what matlab say is:
"file may be corrupt"

Regards

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[R] help on wavelet

2008-09-09 Thread giov

Hi,
I have little experience using wavelet and I would like to know if it is
possible,using R wavelet package, to have a plot of frequency versus time. 

thank you

giov
-- 
View this message in context: 
http://www.nabble.com/help-on-wavelet-tp19395583p19395583.html
Sent from the R help mailing list archive at Nabble.com.

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[R] binomial(link="inverse")

2008-09-09 Thread Ben Bolker

  this may be a better question for r-devel, but ...

  Is there a particular reason (and if so, what is it) that
the inverse link is not in the list of allowable link functions
for the binomial family?  I initially thought this might
have something to do with the properties of canonical
vs non-canonical link functions, but since other link functions
(probit, cloglog, cauchit, log) are allowed, I can't think
of any good reason.  In fact, it's sort of a mystery to me
why the sets of link functions for each family are restricted.
Is this from painful experience that some link functions just
don't work well?

  I can go ahead and hack my own version that allows inverse
link, but it would be nice to know if I'm doing something dumb.

  (The reason I want to do this is that the inverse link
linearizes the Michaelis-Menten function, y = a*x/(b+x) ...)

  cheers
Ben Bolker




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Re: [R] printing all rows

2008-09-09 Thread Adam D. I. Kramer

options("max.print")

$max.print
[1] 9


options(max.print=10)
options("max.print")

$max.print
[1] 1e+05

...so check what your max.print is, and figure out whether you need to
set it to nrow, ncol, or nrow*ncol of your data frame...then do so...though
of course, this is a global variable, so everything you print from then on
will just keep printing and printing.

Really, though, you might get more utility out of write.table and then using
a word processor to read the data in your table.

--Adam

On Tue, 9 Sep 2008, ANJAN PURKAYASTHA wrote:


Hi,
my data table has 38939 rows. R prints  the first 1 columns and then
prints an error message:[ reached getOption("max.print") -- omitted 27821
rows ]].
is it possible to set the maxprint parameter so that R prints all the rows?

tia,
anjan

--
=
anjan purkayastha, phd
bioinformatics analyst
whitehead institute for biomedical research
nine cambridge center
cambridge, ma 02142

purkayas [at] wi [dot] mit [dot] edu
703.740.6939

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Re: [R] help on wavelet

2008-09-09 Thread stephen sefick
It depends on what you want to do.  In wavelet speak frequency is scale.
these are the libraries:
wmtsa - wavCWT (make sure that you pick the wavelet.  I suggest morlet
because it is compactly supported (disappears to zero quickly))
I would also suggest the fields packages for the tim.colors function
which produces the familiar red to blue color scheme.
sowas- more complex stuff here take a look very interesting if you are
trying to tell if two signals are coherent.

hope this helps

stephen

On Tue, Sep 9, 2008 at 12:03 PM, giov <[EMAIL PROTECTED]> wrote:
>
> Hi,
> I have little experience using wavelet and I would like to know if it is
> possible,using R wavelet package, to have a plot of frequency versus time.
>
> thank you
>
> giov
> --
> View this message in context: 
> http://www.nabble.com/help-on-wavelet-tp19395583p19395583.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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[R] creating table of averages

2008-09-09 Thread Lawrence Hanser
Dear Colleagues,

I have a dataframe with variables:

  [1] "ID" "category"   "a11""a12"
"a13""a21"
  [7] "a22""a23""a31""a32"
"b11""b12"
 [13] "b13""b21""b31""b32"
"b33""b41"
 [19] "b42""c11""c12""c21"
"c22""c23"
 [25] "c31""c32""c33""d11"
"d12""d13"
 [31] "d14""d21""d22""d23"
"d24""d25"
 [37] "d31""d32""d33""e11"
"e12""e13"
 [43] "e21""e22""e23""e31"
"e32""e33"
 [49] "f11""f12""f13""f14"
"f21""f22"
 [55] "f23""f24""g11""g12"
"g13""g14"
 [61] "g21""g22""g23""g24"
"g31""g32"
 [67] "g33""g41""g42""g43"
"h11""h12"
 [73] "h13""h21""h22""h23"
"C1.Employ"  "SC11.Ops"
 [79] "SC12.Unit"  "SC13.Nonadvers" "C2.Enterprise"  "SC21.Structure"
"SC22.Gov"   "SC23.Culture"
 [85] "SC24.Stratcomm" "C3.Manage"  "SC31.Resource"  "SC32.Change"
"SC33.Continue"  "C4.Stratthink"
 [91] "SC41.Vision""SC42.Decision"  "SC43.Adapt" "C5.Lead"
"SC51.Develop"   "SC52.Care"
 [97] "SC53.Diversity" "C6.Foster"  "SC61.Teams" "SC62.Negotiate"
"C7.Embody"  "SC71.Ethical"
[103] "SC72.Follower"  "SC73.Warrior"   "SC74.Develop"   "C8.Comm"
"C81.Speak"  "C82.Listen"
[109] "OverallImp"

The variable "category" has four values: Regular, CCM, CFM, and Other

I'd like to create a table like this to feed into barplot2:

row.name  C1.Employ C2.Enterprise  C3.Manage  C4.Stratthink  C5.Lead
C6.Foster  C7.Embody  C8.Comm
Regular 3.68  4.27 3.22
etc..
CCM 4.32  4.56  etc.
CFM  etc.
Other etc.

So far, I have been able to get this far:

 >
mean(subset(impchiefs08,category=="Regular",select=c(C1.Employ,C2.Enterprise,C3.Manage,C4.Stratthink,C5.Lead,C6.Foster,C7.Embody,C8.Comm
)))
C1.Employ C2.Enterprise C3.Manage C4.Stratthink   C5.Lead
C6.Foster C7.Embody   C8.Comm
 3.60  3.85  4.48  4.346667  4.608889
4.44  4.60  4.49
>

But I am stumped as to how to get what I want.

Thanks in advance.

Larry

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[R] randomForest

2008-09-09 Thread Kate Behrman
I am combining many different random forest objects run on the same data set
using the combine ( ) function. After combining the forests I am not sure
whether the variable importance, local importance, and rsq predictors are
recalculated for the new random forest object  or are calculated
individually for each tree ensemble? Is it  possible to  calculate these
predictors for the new random forest object after calling the combine
function? Any help would be greatly apprecaited. Thanks, Kate

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Re: [R] Compiling date

2008-09-09 Thread Dr Eberhard W Lisse

Is this Month-Day or Day-Month or a mixture of both?

I still think using the Format -> Cell -> Date will work
much better...

el


On 09 Sep 2008, at 11:21 , David Scott wrote:


On Mon, 8 Sep 2008, Megh Dal wrote:


Hi,

I have following kind of dataset (all are dates) in my Excel sheet.

09/08/08
09/05/08
09/04/08
09/02/08
09/01/08
29/08/2008
28/08/2008
27/08/2008
26/08/2008
25/08/2008
22/08/2008
21/08/2008
20/08/2008
18/08/2008
14/08/2008
13/08/2008
08/12/08
08/11/08
08/08/08
08/07/08

However I want to use R to compile those data to make all dates in  
same format. Can anyone please tell me any automated way for doing  
that?




Well you have to read them in as character first. Then use sub to  
make the two digit years into four digits. The following could  
probably be improved by a regular expression whiz, but works:



strngs <- c("06/05/08","23/11/2008")
sub("([0-9][0-9]/[0-9][0-9]/)([0-9][0-9]$)","\\120\\2",strngs)

[1] "06/05/2008" "23/11/2008"


David Scott


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Re: [R] creating table of averages

2008-09-09 Thread Adam D. I. Kramer

Maybe something like this:

by(df[,c(77,81,86,90,94,98,101,106)],df$category,apply,2,mean)

...which would then need to be reformatted into a data frame (there is
probably an easy way to do this which I don't know).

aggregate seems like a more reasonable choice, but the function for
aggregate must return scalars, not rows...tapply doesn't take data.frame
inputs. Maybe someone else has a suggestion?

--Adam

On Tue, 9 Sep 2008, Lawrence Hanser wrote:


Dear Colleagues,

I have a dataframe with variables:

 [1] "ID" "category"   "a11""a12"
"a13""a21"
 [7] "a22""a23""a31""a32"
"b11""b12"
[13] "b13""b21""b31""b32"
"b33""b41"
[19] "b42""c11""c12""c21"
"c22""c23"
[25] "c31""c32""c33""d11"
"d12""d13"
[31] "d14""d21""d22""d23"
"d24""d25"
[37] "d31""d32""d33""e11"
"e12""e13"
[43] "e21""e22""e23""e31"
"e32""e33"
[49] "f11""f12""f13""f14"
"f21""f22"
[55] "f23""f24""g11""g12"
"g13""g14"
[61] "g21""g22""g23""g24"
"g31""g32"
[67] "g33""g41""g42""g43"
"h11""h12"
[73] "h13""h21""h22""h23"
"C1.Employ"  "SC11.Ops"
[79] "SC12.Unit"  "SC13.Nonadvers" "C2.Enterprise"  "SC21.Structure"
"SC22.Gov"   "SC23.Culture"
[85] "SC24.Stratcomm" "C3.Manage"  "SC31.Resource"  "SC32.Change"
"SC33.Continue"  "C4.Stratthink"
[91] "SC41.Vision""SC42.Decision"  "SC43.Adapt" "C5.Lead"
"SC51.Develop"   "SC52.Care"
[97] "SC53.Diversity" "C6.Foster"  "SC61.Teams" "SC62.Negotiate"
"C7.Embody"  "SC71.Ethical"
[103] "SC72.Follower"  "SC73.Warrior"   "SC74.Develop"   "C8.Comm"
"C81.Speak"  "C82.Listen"
[109] "OverallImp"

The variable "category" has four values: Regular, CCM, CFM, and Other

I'd like to create a table like this to feed into barplot2:

row.name  C1.Employ C2.Enterprise  C3.Manage  C4.Stratthink  C5.Lead
C6.Foster  C7.Embody  C8.Comm
Regular 3.68  4.27 3.22
etc..
CCM 4.32  4.56  etc.
CFM  etc.
Other etc.

So far, I have been able to get this far:

>
mean(subset(impchiefs08,category=="Regular",select=c(C1.Employ,C2.Enterprise,C3.Manage,C4.Stratthink,C5.Lead,C6.Foster,C7.Embody,C8.Comm
)))
   C1.Employ C2.Enterprise C3.Manage C4.Stratthink   C5.Lead
C6.Foster C7.Embody   C8.Comm
3.60  3.85  4.48  4.346667  4.608889
4.44  4.60  4.49




But I am stumped as to how to get what I want.

Thanks in advance.

Larry

[[alternative HTML version deleted]]

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Re: [R] creating table of averages

2008-09-09 Thread Duncan Murdoch

On 9/9/2008 2:12 PM, Adam D. I. Kramer wrote:

Maybe something like this:

by(df[,c(77,81,86,90,94,98,101,106)],df$category,apply,2,mean)

...which would then need to be reformatted into a data frame (there is
probably an easy way to do this which I don't know).


sparseby() in the reshape package is more flexible than by(). If the 
function returns a vector with a consistent length, you'll get a 
dataframe with columns corresponding to its entries.


Duncan Murdoch



aggregate seems like a more reasonable choice, but the function for
aggregate must return scalars, not rows...tapply doesn't take data.frame
inputs. Maybe someone else has a suggestion?

--Adam

On Tue, 9 Sep 2008, Lawrence Hanser wrote:


Dear Colleagues,

I have a dataframe with variables:

 [1] "ID" "category"   "a11""a12"
"a13""a21"
 [7] "a22""a23""a31""a32"
"b11""b12"
[13] "b13""b21""b31""b32"
"b33""b41"
[19] "b42""c11""c12""c21"
"c22""c23"
[25] "c31""c32""c33""d11"
"d12""d13"
[31] "d14""d21""d22""d23"
"d24""d25"
[37] "d31""d32""d33""e11"
"e12""e13"
[43] "e21""e22""e23""e31"
"e32""e33"
[49] "f11""f12""f13""f14"
"f21""f22"
[55] "f23""f24""g11""g12"
"g13""g14"
[61] "g21""g22""g23""g24"
"g31""g32"
[67] "g33""g41""g42""g43"
"h11""h12"
[73] "h13""h21""h22""h23"
"C1.Employ"  "SC11.Ops"
[79] "SC12.Unit"  "SC13.Nonadvers" "C2.Enterprise"  "SC21.Structure"
"SC22.Gov"   "SC23.Culture"
[85] "SC24.Stratcomm" "C3.Manage"  "SC31.Resource"  "SC32.Change"
"SC33.Continue"  "C4.Stratthink"
[91] "SC41.Vision""SC42.Decision"  "SC43.Adapt" "C5.Lead"
"SC51.Develop"   "SC52.Care"
[97] "SC53.Diversity" "C6.Foster"  "SC61.Teams" "SC62.Negotiate"
"C7.Embody"  "SC71.Ethical"
[103] "SC72.Follower"  "SC73.Warrior"   "SC74.Develop"   "C8.Comm"
"C81.Speak"  "C82.Listen"
[109] "OverallImp"

The variable "category" has four values: Regular, CCM, CFM, and Other

I'd like to create a table like this to feed into barplot2:

row.name  C1.Employ C2.Enterprise  C3.Manage  C4.Stratthink  C5.Lead
C6.Foster  C7.Embody  C8.Comm
Regular 3.68  4.27 3.22
etc..
CCM 4.32  4.56  etc.
CFM  etc.
Other etc.

So far, I have been able to get this far:

>
mean(subset(impchiefs08,category=="Regular",select=c(C1.Employ,C2.Enterprise,C3.Manage,C4.Stratthink,C5.Lead,C6.Foster,C7.Embody,C8.Comm
)))
   C1.Employ C2.Enterprise C3.Manage C4.Stratthink   C5.Lead
C6.Foster C7.Embody   C8.Comm
3.60  3.85  4.48  4.346667  4.608889
4.44  4.60  4.49




But I am stumped as to how to get what I want.

Thanks in advance.

Larry

[[alternative HTML version deleted]]

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Re: [R] creating table of averages

2008-09-09 Thread Lawrence Hanser
Perfect!

Thanks.


On Tue, Sep 9, 2008 at 11:27 AM, Duncan Murdoch <[EMAIL PROTECTED]>wrote:

> On 9/9/2008 2:12 PM, Adam D. I. Kramer wrote:
>
>> Maybe something like this:
>>
>> by(df[,c(77,81,86,90,94,98,101,106)],df$category,apply,2,mean)
>>
>> ...which would then need to be reformatted into a data frame (there is
>> probably an easy way to do this which I don't know).
>>
>
> sparseby() in the reshape package is more flexible than by(). If the
> function returns a vector with a consistent length, you'll get a dataframe
> with columns corresponding to its entries.
>
> Duncan Murdoch
>
>
>
>> aggregate seems like a more reasonable choice, but the function for
>> aggregate must return scalars, not rows...tapply doesn't take data.frame
>> inputs. Maybe someone else has a suggestion?
>>
>> --Adam
>>
>> On Tue, 9 Sep 2008, Lawrence Hanser wrote:
>>
>>  Dear Colleagues,
>>>
>>> I have a dataframe with variables:
>>>
>>>  [1] "ID" "category"   "a11""a12"
>>> "a13""a21"
>>>  [7] "a22""a23""a31""a32"
>>> "b11""b12"
>>> [13] "b13""b21""b31""b32"
>>> "b33""b41"
>>> [19] "b42""c11""c12""c21"
>>> "c22""c23"
>>> [25] "c31""c32""c33""d11"
>>> "d12""d13"
>>> [31] "d14""d21""d22""d23"
>>> "d24""d25"
>>> [37] "d31""d32""d33""e11"
>>> "e12""e13"
>>> [43] "e21""e22""e23""e31"
>>> "e32""e33"
>>> [49] "f11""f12""f13""f14"
>>> "f21""f22"
>>> [55] "f23""f24""g11""g12"
>>> "g13""g14"
>>> [61] "g21""g22""g23""g24"
>>> "g31""g32"
>>> [67] "g33""g41""g42""g43"
>>> "h11""h12"
>>> [73] "h13""h21""h22""h23"
>>> "C1.Employ"  "SC11.Ops"
>>> [79] "SC12.Unit"  "SC13.Nonadvers" "C2.Enterprise"  "SC21.Structure"
>>> "SC22.Gov"   "SC23.Culture"
>>> [85] "SC24.Stratcomm" "C3.Manage"  "SC31.Resource"  "SC32.Change"
>>> "SC33.Continue"  "C4.Stratthink"
>>> [91] "SC41.Vision""SC42.Decision"  "SC43.Adapt" "C5.Lead"
>>> "SC51.Develop"   "SC52.Care"
>>> [97] "SC53.Diversity" "C6.Foster"  "SC61.Teams" "SC62.Negotiate"
>>> "C7.Embody"  "SC71.Ethical"
>>> [103] "SC72.Follower"  "SC73.Warrior"   "SC74.Develop"   "C8.Comm"
>>> "C81.Speak"  "C82.Listen"
>>> [109] "OverallImp"
>>>
>>> The variable "category" has four values: Regular, CCM, CFM, and Other
>>>
>>> I'd like to create a table like this to feed into barplot2:
>>>
>>> row.name  C1.Employ C2.Enterprise  C3.Manage  C4.Stratthink  C5.Lead
>>> C6.Foster  C7.Embody  C8.Comm
>>> Regular 3.68  4.27 3.22
>>> etc..
>>> CCM 4.32  4.56  etc.
>>> CFM  etc.
>>> Other etc.
>>>
>>> So far, I have been able to get this far:
>>>
>>> >
>>>
>>> mean(subset(impchiefs08,category=="Regular",select=c(C1.Employ,C2.Enterprise,C3.Manage,C4.Stratthink,C5.Lead,C6.Foster,C7.Embody,C8.Comm
>>> )))
>>>   C1.Employ C2.Enterprise C3.Manage C4.Stratthink   C5.Lead
>>> C6.Foster C7.Embody   C8.Comm
>>>3.60  3.85  4.48  4.346667  4.608889
>>> 4.44  4.60  4.49
>>>


>>> But I am stumped as to how to get what I want.
>>>
>>> Thanks in advance.
>>>
>>> Larry
>>>
>>>[[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] Hardwarefor R cpu 64 vs 32, dual vs quad

2008-09-09 Thread Henrik Bengtsson
On Tue, Sep 9, 2008 at 6:31 AM, Nic Larson <[EMAIL PROTECTED]> wrote:
> Need to buy fast computer for running R on. Today we use 2,8 MHz intel D cpu
> and the calculations takes around 15 days. Is it possible to get the same
> calculations down to minutes/hours by only changing the hardware?
> Should I go for an really fast dual 32 bit cpu and run R over linux or xp or
> go for an quad core / 64 bit cpu?
> Is it effective to run R on 64 bit (and problem free
> (running/installing))???
> Have around 2000-3000 euro to spend

Faster machines won't do that much.  Without knowing what methods and
algorithms you are running, I bet you a beer that it can be made twice
as fast by just optimizing the code.  My claim applies recursively.
In other words, by optimizing the algorithms/code you can speed up
things quite a bit.  From experience, it is not unlikely to find
bottlenecks in generic algorithms that can be made 10-100 times
faster.  Here is *one* example illustrating that even when you think
the code is "fully optimized" you can still squeeze out more:

  http://wiki.r-project.org/rwiki/doku.php?id=tips:programming:code_optim2

So, start profiling your code to narrow down the parts that takes most
of the CPU time.  help(Rprof) is a start.  There is also a Section
'Profiling R code for speed' in 'Writing R Extensions'.  Good old
verbose print out of system.time() also helps.

My $.02 ...or 2000-3000USD if it was bounty?! ;)

/Henrik

> Thanx for any tip
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
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> and provide commented, minimal, self-contained, reproducible code.
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Re: [R] cluster/snow question

2008-09-09 Thread tolga . i . uzuner
Hi Markus,

Many thanks. Is the "cluster" variable you mention below available in the 
environment of the nodes ? Specifically, within that environment, how 
could one identify the rank of that specific node ?

My code would use that information to partition the problem.

Thanks,
Tolga




Markus Schmidberger <[EMAIL PROTECTED]> 
09/09/2008 07:11
Please respond to
[EMAIL PROTECTED]


To
[EMAIL PROTECTED]
cc
r-help@r-project.org
Subject
Re: [R] cluster/snow question






Hi Tolga,

in SNOW you have to start a cluster with the command

 > library(snow)
 > cluster <- makeCluster(#nodes)

The object "cluster" is a list with an object for each node and each 
object again is a list with all informations (rank, comm, tags)
The size of the cluster is the length of the list.

 > #nodes == length(cluster)

E.g. the rank for node one you can get by
 > cluster[[1]]$rank

Best
Markus

[EMAIL PROTECTED] schrieb:
> Dear R Users,
>
> I am attempting to use the snow package for clustering. Is there a way 
to 
> identfy, in the environment of each node, a rank for that node and also, 

> the total size of the cluster ? 
>
> By way of analogy, I am looking for the functions in snow equivalent to 
> mpi.comm.rank() and mpi.comm.size() from RMPI, in case that makes things 

> clearer.
>
> Thanks in advance,
> Tolga
>
> Generally, this communication is for informational purposes only
> and it is not intended as an offer or solicitation for the purchase
> or sale of any financial instrument or as an official confirmation
> of any transaction. In the event you are receiving the offering
> materials attached below related to your interest in hedge funds or
> private equity, this communication may be intended as an offer or
> solicitation for the purchase or sale of such fund(s).  All market
> prices, data and other information are not warranted as to
> completeness or accuracy and are subject to change without notice.
> Any comments or statements made herein do not necessarily reflect
> those of JPMorgan Chase & Co., its subsidiaries and affiliates.
>
> This transmission may contain information that is privileged,
> confidential, legally privileged, and/or exempt from disclosure
> under applicable law. If you are not the intended recipient, you
> are hereby notified that any disclosure, copying, distribution, or
> use of the information contained herein (including any reliance
> thereon) is STRICTLY PROHIBITED. Although this transmission and any
> attachments are believed to be free of any virus or other defect
> that might affect any computer system into which it is received and
> opened, it is the responsibility of the recipient to ensure that it
> is virus free and no responsibility is accepted by JPMorgan Chase &
> Co., its subsidiaries and affiliates, as applicable, for any loss
> or damage arising in any way from its use. If you received this
> transmission in error, please immediately contact the sender and
> destroy the material in its entirety, whether in electronic or hard
> copy format. Thank you.
> Please refer to http://www.jpmorgan.com/pages/disclosures for
> disclosures relating to UK legal entities.
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 


-- 
Dipl.-Tech. Math. Markus Schmidberger

Ludwig-Maximilians-Universität München
IBE - Institut für medizinische Informationsverarbeitung,
Biometrie und Epidemiologie
Marchioninistr. 15, D-81377 Muenchen
URL: http://ibe.web.med.uni-muenchen.de 
Mail: Markus.Schmidberger [at] ibe.med.uni-muenchen.de
Tel: +49 (089) 7095 - 4599





Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, distribution, or
use of the information contained herein (including any reliance
thereon) is STRICTLY PROHIBITED. Although this transmission and any
attachments are believed to be free of any virus or other defec

[R] Information on the number of CPU's

2008-09-09 Thread tolga . i . uzuner
Dear R Users,
I am on Windows XP SP2 platform, using R version 2.7.2 . I was wondering 
if there is a way to find out, within R, the number of CPU's on my machine 
? I would use this information to set the number of nodes in a cluster, 
depending on the machine. Sys.info() and .Platform do not carry this 
information.
Thanks in advance,
Tolga Uzuner

Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, distribution, or
use of the information contained herein (including any reliance
thereon) is STRICTLY PROHIBITED. Although this transmission and any
attachments are believed to be free of any virus or other defect
that might affect any computer system into which it is received and
opened, it is the responsibility of the recipient to ensure that it
is virus free and no responsibility is accepted by JPMorgan Chase &
Co., its subsidiaries and affiliates, as applicable, for any loss
or damage arising in any way from its use. If you received this
transmission in error, please immediately contact the sender and
destroy the material in its entirety, whether in electronic or hard
copy format. Thank you.
Please refer to http://www.jpmorgan.com/pages/disclosures for
disclosures relating to UK legal entities.
[[alternative HTML version deleted]]

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


Re: [R] : writeMat

2008-09-09 Thread Henrik Bengtsson
Hi,

On Tue, Sep 9, 2008 at 9:53 AM, erola pairo <[EMAIL PROTECTED]> wrote:
> I write a .mat file using the writeMat() command, but when i try to load it
> in Matlab it says that  file may be corrupt. I did it a month ago and it
> worked. It exists any option that I can change for making the file readable
> to Matlab?
>
>> A  <-  c(1:10)
>> dim(A) <- c(2,5)
>> library(R.matlab)
>> writeMat('A.mat', A=A)
>
> And what matlab say is:
> "file may be corrupt"

you need to provide much more information that this.  Please see the
posting guides for the list (see footer), and provide sessionInfo() so
we know what version of R and R.matlab you are using.  What version of
Matlab are you using?  If it worked a month ago, what have you changed
since then?

Cheers

Henrik
(author of R.matlab)

>
> Regards
>
>[[alternative HTML version deleted]]
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
>

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[R] Binning

2008-09-09 Thread Felipe Carrillo
Dear List:
I have a dataset with over 5000 records and I would like to put the Count in 
bins
 based on the ForkLength. e.g.
  Forklength   Count
 32-34?
 35-37?
 38-40?
 and so on...
 and lastly I would like to plot (scatterplot) including the SampleDate
 along the X axis and ForkLength along the Y axis. I recently saw an
  example similar to this one here but I don't want a histogram I  just 
want to see the ForkLength ranges with different colors. For example:
  ForkLength 32-34---green
  ForkLength 35-37---red
  ForkLength 38-40--Orange
  Thanks in advance
 
 SampleDate   ForkLength Count
112/4/2007 32 2
212/6/2007 33 1
312/7/2007 33 2
412/7/2007 33 2
512/7/2007 34 1
612/9/2007 31 1
712/9/2007 33 2
8   12/10/2007 33 5
9   12/10/2007 34 1
10  12/11/2007 33 2
11  12/15/2007 34 1
12  12/16/2007 33 2
13  12/17/2007 35 1
14  12/19/2007 33 1
15  12/19/2007 35 1
16  12/20/2007 31 1
17  12/20/2007 32 1
18  12/20/2007 33 1
19  12/20/2007 34 3
20  12/21/2007 31 1
21  12/21/2007 32 3
22  12/21/2007 33 4
23  12/21/2007 3411
24  12/21/2007 3516
25  12/21/2007 36 3
26  12/21/2007 37 1
27  12/22/2007 32 1
28  12/22/2007 33 3
29  12/22/2007 34 1
30  12/22/2007 35 2
31  12/23/2007 32 1
32  12/23/2007 35 1
33  12/25/2007 32 1
34  12/25/2007 36 1
35  12/26/2007 34 1
36  12/26/2007 35 2
37  12/26/2007 36 1
38  12/27/2007 34 4
39  12/27/2007 35 2
40  12/27/2007 36 2
41  12/28/2007 32 1
42  12/28/2007 33 1
43  12/28/2007 34 1
44  12/28/2007 35 3
45  12/28/2007 36 4
46  12/28/2007 37 6
47  12/28/2007 38 2
48  12/28/2007 39 2
49  12/29/2007 34 1
50  12/29/2007 35 5
51  12/29/2007 36 2
52  12/29/2007 37 1
53  12/30/2007 33 3
54  12/30/2007 3410
55  12/30/2007 3510
56  12/30/2007 36 6
57  12/30/2007 3715
58  12/30/2007 38 3
59  12/31/2007 33 3
60  12/31/2007 34 8
61  12/31/2007 35 9
62  12/31/2007 36 6
63  12/31/2007 37 3
64  12/31/2007 38 1
651/1/2008 34 6
661/1/2008 35 6
671/1/2008 35 1
681/1/2008 36 6
691/1/2008 37 9
701/1/2008 38 1
711/2/2008 34 2
721/2/2008 34 1
731/2/2008 35 2
741/2/2008 36 2
751/2/2008 37 2
761/2/2008 39 1
771/3/2008 34 3
781/3/2008 35 3
791/3/2008 36 2
801/3/2008 37 3
811/8/2008 32 1
821/8/2008 33 7
831/8/2008 34 6
841/8/2008 3510
851/8/2008 3616
861/8/2008 37 7
871/8/2008 38 1
881/8/2008 39 1
891/9/2008 33 1
901/9/2008 3420
911/9/2008 3549
921/9/2008 3649
931/9/2008 3739
941/9/2008 37 1
951/9/2008 3818
961/9/2008 39 1
971/9/2008 40 1
98   1/10/2008 32 3
99   1/10/2008 3313
100  1/10/2008 3456
101  1/10/2008 3533
102  1/10/2008 3624
103  1/10/2008 3718
104  1/10/2008 39 1
105  1/11/2008 33 7
106  1/11/2008 3446
107  1/11/2008 3541
108  1/11/2008 3628
109  1/11/2008 3729

Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA

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Re: [R] cluster/snow question

2008-09-09 Thread Luke Tierney

On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


Hi Markus,

Many thanks. Is the "cluster" variable you mention below available in the
environment of the nodes ? Specifically, within that environment, how
could one identify the rank of that specific node ?


No -- that isn't the way snow works.  With snow the partitioning is
done on the master. If you need a node to know how many other nodes
there are or which index it represents in a clusterApply call then you
need to pass that information in the arguments.

luke



My code would use that information to partition the problem.

Thanks,
Tolga




Markus Schmidberger <[EMAIL PROTECTED]>
09/09/2008 07:11
Please respond to
[EMAIL PROTECTED]


To
[EMAIL PROTECTED]
cc
r-help@r-project.org
Subject
Re: [R] cluster/snow question






Hi Tolga,

in SNOW you have to start a cluster with the command

> library(snow)
> cluster <- makeCluster(#nodes)

The object "cluster" is a list with an object for each node and each
object again is a list with all informations (rank, comm, tags)
The size of the cluster is the length of the list.

> #nodes == length(cluster)

E.g. the rank for node one you can get by
> cluster[[1]]$rank

Best
Markus

[EMAIL PROTECTED] schrieb:

Dear R Users,

I am attempting to use the snow package for clustering. Is there a way

to

identfy, in the environment of each node, a rank for that node and also,



the total size of the cluster ?

By way of analogy, I am looking for the functions in snow equivalent to
mpi.comm.rank() and mpi.comm.size() from RMPI, in case that makes things



clearer.

Thanks in advance,
Tolga

Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, distribution, or
use of the information contained herein (including any reliance
thereon) is STRICTLY PROHIBITED. Although this transmission and any
attachments are believed to be free of any virus or other defect
that might affect any computer system into which it is received and
opened, it is the responsibility of the recipient to ensure that it
is virus free and no responsibility is accepted by JPMorgan Chase &
Co., its subsidiaries and affiliates, as applicable, for any loss
or damage arising in any way from its use. If you received this
transmission in error, please immediately contact the sender and
destroy the material in its entirety, whether in electronic or hard
copy format. Thank you.
Please refer to http://www.jpmorgan.com/pages/disclosures for
disclosures relating to UK legal entities.
   [[alternative HTML version deleted]]

__
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PLEASE do read the posting guide

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

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




--
Dipl.-Tech. Math. Markus Schmidberger

Ludwig-Maximilians-Universit?? M??chen
IBE - Institut f?? medizinische Informationsverarbeitung,
Biometrie und Epidemiologie
Marchioninistr. 15, D-81377 Muenchen
URL: http://ibe.web.med.uni-muenchen.de
Mail: Markus.Schmidberger [at] ibe.med.uni-muenchen.de
Tel: +49 (089) 7095 - 4599





Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, 

[R] splitting time vector into days

2008-09-09 Thread Alexy Khrabrov
Greetings -- I have a dataframe a with one element a vector, time, of  
POSIXct values.  What's a good way to split the data frame into  
periods of a$time, e.g. days, and apply a function, e.g. mean, to some  
other column of the dataframe, e.g. a$value?


Cheers,
Alexy

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Re: [R] Modality Test

2008-09-09 Thread roger koenker

the diptest package, perhaps?


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 Sep 9, 2008, at 11:23 AM, Amin W. Mugera wrote:



Dear Readers:

I have two issues in nonparametric statistical analysis that i need
help:

First, does R have a package that can implement the multimodality  
test,

e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
an earlier thread (sometime in 2003) where someone was trying to write
a code for the Silverman test of multimodality. Is there any other
tests that can enable me to know how many modes are in a distribution?

Second, i would like to test whether two distributions are equal.  
Does R

have a  package than can implement the Li (1996) test of the equality
of two distributions? Is there any other test i can use rather than  
the

Li test?

Thank you in advance for your help.

Amin Mugera
Graduate Student
AgEcon Dept. Kansas State University

__
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Re: [R] cluster/snow question

2008-09-09 Thread tolga . i . uzuner
Understood, that's what I'll do. I'm thinking of exporting the number of 
nodes to all nodes and passing in the node rank as 1:nonodes through 
clusterApply.
Thanks all,
Tolga




Luke Tierney <[EMAIL PROTECTED]> 
09/09/2008 20:11

To
[EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@r-project.org
Subject
Re: [R] cluster/snow question






On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:

> Hi Markus,
>
> Many thanks. Is the "cluster" variable you mention below available in 
the
> environment of the nodes ? Specifically, within that environment, how
> could one identify the rank of that specific node ?

No -- that isn't the way snow works.  With snow the partitioning is
done on the master. If you need a node to know how many other nodes
there are or which index it represents in a clusterApply call then you
need to pass that information in the arguments.

luke

>
> My code would use that information to partition the problem.
>
> Thanks,
> Tolga
>
>
>
>
> Markus Schmidberger <[EMAIL PROTECTED]>
> 09/09/2008 07:11
> Please respond to
> [EMAIL PROTECTED]
>
>
> To
> [EMAIL PROTECTED]
> cc
> r-help@r-project.org
> Subject
> Re: [R] cluster/snow question
>
>
>
>
>
>
> Hi Tolga,
>
> in SNOW you have to start a cluster with the command
>
> > library(snow)
> > cluster <- makeCluster(#nodes)
>
> The object "cluster" is a list with an object for each node and each
> object again is a list with all informations (rank, comm, tags)
> The size of the cluster is the length of the list.
>
> > #nodes == length(cluster)
>
> E.g. the rank for node one you can get by
> > cluster[[1]]$rank
>
> Best
> Markus
>
> [EMAIL PROTECTED] schrieb:
>> Dear R Users,
>>
>> I am attempting to use the snow package for clustering. Is there a way
> to
>> identfy, in the environment of each node, a rank for that node and 
also,
>
>> the total size of the cluster ?
>>
>> By way of analogy, I am looking for the functions in snow equivalent to
>> mpi.comm.rank() and mpi.comm.size() from RMPI, in case that makes 
things
>
>> clearer.
>>
>> Thanks in advance,
>> Tolga
>>
>> Generally, this communication is for informational purposes only
>> and it is not intended as an offer or solicitation for the purchase
>> or sale of any financial instrument or as an official confirmation
>> of any transaction. In the event you are receiving the offering
>> materials attached below related to your interest in hedge funds or
>> private equity, this communication may be intended as an offer or
>> solicitation for the purchase or sale of such fund(s).  All market
>> prices, data and other information are not warranted as to
>> completeness or accuracy and are subject to change without notice.
>> Any comments or statements made herein do not necessarily reflect
>> those of JPMorgan Chase & Co., its subsidiaries and affiliates.
>>
>> This transmission may contain information that is privileged,
>> confidential, legally privileged, and/or exempt from disclosure
>> under applicable law. If you are not the intended recipient, you
>> are hereby notified that any disclosure, copying, distribution, or
>> use of the information contained herein (including any reliance
>> thereon) is STRICTLY PROHIBITED. Although this transmission and any
>> attachments are believed to be free of any virus or other defect
>> that might affect any computer system into which it is received and
>> opened, it is the responsibility of the recipient to ensure that it
>> is virus free and no responsibility is accepted by JPMorgan Chase &
>> Co., its subsidiaries and affiliates, as applicable, for any loss
>> or damage arising in any way from its use. If you received this
>> transmission in error, please immediately contact the sender and
>> destroy the material in its entirety, whether in electronic or hard
>> copy format. Thank you.
>> Please refer to http://www.jpmorgan.com/pages/disclosures for
>> disclosures relating to UK legal entities.
>>[[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
> --
> Dipl.-Tech. Math. Markus Schmidberger
>
> Ludwig-Maximilians-Universit?? M??chen
> IBE - Institut f?? medizinische Informationsverarbeitung,
> Biometrie und Epidemiologie
> Marchioninistr. 15, D-81377 Muenchen
> URL: http://ibe.web.med.uni-muenchen.de
> Mail: Markus.Schmidberger [at] ibe.med.uni-muenchen.de
> Tel: +49 (089) 7095 - 4599
>
>
>
>
>
> Generally, this communication is for informational purposes only
> and it is not intended as an offer or solicitation for the purchase
> or sale of any financial instrument or as an official confirmation
> of any transaction. In the event you are receiving the offering
> materials attached below related to your interest in hedge funds or
> private equity, this communicatio

Re: [R] Information on the number of CPU's

2008-09-09 Thread Prof Brian Ripley

On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


Dear R Users,
I am on Windows XP SP2 platform, using R version 2.7.2 . I was wondering
if there is a way to find out, within R, the number of CPU's on my machine
? I would use this information to set the number of nodes in a cluster,
depending on the machine. Sys.info() and .Platform do not carry this
information.


Correct, since

a) R does not make use of more than 1.

b) It is really not portable, and not even well-defined.  (How many CPUs 
has a hyperthreaded dual Xeon?  Some say 2, some say 4.  Do you want 
CPUs or cores?  If this is a virtualized OS, is the physical number or the 
logical number?)


In the case of Windows, how depends on the Windows version.  The w32api 
(XP or later) call GetNativeSystemInfo will tell you the number of CPUs, 
for some (unstated) definition of 'CPU'.  Later versions have 
GetLogicalProcessorInformation, which can give the number of cores.



Thanks in advance,
Tolga Uzuner


--
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

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


Re: [R] Information on the number of CPU's

2008-09-09 Thread tolga . i . uzuner

Many thanks, that's very helpful.
Regards,
Tolga


- Original Message -
From: Prof Brian Ripley [EMAIL PROTECTED]
Sent: 09/09/2008 20:57 CET
To: Tolga Uzuner
Cc: r-help@r-project.org
Subject: Re: [R] Information on the number of CPU's



On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


Dear R Users,
I am on Windows XP SP2 platform, using R version 2.7.2 . I was wondering
if there is a way to find out, within R, the number of CPU's on my machine
? I would use this information to set the number of nodes in a cluster,
depending on the machine. Sys.info() and .Platform do not carry this
information.


Correct, since

a) R does not make use of more than 1.

b) It is really not portable, and not even well-defined.  (How many CPUs 
has a hyperthreaded dual Xeon?  Some say 2, some say 4.  Do you want 
CPUs or cores?  If this is a virtualized OS, is the physical number or the 
logical number?)


In the case of Windows, how depends on the Windows version.  The w32api 
(XP or later) call GetNativeSystemInfo will tell you the number of CPUs, 
for some (unstated) definition of 'CPU'.  Later versions have 
GetLogicalProcessorInformation, which can give the number of cores.



Thanks in advance,
Tolga Uzuner


--
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
Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, distribution, or
use of the information contained herein (including any reliance
thereon) is STRICTLY PROHIBITED. Although this transmission and any
attachments are believed to be free of any virus or other defect
that might affect any computer system into which it is received and
opened, it is the responsibility of the recipient to ensure that it
is virus free and no responsibility is accepted by JPMorgan Chase &
Co., its subsidiaries and affiliates, as applicable, for any loss
or damage arising in any way from its use. If you received this
transmission in error, please immediately contact the sender and
destroy the material in its entirety, whether in electronic or hard
copy format. Thank you.
Please refer to http://www.jpmorgan.com/pages/disclosures for
disclosures relating to UK legal entities.

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


Re: [R] Modality Test

2008-09-09 Thread Mark Difford

Hi Amin,

>> First, does R have a package that can implement the multimodality test, 
>> e.g., the Silverman test, DIP test, MAP test or Runt test.

Jeremy Tantrum (a Ph.D. student of Werner Steutzle's, c. 2003/04) did some
work on this. There is some useful code on Steutzle's website:

http://www.stat.washington.edu/wxs/Stat593-s03/Code/jeremy-unimodality.R

I used it last year when I was trying to solve the problem of how best to
compare lots of density curves (age distributions of 3 spp. of tree
euphorbias from about very different 35 sites). In particular I had to
ensure that I wasn't creating spurious bimodality at a particular age range
when combining sites.

You might find it useful. Feel free to contact me off list if the code has
gone, as I think I still have it (somewhere).

Regards, Mark.


Amin W. Mugera wrote:
> 
> 
> Dear Readers:
> 
> I have two issues in nonparametric statistical analysis that i need
> help:
> 
> First, does R have a package that can implement the multimodality test,
> e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
> an earlier thread (sometime in 2003) where someone was trying to write
> a code for the Silverman test of multimodality. Is there any other
> tests that can enable me to know how many modes are in a distribution?
> 
> Second, i would like to test whether two distributions are equal. Does R
> have a  package than can implement the Li (1996) test of the equality
> of two distributions? Is there any other test i can use rather than the
> Li test?
> 
> Thank you in advance for your help.
> 
> Amin Mugera
> Graduate Student
> AgEcon Dept. Kansas State University
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
> 

-- 
View this message in context: 
http://www.nabble.com/Modality-Test-tp19396085p19400095.html
Sent from the R help mailing list archive at Nabble.com.

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


Re: [R] Modality Test

2008-09-09 Thread Mark Difford

Whoops! I think that should be Stuetzle --- though I very much doubt that he
reads the list.


Mark Difford wrote:
> 
> Hi Amin,
> 
>>> First, does R have a package that can implement the multimodality test, 
>>> e.g., the Silverman test, DIP test, MAP test or Runt test.
> 
> Jeremy Tantrum (a Ph.D. student of Werner Steutzle's, c. 2003/04) did some
> work on this. There is some useful code on Steutzle's website:
> 
> http://www.stat.washington.edu/wxs/Stat593-s03/Code/jeremy-unimodality.R
> 
> I used it last year when I was trying to solve the problem of how best to
> compare lots of density curves (age distributions of 3 spp. of tree
> euphorbias from about very different 35 sites). In particular I had to
> ensure that I wasn't creating spurious bimodality at a particular age
> range when combining sites.
> 
> You might find it useful. Feel free to contact me off list if the code has
> gone, as I think I still have it (somewhere).
> 
> Regards, Mark.
> 
> 
> Amin W. Mugera wrote:
>> 
>> 
>> Dear Readers:
>> 
>> I have two issues in nonparametric statistical analysis that i need
>> help:
>> 
>> First, does R have a package that can implement the multimodality test,
>> e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
>> an earlier thread (sometime in 2003) where someone was trying to write
>> a code for the Silverman test of multimodality. Is there any other
>> tests that can enable me to know how many modes are in a distribution?
>> 
>> Second, i would like to test whether two distributions are equal. Does R
>> have a  package than can implement the Li (1996) test of the equality
>> of two distributions? Is there any other test i can use rather than the
>> Li test?
>> 
>> Thank you in advance for your help.
>> 
>> Amin Mugera
>> Graduate Student
>> AgEcon Dept. Kansas State University
>> 
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>> 
>> 
> 
> 

-- 
View this message in context: 
http://www.nabble.com/Modality-Test-tp19396085p19400138.html
Sent from the R help mailing list archive at Nabble.com.

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


Re: [R] Information on the number of CPU's

2008-09-09 Thread Luke Tierney

The wmic command line utility can also be used to query this; on a
dual-core Vista laptop I get

C:\Users\luke>wmic cpu get NumberOfCores,NumberOfLogicalProcessors
NumberOfCores  NumberOfLogicalProcessors
2  2

luke

--

Luke Tierney
University of Iowa  Phone: 319-335-3386
Department of Statistics andFax:   319-335-3017
   Actuarial Science
241 Schaeffer Hall  email:  [EMAIL PROTECTED]
Iowa City, IA 52242 WWW:  http://www.stat.uiowa.edu

On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


Many thanks, that's very helpful.
Regards,
Tolga


- Original Message -
From: Prof Brian Ripley [EMAIL PROTECTED]
Sent: 09/09/2008 20:57 CET
To: Tolga Uzuner
Cc: r-help@r-project.org
Subject: Re: [R] Information on the number of CPU's



On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


Dear R Users,
I am on Windows XP SP2 platform, using R version 2.7.2 . I was wondering
if there is a way to find out, within R, the number of CPU's on my machine
? I would use this information to set the number of nodes in a cluster,
depending on the machine. Sys.info() and .Platform do not carry this
information.


Correct, since

a) R does not make use of more than 1.

b) It is really not portable, and not even well-defined.  (How many CPUs has 
a hyperthreaded dual Xeon?  Some say 2, some say 4.  Do you want CPUs or 
cores?  If this is a virtualized OS, is the physical number or the logical 
number?)


In the case of Windows, how depends on the Windows version.  The w32api (XP 
or later) call GetNativeSystemInfo will tell you the number of CPUs, for some 
(unstated) definition of 'CPU'.  Later versions have 
GetLogicalProcessorInformation, which can give the number of cores.



Thanks in advance,
Tolga Uzuner


--
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
Generally, this communication is for informational purposes only
and it is not intended as an offer or solicitation for the purchase
or sale of any financial instrument or as an official confirmation
of any transaction. In the event you are receiving the offering
materials attached below related to your interest in hedge funds or
private equity, this communication may be intended as an offer or
solicitation for the purchase or sale of such fund(s).  All market
prices, data and other information are not warranted as to
completeness or accuracy and are subject to change without notice.
Any comments or statements made herein do not necessarily reflect
those of JPMorgan Chase & Co., its subsidiaries and affiliates.

This transmission may contain information that is privileged,
confidential, legally privileged, and/or exempt from disclosure
under applicable law. If you are not the intended recipient, you
are hereby notified that any disclosure, copying, distribution, or
use of the information contained herein (including any reliance
thereon) is STRICTLY PROHIBITED. Although this transmission and any
attachments are believed to be free of any virus or other defect
that might affect any computer system into which it is received and
opened, it is the responsibility of the recipient to ensure that it
is virus free and no responsibility is accepted by JPMorgan Chase &
Co., its subsidiaries and affiliates, as applicable, for any loss
or damage arising in any way from its use. If you received this
transmission in error, please immediately contact the sender and
destroy the material in its entirety, whether in electronic or hard
copy format. Thank you.
Please refer to http://www.jpmorgan.com/pages/disclosures for
disclosures relating to UK legal entities.

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Modality Test

2008-09-09 Thread Mark Difford

Hi Amin,

And I have just remembered that there is a function called curveRep in Frank
Harrell's Hmisc package that might be useful, even if not quite in the
channel of your enquiry. curveRep was added to the package after my
struggles, so I never used it and so don't know how well it performs (quite
well, I would think).

Regards, Mark.


Amin W. Mugera wrote:
> 
> 
> Dear Readers:
> 
> I have two issues in nonparametric statistical analysis that i need
> help:
> 
> First, does R have a package that can implement the multimodality test,
> e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
> an earlier thread (sometime in 2003) where someone was trying to write
> a code for the Silverman test of multimodality. Is there any other
> tests that can enable me to know how many modes are in a distribution?
> 
> Second, i would like to test whether two distributions are equal. Does R
> have a  package than can implement the Li (1996) test of the equality
> of two distributions? Is there any other test i can use rather than the
> Li test?
> 
> Thank you in advance for your help.
> 
> Amin Mugera
> Graduate Student
> AgEcon Dept. Kansas State University
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
> 

-- 
View this message in context: 
http://www.nabble.com/Modality-Test-tp19396085p19400426.html
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[R] NMDS and varimax rotation

2008-09-09 Thread Bernd Panassiti
hello,

subsequently to a NMDS analysis (performed with metaMDS or isoMDS) is 
it possible to 
rotate the axis through a varimax-rotation?

Thanks in advance.

Bernd Panassiti

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[R] csaps in R?

2008-09-09 Thread Dr Carbon
Is there is function in R equivalent to Matlab's csaps? I need a
spline function with the same calculation of the smoothing parameter
in csaps to compare some results. AFAIK, the spar in smooth.spline is
related but not the same.

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[R] tsdiag error

2008-09-09 Thread rkevinburton
Does anyone know why I get the following error when trying tsdiag?

Error in UseMethod("tsdiag") : no applicable method for "tsdiag"

I am invoking it as: tsdiag(mar).

Thank you.

Kevin

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Re: [R] splitting time vector into days

2008-09-09 Thread stephen sefick
?aggregate
?window.zoo
?rollapply

anyway have a look at package zoo

On Tue, Sep 9, 2008 at 3:25 PM, Alexy Khrabrov <[EMAIL PROTECTED]> wrote:
> Greetings -- I have a dataframe a with one element a vector, time, of
> POSIXct values.  What's a good way to split the data frame into periods of
> a$time, e.g. days, and apply a function, e.g. mean, to some other column of
> the dataframe, e.g. a$value?
>
> Cheers,
> Alexy
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] NMDS and varimax rotation

2008-09-09 Thread stephen sefick
have you looked at the vegan viginette- I know there is proscrutes rotation.

On Tue, Sep 9, 2008 at 3:54 PM, Bernd Panassiti
<[EMAIL PROTECTED]> wrote:
> hello,
>
> subsequently to a NMDS analysis (performed with metaMDS or isoMDS) is
> it possible to
> rotate the axis through a varimax-rotation?
>
> Thanks in advance.
>
> Bernd Panassiti
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

__
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[R] building a package that contains S4 classes and methods

2008-09-09 Thread Marie Pierre Sylvestre
Hello R users,

I am trying to make a my first package and I get an error that I can
understand. The package is build out of three files (one for functions, 1
for s4 classes and 1 for s4 methods).

Once I source them I run 

package.skeleton( name="TDC" )

within a R session and I get 

Creating directories ...
Creating DESCRIPTION ...
Creating Read-and-delete-me ...
Saving functions and data ...
Making help files ...
Done.
Further steps are described in './TDC/Read-and-delete-me'.
Warning messages:
1: In dump(internalObjs, file = file.path(code_dir,
sprintf("%s-internal.R",  :
  deparse of an S4 object will not be source()able
2: In dump(internalObjs, file = file.path(code_dir,
sprintf("%s-internal.R",  :
  deparse of an S4 object will not be source()able
3: In dump(internalObjs, file = file.path(code_dir,
sprintf("%s-internal.R",  :
  deparse of an S4 object will not be source()able
4: In dump(internalObjs, file = file.path(code_dir,
sprintf("%s-internal.R",  :
  deparse may be incomplete


I keep going in spite of the warnings with 
R CMD check --no-examples TDC

and I get 
* checking for working pdflatex ... OK
* using log directory
'/home/mariepierre/Packages/PermAlgo/PermAlgo/PermAlgo2/TDC.Rcheck'
* using R version 2.7.1 (2008-06-23)
* using session charset: UTF-8
* checking for file 'TDC/DESCRIPTION' ... OK
* checking extension type ... Package
* this is package 'TDC' version '1.0'
* checking package dependencies ... OK
* checking if this is a source package ... OK
* checking whether package 'TDC' can be installed ... ERROR
Installation failed.

The error file says:
 
* Installing *source* package 'TDC' ...
** R
** preparing package for lazy loading
Error in parse(n = -1, file = file) : unexpected '<' at
102: `.__C__BindArgs` <-
103: <
Calls:  -> code2LazyLoadDB -> sys.source -> parse
Execution halted
ERROR: lazy loading failed for package 'TDC'
** Removing
'/home/mariepierre/Packages/PermAlgo/PermAlgo/PermAlgo2/TDC.Rcheck/TDC'

The problem is with my classes and methods. The respective files contain:

setClass("BindArgs",  signature( "function" ))
setClass("BindArgs2", signature( "function" )) 

and

setMethod("initialize", "BindArgs", function( .Object, f, ... )
  callNextMethod( .Object, function( x ) f( x, ... ) )) 

setMethod("initialize", "BindArgs2", function( .Object, f, ...)
  callNextMethod( .Object, function( x, y ) f( x, y, ... ) )) 

Everything works well within a R session but I can build the package.

If I look at the internal R file that this created I get

`.__C__BindArgs` <-

`.__C__BindArgs2` <-

`.__M__initialize:methods` <-

`.__T__initialize:methods` <-


Well, let just say that I am new to classes so this confuses me greatly. I
have checked the documentation and tried a few things but I reached my
personal limits!

I am using R 2.7.1 on Linux Fedora 8.

Any comments on what is happening and/or help would be greatly appreciated.

MP

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Re: [R] splitting time vector into days

2008-09-09 Thread jim holtman
Here is one way of doing it:

> x <- data.frame(dates=seq(as.POSIXct('2008-09-08'), by='7 hours', length=10),
+ values=1:10)
> # split into days
> x.s <- split(x, format(x$dates, "%Y%m%d"))
> x.s
$`20080908`
dates values
1 2008-09-08 00:00:00  1
2 2008-09-08 07:00:00  2
3 2008-09-08 14:00:00  3
4 2008-09-08 21:00:00  4

$`20080909`
dates values
5 2008-09-09 04:00:00  5
6 2008-09-09 11:00:00  6
7 2008-09-09 18:00:00  7

$`20080910`
 dates values
8  2008-09-10 01:00:00  8
9  2008-09-10 08:00:00  9
10 2008-09-10 15:00:00 10

> lapply(x.s, function(.df) mean(.df$values))
$`20080908`
[1] 2.5

$`20080909`
[1] 6

$`20080910`
[1] 9

>


On Tue, Sep 9, 2008 at 3:25 PM, Alexy Khrabrov <[EMAIL PROTECTED]> wrote:
> Greetings -- I have a dataframe a with one element a vector, time, of
> POSIXct values.  What's a good way to split the data frame into periods of
> a$time, e.g. days, and apply a function, e.g. mean, to some other column of
> the dataframe, e.g. a$value?
>
> Cheers,
> Alexy
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

What is the problem that you are trying to solve?

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Re: [R] Help with 'spectrum'

2008-09-09 Thread Prof Brian Ripley
This is why some help pages have references: please use them (Venables & 
Ripley explain the exact formulae used in R).


On Tue, 9 Sep 2008, [EMAIL PROTECTED] wrote:


For the command 'spectrum' I read:

The spectrum here is defined with scaling 1/frequency(x), following 
S-PLUS. This makes the spectral density a density over the range 
(-frequency(x)/2, +frequency(x)/2], whereas a more common scaling is 2π 
and range (-0.5, 0.5] (e.g., Bloomfield) or 1 and range (-π, π].



Forgive my ignorance but I am having a hard time interpreting this. Does 
this mean that in the spectrum output every element of the $spec array 
is scaled by 1/frequency(x)? I am having a hard time determing what is 
meant by 'frequency'.


So please do look up the help for frequency().

 Say I define a time series for a year with samples 
for every day. I input a 'frequency' of 365 (which in my mind is the 
period).


The point is that your time unit is 1 year, and your measurements are 
every 1/365 year.  That is unrelated to the 'period' (no one mentioned 
periodicity yet).


On the output of 'spectrum' would this mean that every element 
of the $spec array is scaled by 1/365? There is a corresponding 
frequency array on the output from 'spectrum'. If the frequency is 365 
and an element in the frequency array output from 'spectrum' is .1 am I 
to assume that the period is 36.5 and a corresponding sin wave would be 
sin(2 * pi * 36.5/365)?


Hmm, you need a 't' in there (and a phase).  The issue is the units for t. 
A frequency in the 'freq' element of the output of 0.1 corresponds to 10 
cycles per unit of time, and in your example the unit of time is 365 
observations.  So the sine (sic) wave is sin(2*pi*0.1*t + phi), where the 
increments in 't' are 1/365: that gives 10 complete cycles in observations 
at, say, c(1990, 1) ... c(1990, 365), the days of 1990 (not a leap year).



Thank you in advance for helping me clear up some confusion.

Kevin


--
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] Binning

2008-09-09 Thread jim holtman
This should do what you want.

#--x <- read.table('clipboard', header=TRUE, as.is=TRUE)
# convert dates
x$date <- as.POSIXct(strptime(x$SampleDate, "%m/%d/%Y"))
# put ForkLength into bins
x$bins <- cut(x$ForkLength, breaks=c(32, 34, 37, 40), include.lowest=TRUE)
# count the bins
tapply(x$Count, x$bins, sum)
# plot the data
plot(x$date, x$ForkLength, col=c('green', 'red', 'orange')[x$bins])






On Tue, Sep 9, 2008 at 3:12 PM, Felipe Carrillo
<[EMAIL PROTECTED]> wrote:
> Dear List:
> I have a dataset with over 5000 records and I would like to put the Count in 
> bins
>  based on the ForkLength. e.g.
>  Forklength   Count
> 32-34?
> 35-37?
> 38-40?
> and so on...
> and lastly I would like to plot (scatterplot) including the SampleDate
> along the X axis and ForkLength along the Y axis. I recently saw an
>  example similar to this one here but I don't want a histogram I  
> just want to see the ForkLength ranges with different colors. For example:
>  ForkLength 32-34---green
>  ForkLength 35-37---red
>  ForkLength 38-40--Orange
>  Thanks in advance
>
>  SampleDate   ForkLength Count
> 112/4/2007 32 2
> 212/6/2007 33 1
> 312/7/2007 33 2
> 412/7/2007 33 2
> 512/7/2007 34 1
> 612/9/2007 31 1
> 712/9/2007 33 2
> 8   12/10/2007 33 5
> 9   12/10/2007 34 1
> 10  12/11/2007 33 2
> 11  12/15/2007 34 1
> 12  12/16/2007 33 2
> 13  12/17/2007 35 1
> 14  12/19/2007 33 1
> 15  12/19/2007 35 1
> 16  12/20/2007 31 1
> 17  12/20/2007 32 1
> 18  12/20/2007 33 1
> 19  12/20/2007 34 3
> 20  12/21/2007 31 1
> 21  12/21/2007 32 3
> 22  12/21/2007 33 4
> 23  12/21/2007 3411
> 24  12/21/2007 3516
> 25  12/21/2007 36 3
> 26  12/21/2007 37 1
> 27  12/22/2007 32 1
> 28  12/22/2007 33 3
> 29  12/22/2007 34 1
> 30  12/22/2007 35 2
> 31  12/23/2007 32 1
> 32  12/23/2007 35 1
> 33  12/25/2007 32 1
> 34  12/25/2007 36 1
> 35  12/26/2007 34 1
> 36  12/26/2007 35 2
> 37  12/26/2007 36 1
> 38  12/27/2007 34 4
> 39  12/27/2007 35 2
> 40  12/27/2007 36 2
> 41  12/28/2007 32 1
> 42  12/28/2007 33 1
> 43  12/28/2007 34 1
> 44  12/28/2007 35 3
> 45  12/28/2007 36 4
> 46  12/28/2007 37 6
> 47  12/28/2007 38 2
> 48  12/28/2007 39 2
> 49  12/29/2007 34 1
> 50  12/29/2007 35 5
> 51  12/29/2007 36 2
> 52  12/29/2007 37 1
> 53  12/30/2007 33 3
> 54  12/30/2007 3410
> 55  12/30/2007 3510
> 56  12/30/2007 36 6
> 57  12/30/2007 3715
> 58  12/30/2007 38 3
> 59  12/31/2007 33 3
> 60  12/31/2007 34 8
> 61  12/31/2007 35 9
> 62  12/31/2007 36 6
> 63  12/31/2007 37 3
> 64  12/31/2007 38 1
> 651/1/2008 34 6
> 661/1/2008 35 6
> 671/1/2008 35 1
> 681/1/2008 36 6
> 691/1/2008 37 9
> 701/1/2008 38 1
> 711/2/2008 34 2
> 721/2/2008 34 1
> 731/2/2008 35 2
> 741/2/2008 36 2
> 751/2/2008 37 2
> 761/2/2008 39 1
> 771/3/2008 34 3
> 781/3/2008 35 3
> 791/3/2008 36 2
> 801/3/2008 37 3
> 811/8/2008 32 1
> 821/8/2008 33 7
> 831/8/2008 34 6
> 841/8/2008 3510
> 851/8/2008 3616
> 861/8/2008 37 7
> 871/8/2008 38 1
> 881/8/2008 39 1
> 891/9/2008 33 1
> 901/9/2008 3420
> 911/9/2008 3549
> 921/9/2008 3649
> 931/9/2008 3739
> 941/9/2008 37 1
> 951/9/2008 3818
> 961/9/2008 39 1
> 971/9/2008 40 1
> 98   1/10/2008 32 3
> 99   1/10/2008 3313
> 100  1/10/2008 3456
> 101  1/10/2008 3533
> 102  1/10/2008 3624
> 103  1/10/2008 3718
> 104  1/10/2008 39 1
> 105  1/11/2008 33 7
> 106  1/11/2008 3446
> 107  1/11/2008 3541
> 108  1/11/2008 3628
> 109  1/11/2008 3729
>
> Felipe D. Carrillo
> Supervisory Fishery Biologist
> Department of the Interior
> US Fish & Wildlife Service
> Cali

Re: [R] naive variance in GEE

2008-09-09 Thread Thomas Lumley

On Mon, 8 Sep 2008, Qiong Yang wrote:


Hi,

The standard error from logistic regression is slightly different from the 
naive SE from GEE under independence working correlation structure.


Yes


Shouldn't they be identical? Anyone has insight about this?


No, they shouldn't. They are different estimators of the same quantity, 
like the mean and median of a symmetric distribution.


-thomas





Thanks,
Qiong

a<-rbinom(1000,1)
b<-rbinom(1000,2,0.1)
c<-rbinom(1000,10,0.5)
summary(gee(a~b, id=c,family="binomial",corstr="independence"))$coef
summary(glm(a~b,family="binomial"))

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



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

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Re: [R] naive variance in GEE

2008-09-09 Thread Thomas Lumley


Sorry, I misread your message. Prof Ripley is right, as usual -- the 
estimates use different stopping criteria and so are just numerically 
different.


-thomas

On Tue, 9 Sep 2008, Thomas Lumley wrote:


On Mon, 8 Sep 2008, Qiong Yang wrote:


Hi,

The standard error from logistic regression is slightly different from the 
naive SE from GEE under independence working correlation structure.


Yes


Shouldn't they be identical? Anyone has insight about this?


No, they shouldn't. They are different estimators of the same quantity, like 
the mean and median of a symmetric distribution.


-thomas





Thanks,
Qiong

a<-rbinom(1000,1)
b<-rbinom(1000,2,0.1)
c<-rbinom(1000,10,0.5)
summary(gee(a~b, id=c,family="binomial",corstr="independence"))$coef
summary(glm(a~b,family="binomial"))

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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.



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



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

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[R] [R-pkgs] survey package

2008-09-09 Thread Thomas Lumley


Version 3.9 of the survey package is now on CRAN.  Since the last 
announcement (version 3.6-11, about a year ago) the main changes are
 - Database-backed survey objects: the data can live in a SQLite (or other 
DBI-compatible) database and be loaded as needed.

 - Ordinal logistic regression
 - Support for the 'mitools' package and multiply-imputed data
 - Conditioning plots, transparent scatterplots, survival and CDF plots.

There is more information on the package web page at
http://faculty.washington.edu/tlumley/survey/

-thomas

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

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Re: [R] Modality Test

2008-09-09 Thread Bert Gunter
Here is a function that tests for equality of however many distributions as
you like:

equaldist <- function(...){
## ... numeric sample vectors from the possibly different distributions to
be tested
## returns TRUE only if the distributions are the same
FALSE
}

;-)

-- Bert Gunter
Genentech 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Mark Difford
Sent: Tuesday, September 09, 2008 1:23 PM
To: r-help@r-project.org
Subject: Re: [R] Modality Test


Hi Amin,

And I have just remembered that there is a function called curveRep in Frank
Harrell's Hmisc package that might be useful, even if not quite in the
channel of your enquiry. curveRep was added to the package after my
struggles, so I never used it and so don't know how well it performs (quite
well, I would think).

Regards, Mark.


Amin W. Mugera wrote:
> 
> 
> Dear Readers:
> 
> I have two issues in nonparametric statistical analysis that i need
> help:
> 
> First, does R have a package that can implement the multimodality test,
> e.g., the Silverman test, DIP test, MAP test or Runt test. I have seen
> an earlier thread (sometime in 2003) where someone was trying to write
> a code for the Silverman test of multimodality. Is there any other
> tests that can enable me to know how many modes are in a distribution?
> 
> Second, i would like to test whether two distributions are equal. Does R
> have a  package than can implement the Li (1996) test of the equality
> of two distributions? Is there any other test i can use rather than the
> Li test?
> 
> Thank you in advance for your help.
> 
> Amin Mugera
> Graduate Student
> AgEcon Dept. Kansas State University
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 
> 

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[R] "figure margins too large" issue

2008-09-09 Thread Hui-Yi Chu
Hi everyone,

I updated R from 2.6.2 to 2.7.2 recently but keep getting the error "figure
margins too large" when plot pictures which never happened to me when using
2.6.2. After googling and searching the mailing list, I still have no idea
how to optimize it. One of my packages fixed bugs in 2.7.2 version, so that
I have to use it. Any suggestions are really appreciated!!!

Best regards,
Hui-Yi Chu

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[R] test for a single variance

2008-09-09 Thread Edna Bell
Dear R Gurus:

Is there a test for a single variance available in R, please?

Thanks,
Edna Bell

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[R] New member with question on multiple comparisons in mixed effects models

2008-09-09 Thread Giuliano Matessi
Dear fellow R.users/.lovers,
 
I am very new to both R and this list, so I hope you will be patient with me in 
the beginning if my enquiries are inappropriate/unclear.
 
I am trying to perform some rather complex statistical modelling using 
mixed-effects models.
 
I have, after a rather difficult beginning, finally boiled down my model (using 
the lme function in nlme) to a couple of fixed effects and their interactions, 
plus a couple of nested random effects. 
 
For the moment I will not include the details because the dataset is rather 
large and without it is difficult to be more specific, but I can provide them 
if requested.
 
My question is maybe a bit trivial, but I would like to get out of the model 
the pairwise comparisons for all the factor levels using TukeyHSD, but when I 
try to do this using the "glht" method in the multcomp library I get an error 
message which seems to indicate a wrong synthax, yet I think I am following the 
instructions given on the package documentation. Below are the code lines and 
the R reply.
 
> glht(inner.exch.M9.lme,linfct=mcp(Personality:Day = "Tukey"))
Error: unexpected '=' in "glht(inner.exch.M9.lme,linfct=mcp(Personality:Day ="

(inner.exch.M9.lme is the model object, Personality and Day are the two fixed 
factors the levels of which I want to compare)
 
I also tried without the interaction symbol and only one factor, with the same 
result.
 
As I understand it, the multcomp package is new so maybe the method is not 
implemented for complex designs yet.
 
Any suggestion will be appreciated (including such as "read the instructions" 
if they can point me to such instructions!).
 
 
Thank you very much in advance and all the best
 
Giuliano
 
 
 

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[R] Package installation in Windows

2008-09-09 Thread boxyzzy
As a sys admin, how do I install packages so that there is one common
library for all users of the MS Windows computer, instead of the
default individual location for each user?

I've done this for Linux.

Thanks, in advance, for your help.

Mike

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