Re: [R] Re move row.names column in dataframe

[Bill.Venables]

- There is no column in the data frame called "row.names".  It only
looks that way when you use the viewer.

- All data frames must have row and column names.  This is not
negotiable.  Also, the row.names will not interfere with any merging
operation because they are not a column of the data frame: they are the
row names.


Bill Venables
CSIRO Laboratories
mailto:[EMAIL PROTECTED]
http://www.cmis.csiro.au/bill.venables/ 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of tonyxv
Sent: Sunday, 22 June 2008 11:15 AM
To: r-help@r-project.org
Subject: [R] Re move row.names column in dataframe



Hello,

aa<-c(1,1,2,2,3,3,4,4,5,5,6,6)
bb<-c(56,56,33,33,53,53,20,20,63,63,9,9) 
cc<-data.frame(aa,bb) 
uniquedf <- unique(cc)
View(uniquedf)


Why does the column "row.names" appear in the new dataframe and how do I
get
rid of it.

"uniquedf$row.names <- NULL" does not seem to work.

For what I'm doing this is a useless column and can cause confusion
and/or
problems for subsequent merging/filtering etc.


Thanks.
-- 
View this message in context:
http://www.nabble.com/Remove-row.names-column-in-dataframe-tp18050179p18
050179.html
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Re: [R] boxplot colors

[HBaize]

You are drawing four box plots, not two. Two of them are just the number 5.
The two box plots that are only "5" don't have a box, so you can't see that
they're red. Try this:

boxplot(1:19,20:39,col=c("red","blue"))


Paul Adams-8 wrote:
> 
> Hello everyone,
> 
> 
> I am trying to color two boxplots on the same graph two 
> different colors.
> I have tried the following
> 
> boxplot(5,1:19,5,20:39,col=c("red","blue"))
> but all I get are two blue boxes.
> Any help would be appreciated
> Paul
> 
> 
> 
>   
>   [[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] boxplot problem

[HBaize]

The problem is dat is a data object, not a function. You used the syntax for
a function "dat(1:19)" 
What you probably want is: 

gc <- dat[,1:19]
act <- dat[,20:39]

That will select columns 1 through 19 and put them into the object gc, and
act will get columns 20 through 39. Is that what you want? It is hard to
tell because we don't have access to the data file read into the object
"dat" using the read.table function. 

BTW, is this a homework assignment?  ;-)



Paul Adams-8 wrote:
> 
> Hello everyone,
> I am trying to plot a boxplot but am coming up with the error :
> "could not find function dat" I have used the following code.
> dat<-read.table(file="C:\\Documents and
> Settings\\...txt",header=T,row.names=1,blank.lines.skip=F,na.strings="NA")
> file.show(file="C:\\Documents and Settings\\..txt")
> data<-read.table(file="C:\\Documents and Settings\\.txt",header=T)
> file.show(file="C:\\Documentstxt")
> 
> gc<-dat(1:19)
> act<-dat(20:39)
> x<-as.numeric(dat(100,gc))
> y<-as.numeric(dat(100,act))
> x<-x(!is.na(x))
> y<-y(!is.na(y))
> xy.list<-list(x,y)
> boxplot(xy.list,col=c("red","blue"),main="Gene 100")
> I can not figure out what I am doing wrong with function dat. Any help
> would be
> appreciated.
> Paul
> 
> 
> 
>   
>   [[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] R vs. Bugs

[Prof Brian Ripley]

Your request is too vague for us to be very helpful.  However OpenBUGS 
runs without very frequent crashes only on some ix86 Linux machines -- and 
what those are is unclear and Uwe Ligges and I (working on BRugs) have 
been unable to find one recently.


There are dozens of Bayesian MCMC packages on CRAN (look at its Bayesian 
task view).  Most are less general and faster than BUGS.


There is no 'AMCMC package in R'.  There is at least one third-party 
effort of that name, not on CRAN and explicitly claiming


   The R function amcmc() will tend to run rather _slowly_,

(his emphasis) so perhaps that is not the one you mean.

On Sun, 22 Jun 2008, Peter Muhlberg wrote:


A naive question from a non-statistician:  I'm looking into running a
Bayesian analysis of a model with high dimensionality.  It's not a
standard model (the likelihood requires a lot of code to implement),


'code' in what language?  Note that MCMC does *not* require the likelihood 
to be calculated, and its renaissance in statistics ca 30 years ago was 
for models for which the likelihood is not even known completely.



and I'm using a Linux machine.  Was wondering if someone
has any thoughts on what the advantages of OpenBugs are as
opposed to just R (or should I be thinking WinBUGS under Wine?)?  The
AMCMC package in R promises to run MCMC's very rapidly.  Have read
that OpenBugs as a project was 'stalling' in 2007.

Peter


--
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] I can't see plots

[Josep Lluís Figueras]

Hi,

I use RSPerl on Ubuntu and Apache2 web server. It is my first experience
with R language :-)

I have the next code embedded into a Perl CGI:

use R;
use RReferences;

my @x;
&R::initR("--silent","--no-save");
&R::library("RSPerl");
@x = &R::call("rnorm", 10);
&R::call("plot", [EMAIL PROTECTED]);

The CGI works but no plot is shown.

Apache2 web server log

[[alternative HTML version deleted]]

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

[R] No HTML package list update in 2.7.0?

[Jim Lemon]

Hi all,
As I was uploading a couple of little fixes to prettyR, I remembered
that there was a question I wanted to ask. I remember the package list
being updated whenever I installed a new package (at least on Linux, oh
yes, Fedora Core 9 now), but not on Windows (I just used to unzip the
packages and manually update the package list). Since I have upgraded to
2.7.0, the HTML package list isn't updated. I couldn't find anything
about this in the install stuff for the new features (what is a tangled
vignette anyway?). Is this just something that has slipped beneath the
legendary radar of the team? Am I really the first to notice this?

Jim

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[R] I can't see plots

[Josep Lluís Figueras]

Hi,

I am using RSPerl package on Ubuntu and Apache2 web server. It is my first
experience with R language :-)

I have the next code embedded into a Perl CGI:

use R;
use RReferences;

my @x;
&R::initR("--silent","--no-save");
&R::library("RSPerl");
@x = &R::call("rnorm", 10);
&R::call("plot", [EMAIL PROTECTED]);

The CGI works but no plot is shown.

Apache2 web server log says:

[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Error in pdf() :
unable to start device pdf, referer: http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In addition: Warning
message:, referer: http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In pdf() : cannot open
'pdf' file argument 'Rplots.pdf', referer:
http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Caught error in
R::call(), referer: http://localhost/html/proveta.html

However, if I use R directly plots are correctly shown.

Can anybody help me?

Thanks!

Josep Lluís Figueras
(from Barcelona)

[[alternative HTML version deleted]]

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

Re: [R] passing arguments to a function problem

[Gabor Grothendieck]

See orderBy in the doBy package.

On Sat, Jun 21, 2008 at 7:58 PM, Jiří Voller <[EMAIL PROTECTED]> wrote:
> Dear R-users,
> is there some way how to pass various colnames to the following code for
> sorting data.frames?
>
> mydf.ordered<-with(mydf, mydf[order(colname1,colname2, colnameX, decreasing
> = TRUE), ])
>
>
>
> I was trying something like
>
> Afunction<-function (mydf,colnames,decreasing=T){
>
> mydf.ordered<-with(mydf, mydf[order(colnames, decreasing = decreasing),
> ])
>
>  }
>
> but it didnt work
> please help
> thank you
>
>
> Jiri Voller
>
>[[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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] No HTML package list update in 2.7.0?

[Jonathan Baron]

See

https://bugzilla.redhat.com/show_bug.cgi?id=442727

Right now I seem to be using
/usr/share/doc/R-2.7.0/html
to get the list (on finzi.psych.upenn.edu)

On 06/22/08 20:43, Jim Lemon wrote:
> Hi all,
> As I was uploading a couple of little fixes to prettyR, I remembered
> that there was a question I wanted to ask. I remember the package list
> being updated whenever I installed a new package (at least on Linux, oh
> yes, Fedora Core 9 now), but not on Windows (I just used to unzip the
> packages and manually update the package list). Since I have upgraded to
> 2.7.0, the HTML package list isn't updated. I couldn't find anything
> about this in the install stuff for the new features (what is a tangled
> vignette anyway?). Is this just something that has slipped beneath the
> legendary radar of the team? Am I really the first to notice this?
> 
> Jim

-- 
Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron

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Re: [R] Coloring Stripchart Points, or Better, Lattice Equivalent

[Bryan Hanson]

Below is a revised set of code that demonstrates my question a little more
clearly, I hope.

When plotting all the data (5th panel), col & sym don't seem to be passed
correctly, as the (random) first value for col & sym are used for all points
(run the code, then run it again, you'll see how the 5th panel changes
depending upon col & sym for the first data point).  The 5th panel should
ideally be the "sum" of the 4 panels above, keeping col & sym intact.

Also, I would rather have this in lattice or ggplot2, if anyone sees how to
convert it.

Thanks once again, several of you have made very useful suggestions off
list.  Bryan

samples <- 100 # must be even
index <- round(runif(samples, 1, 100)) # set up data
resp <- rbinom(samples, 1, 0.5)
yr <- rep(c("2005", "2006"), samples/2)
all <- data.frame(index, resp, yr)
all$sym <- ifelse(all$resp == 1, 1, 3)
all$col <- ifelse(all$yr == 2005, "red", "blue")
all$count <- rep(1, length(all$index))
all <- all[order(all$index, all$yr, all$resp),] # for easier inspection
row.names(all) <- c(1:samples) # for easier inspection

one <- all[(all$yr == 2005 & all$resp == 0),] # First 2005/0 at top
two <- all[(all$yr == 2005 & all$resp == 1),] # Then 2005/1
three <- all[(all$yr == 2006 & all$resp == 0),] # Now 2006/0
four <- all[(all$yr == 2006 & all$resp == 1),] # Finally 2006/1

par(mfrow = c(5, 1))
par(plt = c(0.1, 0.9, 0.25, 0.75))
stripchart(one$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
= one$col, pch = one$sym)
mtext("2005/0", side = 3)
stripchart(two$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
= two$col, pch = two$sym)
mtext("2005/1", side = 3)
stripchart(three$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
col = three$col, pch = three$sym)
mtext("2006/0", side = 3)
stripchart(four$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
col = four$col, pch = four$sym)
mtext("2006/1", side = 3)
stripchart(all$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
= all$col, pch = all$sym)
mtext("col & sym always taken from 1st data point when all data is
plotted!", side = 3)

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Re: [R] I can't see plots

[Duncan Murdoch]

Josep Lluís Figueras wrote:

Hi,

I am using RSPerl package on Ubuntu and Apache2 web server. It is my first
experience with R language :-)

I have the next code embedded into a Perl CGI:

use R;
use RReferences;

my @x;
&R::initR("--silent","--no-save");
&R::library("RSPerl");
@x = &R::call("rnorm", 10);
&R::call("plot", [EMAIL PROTECTED]);

The CGI works but no plot is shown.

Apache2 web server log says:

[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Error in pdf() :
unable to start device pdf, referer: http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In addition: Warning
message:, referer: http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In pdf() : cannot open
'pdf' file argument 'Rplots.pdf', referer:
http://localhost/html/proveta.html
[Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Caught error in
R::call(), referer: http://localhost/html/proveta.html

However, if I use R directly plots are correctly shown.

  
Because you aren't running R interactively, it's trying to open a pdf 
device.  This is failing (file permissions maybe?).


You should probably specify exactly what device to open, and if it's 
something that writes a file, where to write it.


Duncan Murdoch

Can anybody help me?

Thanks!

Josep Lluís Figueras
(from Barcelona)

[[alternative HTML version deleted]]

  



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



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

Re: [R] I can't see plots

[milton ruser]

Hi there,

try to open a graphic device with x11(), may be.

cheers,

miltinho

On 6/22/08, Josep Lluís Figueras <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I am using RSPerl package on Ubuntu and Apache2 web server. It is my first
> experience with R language :-)
>
> I have the next code embedded into a Perl CGI:
>
> use R;
> use RReferences;
>
> my @x;
> &R::initR("--silent","--no-save");
> &R::library("RSPerl");
> @x = &R::call("rnorm", 10);
> &R::call("plot", [EMAIL PROTECTED]);
>
> The CGI works but no plot is shown.
>
> Apache2 web server log says:
>
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Error in pdf() :
> unable to start device pdf, referer: http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In addition: Warning
> message:, referer: http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In pdf() : cannot
> open
> 'pdf' file argument 'Rplots.pdf', referer:
> http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Caught error in
> R::call(), referer: http://localhost/html/proveta.html
>
> However, if I use R directly plots are correctly shown.
>
> Can anybody help me?
>
> Thanks!
>
> Josep Lluís Figueras
> (from Barcelona)
>
>[[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.
>
>

[[alternative HTML version deleted]]

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

Re: [R] I can't see plots

[Daniel Cegielka]

Maybe this help you

http://www.rforge.net/Cairo/index.html

http://www.rosuda.org/R/GDD/

daniel cegielka


-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of milton ruser
Sent: Sunday, June 22, 2008 5:56 PM
To: Josep Lluís Figueras
Cc: r-help@r-project.org
Subject: Re: [R] I can't see plots

Hi there,

try to open a graphic device with x11(), may be.

cheers,

miltinho

On 6/22/08, Josep Llums Figueras <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I am using RSPerl package on Ubuntu and Apache2 web server. It is my first
> experience with R language :-)
>
> I have the next code embedded into a Perl CGI:
>
> use R;
> use RReferences;
>
> my @x;
> &R::initR("--silent","--no-save");
> &R::library("RSPerl");
> @x = &R::call("rnorm", 10);
> &R::call("plot", [EMAIL PROTECTED]);
>
> The CGI works but no plot is shown.
>
> Apache2 web server log says:
>
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Error in pdf() :
> unable to start device pdf, referer: http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In addition: Warning
> message:, referer: http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] In pdf() : cannot
> open
> 'pdf' file argument 'Rplots.pdf', referer:
> http://localhost/html/proveta.html
> [Sun Jun 22 12:01:41 2008] [error] [client 127.0.0.1] Caught error in
> R::call(), referer: http://localhost/html/proveta.html
>
> However, if I use R directly plots are correctly shown.
>
> Can anybody help me?
>
> Thanks!
>
> Josep Llums Figueras
> (from Barcelona)
>
>[[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.
>
>

[[alternative HTML version deleted]]

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

Re: [R] Coloring Stripchart Points, or Better, Lattice Equivalent

[Gabor Grothendieck]

Try this:

library(lattice)
all$resp <- as.factor(all$resp)
stripplot(~ index | resp * yr, all, col = all$col, pch = all$sym,
layout = c(1, 4))


On Sun, Jun 22, 2008 at 10:43 AM, Bryan Hanson <[EMAIL PROTECTED]> wrote:
> Below is a revised set of code that demonstrates my question a little more
> clearly, I hope.
>
> When plotting all the data (5th panel), col & sym don't seem to be passed
> correctly, as the (random) first value for col & sym are used for all points
> (run the code, then run it again, you'll see how the 5th panel changes
> depending upon col & sym for the first data point).  The 5th panel should
> ideally be the "sum" of the 4 panels above, keeping col & sym intact.
>
> Also, I would rather have this in lattice or ggplot2, if anyone sees how to
> convert it.
>
> Thanks once again, several of you have made very useful suggestions off
> list.  Bryan
>
> samples <- 100 # must be even
> index <- round(runif(samples, 1, 100)) # set up data
> resp <- rbinom(samples, 1, 0.5)
> yr <- rep(c("2005", "2006"), samples/2)
> all <- data.frame(index, resp, yr)
> all$sym <- ifelse(all$resp == 1, 1, 3)
> all$col <- ifelse(all$yr == 2005, "red", "blue")
> all$count <- rep(1, length(all$index))
> all <- all[order(all$index, all$yr, all$resp),] # for easier inspection
> row.names(all) <- c(1:samples) # for easier inspection
>
> one <- all[(all$yr == 2005 & all$resp == 0),] # First 2005/0 at top
> two <- all[(all$yr == 2005 & all$resp == 1),] # Then 2005/1
> three <- all[(all$yr == 2006 & all$resp == 0),] # Now 2006/0
> four <- all[(all$yr == 2006 & all$resp == 1),] # Finally 2006/1
>
> par(mfrow = c(5, 1))
> par(plt = c(0.1, 0.9, 0.25, 0.75))
> stripchart(one$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
> = one$col, pch = one$sym)
> mtext("2005/0", side = 3)
> stripchart(two$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
> = two$col, pch = two$sym)
> mtext("2005/1", side = 3)
> stripchart(three$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
> col = three$col, pch = three$sym)
> mtext("2006/0", side = 3)
> stripchart(four$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
> col = four$col, pch = four$sym)
> mtext("2006/1", side = 3)
> stripchart(all$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
> = all$col, pch = all$sym)
> mtext("col & sym always taken from 1st data point when all data is
> plotted!", side = 3)
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] R vs. Bugs

[Peter Muhlberg]

I've done some looking around in R and elsewhere to answer my question
on the value of R vs. Bugs for MCMC.  So, for anyone who is curious,
here's what I think I've found:  Bugs compiles its code, which should
make it much faster than a pure R program.  Packages such as AMCMC run
MCMC in R, potentially with a user-defined C function for the
density--which should make it comparable in speed to Bugs.  The
packages MCMCpack  (MCMCmetrop1R function) and mcmc seem designed to
run w/ a density function written in R.   MCMCpack does have functions
that use precompiled C code from the Scythe library (which looks
nice), but I see no simple way to add a C density function.  AMCMC and
Bugs seem to use adaptive MCMC, but the other R packages don't appear
to do so, which may mean another performance reduction.

I see no way to insert my own proposal density in the R functions.
JAG, a Java-based version of BUGS, apparently allows users to create
their own samplers, which might be a way to insert a different
proposal density.  Details about how to install a sampler are not
given in the manual, which, incidentally, is nevertheless much better
than the Bugs manual.  Also, the proposal density I'd want would
probably treat different variables differently, so I may need
Metropolis within Gibbs, not standard Gibbs sampling.  Can't get a
clear picture of what JAG's algorithm(s) are--the manual doesn't
mention Metropolis.

WinBugs and OpenBugs can't be made to run easily on Linux.  It looks
like WinBugs running under WINE might be the simplest viable
configuration, though I don't know how well or quickly it runs under
WINE or how much memory WINE ends up consuming.

Given all this, it may be easiest for my purposes to try to tweak the
AMCMC code to allow a different proposal density.  Maybe.

Peter

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Re: [R] Coloring Stripchart Points, or Better, Lattice Equivalent

[Gabor Grothendieck]

Actually I am not sure if my prior answer was correct.  I think its ok
with one panel but
you might have to use a panel function is there are several. With one
panel it seems
ok:

stripplot(~ index, all, col = all$col, pch = all$sym)

On Sun, Jun 22, 2008 at 12:28 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Try this:
>
> library(lattice)
> all$resp <- as.factor(all$resp)
> stripplot(~ index | resp * yr, all, col = all$col, pch = all$sym,
> layout = c(1, 4))
>
>
> On Sun, Jun 22, 2008 at 10:43 AM, Bryan Hanson <[EMAIL PROTECTED]> wrote:
>> Below is a revised set of code that demonstrates my question a little more
>> clearly, I hope.
>>
>> When plotting all the data (5th panel), col & sym don't seem to be passed
>> correctly, as the (random) first value for col & sym are used for all points
>> (run the code, then run it again, you'll see how the 5th panel changes
>> depending upon col & sym for the first data point).  The 5th panel should
>> ideally be the "sum" of the 4 panels above, keeping col & sym intact.
>>
>> Also, I would rather have this in lattice or ggplot2, if anyone sees how to
>> convert it.
>>
>> Thanks once again, several of you have made very useful suggestions off
>> list.  Bryan
>>
>> samples <- 100 # must be even
>> index <- round(runif(samples, 1, 100)) # set up data
>> resp <- rbinom(samples, 1, 0.5)
>> yr <- rep(c("2005", "2006"), samples/2)
>> all <- data.frame(index, resp, yr)
>> all$sym <- ifelse(all$resp == 1, 1, 3)
>> all$col <- ifelse(all$yr == 2005, "red", "blue")
>> all$count <- rep(1, length(all$index))
>> all <- all[order(all$index, all$yr, all$resp),] # for easier inspection
>> row.names(all) <- c(1:samples) # for easier inspection
>>
>> one <- all[(all$yr == 2005 & all$resp == 0),] # First 2005/0 at top
>> two <- all[(all$yr == 2005 & all$resp == 1),] # Then 2005/1
>> three <- all[(all$yr == 2006 & all$resp == 0),] # Now 2006/0
>> four <- all[(all$yr == 2006 & all$resp == 1),] # Finally 2006/1
>>
>> par(mfrow = c(5, 1))
>> par(plt = c(0.1, 0.9, 0.25, 0.75))
>> stripchart(one$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>> = one$col, pch = one$sym)
>> mtext("2005/0", side = 3)
>> stripchart(two$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>> = two$col, pch = two$sym)
>> mtext("2005/1", side = 3)
>> stripchart(three$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
>> col = three$col, pch = three$sym)
>> mtext("2006/0", side = 3)
>> stripchart(four$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
>> col = four$col, pch = four$sym)
>> mtext("2006/1", side = 3)
>> stripchart(all$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>> = all$col, pch = all$sym)
>> mtext("col & sym always taken from 1st data point when all data is
>> plotted!", side = 3)
>>
>> __
>> 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] Coloring Stripchart Points, or Better, Lattice Equivalent

[Bryan Hanson]

Thanks Gabor, I'm getting closer.

Is there a way to spread out resp values vertically for a given value of
index?  In base graphics, stripchart does this with method = "stack".  But
in lattice, stack = TRUE does something rather different, and I don't see a
combination of lattice arguments that does it like base graphics.

Thanks, Bryan


On 6/22/08 12:48 PM, "Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:

> Actually I am not sure if my prior answer was correct.  I think its ok
> with one panel but
> you might have to use a panel function is there are several. With one
> panel it seems
> ok:
> 
> stripplot(~ index, all, col = all$col, pch = all$sym)
> 
> On Sun, Jun 22, 2008 at 12:28 PM, Gabor Grothendieck
> <[EMAIL PROTECTED]> wrote:
>> Try this:
>> 
>> library(lattice)
>> all$resp <- as.factor(all$resp)
>> stripplot(~ index | resp * yr, all, col = all$col, pch = all$sym,
>> layout = c(1, 4))
>> 
>> 
>> On Sun, Jun 22, 2008 at 10:43 AM, Bryan Hanson <[EMAIL PROTECTED]> wrote:
>>> Below is a revised set of code that demonstrates my question a little more
>>> clearly, I hope.
>>> 
>>> When plotting all the data (5th panel), col & sym don't seem to be passed
>>> correctly, as the (random) first value for col & sym are used for all points
>>> (run the code, then run it again, you'll see how the 5th panel changes
>>> depending upon col & sym for the first data point).  The 5th panel should
>>> ideally be the "sum" of the 4 panels above, keeping col & sym intact.
>>> 
>>> Also, I would rather have this in lattice or ggplot2, if anyone sees how to
>>> convert it.
>>> 
>>> Thanks once again, several of you have made very useful suggestions off
>>> list.  Bryan
>>> 
>>> samples <- 100 # must be even
>>> index <- round(runif(samples, 1, 100)) # set up data
>>> resp <- rbinom(samples, 1, 0.5)
>>> yr <- rep(c("2005", "2006"), samples/2)
>>> all <- data.frame(index, resp, yr)
>>> all$sym <- ifelse(all$resp == 1, 1, 3)
>>> all$col <- ifelse(all$yr == 2005, "red", "blue")
>>> all$count <- rep(1, length(all$index))
>>> all <- all[order(all$index, all$yr, all$resp),] # for easier inspection
>>> row.names(all) <- c(1:samples) # for easier inspection
>>> 
>>> one <- all[(all$yr == 2005 & all$resp == 0),] # First 2005/0 at top
>>> two <- all[(all$yr == 2005 & all$resp == 1),] # Then 2005/1
>>> three <- all[(all$yr == 2006 & all$resp == 0),] # Now 2006/0
>>> four <- all[(all$yr == 2006 & all$resp == 1),] # Finally 2006/1
>>> 
>>> par(mfrow = c(5, 1))
>>> par(plt = c(0.1, 0.9, 0.25, 0.75))
>>> stripchart(one$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>>> = one$col, pch = one$sym)
>>> mtext("2005/0", side = 3)
>>> stripchart(two$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>>> = two$col, pch = two$sym)
>>> mtext("2005/1", side = 3)
>>> stripchart(three$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
>>> col = three$col, pch = three$sym)
>>> mtext("2006/0", side = 3)
>>> stripchart(four$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
>>> col = four$col, pch = four$sym)
>>> mtext("2006/1", side = 3)
>>> stripchart(all$index, method = "stack", ylim = c(0,10), xlim = c(1,100), col
>>> = all$col, pch = all$sym)
>>> mtext("col & sym always taken from 1st data point when all data is
>>> plotted!", side = 3)
>>> 
>>> __
>>> 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] Redefining the for loop

[ast_shopping]

Hi,

I have the following problem:

I want to redefine the for loop (or define a similar statement) to change its 
behavior under some circumstances. So, I would like sthg. like

"For" = function ( var, vec, statement ) {
 
  if ( ... ) {
/* my implementation of for */
  }
  else {
/* call R's for loop */
  }
}

I tried to manipulate the elements of the list one gets by

e = quote( for ( i in (1:10) ) print("*") )

That means, storing the variable in e[[2]] (my trials didn't work), and the 
statement in e[[4]], but I failed.

How can one do this in R? It would be best, if I could redefine "for" directly, 
so that one can use the same syntax, i.e., with the statement after the 
function call.

Can somebody help me?


Thanks!







{
-- 

Jetzt dabei sein: http://www.shortview.de/[EMAIL PROTECTED]

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[R] readLines problem in iteration

[Yasin Hajizadeh]

Hi there
  My script at each iteration produces a .dat file which I save in a directory 
with write.table. Then I need to read each line of this file and create a 
directory for each line and put elements of that row in that directory. I do it 
with the following script which I have inserted somewhere between my outer 
iteration loop, it works fine for the first iteration, but the problem is that 
it does nothing at next iterations. How can I solve this problem, is there 
anything wrong with my code?
  Thanks in advance for your help.
   
   
write.table(X, file = "model.dat", append = FALSE, sep = " ", eol = "\n", 
na = "NA", dec = ".", row.names = FALSE, col.names = FALSE, qmethod = 
c("escape", "double")) 
  
  mother<-getwd()
  
  
  count <- 0;
  
  con<- file("model.dat", open="r")

   
  while (length(line) > 0) {
   
  line<-readLines(con,n=1)
  count <- count +1
  cat("read row no", count, ",\n", sep="");
  print(line)
  
  foldername <- paste("Iteration",c, "ant", count)
  dir.create (foldername)
  
  setwd(foldername)
  write.table(line, file = "params.in", append = FALSE, sep = " ", eol = "\n", 
na = "NA", dec = ".", row.names = FALSE, col.names = FALSE, qmethod = 
c("escape", "double"))
  
  #system(run_misf_calc.py)
  
  setwd(mother)
   
   
  }
  
  
  close(con)

   
[[alternative HTML version deleted]]

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Re: [R] Coloring Stripchart Points, or Better, Lattice Equivalent

[Gabor Grothendieck]

I tried it with a simple example to make it easier to check the
result and it does seem we need a panel function, pnl:

library(lattice)
DF <- data.frame(x = 1:5, col = c(1:4, 1), g = c(1, 1, 2, 2, 2), pch =
c(1:3, 2:1))
pnl <- function(x, y, subscripts, col, pch, ...) panel.stripplot(x, y,
col = col[subscripts], pch = pch[subscripts], ...)
stripplot(~ x | g, DF, col = DF$col, pch = DF$pch, panel = pnl)

Using the same panel function, pnl, we can spread out resp into two different
panels like this:

all$resp <- as.factor(all$resp) # as before
stripplot(~ index | resp, all, col = all$col, pch = all$sym, panel =
pnl, layout = 1:2)

On Sun, Jun 22, 2008 at 1:09 PM, Bryan Hanson <[EMAIL PROTECTED]> wrote:
> Thanks Gabor, I'm getting closer.
>
> Is there a way to spread out resp values vertically for a given value of
> index?  In base graphics, stripchart does this with method = "stack".  But
> in lattice, stack = TRUE does something rather different, and I don't see a
> combination of lattice arguments that does it like base graphics.
>
> Thanks, Bryan
>
>
> On 6/22/08 12:48 PM, "Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
>
>> Actually I am not sure if my prior answer was correct.  I think its ok
>> with one panel but
>> you might have to use a panel function is there are several. With one
>> panel it seems
>> ok:
>>
>> stripplot(~ index, all, col = all$col, pch = all$sym)
>>
>> On Sun, Jun 22, 2008 at 12:28 PM, Gabor Grothendieck
>> <[EMAIL PROTECTED]> wrote:
>>> Try this:
>>>
>>> library(lattice)
>>> all$resp <- as.factor(all$resp)
>>> stripplot(~ index | resp * yr, all, col = all$col, pch = all$sym,
>>> layout = c(1, 4))
>>>
>>>
>>> On Sun, Jun 22, 2008 at 10:43 AM, Bryan Hanson <[EMAIL PROTECTED]> wrote:
 Below is a revised set of code that demonstrates my question a little more
 clearly, I hope.

 When plotting all the data (5th panel), col & sym don't seem to be passed
 correctly, as the (random) first value for col & sym are used for all 
 points
 (run the code, then run it again, you'll see how the 5th panel changes
 depending upon col & sym for the first data point).  The 5th panel should
 ideally be the "sum" of the 4 panels above, keeping col & sym intact.

 Also, I would rather have this in lattice or ggplot2, if anyone sees how to
 convert it.

 Thanks once again, several of you have made very useful suggestions off
 list.  Bryan

 samples <- 100 # must be even
 index <- round(runif(samples, 1, 100)) # set up data
 resp <- rbinom(samples, 1, 0.5)
 yr <- rep(c("2005", "2006"), samples/2)
 all <- data.frame(index, resp, yr)
 all$sym <- ifelse(all$resp == 1, 1, 3)
 all$col <- ifelse(all$yr == 2005, "red", "blue")
 all$count <- rep(1, length(all$index))
 all <- all[order(all$index, all$yr, all$resp),] # for easier inspection
 row.names(all) <- c(1:samples) # for easier inspection

 one <- all[(all$yr == 2005 & all$resp == 0),] # First 2005/0 at top
 two <- all[(all$yr == 2005 & all$resp == 1),] # Then 2005/1
 three <- all[(all$yr == 2006 & all$resp == 0),] # Now 2006/0
 four <- all[(all$yr == 2006 & all$resp == 1),] # Finally 2006/1

 par(mfrow = c(5, 1))
 par(plt = c(0.1, 0.9, 0.25, 0.75))
 stripchart(one$index, method = "stack", ylim = c(0,10), xlim = c(1,100), 
 col
 = one$col, pch = one$sym)
 mtext("2005/0", side = 3)
 stripchart(two$index, method = "stack", ylim = c(0,10), xlim = c(1,100), 
 col
 = two$col, pch = two$sym)
 mtext("2005/1", side = 3)
 stripchart(three$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
 col = three$col, pch = three$sym)
 mtext("2006/0", side = 3)
 stripchart(four$index, method = "stack", ylim = c(0,10), xlim = c(1,100),
 col = four$col, pch = four$sym)
 mtext("2006/1", side = 3)
 stripchart(all$index, method = "stack", ylim = c(0,10), xlim = c(1,100), 
 col
 = all$col, pch = all$sym)
 mtext("col & sym always taken from 1st data point when all data is
 plotted!", side = 3)

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Re: [R] error related to approxfun in R 2.7.0

[Karl Marx]

Fantastic, works perfectly!
Would you have any pointers to where I can learn more about why/how this works?
Thanks!

Regarding the P.S., yes...  it seemed appropriate at the time :-)

2008/6/21, Martin Maechler <[EMAIL PROTECTED]>:
>> "KM" == Karl Marx <[EMAIL PROTECTED]>
>> on Thu, 19 Jun 2008 23:01:31 +0200 writes:
>
> KM> I'm testing R version 2.7.0 on windows and there seems
> KM> to be a compatibility issue with objects that were
> KM> created by "approxfun" in older versions. As long as the
> KM> objects were created in version 2.7.0 things work ok,
> KM> but calling the interpolated functions from R version
> KM> 2.0.1 causes this error:
>
> KM> Error in .C("R_approx", as.double(x), as.double(y),
> KM> as.integer(n), xout = as.double(v), : C symbol name
> KM> "R_approx" not in DLL for package "base"
>
> KM> What's the problem and how can t be resolved (need to be
> KM> compatible with old data)?
>
> approxfun() wa moved from "base" to "stats".
>
> If   f <- approxfun(...)  in R 2.0.1
> and you save f , load it in R 2.7.0,
> you can do
>
> body(f)[[2]][["PACKAGE"]] <- "stats"
>
> and then use f(..) in R 2.7.0.
>
> Martin Maechler, ETH Zurich
>
> PS: Do I guess correctly, that you are Swiss?
>

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[R] two newbie questions

[Donald Braman]

# I've tried to make this easy to paste into R, though it's probably
so simple you won't need to.
# I have some data (there are many more variables, but this is a
reasonable approximation of it)

# here's a fabricated data frame that is similar in form to mine:
my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1
var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
"intv1", "intv2")
names(my.df) <- var.list

# I have some are DVs:
dvs <- c("dv1", "dv2", "dv3")

# some IVs:
ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")

# and some binary interaction variables:
intvs <- c("intv1", "intv2")
library(car)
my.df[intvs] <- lapply(my.df[intvs], function(x)
 recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))

# now I loop through a series of interactions using the vector numbers:
for(dv in 1:3) {
 for(iv in 4:8) {
  for (intv in 9:10) {
   jpeg(paste(names(my.df[iv]), names(my.df[dv]), names(my.df[intv]),
".jpg", sep="_"))
   with(data.frame(my.df), {
my.fit <- lm( my.df[[dv]] ~ my.df[[iv]] + my.df[[intv]] +
my.df[[iv]]:my.df[[intv]])
colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
plot(my.df[[iv]], my.df[[dv]], xlab=names(my.df[iv]),
ylab=names(my.df[dv]), col=colors, pch=".")
curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
   })
   dev.off()
  }
 }
}


# Question1: Works fine, but using the vector numbers feels kludgy --
especially if the variables in question aren't consecutive.
# Is there a more elegant way of doing this with my lists of variable
names? Something like this, for example:
for(dv in dvs) {
 for(iv in ivs) {
  for (intv in intvs) {
   jpeg(paste(dv, iv, intv, ".jpg", sep="_"))
   with(data.frame(my.df), {
my.fit <- lm(my.df[dv] ~ my.df[iv] + my.df[intv] + my.df[iv]:my.df[intv])
colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
plot(my.df[iv], my.df[dv], xlab=iv, ylab=names(dv), col=colors, pch=".")
curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
   })
   dev.off()
  }
 }
}

# Clearly that's wrong -- why it's wrong is obscure to me, though!
Please educate me!

# Question2: Could this could be done by using "apply" rather than a loop?
# Or is looping better here bc there are several actions performed at
each iteration?
# I'm still trying to get my head around all the ways to ditch looping in R.


Donald Braman
http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
http://research.yale.edu/culturalcognition
http://ssrn.com/author=286206

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

Re: [R] two newbie questions

[jim holtman]

This does away with the 'for' loops and uses 'expand.grid' to create
the combinations.  I think I got the right variables substituted:

my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1
var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
"intv1", "intv2")
names(my.df) <- var.list

# I have some are DVs:
dvs <- c("dv1", "dv2", "dv3")

# some IVs:
ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")

# and some binary interaction variables:
intvs <- c("intv1", "intv2")
library(car)
my.df[intvs] <- lapply(my.df[intvs], function(x)
 recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))

# now I loop through a series of interactions using the vector numbers:
# create a dataframe of values to check
xpnd <- expand.grid(dvs, ivs, intvs) # create combinations
invisible(apply(xpnd, 1, function(.row) {
  jpeg(paste(paste(.row, collapse="_"),".jpg", sep=''))

   my.fit <- lm( my.df[[.row[1]]] ~ my.df[[.row[2]]] + my.df[[.row[3]]] +
my.df[[.row[2]]]:my.df[[.row[3]]])
   colors <- ifelse (my.df[[.row[3]]] == 1, "black", "grey")
   plot(my.df[[.row[2]]], my.df[[.row[1]]], xlab=.row[2],
ylab=.row[1], col=colors, pch=".")
   curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
   curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")

  dev.off()
 }
))


On Sun, Jun 22, 2008 at 6:26 PM, Donald Braman <[EMAIL PROTECTED]> wrote:
> # I've tried to make this easy to paste into R, though it's probably
> so simple you won't need to.
> # I have some data (there are many more variables, but this is a
> reasonable approximation of it)
>
> # here's a fabricated data frame that is similar in form to mine:
> my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1
> var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
> "intv1", "intv2")
> names(my.df) <- var.list
>
> # I have some are DVs:
> dvs <- c("dv1", "dv2", "dv3")
>
> # some IVs:
> ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")
>
> # and some binary interaction variables:
> intvs <- c("intv1", "intv2")
> library(car)
> my.df[intvs] <- lapply(my.df[intvs], function(x)
>  recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))
>
> # now I loop through a series of interactions using the vector numbers:
> for(dv in 1:3) {
>  for(iv in 4:8) {
>  for (intv in 9:10) {
>   jpeg(paste(names(my.df[iv]), names(my.df[dv]), names(my.df[intv]),
> ".jpg", sep="_"))
>   with(data.frame(my.df), {
>my.fit <- lm( my.df[[dv]] ~ my.df[[iv]] + my.df[[intv]] +
> my.df[[iv]]:my.df[[intv]])
>colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
>plot(my.df[[iv]], my.df[[dv]], xlab=names(my.df[iv]),
> ylab=names(my.df[dv]), col=colors, pch=".")
>curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
>curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
>   })
>   dev.off()
>  }
>  }
> }
>
>
> # Question1: Works fine, but using the vector numbers feels kludgy --
> especially if the variables in question aren't consecutive.
> # Is there a more elegant way of doing this with my lists of variable
> names? Something like this, for example:
> for(dv in dvs) {
>  for(iv in ivs) {
>  for (intv in intvs) {
>   jpeg(paste(dv, iv, intv, ".jpg", sep="_"))
>   with(data.frame(my.df), {
>my.fit <- lm(my.df[dv] ~ my.df[iv] + my.df[intv] + my.df[iv]:my.df[intv])
>colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
>plot(my.df[iv], my.df[dv], xlab=iv, ylab=names(dv), col=colors, pch=".")
>curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
>curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
>   })
>   dev.off()
>  }
>  }
> }
>
> # Clearly that's wrong -- why it's wrong is obscure to me, though!
> Please educate me!
>
> # Question2: Could this could be done by using "apply" rather than a loop?
> # Or is looping better here bc there are several actions performed at
> each iteration?
> # I'm still trying to get my head around all the ways to ditch looping in R.
>
>
> Donald Braman
> http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
> http://research.yale.edu/culturalcognition
> http://ssrn.com/author=286206
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

What is the problem you are trying to solve?

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Re: [R] readLines problem in iteration

[jim holtman]

What is the value of 'line' on the first iteration?  It appears to be
undefined.  You probably want something like this:

while (TRUE){
line <- readLines(con, n=1)
if (length(line) == 0) break
.your normal statements.
}

On Sun, Jun 22, 2008 at 3:29 PM, Yasin Hajizadeh
<[EMAIL PROTECTED]> wrote:
> Hi there
>  My script at each iteration produces a .dat file which I save in a directory 
> with write.table. Then I need to read each line of this file and create a 
> directory for each line and put elements of that row in that directory. I do 
> it with the following script which I have inserted somewhere between my outer 
> iteration loop, it works fine for the first iteration, but the problem is 
> that it does nothing at next iterations. How can I solve this problem, is 
> there anything wrong with my code?
>  Thanks in advance for your help.
>
>
>write.table(X, file = "model.dat", append = FALSE, sep = " ", eol = "\n", 
> na = "NA", dec = ".", row.names = FALSE, col.names = FALSE, qmethod = 
> c("escape", "double"))
>
>  mother<-getwd()
>
>
>  count <- 0;
>
>  con<- file("model.dat", open="r")
>
>
>  while (length(line) > 0) {
>
>  line<-readLines(con,n=1)
>  count <- count +1
>  cat("read row no", count, ",\n", sep="");
>  print(line)
>
>  foldername <- paste("Iteration",c, "ant", count)
>  dir.create (foldername)
>
>  setwd(foldername)
>  write.table(line, file = "params.in", append = FALSE, sep = " ", eol = "\n", 
> na = "NA", dec = ".", row.names = FALSE, col.names = FALSE, qmethod = 
> c("escape", "double"))
>
>  #system(run_misf_calc.py)
>
>  setwd(mother)
>
>
>  }
>
>
>  close(con)
>
>
>[[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.
>



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

What is the problem you are trying to solve?

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

Re: [R] two newbie questions

[Donald Braman]

Wow -- many thanks for the mind-*expanding* help!  I'm really impressed by
R's ability to handle this so concisely  It's going to take me a while to
get used to applying things to vectors, but the more I understand, the nicer
R looks.


On Sun, Jun 22, 2008 at 6:59 PM, jim holtman <[EMAIL PROTECTED]> wrote:

> This does away with the 'for' loops and uses 'expand.grid' to create
> the combinations.  I think I got the right variables substituted:
>
> my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1
> var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
> "intv1", "intv2")
> names(my.df) <- var.list
>
> # I have some are DVs:
> dvs <- c("dv1", "dv2", "dv3")
>
> # some IVs:
> ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")
>
> # and some binary interaction variables:
> intvs <- c("intv1", "intv2")
> library(car)
> my.df[intvs] <- lapply(my.df[intvs], function(x)
>  recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))
>
> # now I loop through a series of interactions using the vector numbers:
> # create a dataframe of values to check
> xpnd <- expand.grid(dvs, ivs, intvs) # create combinations
> invisible(apply(xpnd, 1, function(.row) {
>  jpeg(paste(paste(.row, collapse="_"),".jpg", sep=''))
>
>   my.fit <- lm( my.df[[.row[1]]] ~ my.df[[.row[2]]] + my.df[[.row[3]]] +
>my.df[[.row[2]]]:my.df[[.row[3]]])
>   colors <- ifelse (my.df[[.row[3]]] == 1, "black", "grey")
>   plot(my.df[[.row[2]]], my.df[[.row[1]]], xlab=.row[2],
>ylab=.row[1], col=colors, pch=".")
>curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
>   curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
>
>  dev.off()
>  }
> ))
>
>
> On Sun, Jun 22, 2008 at 6:26 PM, Donald Braman <[EMAIL PROTECTED]>
> wrote:
> > # I've tried to make this easy to paste into R, though it's probably
> > so simple you won't need to.
> > # I have some data (there are many more variables, but this is a
> > reasonable approximation of it)
> >
> > # here's a fabricated data frame that is similar in form to mine:
> > my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1
> > var.list <- c("dv1", "dv2", "dv3", "iv1", "iv2", "iv3", "iv4", "iv5",
> > "intv1", "intv2")
> > names(my.df) <- var.list
> >
> > # I have some are DVs:
> > dvs <- c("dv1", "dv2", "dv3")
> >
> > # some IVs:
> > ivs <- c("iv1", "iv2", "iv3", "iv4", "iv5")
> >
> > # and some binary interaction variables:
> > intvs <- c("intv1", "intv2")
> > library(car)
> > my.df[intvs] <- lapply(my.df[intvs], function(x)
> >  recode(x, recodes = "lo:3.5=0; 3.5:hi=1; ",as.factor.result = FALSE))
> >
> > # now I loop through a series of interactions using the vector numbers:
> > for(dv in 1:3) {
> >  for(iv in 4:8) {
> >  for (intv in 9:10) {
> >   jpeg(paste(names(my.df[iv]), names(my.df[dv]), names(my.df[intv]),
> > ".jpg", sep="_"))
> >   with(data.frame(my.df), {
> >my.fit <- lm( my.df[[dv]] ~ my.df[[iv]] + my.df[[intv]] +
> > my.df[[iv]]:my.df[[intv]])
> >colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
> >plot(my.df[[iv]], my.df[[dv]], xlab=names(my.df[iv]),
> > ylab=names(my.df[dv]), col=colors, pch=".")
> >curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
> >curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
> >   })
> >   dev.off()
> >  }
> >  }
> > }
> >
> >
> > # Question1: Works fine, but using the vector numbers feels kludgy --
> > especially if the variables in question aren't consecutive.
> > # Is there a more elegant way of doing this with my lists of variable
> > names? Something like this, for example:
> > for(dv in dvs) {
> >  for(iv in ivs) {
> >  for (intv in intvs) {
> >   jpeg(paste(dv, iv, intv, ".jpg", sep="_"))
> >   with(data.frame(my.df), {
> >my.fit <- lm(my.df[dv] ~ my.df[iv] + my.df[intv] +
> my.df[iv]:my.df[intv])
> >colors <- ifelse (my.df[[intv]] == 1, "black", "grey")
> >plot(my.df[iv], my.df[dv], xlab=iv, ylab=names(dv), col=colors,
> pch=".")
> >curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
> >curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
> >   })
> >   dev.off()
> >  }
> >  }
> > }
> >
> > # Clearly that's wrong -- why it's wrong is obscure to me, though!
> > Please educate me!
> >
> > # Question2: Could this could be done by using "apply" rather than a
> loop?
> > # Or is looping better here bc there are several actions performed at
> > each iteration?
> > # I'm still trying to get my head around all the ways to ditch looping in
> R.
> >
> >
> > Donald Braman
> > http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
> > http://research.yale.edu/culturalcognition
> > http://ssrn.com/author=286206
> >
> > __
> > 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, mini

[R] ggplot2-barplot

[Felipe Carrillo]

Hi all:
I have been using ggplot2 graphics for quite some time now and I really like 
it. However, I haven't used barplots enough to understand the arquitecture 
behind it. Can someone show me how to make this simple barplot graph with 
ggplot2? I want "PondName" along the x axis and "avgWt" along the Y axis which 
represents the avgWt by each Pond. 

PondNameavgWt
Pond01  21.5
Pond02  17.8
Pond03  17.8
Pond04  17.8
Pond05  16.375
Pond06  21.5
Pond07  21.5

Using sigmaplot or Excel is a simple task but I can't get it to work with 
ggplot2. I was trying:

qplot(PondName,data=mydata,geom="bar") + geom_bar(fill=factor(PondName))
but I don't get the desired results. Thanks

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Re: [R] R vs. Bugs

[Paul Johnson]

Hey, good topic for a thread.  I've wrestled with this over the years.
I think there's some user confusion about what WinBUGS does.  People
who did not see BUGS before WinBUGS tend not to understand this in the
same way...

The unique / important contributions from WinBUGS  are the collection
of working examples and the Doodle code generator thing.  All of the
rest of it can be easily replaced by open source tools that exist or
could easily exist.


On Sun, Jun 22, 2008 at 11:34 AM, Peter Muhlberg
<[EMAIL PROTECTED]> wrote:
> I've done some looking around in R and elsewhere to answer my question
> on the value of R vs. Bugs for MCMC.  So, for anyone who is curious,
> here's what I think I've found:  Bugs compiles its code, which should
> make it much faster than a pure R program.  Packages such as AMCMC run
> MCMC in R, potentially with a user-defined C function for the
> density--which should make it comparable in speed to Bugs.  The
> packages MCMCpack  (MCMCmetrop1R function) and mcmc seem designed to
> run w/ a density function written in R.   MCMCpack does have functions
> that use precompiled C code from the Scythe library (which looks
> nice), but I see no simple way to add a C density function.  AMCMC and
> Bugs seem to use adaptive MCMC, but the other R packages don't appear
> to do so, which may mean another performance reduction.

Think of "performance" as a combination of development time and run time.

Andrew Martin's MCMCpack reduces development time by giving people
some pre-packaged Gibbs sampling fitters for standard models, such as
logistic regression.  It still uses the same iterative sampling
routines "under the hood" as most other MCMC approaches.  The only
difference there is that it has routines formatted in a way that will
be familiar to R users.  I do not believe a simulation model conducted
in MCMCpack will take a different amount of "run time" than a well
coded, custom version of the same that is prepared in WinBUGS or any
other GIBBS sampler.  The big difference is that MCMCpack offers
routines for familiar models, and if one wants to "thrown in" some
random parameter here or there, then MCMCpack won't be able to handle
it.

The U. Chicago professors provide the package bayesm which is roughly
the same kind of thing.  For pre-existing model types, an MCMC model
can be conducted.

Students ask "Why learn BUGS when I can use MCMCpack (or bayesm)?"
Answer: no reason, unless you want to propose a model that is not
already coded up by the package.

>
> I see no way to insert my own proposal density in the R functions.
> JAG, a Java-based version of BUGS, apparently allows users to create

If you mean to refer to Martyn Plummer's project JAGS:

http://www-ice.iarc.fr/~martyn/software/jags/

JAGS is "Just Another Gibbs Sampler", and it is decidedly not written in Java.
It is, rather, written in C++.

JAGS is proposed as a more-or-less complete drop in replacement for
BUGS, so the user can write up a BUGS style model and then hand it
over to JAGS for processing, the same way we used BUGS before the
WinBugs came along and tried to make it a pointy-clicky experience.
JAGS has versions of the classic BUGS examples, and I think it is very
well done.   If you want to run a Gibbs Sampling exercise in Linux, I
suggest you seriously consider using JAGS.  You might use WinBUGS
Doodle thingie to sketch out the code for your model, but then
translate it to JAGS. (I put a bit of thought into packaging an RPM
for Fedora users a few months ago, but ran into a few little packaging
errors that put me off of it.  Now I'm running Ubuntu and packaging
for that is harder, at least for me...)

I notice there are some R packages that are providing pre-packaged
estimators for common models through JAGS.  Check witness:

bayescount  Bayesian analysis of count distributions with JAGS
bayesmixBayesian Mixture Models with JAGS

It is disappointing/frustrating to me that the source code for
WinBUGS/OpenBugs is kept in secret, because here's what I'd really
like.

1. Make a version of Doodle for sketching out models.  As far as I can
see, Doodle is the only truly uniquely valuable component in Win/Open
BUGS.  It helps people get started by providing a code template.

2. Create an avenue for that Doodle code to travel to JAGS.  rJAGS
exists as an interface between R and jags.


-- 
Paul E. Johnson
Professor, Political Science
1541 Lilac Lane, Room 504
University of Kansas

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[R] Simulating Gaussian Mixture Models

[Peng Jiang]

Hi,
  Is there any package that I can use to simulate the Gaussian  
Mixture Model , which is a mixture modeling method that is widely used  
in statistical learning theory.
 I know there is a mclust, however, I think it is a little bit  
different from my problem.

Thanks very much..

 regards.








--
Peng Jiang
江鹏
Ph.D. Candidate

Antai College of Economics & Management
安泰经济管理学院
Department of Mathematics
数学系
Shanghai Jiaotong University (Minhang Campus)
800 Dongchuan Road
200240 Shanghai
P. R. China

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[R] problem in R for Linear mixed model~

[Manli Yan]

   Dear R users:
  I just got confused some R code used in linear mixed model~
  example,two factors,A, B,C,A is fixed ,B,C are random,and B is nested in
C,if I wannt to use linear mixed model,are the following code correct for
each case?
case1:want to know random effect of B,
   case1<-lme(y~A*B*C,random=~B|C) where "B|C" stand for what?,mean B is
nested in C?

case2: how to wirte random effect of C?
   case2<-lme(y~A*B*C,random=~C)? this doesnt work out,it seem it must have
somehing like #|$

case3.omitting the random effect for B from case1
  case3<-update(case1,random=~1|C),so I just type 1,so the random effect of
B will be removed from the model,there only left random effect of c ,the
random effect I removed ,which include both random intercept and slope
,correct??

   case4:omitting the random  intercept
  case4<-update(case1,random=B-1|C)
  this code I got from some paper,it said by inputing B-1|C,then the random
intercept  is removed,so,if I want to remove random slode,I input B-2|C,it
doesnt work out.

   case5 :how to know the both random effect of B,and C,I dont know how to
wirtie this in R
.
  And I am a little confused of these R code,especially the #|#
part,what deos this syntax really mean in LME package,

  Thank you for your time~

[[alternative HTML version deleted]]

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Re: [R] Simulating Gaussian Mixture Models

[Tatiana Benaglia]

Take a look at mixtools.

Tatiana



On Jun 22, 2008, at 9:23 PM, Peng Jiang wrote:




Hi,
 Is there any package that I can use to simulate the Gaussian  
Mixture Model , which is a mixture modeling method that is widely  
used in statistical learning theory.
I know there is a mclust, however, I think it is a little bit  
different from my problem.

Thanks very much..

regards.








--
Peng Jiang
江鹏
Ph.D. Candidate

Antai College of Economics & Management
安泰经济管理学院
Department of Mathematics
数学系
Shanghai Jiaotong University (Minhang Campus)
800 Dongchuan Road
200240 Shanghai
P. R. China

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


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

Re: [R] Simulating Gaussian Mixture Models

[Wensui Liu]

Hi, Peng,
I had a piece of SAS code for 2-class gaussian mixture from my blog.
You might convert it to R code.

2-Class Gaussian Mixture Model in SAS
data d1;
  do i = 1 to 100;
x = ranuni(1);
e = rannor(1);
y = 5 * x + e;
output;
  end;
run;

data d2;
  do i = 1 to 100;
x = ranuni(2);
e = rannor(2);
y = 15 + 10 * x - 5 * x ** 2 + e;
output;
  end;
run;

data data;
  set d1 d2;
run;

proc nlmixed data = data tech = quanew maxiter = 1000;
  parms b10 = 0 b11 = 5 b12 = 0 b20 = 15 b21 = 10 b22 = -5
prior = 0.1 to 0.9 by 0.01 sigma = 1;

  mu1 = b10 + b11 * x + b12 * x * x;
  mu2 = b20 + b21 * x + b22 * x * x;
  pi = constant('pi');

  P1 = 1 / (((2 * pi) ** 0.5) * sigma) * exp(-0.5 * ((y - mu1) / sigma) ** 2);
  P2 = 1 / (((2 * pi) ** 0.5) * sigma) * exp(-0.5 * ((y - mu2) / sigma) ** 2);

  LH = P1 * prior + P2 * (1 - prior);
  LL = log(LH);

  model y ~ general(LL);
run;

/*
  Parameter Estimates

  Standard
 Parameter  Estimate ErrorDF  t Value  Pr > |t|   Alpha
 b10 -0.17440.3450   200-0.510.61370.05
 b11  5.34261.5040   200 3.550.00050.05
 b12-0.064541.4334   200-0.050.96410.05
 b20 15.36520.3099   20049.57<.00010.05
 b21  9.62971.4970   200 6.43<.00010.05
 b22 -5.47951.4776   200-3.710.00030.05
 prior0.5000   0.03536   20014.14<.00010.05
 sigma1.0049   0.05025   20020.00<.00010.05
*/
On 6/22/08, Peng Jiang <[EMAIL PROTECTED]> wrote:
>
>
>  Hi,
>   Is there any package that I can use to simulate the Gaussian Mixture Model
> , which is a mixture modeling method that is widely used in statistical
> learning theory.
>   I know there is a mclust, however, I think it is a little bit different
> from my problem.
>  Thanks very much..
>
>   regards.
>
>
>
>
>
>
>
>
>  --
>  Peng Jiang
>  江鹏
>  Ph.D. Candidate
>
>  Antai College of Economics & Management
>  安泰经济管理学院
>  Department of Mathematics
>  数学系
>  Shanghai Jiaotong University (Minhang Campus)
>  800 Dongchuan Road
>  200240 Shanghai
>  P. R. China
>
>  __
>  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.
>


-- 
===
WenSui Liu
Acquisition Risk, Chase
Email : [EMAIL PROTECTED]
Blog   : statcompute.spaces.live.com
===
__
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and provide commented, minimal, self-contained, reproducible code.

[R] Getting only label column of a data frame

[Gundala Viswanath]

Hi,

How can I extract the label only from a given data frame.

Fore example from this data frame.

> print(dataf)
  V1  V2  V3  V4  V5  V6  V7  V8  V9
1114514.317.131.241.745.849.868.670.672.9
3545 10.215.620.923.231.431.736.248.451.9
8951 15.217.520.021.432.449.751.358.358.9
1109759.565.9   117.5   118.0   118.9   122.5   126.3   156.5   157.0

I want to get the label (first) column only:

11145
3545
8951
11097

Is there a quick way to achieve that?
I tried this but fail:

mylable <- dataf[,1]

Please advice.

- Gundala Viswanath
Jakarta - Indonesia

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

Re: [R] Getting only label column of a data frame

[jim holtman]

Is this what you want:

> x
V1   V2V3V4V5V6V7V8V9
11145 14.3 17.1  31.2  41.7  45.8  49.8  68.6  70.6  72.9
3545  10.2 15.6  20.9  23.2  31.4  31.7  36.2  48.4  51.9
8951  15.2 17.5  20.0  21.4  32.4  49.7  51.3  58.3  58.9
11097 59.5 65.9 117.5 118.0 118.9 122.5 126.3 156.5 157.0
> x[,1]
[1] 14.3 10.2 15.2 59.5
> x[,1,drop=FALSE]
V1
11145 14.3
3545  10.2
8951  15.2
11097 59.5
>

You probably need the 'drop=FALSE';  see ?'['

On Sun, Jun 22, 2008 at 10:16 PM, Gundala Viswanath <[EMAIL PROTECTED]> wrote:
> Hi,
>
> How can I extract the label only from a given data frame.
>
> Fore example from this data frame.
>
>> print(dataf)
>  V1  V2  V3  V4  V5  V6  V7  V8  V9
> 1114514.317.131.241.745.849.868.670.672.9
> 3545 10.215.620.923.231.431.736.248.451.9
> 8951 15.217.520.021.432.449.751.358.358.9
> 1109759.565.9   117.5   118.0   118.9   122.5   126.3   156.5   157.0
>
> I want to get the label (first) column only:
>
> 11145
> 3545
> 8951
> 11097
>
> Is there a quick way to achieve that?
> I tried this but fail:
>
> mylable <- dataf[,1]
>
> Please advice.
>
> - Gundala Viswanath
> Jakarta - Indonesia
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

What is the problem you are trying to solve?

__
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] Getting only label column of a data frame

[jim holtman]

Wait.  The first column is really the rowname.  So you want this:

> row.names(x)
[1] "11145" "3545"  "8951"  "11097"


On Sun, Jun 22, 2008 at 10:16 PM, Gundala Viswanath <[EMAIL PROTECTED]> wrote:
> Hi,
>
> How can I extract the label only from a given data frame.
>
> Fore example from this data frame.
>
>> print(dataf)
>  V1  V2  V3  V4  V5  V6  V7  V8  V9
> 1114514.317.131.241.745.849.868.670.672.9
> 3545 10.215.620.923.231.431.736.248.451.9
> 8951 15.217.520.021.432.449.751.358.358.9
> 1109759.565.9   117.5   118.0   118.9   122.5   126.3   156.5   157.0
>
> I want to get the label (first) column only:
>
> 11145
> 3545
> 8951
> 11097
>
> Is there a quick way to achieve that?
> I tried this but fail:
>
> mylable <- dataf[,1]
>
> Please advice.
>
> - Gundala Viswanath
> Jakarta - Indonesia
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

What is the problem you are trying to solve?

__
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] problem in R for Linear mixed model~

[Daniel Malter]

Hi,

random=~1|B/C
C is nested in B

##Example
data=Oats
regress=lme(yield~nitro*Variety,data=Oats,random=~1|Block/Variety)
##i.e. variety is nested in block
summary(regress)
##End of example

Best,
Daniel 


-
cuncta stricte discussurus
-

-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Manli Yan
Gesendet: Sunday, June 22, 2008 9:30 PM
An: r-help@r-project.org
Betreff: [R] problem in R for Linear mixed model~

   Dear R users:
  I just got confused some R code used in linear mixed model~
  example,two factors,A, B,C,A is fixed ,B,C are random,and B is nested in
C,if I wannt to use linear mixed model,are the following code correct for
each case?
case1:want to know random effect of B,
   case1<-lme(y~A*B*C,random=~B|C) where "B|C" stand for what?,mean B is
nested in C?

case2: how to wirte random effect of C?
   case2<-lme(y~A*B*C,random=~C)? this doesnt work out,it seem it must have
somehing like #|$

case3.omitting the random effect for B from case1
  case3<-update(case1,random=~1|C),so I just type 1,so the random effect of
B will be removed from the model,there only left random effect of c ,the
random effect I removed ,which include both random intercept and slope
,correct??

   case4:omitting the random  intercept
  case4<-update(case1,random=B-1|C)
  this code I got from some paper,it said by inputing B-1|C,then the random
intercept  is removed,so,if I want to remove random slode,I input B-2|C,it
doesnt work out.

   case5 :how to know the both random effect of B,and C,I dont know how to
wirtie this in R .
  And I am a little confused of these R code,especially the #|# part,what
deos this syntax really mean in LME package,

  Thank you for your time~

[[alternative HTML version deleted]]

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

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

Re: [R] Getting only label column of a data frame

[Gundala Viswanath]

Thanks so much Jim.

- Gundala Viswanath
Jakarta - Indonesia


On Mon, Jun 23, 2008 at 11:22 AM, jim holtman <[EMAIL PROTECTED]> wrote:
> Wait.  The first column is really the rowname.  So you want this:
>
>> row.names(x)
> [1] "11145" "3545"  "8951"  "11097"
>
>
> On Sun, Jun 22, 2008 at 10:16 PM, Gundala Viswanath <[EMAIL PROTECTED]> wrote:
>> Hi,
>>
>> How can I extract the label only from a given data frame.
>>
>> Fore example from this data frame.
>>
>>> print(dataf)
>>  V1  V2  V3  V4  V5  V6  V7  V8  V9
>> 1114514.317.131.241.745.849.868.670.672.9
>> 3545 10.215.620.923.231.431.736.248.451.9
>> 8951 15.217.520.021.432.449.751.358.358.9
>> 1109759.565.9   117.5   118.0   118.9   122.5   126.3   156.5   157.0
>>
>> I want to get the label (first) column only:
>>
>> 11145
>> 3545
>> 8951
>> 11097
>>
>> Is there a quick way to achieve that?
>> I tried this but fail:
>>
>> mylable <- dataf[,1]
>>
>> Please advice.
>>
>> - Gundala Viswanath
>> Jakarta - Indonesia
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>

__
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] One-to-one matching?

[Alec.Zwart]

Hi folks, 

Can anyone suggest an efficient way to do "matching without
replacement", or "one-to-one matching"?  pmatch() doesn't quite provide
what I need...

For example, 

lookupTable <- c("a","b","c","d","e","f")
matchSample <- c("a","a","b","d")
##Normal match() behaviour:
match(matchSample,lookupTable)
[1] 1 1 2 4

My problem here is that both "a"s in matchSample are matched to the same
"a" in the lookup table.  I need the elements of the lookup table to be
excluded from the table as they are matched, so that no match can be
found for the second "a".  

Function pmatch() comes close to what I need:

pmatch(matchSample,lookupTable)
[1] 1 NA 2 4

Yep!  However, pmatch() incorporates partial matching, which I
definitely don't want:

lookupTable <- c("a","b","c","d","e","f") 
matchSample <- c("a","a","b","d")
pmatch(matchSample,lookupTable)
[1] 1 6 2 4
## i.e. the second "a", matches "f" - I don't want this.

Of course, when identical items ARE duplicated in both sample and lookup
table, I need the matching to reflect this:

lookupTable <- c("a","a","c","d","e","f")
matchSample <- c("a","a","c","d")
##Normal match() behaviour
match(matchSample,lookupTable)
[1] 1 1 3 4

No good - pmatch() is better:

lookupTable <- c("a","a","c","d","e","f")
matchSample <- c("a","a","c","d")
pmatch(matchSample,lookupTable)
[1] 1 2 3 4

...but we still have the partial matching issue...

##And of course, as per the usual behaviour of match(), sample elements
missing from the lookup table should return NA:

matchSample <- c("a","frog","e","d") ; print(matchSample)
match(matchSample,lookupTable)

Is there a nifty way to get what I'm after without resorting to a for
loop? (my code's already got too blasted many of those...)

Thanks, 

Alec Zwart
CMIS CSIRO
[EMAIL PROTECTED]

__
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] problem in R for Linear mixed model~

[Manli Yan]

  Hi:
  Thank you very much,and for your case
  if I wirte as
 regress=lme(yield~nitro*Variety,data=Oats,random=~Variety|Block)

what this mean?
 Regards


2008/6/22 Daniel Malter <[EMAIL PROTECTED]>:

> Hi,
>
> random=~1|B/C
> C is nested in B
>
> ##Example
> data=Oats
> regress=lme(yield~nitro*Variety,data=Oats,random=~1|Block/Variety)
> ##i.e. variety is nested in block
> summary(regress)
> ##End of example
>
> Best,
> Daniel
>
>
> -
> cuncta stricte discussurus
> -
>
> -Ursprüngliche Nachricht-
> Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
> Auftrag von Manli Yan
> Gesendet: Sunday, June 22, 2008 9:30 PM
> An: r-help@r-project.org
> Betreff: [R] problem in R for Linear mixed model~
>
>   Dear R users:
>  I just got confused some R code used in linear mixed model~
>  example,two factors,A, B,C,A is fixed ,B,C are random,and B is nested in
> C,if I wannt to use linear mixed model,are the following code correct for
> each case?
> case1:want to know random effect of B,
>   case1<-lme(y~A*B*C,random=~B|C) where "B|C" stand for what?,mean B is
> nested in C?
>
> case2: how to wirte random effect of C?
>   case2<-lme(y~A*B*C,random=~C)? this doesnt work out,it seem it must have
> somehing like #|$
>
> case3.omitting the random effect for B from case1
>  case3<-update(case1,random=~1|C),so I just type 1,so the random effect of
> B will be removed from the model,there only left random effect of c ,the
> random effect I removed ,which include both random intercept and slope
> ,correct??
>
>   case4:omitting the random  intercept
>  case4<-update(case1,random=B-1|C)
>  this code I got from some paper,it said by inputing B-1|C,then the random
> intercept  is removed,so,if I want to remove random slode,I input B-2|C,it
> doesnt work out.
>
>   case5 :how to know the both random effect of B,and C,I dont know how to
> wirtie this in R .
>  And I am a little confused of these R code,especially the #|# part,what
> deos this syntax really mean in LME package,
>
>  Thank you for your time~
>
>[[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.
>
>

[[alternative HTML version deleted]]

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

[R] Columnwise Alternate Subsetting of a Data Frame

[Gundala Viswanath]

Hi,

Given this data frame:

> print(repo.dat)
   V1 V2  V3  V4
  V5    (can be more)
1  1007_s_at   DDR1  2865.1  2901.3  1978.3   300.2
2   1053_at  RFC2   103.681.6   108.0  101.3

I want to split them into two data frames yielding:

   V1 V2 V3  ... etc
1  1007_s_at   DDR1  2865.11978.3
2   1053_at  RFC2   103.6  108.0

(V2 and V3 is obtain from V2 and V4 of original repo.dat)

and

   V1V2   V3   ...etc
1  1007_s_at   DDR12901.3 300.2
2   1053_at  RFC2  81.6101.3


(V2 and V3 is obtain from V3 and V5 of original repo.dat)

In principle what we desire to do is to:
1. Given original data frame with V1, V2, V3, V4
2. Split the data frame into two based on V2,V4,V6 ... (even) and V3,
V5, V7 ..(odd) in the original data frame
3. Meanwhile keep row names and V1 in two newly formed data frames.

Is there a compact way to do it?

- Gundala Viswanath
Jakarta - Indonesia

__
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] Simulating Gaussian Mixture Models

[Xiaohui Chen]

First, simulate a uniform r.v on [0,1] and then cast it to binary label 
according to your underlying mixing probability;

Second, simulate a Gaussian r.v. in above selected component.

Of course, you can vecterize the two steps to simply your code.

X

Peng Jiang 写道:



Hi,
  Is there any package that I can use to simulate the Gaussian Mixture 
Model , which is a mixture modeling method that is widely used in 
statistical learning theory.
 I know there is a mclust, however, I think it is a little bit 
different from my problem.

Thanks very much..

 regards.








--
Peng Jiang
江鹏
Ph.D. Candidate

Antai College of Economics & Management
安泰经济管理学院
Department of Mathematics
数学系
Shanghai Jiaotong University (Minhang Campus)
800 Dongchuan Road
200240 Shanghai
P. R. China

__
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.

Re: [R] One-to-one matching?

[Gabor Grothendieck]

Try this.  apseq() sorts the input and appends a
sequence number: 0, 1, ... to successive
occurrences of each value.  Apply that to both
vectors transforms it into a problem that works
with ordinary match:

> lookupTable <- c("a", "a","b","c","d","e","f")
> matchSample <- c("a", "a","a","b","d")
>
> # sort and append sequence no
> apseq <- function(x) {
+ x <- sort(x)
+ s <- cumsum(!duplicated(x))
+ paste(x, seq(s) - match(s, s))
+ }
>
> match(apseq(matchSample), apseq(lookupTable))
[1]  1  2 NA  3  5


On Sun, Jun 22, 2008 at 10:57 PM,  <[EMAIL PROTECTED]> wrote:
> Hi folks,
>
> Can anyone suggest an efficient way to do "matching without
> replacement", or "one-to-one matching"?  pmatch() doesn't quite provide
> what I need...
>
> For example,
>
> lookupTable <- c("a","b","c","d","e","f")
> matchSample <- c("a","a","b","d")
> ##Normal match() behaviour:
> match(matchSample,lookupTable)
> [1] 1 1 2 4
>
> My problem here is that both "a"s in matchSample are matched to the same
> "a" in the lookup table.  I need the elements of the lookup table to be
> excluded from the table as they are matched, so that no match can be
> found for the second "a".
>
> Function pmatch() comes close to what I need:
>
> pmatch(matchSample,lookupTable)
> [1] 1 NA 2 4
>
> Yep!  However, pmatch() incorporates partial matching, which I
> definitely don't want:
>
> lookupTable <- c("a","b","c","d","e","f")
> matchSample <- c("a","a","b","d")
> pmatch(matchSample,lookupTable)
> [1] 1 6 2 4
> ## i.e. the second "a", matches "f" - I don't want this.
>
> Of course, when identical items ARE duplicated in both sample and lookup
> table, I need the matching to reflect this:
>
> lookupTable <- c("a","a","c","d","e","f")
> matchSample <- c("a","a","c","d")
> ##Normal match() behaviour
> match(matchSample,lookupTable)
> [1] 1 1 3 4
>
> No good - pmatch() is better:
>
> lookupTable <- c("a","a","c","d","e","f")
> matchSample <- c("a","a","c","d")
> pmatch(matchSample,lookupTable)
> [1] 1 2 3 4
>
> ...but we still have the partial matching issue...
>
> ##And of course, as per the usual behaviour of match(), sample elements
> missing from the lookup table should return NA:
>
> matchSample <- c("a","frog","e","d") ; print(matchSample)
> match(matchSample,lookupTable)
>
> Is there a nifty way to get what I'm after without resorting to a for
> loop? (my code's already got too blasted many of those...)
>
> Thanks,
>
> Alec Zwart
> CMIS CSIRO
> [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.
>

__
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] One-to-one matching?

[Moshe Olshansky]

How about

x <- lookupTable
x[!(x %in% matchSample)] <- NA
pmatch(matchSample,x)


Regards,

Moshe.

--- On Mon, 23/6/08, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:

> From: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
> Subject: [R] One-to-one matching?
> To: r-help@r-project.org
> Received: Monday, 23 June, 2008, 12:57 PM
> Hi folks, 
> 
> Can anyone suggest an efficient way to do "matching
> without
> replacement", or "one-to-one matching"? 
> pmatch() doesn't quite provide
> what I need...
> 
> For example, 
> 
> lookupTable <-
> c("a","b","c","d","e","f")
> matchSample <-
> c("a","a","b","d")
> ##Normal match() behaviour:
> match(matchSample,lookupTable)
> [1] 1 1 2 4
> 
> My problem here is that both "a"s in matchSample
> are matched to the same
> "a" in the lookup table.  I need the elements of
> the lookup table to be
> excluded from the table as they are matched, so that no
> match can be
> found for the second "a".  
> 
> Function pmatch() comes close to what I need:
> 
> pmatch(matchSample,lookupTable)
> [1] 1 NA 2 4
> 
> Yep!  However, pmatch() incorporates partial matching,
> which I
> definitely don't want:
> 
> lookupTable <-
> c("a","b","c","d","e","f")
> 
> matchSample <-
> c("a","a","b","d")
> pmatch(matchSample,lookupTable)
> [1] 1 6 2 4
> ## i.e. the second "a", matches
> "f" - I don't want this.
> 
> Of course, when identical items ARE duplicated in both
> sample and lookup
> table, I need the matching to reflect this:
> 
> lookupTable <-
> c("a","a","c","d","e","f")
> matchSample <-
> c("a","a","c","d")
> ##Normal match() behaviour
> match(matchSample,lookupTable)
> [1] 1 1 3 4
> 
> No good - pmatch() is better:
> 
> lookupTable <-
> c("a","a","c","d","e","f")
> matchSample <-
> c("a","a","c","d")
> pmatch(matchSample,lookupTable)
> [1] 1 2 3 4
> 
> ...but we still have the partial matching issue...
> 
> ##And of course, as per the usual behaviour of match(),
> sample elements
> missing from the lookup table should return NA:
> 
> matchSample <-
> c("a","frog","e","d")
> ; print(matchSample)
> match(matchSample,lookupTable)
> 
> Is there a nifty way to get what I'm after without
> resorting to a for
> loop? (my code's already got too blasted many of
> those...)
> 
> Thanks, 
> 
> Alec Zwart
> CMIS CSIRO
> [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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Re: [R] One-to-one matching?

[Charles C. Berry]

Or use make.unique() in place of apseq()

On Sun, 22 Jun 2008, Gabor Grothendieck wrote:


Try this.  apseq() sorts the input and appends a
sequence number: 0, 1, ... to successive
occurrences of each value.  Apply that to both
vectors transforms it into a problem that works
with ordinary match:


lookupTable <- c("a", "a","b","c","d","e","f")
matchSample <- c("a", "a","a","b","d")

# sort and append sequence no
apseq <- function(x) {

+ x <- sort(x)
+ s <- cumsum(!duplicated(x))
+ paste(x, seq(s) - match(s, s))
+ }


match(apseq(matchSample), apseq(lookupTable))

[1]  1  2 NA  3  5


On Sun, Jun 22, 2008 at 10:57 PM,  <[EMAIL PROTECTED]> wrote:

Hi folks,

Can anyone suggest an efficient way to do "matching without
replacement", or "one-to-one matching"?  pmatch() doesn't quite provide
what I need...

For example,

lookupTable <- c("a","b","c","d","e","f")
matchSample <- c("a","a","b","d")
##Normal match() behaviour:
match(matchSample,lookupTable)
[1] 1 1 2 4

My problem here is that both "a"s in matchSample are matched to the same
"a" in the lookup table.  I need the elements of the lookup table to be
excluded from the table as they are matched, so that no match can be
found for the second "a".

Function pmatch() comes close to what I need:

pmatch(matchSample,lookupTable)
[1] 1 NA 2 4

Yep!  However, pmatch() incorporates partial matching, which I
definitely don't want:

lookupTable <- c("a","b","c","d","e","f")
matchSample <- c("a","a","b","d")
pmatch(matchSample,lookupTable)
[1] 1 6 2 4
## i.e. the second "a", matches "f" - I don't want this.

Of course, when identical items ARE duplicated in both sample and lookup
table, I need the matching to reflect this:

lookupTable <- c("a","a","c","d","e","f")
matchSample <- c("a","a","c","d")
##Normal match() behaviour
match(matchSample,lookupTable)
[1] 1 1 3 4

No good - pmatch() is better:

lookupTable <- c("a","a","c","d","e","f")
matchSample <- c("a","a","c","d")
pmatch(matchSample,lookupTable)
[1] 1 2 3 4

...but we still have the partial matching issue...

##And of course, as per the usual behaviour of match(), sample elements
missing from the lookup table should return NA:

matchSample <- c("a","frog","e","d") ; print(matchSample)
match(matchSample,lookupTable)

Is there a nifty way to get what I'm after without resorting to a for
loop? (my code's already got too blasted many of those...)

Thanks,

Alec Zwart
CMIS CSIRO
[EMAIL PROTECTED]

__
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__
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]  UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

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Re: [R] Columnwise Alternate Subsetting of a Data Frame

[Moshe Olshansky]

You could do something like:

k <- dim(repo.dat)[2]
odd <- seq(from=1,to=k,by=2)
even <- seq(from=2,to=k,by=2)
even<-c(1,even)
df1 <- repo.dat[,odd)
colnames(df1) <- paste("V",odd,sep="")
df2 <- repo.dat[,even)
colnames(df2) <- paste("V",even,sep="")


--- On Mon, 23/6/08, Gundala Viswanath <[EMAIL PROTECTED]> wrote:

> From: Gundala Viswanath <[EMAIL PROTECTED]>
> Subject: [R] Columnwise Alternate Subsetting of a Data Frame
> To: [EMAIL PROTECTED]
> Received: Monday, 23 June, 2008, 1:03 PM
> Hi,
> 
> Given this data frame:
> 
> > print(repo.dat)
>V1 V2  V3   
>   V4
>   V5    (can be more)
> 1  1007_s_at   DDR1  2865.1  2901.3  1978.3   300.2
> 2   1053_at  RFC2   103.681.6   108.0 
> 101.3
> 
> I want to split them into two data frames yielding:
> 
>V1 V2 V3
>  ... etc
> 1  1007_s_at   DDR1  2865.11978.3
> 2   1053_at  RFC2   103.6  108.0
> 
> (V2 and V3 is obtain from V2 and V4 of original repo.dat)
> 
> and
> 
>V1V2
>   V3   ...etc
> 1  1007_s_at   DDR12901.3 300.2
> 2   1053_at  RFC2  81.6101.3
> 
> 
> (V2 and V3 is obtain from V3 and V5 of original repo.dat)
> 
> In principle what we desire to do is to:
> 1. Given original data frame with V1, V2, V3, V4
> 2. Split the data frame into two based on V2,V4,V6 ...
> (even) and V3,
> V5, V7 ..(odd) in the original data frame
> 3. Meanwhile keep row names and V1 in two newly formed data
> frames.
> 
> Is there a compact way to do it?
> 
> - Gundala Viswanath
> Jakarta - Indonesia
> 
> __
> 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] Columnwise Alternate Subsetting of a Data Frame

[Charles C. Berry]

See

?split

help.search() is (or should be) your friend.

HTH,

Chuck

On Mon, 23 Jun 2008, Gundala Viswanath wrote:


Hi,

Given this data frame:


print(repo.dat)

  V1 V2  V3  V4
 V5    (can be more)
1  1007_s_at   DDR1  2865.1  2901.3  1978.3   300.2
2   1053_at  RFC2   103.681.6   108.0  101.3

I want to split them into two data frames yielding:

  V1 V2 V3  ... etc
1  1007_s_at   DDR1  2865.11978.3
2   1053_at  RFC2   103.6  108.0

(V2 and V3 is obtain from V2 and V4 of original repo.dat)

and

  V1V2   V3   ...etc
1  1007_s_at   DDR12901.3 300.2
2   1053_at  RFC2  81.6101.3


(V2 and V3 is obtain from V3 and V5 of original repo.dat)

In principle what we desire to do is to:
1. Given original data frame with V1, V2, V3, V4
2. Split the data frame into two based on V2,V4,V6 ... (even) and V3,
V5, V7 ..(odd) in the original data frame
3. Meanwhile keep row names and V1 in two newly formed data frames.

Is there a compact way to do it?

- Gundala Viswanath
Jakarta - Indonesia

__
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.



Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]  UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

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

[R] Pairwise Partitioning of a Vector

[Gundala Viswanath]

Hi,

How can I partitioned an example vector like this

> print(myvector)
 [1]   30.9   60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9

into the following pairwise partition:

PAIR1
part1  = 30.9
part2  = 60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9

PAIR2
part1 = 30.9   60.1
part2 =  70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9



PAIR9
part1 = 30.9   60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8
part2 = 113.9


I'm stuck with this kind of loop:

__BEGIN__

# gexp is a Vector

process_two_partition <- function(gexp) {
sort.gexp <- sort(as.matrix(gexp))
print(sort.gexp)

for (posb in 1:ncol(gexp)) {
for (pose in 1:ncol(gexp)) {

  sp_b <- pose+1
  sp_e <- ncol(gexp)

  # This two doesn't do what I want
  part1 <- sort.gexp[posb:pose]
  part2 <- sort.gexp[sp_b:sp_e]

 # we later want to process part1 and part2 separately

}
}

}

__END__

- Gundala Viswanath
Jakarta - Indonesia

__
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Re: [R] Pairwise Partitioning of a Vector

[Peter Dalgaard]

Gundala Viswanath wrote:

Hi,

How can I partitioned an example vector like this

  

print(myvector)


 [1]   30.9   60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9

into the following pairwise partition:

PAIR1
part1  = 30.9
part2  = 60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9

PAIR2
part1 = 30.9   60.1
part2 =  70.0   73.0   75.0   83.9   93.1   97.6   98.8  113.9



PAIR9
part1 = 30.9   60.1   70.0   73.0   75.0   83.9   93.1   97.6   98.8
part2 = 113.9


I'm stuck with this kind of loop:
  
Your code seems strangely out of sync with your example. Why does gexp 
want to me a matrix and what is the sorting for??


How about

lapply(1:9, function(n) {ix <- seq_len(n); list(part1=myvector[ix], 
part2=myvector[-ix])})


?


__BEGIN__

# gexp is a Vector

process_two_partition <- function(gexp) {
sort.gexp <- sort(as.matrix(gexp))
print(sort.gexp)

for (posb in 1:ncol(gexp)) {
for (pose in 1:ncol(gexp)) {

  sp_b <- pose+1
  sp_e <- ncol(gexp)

  # This two doesn't do what I want
  part1 <- sort.gexp[posb:pose]
  part2 <- sort.gexp[sp_b:sp_e]

 # we later want to process part1 and part2 separately

}
}

}

__END__

- Gundala Viswanath
Jakarta - Indonesia

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

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Re: [R] Pairwise Partitioning of a Vector

[Moshe Olshansky]

One possibility is:

x <- c(30.9, 60.1  , 70.0 ,  73.0 ,  75.0 ,  83.9 ,  93.1 ,  97.6 ,  98.8 , 
113.9)
for (i in 1:9) assign(paste("PAIR",i,sep=""),list(part1 = x[1:i],part2 = 
x[-(1:i)]))




--- On Mon, 23/6/08, Gundala Viswanath <[EMAIL PROTECTED]> wrote:

> From: Gundala Viswanath <[EMAIL PROTECTED]>
> Subject: [R] Pairwise Partitioning of a Vector
> To: [EMAIL PROTECTED]
> Received: Monday, 23 June, 2008, 4:13 PM
> Hi,
> 
> How can I partitioned an example vector like this
> 
> > print(myvector)
>  [1]   30.9   60.1   70.0   73.0   75.0   83.9   93.1  
> 97.6   98.8  113.9
> 
> into the following pairwise partition:
> 
> PAIR1
> part1  = 30.9
> part2  = 60.1   70.0   73.0   75.0   83.9   93.1   97.6  
> 98.8  113.9
> 
> PAIR2
> part1 = 30.9   60.1
> part2 =  70.0   73.0   75.0   83.9   93.1   97.6   98.8 
> 113.9
> 
> 
> 
> PAIR9
> part1 = 30.9   60.1   70.0   73.0   75.0   83.9   93.1  
> 97.6   98.8
> part2 = 113.9
> 
> 
> I'm stuck with this kind of loop:
> 
> __BEGIN__
> 
> # gexp is a Vector
> 
> process_two_partition <- function(gexp) {
> sort.gexp <- sort(as.matrix(gexp))
> print(sort.gexp)
> 
> for (posb in 1:ncol(gexp)) {
> for (pose in 1:ncol(gexp)) {
> 
>   sp_b <- pose+1
>   sp_e <- ncol(gexp)
> 
>   # This two doesn't do what I want
>   part1 <- sort.gexp[posb:pose]
>   part2 <- sort.gexp[sp_b:sp_e]
> 
>  # we later want to process part1 and part2
> separately
> 
> }
> }
> 
> }
> 
> __END__
> 
> - Gundala Viswanath
> Jakarta - Indonesia
> 
> __
> 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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