date:20080616


If you want a new device opened you can call dev.new() In current R.

X11() does have xpos and ypos, *but* there is no guarantee that any value 
you choose will not be the same as the default.  Indeed, with
windows(5,5,xpos=0) you will put your second window on top of the first 
one in my setup.


I entirely agree with Patrick, and would add 'why do you think a 5 square 
window is appropriate on both 8 and 30 monitors?'.  Please curb your 
enthusiasm to make decisions for your package users.


On Mon, 16 Jun 2008, Christophe Genolini wrote:




Generally, I like to use a pdf device which can have any number of
pages.
I am proposing a graphical way a selecting a clusterization based a quality 
criterion. In order to avoid local maximum, you can build something like 10 
000 clusterization. Exporting all of them has no sence. So I open two graph, 
on the left you can see all the quality criterion. On the rigth, you can see 
the clusterization that you are curently selecting on the left graph. The 
idea is not to export all the clusterization but to chose one. So you can 
change the selected clusterization by using the arrow. This is possible only 
on a screen.



But an X11 device will also work for single plots and there's no need
to specifically set it in the package.


Thanks a lot.
Is there a way to set the position at which the x11 windows shall open ? I do 
not want to be very precise, I just want the two windows to not open one on 
the top of the other...




Why not let users set the device they wish to use?
The function that is calling windows(5,5,xpos=0) is called choice(). So 
users that do not like the output that I propose are free to not call 
choice(), others are free to use it...



 Requests to the maintainer gets no response.


Lol ! I am so proud when someone is using one of my two package that I answer 
very fast... ;-)

May be it will change when I will have ten to maintain ?

Christophe

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



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

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[R] Display a jpeg inside a widget which already has text

2008-06-16 Thread danzam


Hi 

I am trying to insert a jpeg into a widget I have created. 
I have used this link
http://bioinf.wehi.edu.au/~wettenhall/RTclTkExamples/showImage.html to
display a jpeg inside a newly created widget but would like some info on how
to insert into a pre existing widget. 
This is my code below for my first widget. 

tt-tktoplevel()
fontHeading2 - tkfont.create(family=arial,size=16, slant=italic)
fontHeading3 - tkfont.create(family=arial,size=12)
fontTextLabel - tkfont.create(family=arial,size=12)
fontlabel - tkfont.create(family=arial,size=10,weight=bold)
tkgrid(tklabel(tt,height=2,width=50, text=))

tkgrid(tklabel(tt,text=))
tkgrid(tklabel(tt,text='Enhance Your Decision
Making',font=fontHeading2,foreground=black))
tkgrid(tklabel(tt,text= v 4.0, font=fontTextLabel))
tkgrid(tklabel(tt,height=2,width=50,text=))
tkgrid(tklabel(tt,text=))

##I would like to insert an image into this widget


Any help would be greatly appreciated.

cheers
Daniel 


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[R] error in dat

2008-06-16 Thread Paul Adams

Hello everyone,
I have the following code which keeps giving me an error.
The code is:
dat-read.table(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisen.txt,header=T,row.names=1,blank.lines.skip=F,na..strings=NA)
dimnames(dat)((1)) -as.character(dat(,1))
dat-dat(,-1)
dat-as.data.frame(dat)
file.show(file=C:\\Documents and Settings\\Owner\\My Documents\\eisen.txt)
ann-read.table(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisenClasses.txt,header=T)
file.show(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisenClasses.txt)
cl-as.character(ann[,2])
dat-dat[,cl]
gc-cl(1:19)
act-cl(20:39)
x-as.numeric(dat(2000,gc))
y-as.numeric(dat(2000,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 2000)
The error is:   error in eval .with.vis(expr, envir, enclos)  :
could not find function dat
I do not understand what this error is? Also is there a 
command to get a more verbose explanation of the error 
Any help would be appreciated
Paul



  
[[alternative HTML version deleted]]

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Re: [R] randomForest, 'No forest component...' error while calling Predict()

2008-06-16 Thread Jim_S


Now I read it in the manual If xtest is given, defaults to FALSE. 
Thanks for your help!

Jim


Moshe Olshansky-2 wrote:
 
 You must use randomForest with keep.forest=TRUE - otherwise the Forest
 object is not saved and so no prediction can be made.
 
 
 --- On Mon, 16/6/08, Jim_S [EMAIL PROTECTED] wrote:
 
 From: Jim_S [EMAIL PROTECTED]
 Subject: [R]  randomForest, 'No forest component...' error while calling
 Predict()
 To: r-help@r-project.org
 Received: Monday, 16 June, 2008, 8:13 AM
 Dear R-users,
 
 While making a prediction using the randomForest function
 (package
 randomForest) I'm getting the following error message: 
 
 Error in predict.randomForest(model, newdata = CV) : 
 No forest component
 in the object
 Here's my complete code. For reproducing this task,
 please find my 2 data
 sets attached (
 http://www.nabble.com/file/p17855119/data.rar data.rar ).
 
 Thanks in advance for any help!
 
  - Jim
 
 CT -
 read.table(CT.txt,header=TRUE,sep=\t)
 CV -
 read.table(CV.txt,header=TRUE,sep=\t)
 # Both CT  CV have the syntaxis X1, X2,...,X97,Y where
 all variables are
 numeric 
 x - CT[,-98]
 y - CT[,98]
 xtest - CV[,-98]
 ytest - CV[,98]
 library(randomForest)
 model - randomForest(x ,y , xtest,
 ytest,ntree=500,mtry=32,nodesize=5,nPerm=2)
 model
 
 #Call:
 # randomForest(x = x, y = y, xtest = xtest, ytest = ytest,
 ntree = 500, 
 mtry = 32, nodesize = 5, 
 # nPerm = 2) 
 #   Type of random forest: regression
 #Number of trees: 500
 #No. of variables tried at each split: 32
 #
 #  Mean of squared residuals: 0.0022117
 #% Var explained: -1.01
 #   Test set MSE: 0
 #% Var explained: 2.61
  
 
 prediction - predict(model, newdata= CT)
 #Error in predict.randomForest(model, newdata = CV) : 
 #  No forest component in the object
 # 
 -- 
 View this message in context:
 http://www.nabble.com/randomForest%2C-%27No-forest-component...%27-error-while-calling-Predict%28%29-tp17855119p17855119.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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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.
 
 __
 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] Superimposing Line over Histogram in Density Plot

2008-06-16 Thread Dimitris Rizopoulos


try something like this:

x - rnorm(200)
hist(x, col = blue, freq = FALSE)
lines(density(x), col = red, lwd = 2)


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven

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


- Original Message - 
From: Gundala Viswanath [EMAIL PROTECTED]

To: [EMAIL PROTECTED]
Sent: Monday, June 16, 2008 8:25 AM
Subject: [R] Superimposing Line over Histogram in Density Plot



Hi,

Currently I have a density plot generated with this
snippet.

Is there a way I can add a line curve on top of it?
I mean in one figure


__BEGIN__
  myhist - hist(x
   col=blue,
   main = Density Plot,
   xlab = Exp Level,
   )
__END__


- Gundala Viswanath
Jakarta - Indonesia

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

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Re: [R] package under unix

2008-06-16 Thread cgenolin


Hi

I entirely agree with Patrick, and would add 'why do you think a 5 
square window is appropriate on both 8 and 30 monitors?'.


I do not realy care about 5, 8 or 20, I use it when I devellop, I 
will remove it for the final package.
I just want the two windows to not be one on the top of the other. I do 
not care if one in on right, the other left, or one at the top of the 
screen, the other at the bottom...



X11() does have xpos and ypos,


I am using R 2.7 but I do not manage to deal with X11 and xpos :
X11(xpos=0) give an error
windows.options(xpos=0) ; X11() open a windows but it does not change 
the place.

Am I misunderstanding something ?

Christophe


curb your enthusiasm to make decisions for your package users.

On Mon, 16 Jun 2008, Christophe Genolini wrote:




Generally, I like to use a pdf device which can have any number of
pages.
I am proposing a graphical way a selecting a clusterization based a 
quality criterion. In order to avoid local maximum, you can build 
something like 10 000 clusterization. Exporting all of them has no 
sence. So I open two graph, on the left you can see all the quality 
criterion. On the rigth, you can see the clusterization that you are 
curently selecting on the left graph. The idea is not to export all 
the clusterization but to chose one. So you can change the selected 
clusterization by using the arrow. This is possible only on a screen.



But an X11 device will also work for single plots and there's no need
to specifically set it in the package.


Thanks a lot.
Is there a way to set the position at which the x11 windows shall 
open ? I do not want to be very precise, I just want the two windows 
to not open one on the top of the other...




Why not let users set the device they wish to use?
The function that is calling windows(5,5,xpos=0) is called 
choice(). So users that do not like the output that I propose are 
free to not call choice(), others are free to use it...



 Requests to the maintainer gets no response.


Lol ! I am so proud when someone is using one of my two package that 
I answer very fast... ;-)

May be it will change when I will have ten to maintain ?

Christophe

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



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



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

Re: [R] package under unix

X11() is the standard Unix device: see your subject line.  Testing x11() 
under Windows isn't relevant to your subject line (which in any case is 
not helpful -- the posting guide asks you to use informative subject 
lines).


There is no 'R 2.7', and you are explicitly asked in the posting guide not 
to be inaccurate about this.


On Mon, 16 Jun 2008, [EMAIL PROTECTED] wrote:


Hi

I entirely agree with Patrick, and would add 'why do you think a 5 square 
window is appropriate on both 8 and 30 monitors?'.


I do not realy care about 5, 8 or 20, I use it when I devellop, I will 
remove it for the final package.
I just want the two windows to not be one on the top of the other. I do not 
care if one in on right, the other left, or one at the top of the screen, the 
other at the bottom...



X11() does have xpos and ypos,


I am using R 2.7 but I do not manage to deal with X11 and xpos :
X11(xpos=0) give an error
windows.options(xpos=0) ; X11() open a windows but it does not change the 
place.

Am I misunderstanding something ?

Christophe


curb your enthusiasm to make decisions for your package users.

On Mon, 16 Jun 2008, Christophe Genolini wrote:




Generally, I like to use a pdf device which can have any number of
pages.
I am proposing a graphical way a selecting a clusterization based a 
quality criterion. In order to avoid local maximum, you can build 
something like 10 000 clusterization. Exporting all of them has no sence. 
So I open two graph, on the left you can see all the quality criterion. On 
the rigth, you can see the clusterization that you are curently selecting 
on the left graph. The idea is not to export all the clusterization but to 
chose one. So you can change the selected clusterization by using the 
arrow. This is possible only on a screen.



But an X11 device will also work for single plots and there's no need
to specifically set it in the package.


Thanks a lot.
Is there a way to set the position at which the x11 windows shall open ? I 
do not want to be very precise, I just want the two windows to not open 
one on the top of the other...




Why not let users set the device they wish to use?
The function that is calling windows(5,5,xpos=0) is called choice(). So 
users that do not like the output that I propose are free to not call 
choice(), others are free to use it...



 Requests to the maintainer gets no response.


Lol ! I am so proud when someone is using one of my two package that I 
answer very fast... ;-)

May be it will change when I will have ten to maintain ?

Christophe

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

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



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






Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre





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

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

Re: [R] error in dat

2008-06-16 Thread Wacek Kusnierczyk

Paul Adams wrote:
 Hello everyone,
 I have the following code which keeps giving me an error.
 The code is:
 dat-read.table(file=C:\\Documents and Settings\\Owner\\My 
 Documents\\eisen.txt,header=T,row.names=1,blank.lines.skip=F,na..strings=NA)
 dimnames(dat)((1)) -as.character(dat(,1))
 dat-dat(,-1)
 dat-as.data.frame(dat)
 file.show(file=C:\\Documents and Settings\\Owner\\My Documents\\eisen.txt)
 ann-read.table(file=C:\\Documents and Settings\\Owner\\My 
 Documents\\eisenClasses.txt,header=T)
 file.show(file=C:\\Documents and Settings\\Owner\\My 
 Documents\\eisenClasses.txt)
 cl-as.character(ann[,2])
 dat-dat[,cl]
 gc-cl(1:19)
 act-cl(20:39)
 x-as.numeric(dat(2000,gc))
 y-as.numeric(dat(2000,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 2000)
 The error is:   error in eval .with.vis(expr, envir, enclos)  :
 could not find function dat

you misuse the syntax, check the docs.  with 'dat(...)' r tries to apply
dat, but dat is a data frame, and is thus not applicable.  what you want
is dat[...].

you can argue that the error message is misleading;  unless you defined
one, r cannot find a function named 'dat', but it does find your data
frame, and it should complain about its non-applicability. 

vQ

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

Re: [R] error in dat


On Mon, 16 Jun 2008, Wacek Kusnierczyk wrote:


Paul Adams wrote:

Hello everyone,
I have the following code which keeps giving me an error.
The code is:
dat-read.table(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisen.txt,header=T,row.names=1,blank.lines.skip=F,na..strings=NA)
dimnames(dat)((1)) -as.character(dat(,1))
dat-dat(,-1)
dat-as.data.frame(dat)
file.show(file=C:\\Documents and Settings\\Owner\\My Documents\\eisen.txt)
ann-read.table(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisenClasses.txt,header=T)
file.show(file=C:\\Documents and Settings\\Owner\\My 
Documents\\eisenClasses.txt)
cl-as.character(ann[,2])
dat-dat[,cl]
gc-cl(1:19)
act-cl(20:39)
x-as.numeric(dat(2000,gc))
y-as.numeric(dat(2000,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 2000)
The error is:   error in eval .with.vis(expr, envir, enclos)  :
could not find function dat


you misuse the syntax, check the docs.  with 'dat(...)' r tries to apply
dat, but dat is a data frame, and is thus not applicable.  what you want
is dat[...].

you can argue that the error message is misleading;  unless you defined
one, r cannot find a function named 'dat', but it does find your data
frame, and it should complain about its non-applicability.


It is this explanation which is misleading.  R (not r) looks for a 
function named 'dat': it does not find the data frame when looking for a 
function.  To be explicit, when R encounters foo() it looks in the current 
environment for a function named 'foo' and ignores all other objects named 
'foo' even if they are higher on the search path.  This was not the 
behaviour of Blue-Book S, but it has been the behvaviour of S and R for 
many years.


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

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

[R] How can I stop strwrap removing escape characters? (and solution to the screen wrapping question from 11/6)

2008-06-16 Thread Toby Marthews

After the helpful reply from Greg Snow (below), I've written a function
printtoscreen which does the screen wrapping I need (although not in a
very elegant way). This works fine for strings without any escape
characters in them, e.g.

 printtoscreen=function(str) {
+ str2=strwrap(str,width=getOption(width))
+ if (length(str2)1)
{str2[2:length(str2)]=paste(\b\b,str2[2:length(str2)],sep=)} #to
remove the , s toString puts in
+ str2=paste(str2,\n,sep=)
+ cat(str2)
+ }
 getOption(width)
[1] 59

 cat(Measured lengths are (,toString(1:20/100),) mm.\n)
Measured lengths are ( 0.01, 0.02, 0.03, 0.04, 0.05, 0.06,$

 printtoscreen(paste(Measured lengths are (,toString(1:20/100),)
mm.\n,sep=))
Measured lengths are (0.01, 0.02, 0.03, 0.04, 0.05, 0.06,
0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16,
0.17, 0.18, 0.19, 0.2) mm.


HOWEVER, if you use any escape characters in your string, it fails because
strwrap removes all of them:


 cat(\n\tHello.\n)

Hello.
 printtoscreen(\n\tHello.\n)
Hello.


Is there any way to prevent strwrap doing this? I haven't found any
options for this.

Thanks very much (and thanks to Greg Snow for the tip about strwrap before).

Toby Marthews


==
Le Mer 11 juin 2008 18:00, Greg Snow a écrit :
 You could try passing your character string to the strwrap function
first,
 then use cat on the result.

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



 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Toby Marthews Sent:
Wednesday, June 11, 2008 6:52 AM
 To: r-help@r-project.org
 Subject: [R] Word wrapping for character objects (WINDOWS R ONLY) Can
anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM
DOESN'T OCCUR ON LINUX **
 I want to print a long character to screen:
  getOption(width)
 [1] 60
  z=(1:20)/10#z is a vector of length between 20 and 30 (depending
on user options) containing lengths in mm (i.e. each element is 1-5
characters long)
  str1=paste(The depths chosen are (,toString(z),) mm, and more text
...\n)
  cat(str1)
 The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,
1.1, 1.2, 1.3, 1.$
 
 The problem is that on R for Windows the string is cropped by the
window size (hence the $). On R for Linux, this doesn't happen and the
text is word wrapped (the default for the shell, I guess):
  cat(str1)
 The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,
1.1, 1.2, 1.3,
 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2 ) mm, and more text ...
 
 I can't find any option for word wrapping in the cat command
(fill=TRUE has no effect). I also checked the menu Edit - GUI
preferences..., but there doesn't seem to be a Word Wrap option there
either.
 How do I get word wrapping like this in Windows?
 THANKS FOR ANY HELP!
 Toby Marthews
 Previous relevant posts:
 - The post from 2006 about Screen Wrapping
(http://tolstoy.newcastle.edu.au/R/help/06/05/26673.html) which Brian
Ripley answered was about controlling how long vectors are cropped to the
screen. Unfortunately, the width option in options() does not
affect character objects, so I can't use that control here.
 - I sent the same question to [EMAIL PROTECTED] in Oct 2007,
but noone there could help me with it.
 - Try the following command on Windows R with a small window
(getOption(width)117) and a large window (getOption(width)117) and
you'll see you get extra nonexistent options in the menu:
 a=c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec);menu(a)
I guess this is a related problem?

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] NLME: Prediction intervals for random effects

2008-06-16 Thread Marc Bernard

Dear all, 
nbsp;
Is there a function to calculate thenbsp; prediction intervals for random 
effects in non-linear mixed models? I found a way to do it for linear mixed 
models but not for non-linearnbsp;mixed one.
nbsp;
Many thnaks
nbsp;
Bernard
nbsp;
nbsp;


  
_ 

o.fr
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Re: [R] How can I stop strwrap removing escape characters? (and solution to the screen wrapping question from 11/6)

2008-06-16 Thread Gavin Simpson

On Mon, 2008-06-16 at 11:33 +0200, Toby Marthews wrote:
 After the helpful reply from Greg Snow (below), I've written a function
 printtoscreen which does the screen wrapping I need (although not in a
 very elegant way). This works fine for strings without any escape
 characters in them, e.g.
 
  printtoscreen=function(str) {
 + str2=strwrap(str,width=getOption(width))
 + if (length(str2)1)
 {str2[2:length(str2)]=paste(\b\b,str2[2:length(str2)],sep=)} #to
 remove the , s toString puts in
 + str2=paste(str2,\n,sep=)
 + cat(str2)
 + }
  getOption(width)
 [1] 59

Hi Toby,

Have you considered writeLines() as a in-built alternative to
printtoscreen:

 writeLines(strwrap(paste(Measured lengths are (, paste(1:20/100,
collapse = , ), ) mm., sep = ), width = 60))
Measured lengths are (0.01, 0.02, 0.03, 0.04, 0.05, 0.06,
0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16,
0.17, 0.18, 0.19, 0.2) mm.

The above is how I would have tackled this previously, but I see
toString is something I should add my battery of useful R manipulation
routines:

 writeLines(strwrap(paste(Measured lengths are (, toString(1:20 /
100), ) mm., sep = ), width = 60))
Measured lengths are (0.01, 0.02, 0.03, 0.04, 0.05, 0.06,
0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16,
0.17, 0.18, 0.19, 0.2) mm.

Note you don't need the trailing \n from your example if using
writeLines (argument sep to writeLines adds this by default).

  cat(Measured lengths are (,toString(1:20/100),) mm.\n)
 Measured lengths are ( 0.01, 0.02, 0.03, 0.04, 0.05, 0.06,$
 
  printtoscreen(paste(Measured lengths are (,toString(1:20/100),)
 mm.\n,sep=))
 Measured lengths are (0.01, 0.02, 0.03, 0.04, 0.05, 0.06,
 0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16,
 0.17, 0.18, 0.19, 0.2) mm.
 
 
 HOWEVER, if you use any escape characters in your string, it fails because
 strwrap removes all of them:
 
 
  cat(\n\tHello.\n)
 
 Hello.
  printtoscreen(\n\tHello.\n)
 Hello.
 
 
 Is there any way to prevent strwrap doing this? I haven't found any
 options for this.

No, not that I can see and this is as documented on ?strwrap, though
consider the prefix argument of strwrap to achieve the same end result
as your simple example.

 writeLines(strwrap(\n\tHello.\n, width = 60, prefix = \n\t))

Hello.

Your actual use case may be more complicated than this, however, but I
don't have a solution to hand if it is.

HTH

G

 
 Thanks very much (and thanks to Greg Snow for the tip about strwrap before).
 
 Toby Marthews
 
 
 ==
 Le Mer 11 juin 2008 18:00, Greg Snow a écrit :
  You could try passing your character string to the strwrap function
 first,
  then use cat on the result.
 
  --
  Gregory (Greg) L. Snow Ph.D.
  Statistical Data Center
  Intermountain Healthcare
  [EMAIL PROTECTED]
  (801) 408-8111
 
 
 
  -Original Message-
  From: [EMAIL PROTECTED]
  [mailto:[EMAIL PROTECTED] On Behalf Of Toby Marthews Sent:
 Wednesday, June 11, 2008 6:52 AM
  To: r-help@r-project.org
  Subject: [R] Word wrapping for character objects (WINDOWS R ONLY) Can
 anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM
 DOESN'T OCCUR ON LINUX **
  I want to print a long character to screen:
   getOption(width)
  [1] 60
   z=(1:20)/10#z is a vector of length between 20 and 30 (depending
 on user options) containing lengths in mm (i.e. each element is 1-5
 characters long)
   str1=paste(The depths chosen are (,toString(z),) mm, and more text
 ...\n)
   cat(str1)
  The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,
 1.1, 1.2, 1.3, 1.$
  
  The problem is that on R for Windows the string is cropped by the
 window size (hence the $). On R for Linux, this doesn't happen and the
 text is word wrapped (the default for the shell, I guess):
   cat(str1)
  The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,
 1.1, 1.2, 1.3,
  1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2 ) mm, and more text ...
  
  I can't find any option for word wrapping in the cat command
 (fill=TRUE has no effect). I also checked the menu Edit - GUI
 preferences..., but there doesn't seem to be a Word Wrap option there
 either.
  How do I get word wrapping like this in Windows?
  THANKS FOR ANY HELP!
  Toby Marthews
  Previous relevant posts:
  - The post from 2006 about Screen Wrapping
 (http://tolstoy.newcastle.edu.au/R/help/06/05/26673.html) which Brian
 Ripley answered was about controlling how long vectors are cropped to the
 screen. Unfortunately, the width option in options() does not
 affect character objects, so I can't use that control here.
  - I sent the same question to [EMAIL PROTECTED] in Oct 2007,
 but noone there could help me with it.
  - Try the following command on Windows R with a small window
 (getOption(width)117) and a large window (getOption(width)117) and
 you'll see you get extra nonexistent options in the menu:
  a=c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec);menu(a)
 I guess this

Re: [R] How can I stop strwrap removing escape characters? (and solution to the screen wrapping question from 11/6)

I think by 'escape characters' you mean 'whitespace characters'.  There 
are no 'escape characters' in your string -- the escape character is \, 
and \n and \t are converted by the parser to characters when the string is 
constructed and so are not escapted in the string.


?strwrap explicitly says 'Whitespace in the input is destroyed.', which 
seems clear enough in your case.  (What is meant by 'whitespace' is not 
completely clear, but space, tab (\t) and LF (\n) are always whitespace 
and it seems it means just those three and not FF, VT nor CR.)


On Mon, 16 Jun 2008, Toby Marthews wrote:


After the helpful reply from Greg Snow (below), I've written a function
printtoscreen which does the screen wrapping I need (although not in a
very elegant way). This works fine for strings without any escape
characters in them, e.g.


printtoscreen=function(str) {

+ str2=strwrap(str,width=getOption(width))
+ if (length(str2)1)
{str2[2:length(str2)]=paste(\b\b,str2[2:length(str2)],sep=)} #to
remove the , s toString puts in
+ str2=paste(str2,\n,sep=)
+ cat(str2)
+ }

getOption(width)

[1] 59


cat(Measured lengths are (,toString(1:20/100),) mm.\n)

Measured lengths are ( 0.01, 0.02, 0.03, 0.04, 0.05, 0.06,$


printtoscreen(paste(Measured lengths are (,toString(1:20/100),)

mm.\n,sep=))
Measured lengths are (0.01, 0.02, 0.03, 0.04, 0.05, 0.06,
0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16,
0.17, 0.18, 0.19, 0.2) mm.




HOWEVER, if you use any escape characters in your string, it fails because
strwrap removes all of them:



cat(\n\tHello.\n)


   Hello.

printtoscreen(\n\tHello.\n)

Hello.




Is there any way to prevent strwrap doing this? I haven't found any
options for this.

Thanks very much (and thanks to Greg Snow for the tip about strwrap before).

Toby Marthews


==
Le Mer 11 juin 2008 18:00, Greg Snow a écrit :

You could try passing your character string to the strwrap function

first,

then use cat on the result.

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




-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Toby Marthews Sent:

Wednesday, June 11, 2008 6:52 AM

To: r-help@r-project.org
Subject: [R] Word wrapping for character objects (WINDOWS R ONLY) Can

anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM
DOESN'T OCCUR ON LINUX **

I want to print a long character to screen:

getOption(width)

[1] 60

z=(1:20)/10#z is a vector of length between 20 and 30 (depending

on user options) containing lengths in mm (i.e. each element is 1-5
characters long)

str1=paste(The depths chosen are (,toString(z),) mm, and more text

...\n)

cat(str1)

The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,

1.1, 1.2, 1.3, 1.$



The problem is that on R for Windows the string is cropped by the

window size (hence the $). On R for Linux, this doesn't happen and the
text is word wrapped (the default for the shell, I guess):

cat(str1)

The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1,

1.1, 1.2, 1.3,

1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2 ) mm, and more text ...



I can't find any option for word wrapping in the cat command

(fill=TRUE has no effect). I also checked the menu Edit - GUI
preferences..., but there doesn't seem to be a Word Wrap option there
either.

How do I get word wrapping like this in Windows?
THANKS FOR ANY HELP!
Toby Marthews
Previous relevant posts:
- The post from 2006 about Screen Wrapping

(http://tolstoy.newcastle.edu.au/R/help/06/05/26673.html) which Brian
Ripley answered was about controlling how long vectors are cropped to the
screen. Unfortunately, the width option in options() does not
affect character objects, so I can't use that control here.

- I sent the same question to [EMAIL PROTECTED] in Oct 2007,

but noone there could help me with it.

- Try the following command on Windows R with a small window

(getOption(width)117) and a large window (getOption(width)117) and
you'll see you get extra nonexistent options in the menu:

a=c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec);menu(a)

I guess this is a related problem?

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

[R] ggplot2: How to remove legend component for geom_errorbar() ?

2008-06-16 Thread Carsten Jaeger

Dear list (probably Hadley),

I'm trying to do a plot like the following, composed of bars and error
bars:

df - data.frame(factor1=gl(2,5), factor2=gl(5,1), y=rnorm(10),
err=0.1) 
ggplot(df, aes(x=factor1, y=y, fill=factor2)) +
geom_bar(position=dodge, stat=identity) + 
geom_errorbar(aes(min=y-err, max=y+err), position=dodge,
width=0.1)

The plot obtained has a legend named factor2, giving as expected the
colours used for the five factor levels of factor2. So far so good.
However, the legend also contains solid horizontal lines, obviously
representing the error bars. While I do not immediately see the sense of
a legend symbol for error bars, is there any easy way to turn these off?

Thanks,
Carsten

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[R] unique-fy colume names

2008-06-16 Thread Daren Tan


 
I have a column containing duplicate entries and need to append numeric 
suffixes to make them unique. How ?
 
e.g., paste(id, c(1:10,1,5,10,10), sep=.)
 [1] id.1  id.2  id.3  id.4  id.5  id.6  id.7  id.8  id.9[10] 
id.10 id.1  id.5  id.10 id.10 
I hope to get
 [1] id.1  id.2  id.3  id.4  id.5  id.6  id.7  id.8  id.9[10] 
id.10 id.1.1  id.5.1  id.10.1 id.10.2
 
_
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and provide commented, minimal, self-contained, reproducible code.

[R] R on an ASUS eee PC, continued - installing packages

2008-06-16 Thread Millo Giovanni

Dear all, 

I just went through the process of installing R on an eeePC 900 running
Linux. As a Windows useR utterly ignorant about Linux, I'd never have
done it without reading your posts and the R Wiki, so first of all:
thank you! 
Next, taking up your thread from some weeks ago, I thought this could be
useful for somebody else too, so here's what I did:

1) I followed wolfgang's step-by-step guide at
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128565.html,
plus 
2) I installed the build dependencies as shown in R Wiki:

## quoting from
http://wiki.r-project.org/rwiki/doku.php?id=getting-started:installation
:debian#build_r ##
How much supporting software you need to install depends on how much
compilation you want to be able to do with R. If you just want to be
able to run R, you can get r-base-core and all the recommended packages
by doing: 

sudo apt-get install r-base

If you want to be able to build and install R packages (including those
from CRAN), you can get all the common header files, as well as
r-base-core by doing: 

sudo apt-get install r-base-dev

If you want to be able to build R from its source code, you can get
build dependencies for R (e.g., compilers, header files) by doing: 

sudo apt-get build-dep r-base
## end quote ##

This way I got all the necessary compilers, headers etc. automagically
installed as well, so that install.packages() works fine now: I
installed the whole Econometrics task view without problems. I still
haven't tried to install Rmetrics in order to save space on the little
machine, because I do not currently need it, but it should go the same
way: maybe this can solve John's problem?

I am aware that probably removing the original OS and installing Debian
as suggested by Dirk Eddelbuettel here
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128605.html would be a
better solution, but I am a complete Linux newbie, never ever tried it
before, so it's good both for me and for family use to retain the
original easy mode GUI while having R running on the same machine.
Full desktop mode (=standard KDE) is also easily enabled by adding
kicker and ksmserver (see http://wiki.eeeuser.com/howto:getkde, I did it
the manual way and it worked; just take heed that kicker and ksmserver
were not found in the '900's repository and I had to take them from the
'701's).

The only thing I wish I were able to do now is to have the graphics
windows defaults changed to a size fitting the small 9'' screen, as now
I have to reduce it and move it to the right by hand every time to
reproduce the results of 'windowstile in Windows. If anybody can
help...

HTH,
Giovanni

Giovanni Millo
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4, 
34131 Trieste (Italy)
tel. +39 040 671184 
fax  +39 040 671160 

# original message: 
From: John C Frain frainj 
Date: Mon, 28 Apr 2008 15:33:31 +0100

Thanks for the step by step guide. It installs a base R and some 
libraries very well. I tries to load some of the Rmetrics libraries, 
using install.packages() nu ran into problems. I used apt-get install 
to install make and gcc from the Xandros repository. gcc also included 
binutils, gcc-4.1 and libsspo. Is there an easy answer to the 
question what I need to install to get the install process working. 
The R Installation and Administration guide says that I just need the 
compilers and tools. Installing robustbase gives the following 
messages. (If it is as simple as R not being able to find the header 
files how do I tell R where to find them?). 


Best Regards 


John 

output in original message suppressed

-- 
John C Frain
Trinity College Dublin
Dublin 2
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:frainj at tcd.ie
mailto:frainj at gmail.com
 
Ai sensi del D.Lgs. 196/2003 si precisa che le informazi...{{dropped:13}}

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Re: [R] package under unix

2008-06-16 Thread Patrick Connolly

On Mon, 16-Jun-2008 at 09:43AM +0200, [EMAIL PROTECTED] wrote:


 I do not realy care about 5, 8 or 20, I use it when I devellop, I
 will remove it for the final package.  I just want the two windows
 to not be one on the top of the other. I do not care if one in on
 right, the other left, or one at the top of the screen, the other at
 the bottom...


 X11() does have xpos and ypos,

 I am using R 2.7 but I do not manage to deal with X11 and xpos :
 X11(xpos=0) give an error
 windows.options(xpos=0) ; X11() open a windows but it does not change the 
 place.

This is a bit of a stab in the dark since I don't use Windows and
can't guess what you've been doing, but I'll try.  Without knowing
what this windows.options is, it's more difficult to answer, but this
might do.  With x11, you could achieve what I think you're aiming at
with

x11(xpos = 1) # makes one on the left
x11(xpos = -1) # makes one on the right

Just how you make a generic way of keeping track of them, you might
have ideas..

HTH

-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~} Great minds discuss ideas
 _( Y )_Middle minds discuss events 
(:_~*~_:)Small minds discuss people  
 (_)-(_)   . Anon
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] R on an ASUS eee PC, continued - installing packages

Millo Giovanni wrote:

 The only thing I wish I were able to do now is to have the graphics
 windows defaults changed to a size fitting the small 9'' screen, as now
 I have to reduce it and move it to the right by hand every time to
 reproduce the results of 'windowstile in Windows. If anybody can
 help...
   
That'll be something like X11.options(width=4,height=4,pointsize=8) in
your ~/.Rprofile

-- 
   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] unique-fy colume names


See ?make.unique

On Mon, 16 Jun 2008, Daren Tan wrote:




I have a column containing duplicate entries and need to append numeric 
suffixes to make them unique. How ?

e.g., paste(id, c(1:10,1,5,10,10), sep=.)
[1] id.1  id.2  id.3  id.4  id.5  id.6  id.7  id.8  id.9[10] id.10 id.1  
id.5  id.10 id.10
I hope to get
[1] id.1  id.2  id.3  id.4  id.5  id.6  id.7  id.8  id.9[10] id.10 id.1.1  
id.5.1  id.10.1 id.10.2

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



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

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

Re: [R] R on an ASUS eee PC, continued - installing packages

2008-06-16 Thread Tom La Bone

I am running R in the konsole of the Kate editor on an eeePC 900 (standard
Xandros OS). When a plot is generated it is initially too large to fit in
the screen, but if I click on the plot it automatically resizes to fit
properly. I have no idea if this is a default behavior.

Tom

Millo Giovanni wrote:

Dear all,

I just went through the process of installing R on an eeePC 900 running
Linux. As a Windows useR utterly ignorant about Linux, I'd never have
done it without reading your posts and the R Wiki, so first of all:
thank you!
Next, taking up your thread from some weeks ago, I thought this could be
useful for somebody else too, so here's what I did:

1) I followed wolfgang's step-by-step guide at
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128565.html,
plus
2) I installed the build dependencies as shown in R Wiki:

## quoting from
http://wiki.r-project.org/rwiki/doku.php?id=getting-started:installation
:debian#build_r ##
How much supporting software you need to install depends on how much
compilation you want to be able to do with R. If you just want to be
able to run R, you can get r-base-core and all the recommended packages
by doing:

sudo apt-get install r-base

If you want to be able to build and install R packages (including those
from CRAN), you can get all the common header files, as well as
r-base-core by doing:

sudo apt-get install r-base-dev

If you want to be able to build R from its source code, you can get
build dependencies for R (e.g., compilers, header files) by doing:

sudo apt-get build-dep r-base
## end quote ##

This way I got all the necessary compilers, headers etc. automagically
installed as well, so that install.packages() works fine now: I
installed the whole Econometrics task view without problems. I still
haven't tried to install Rmetrics in order to save space on the little
machine, because I do not currently need it, but it should go the same
way: maybe this can solve John's problem?

I am aware that probably removing the original OS and installing Debian
as suggested by Dirk Eddelbuettel here
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128605.html would be a
better solution, but I am a complete Linux newbie, never ever tried it
before, so it's good both for me and for family use to retain the
original easy mode GUI while having R running on the same machine.
Full desktop mode (=standard KDE) is also easily enabled by adding
kicker and ksmserver (see http://wiki.eeeuser.com/howto:getkde, I did it
the manual way and it worked; just take heed that kicker and ksmserver
were not found in the '900's repository and I had to take them from the
'701's).

The only thing I wish I were able to do now is to have the graphics
windows defaults changed to a size fitting the small 9'' screen, as now
I have to reduce it and move it to the right by hand every time to
reproduce the results of 'windowstile in Windows. If anybody can
help...

HTH,
Giovanni

Giovanni Millo
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4,
34131 Trieste (Italy)
tel. +39 040 671184
fax +39 040 671160

# original message:
From: John C Frain frainj
Date: Mon, 28 Apr 2008 15:33:31 +0100

Thanks for the step by step guide. It installs a base R and some
libraries very well. I tries to load some of the Rmetrics libraries,
using install.packages() nu ran into problems. I used apt-get install
to install make and gcc from the Xandros repository. gcc also included
binutils, gcc-4.1 and libsspo. Is there an easy answer to the
question what I need to install to get the install process working.
The R Installation and Administration guide says that I just need the
compilers and tools. Installing robustbase gives the following
messages. (If it is as simple as R not being able to find the header
files how do I tell R where to find them?).

Best Regards

John

output in original message suppressed

--
John C Frain
Trinity College Dublin
Dublin 2
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:frainj at tcd.ie
mailto:frainj at gmail.com

Ai sensi del D.Lgs. 196/2003 si precisa che le informazi...{{dropped:13}}

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Re: [R] R on an ASUS eee PC, continued - installing packages


On Mon, 16 Jun 2008, Peter Dalgaard wrote:


Millo Giovanni wrote:


The only thing I wish I were able to do now is to have the graphics
windows defaults changed to a size fitting the small 9'' screen, as now
I have to reduce it and move it to the right by hand every time to
reproduce the results of 'windowstile in Windows. If anybody can
help...


That'll be something like X11.options(width=4,height=4,pointsize=8) in
your ~/.Rprofile


You need grDevices::X11.options, and it is better to use a load hook like

setHook(packageEvent(grDevices, onLoad),
function(...) grDevices::X11.options(width=4,height=4,pointsize=8)
)

There is another way to set the geometry, to set X11 resources as 
described in ?X11.  I have in my ~/.Xresources


R_x11*geometry: 700x700-0+0

(for a 22 90dpi screen, not a 9 one)


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

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Re: [R] package under unix

2008-06-16 Thread cgenolin


x11(xpos = 1) # makes one on the left
x11(xpos = -1) # makes one on the right


Thanks you very much, I'll try that.
Is there any difference between x11 and X11?

Christophe



Patrick Connolly [EMAIL PROTECTED] a Ã©critÂ :

On Mon, 16-Jun-2008 at 09:43AM +0200, [EMAIL PROTECTED] wrote:



I do not realy care about 5, 8 or 20, I use it when I devellop, I
will remove it for the final package.  I just want the two windows
to not be one on the top of the other. I do not care if one in on
right, the other left, or one at the top of the screen, the other at
the bottom...





X11() does have xpos and ypos,


I am using R 2.7 but I do not manage to deal with X11 and xpos :
X11(xpos=0) give an error
windows.options(xpos=0) ; X11() open a windows but it does not change the
place.


This is a bit of a stab in the dark since I don't use Windows and
can't guess what you've been doing, but I'll try.  Without knowing
what this windows.options is, it's more difficult to answer, but this
might do.  With x11, you could achieve what I think you're aiming at
with

x11(xpos = 1) # makes one on the left
x11(xpos = -1) # makes one on the right

Just how you make a generic way of keeping track of them, you might
have ideas..

HTH

--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
  ___Patrick Connolly
{~._.~}  Great minds discuss ideas
_( Y )_ Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_). Anon

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[R] Very puzzled why swapping the columns doesn't swap the column names

2008-06-16 Thread Daren Tan


How can I swap the column names at the same time ?
 
 m - cbind(x=1:3, y=2:4, z=3:5) m x y z[1,] 1 2 3[2,] 2 3 4[3,] 3 4 5
 m[,c(1,2)] - m[,c(2,1)] m x y z[1,] 2 1 3[2,] 3 2 4[3,] 4 3 5
_


[[alternative HTML version deleted]]

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Re: [R] Very puzzled why swapping the columns doesn't swap the column names

2008-06-16 Thread jim holtman

What you are doing with the assignment

m[,c(1,2)] - m[,c(2,1)]

is only changing the values in the matrix; it knows nothing about the
names.  If you want the names of the columns switched, then you have
to do it:

 m - cbind(x=1:3, y=2:4, z=3:5)
 m
 x y z
[1,] 1 2 3
[2,] 2 3 4
[3,] 3 4 5
 m[,c(1,2)] - m[,c(2,1)]
 m
 x y z
[1,] 2 1 3
[2,] 3 2 4
[3,] 4 3 5
 colnames(m)[c(1,2)] - colnames(m)[c(2,1)]
 m
 y x z
[1,] 2 1 3
[2,] 3 2 4
[3,] 4 3 5



On Mon, Jun 16, 2008 at 8:10 AM, Daren Tan [EMAIL PROTECTED] wrote:

 How can I swap the column names at the same time ?

 m - cbind(x=1:3, y=2:4, z=3:5) m x y z[1,] 1 2 3[2,] 2 3 4[3,] 3 4 5
 m[,c(1,2)] - m[,c(2,1)] m x y z[1,] 2 1 3[2,] 3 2 4[3,] 4 3 5
 _


[[alternative HTML version deleted]]

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+1 513 646 9390

What is the problem you are trying to solve?

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Re: [R] Very puzzled why swapping the columns doesn't swap the column names

2008-06-16 Thread Wacek Kusnierczyk

Daren Tan wrote:
 How can I swap the column names at the same time ?
  
   
 m - cbind(x=1:3, y=2:4, z=3:5) m x y z[1,] 1 2 3[2,] 2 3 4[3,] 3 4 5
 m[,c(1,2)] - m[,c(2,1)] m x y z[1,] 2 1 3[2,] 3 2 4[3,] 4 3 5
 
what you do here is to columns 1 and 2 put what is in columns 2 and 1,
respectively. 
what you want is m is columns 2, 1, and 3 from m:

m = m[,c(2,1,3)]

instead of swapping column data and headers separately (you could do
that as well), swap columns.

vQ

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Re: [R] R as chart engine in web-service

2008-06-16 Thread Daniel Cegielka

 Can you put some criteria on 'better'/'faster'.  What throughput are
 you expecting?  How many requests per second will you have to handle?


It's very difficult to tell how many request per second I will have to
handle. If you have normal site it will be max. 2-5 tps (on chart website)
but if you have real time quotes it will be about 300 tps. In RT quotes I
intend use java applet to make RT charts with independent connection.

I thing about service like BigChart.com - put some symbol (GOOG, DJIA etc).

http://bigcharts.marketwatch.com/quickchart/quickchart.asp?symb=djiasid=164
3

and the chart like:

http://www.marketwatch.com/tools/quotes/intchart.asp?submitted=trueintflavo
r=advancedsymb=DJIAorigurl=%2Ftools%2Fquotes%2Fintchart.asptime=8freq=1
startdate=enddate=hiddenTrue=comp=Enter+Symbol(s)%3Acompidx=a~0comp
ind=a~0uf=7168ma=1maval=50lf=1lf2=4lf3=0type=2size=2optstyle=10
13

In this kind of web services its important how faster the chart engine is in
general. If there will be to many request per second I can add more machines
and use load balancer (haproxy or pound).

 What size/type machine will you be running on?

Probably 1-2 Quad Core Intel Xeon 2x6MB Cache, 2.0GHz per machine with 8GB
(or 16GB) RAM. Zope 3 as web server on Gentoo Linux (x86). Memory will be
mounted as a local disk (tmpfs) and python and R binaries will by copy into
RAM. Python and R are compiled with ICC compiler.


 Is data being accessed from a database which
 may influence the timing?

R will connect with Tree Data Server (database):
http://tree.sourceforge.net/
Data can be keep in memcached: http://www.danga.com/memcached/

So I think there is very small influence on the timing.



And I think about quantmod as a R chart package:
http://www.quantmod.com/examples/charting/


The main problem with R is that, R uses x11 server to render images and this
way is very slow and I thing that faster would be the Cairo (or GDD) R
package but I haven't tested this yet.

R engine is very universal and this is the strong of R choice in this kind
of web service... then begin write new (faster?) c/c++ plugin. What you
think?

daniel cegielka



-Original Message-
From: jim holtman [mailto:[EMAIL PROTECTED] 
Sent: Monday, June 16, 2008 1:55 AM
To: Daniel Cegielka
Cc: r-help@r-project.org
Subject: Re: [R] R as chart engine in web-service

Can you put some criteria on 'better'/'faster'.  What throughput are
you expecting?  How many requests per second will you have to handle?
What is the timing of your current method?  What size/type machine
will you be running on?  Is data being accessed from a database which
may influence the timing?  So you need to give some specificity of
what your requirements are.

You can probably get faster (but maybe not better) with a plug-in, but
I would assume that it would also be more effort.

On Sun, Jun 15, 2008 at 6:39 PM, Daniel Cegielka [EMAIL PROTECTED] wrote:
 Hi R-users

 I think about some web service with stock charts and I plan use python as
 web framework with R as mathematical and chart engine. I use rpy to
connect
 R with python. It works good.

 Is it good idea to use R as chart engine? Maybe it's better (faster) to
use
 c/c++ plugin?

 Daniel

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 R-help@r-project.org mailing list
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 PLEASE do read the posting guide
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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Re: [R] R on an ASUS eee PC, continued - installing packages

Prof Brian Ripley wrote:
 On Mon, 16 Jun 2008, Peter Dalgaard wrote:

 Millo Giovanni wrote:

 The only thing I wish I were able to do now is to have the graphics
 windows defaults changed to a size fitting the small 9'' screen, as now
 I have to reduce it and move it to the right by hand every time to
 reproduce the results of 'windowstile in Windows. If anybody can
 help...

 That'll be something like X11.options(width=4,height=4,pointsize=8) in
 your ~/.Rprofile

 You need grDevices::X11.options, and it is better to use a load hook like

 setHook(packageEvent(grDevices, onLoad),
 function(...)
 grDevices::X11.options(width=4,height=4,pointsize=8)
 )

Thanks, Brian.

Obviously, I haven't used this for a while and overlooked the example in
the Examples section of ?Startup. 

I do wonder if there are ways to make this less easily overlooked,
though. Pointers in any or all of ?x11, ?X11.options, and ?Devices could
help, but maybe we need it in an introductory document like R-intro,
along with the discussion of Rprofile files in Customizing the
environment.

 There is another way to set the geometry, to set X11 resources as
 described in ?X11.  I have in my ~/.Xresources

 R_x11*geometry: 700x700-0+0

 (for a 22 90dpi screen, not a 9 one)

That doesn't help with the pointsize, does it? (Quite crucial on low-res
screens, even at 1024x768, 12.3 I find that the default pointsize=12 is
a bit much.)

-p

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

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[R] sizing non-vector point shapes in ggplot2

2008-06-16 Thread mfrumin


Dear all,

With normal plotting, one can size a set of points in a plot using a vector
argument to cex in the points() function.  This works whether you are using
one of the standard R symbols (i.e. 19+) or some ascii symbol, such as '/'

eg: 
plot(1:10, 1:10, type='n');
points(1:10, 1:10, cex = 1:10, pch = '/')

Trying to make the transition to ggplot2, I find that the aesthetic size
mapping does not apply if i do geom_point(..., shape='/') -- the points show
up looking like '/' but they are not sized.  Is there anything to do about
this?  

eg: 
ggplot(data = data.frame(x = 1:10, y = 1:10, size = 1:10), aes(x = x, y =
y)) + geom_point(mapping = aes(size = size), shape = '/')

the plot that I'm making really needs a vertically oriented mark, not a
round-ish point/square/triangle that takes up a lot of area. am I totally
out of luck?  

thanks,
mike
-- 
View this message in context: 
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[R] contrasts using adonis function

2008-06-16 Thread Luis San José García


Hi,

Somebody knows how to make contrasts if i'm using the function adonis?

Thanks.

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Re: [R] R vs SAS and HLM on multilevel analysis- basic question

2008-06-16 Thread Douglas Bates

We may want to move this discussion to the R-SIG-Mixed-Models list,
which I have cc:'d on this reply.

On Sun, Jun 15, 2008 at 6:16 PM, eugen pircalabelu
[EMAIL PROTECTED] wrote:
 Hi R users!

 I am trying to learn some multilevel analysis, but unfortunately i am now 
 very confused. The reason: 
 http://www.ats.ucla.edu/stat/hlm/seminars/hlm_mlm/mlm_hlm_seminar.htm
 http://www.ats.ucla.edu/stat/sas/seminars/sas_mlm/mlm_sas_seminar.htm

 and
 MlmSoftRev. pdf from mlmRev package.

 From what i see, the first two links seem to declare the level one variable 
 as a random part (i don't know sas synthax, but i think i am right ) while 
 Mr. Bates' pdf  says that a grouping variable is the random part of the 
 model, though both models, use roughly the same type of information, some 
 characteristic of the school, along with individual characteristics in 
 explaining individual achivement.

I'm not exactly sure what you are asking.  If you are saying that the
terminology and notation can be confusing, I certainly agree.  I think
those who developed HLM and MLWin have done a tremendous service to
their users in providing them with sophisticated tools for modeling
data.  However, the way that they structure the model is really only
appropriate for models with nested random effects and, to my mind,
introduces many unnecessary and restrictive ways of thinking of the
data and the model.

 Am i mistaken somehow? If not, could they both be valid models (i presume) 
 but each showing something else, in terms of connections between this 
 variables?

As I said, I don't quite understand what you are asking and, rather
than formulate an answer to the wrong question, I'll ask if you can
rephrase your question and perhaps be more explicit about an example.
In particular, you made reference to a school.  Are you referring to
a particular example?

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[R] why does these warning mean?

2008-06-16 Thread Peng Jiang


Hi ,

 I don't know whether it is a Mac specific problem. I hope it does  
not bother you.


When I start R each time I got the following warning , I reinstall for  
many times the problem still exists.


 R(4053,0xa0619fa0) malloc: *** error for object 0x15630350: double  
free

*** set a breakpoint in malloc_error_break to debug
R(4053,0xa0619fa0) malloc: *** error for object 0x15630f10: double free
*** set a breakpoint in malloc_error_break to debug
R(4053,0xa0619fa0) malloc: *** error for object 0x15630350: double free
*** set a breakpoint in malloc_error_break to debug
R(4053,0xa0619fa0) malloc: *** error for object 0x15630f60: double free
*** set a breakpoint in malloc_error_break to debug


My sessionInfo() is
  sessionInfo()
R version 2.7.0 (2008-04-22)
i386-apple-darwin8.10.1

locale:
zh_CN.UTF-8/zh_CN.UTF-8/C/C/zh_CN.UTF-8/zh_CN.UTF-8

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



--
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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Re: [R] Superimposing Line over Histogram in Density Plot

2008-06-16 Thread David Arnold


Look here:

http://msenux.redwoods.edu/mathdept/R/CentralLimit.php

D.

On Jun 15, 2008, at 11:25 PM, Gundala Viswanath wrote:


Hi,

Currently I have a density plot generated with this
snippet.

Is there a way I can add a line curve on top of it?
I mean in one figure


__BEGIN__
   myhist - hist(x
col=blue,
main = Density Plot,
xlab = Exp Level,
)
__END__


- Gundala Viswanath
Jakarta - Indonesia

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[R] R vs SAS and HLM on multilevel analysis- basic question

2008-06-16 Thread Ista Zahn

On Jun 16, 2008, at 6:00 AM, [EMAIL PROTECTED] wrote:

 From: eugen pircalabelu [EMAIL PROTECTED]
 Date: June 15, 2008 7:16:09 PM EDT
 To: R-help [EMAIL PROTECTED]
 Subject: [R] R vs SAS and HLM on multilevel analysis- basic question

 Hi R users!

 I am trying to learn some multilevel analysis, but unfortunately i  
 am now very confused. The 
 reason:http://www.ats.ucla.edu/stat/hlm/seminars/hlm_mlm/mlm_hlm_seminar.htm
 http://www.ats.ucla.edu/stat/sas/seminars/sas_mlm/mlm_sas_seminar.htm

 and
 MlmSoftRev. pdf from mlmRev package.

 From what i see, the first two links seem to declare the level one  
 variable as a random part (i don't know sas synthax, but i think i  
 am right ) while Mr. Bates' pdf  says that a grouping variable is  
 the random part of the model, though both models, use roughly the  
 same type of information, some characteristic of the school, along  
 with individual characteristics in explaining individual achivement.

 Am i mistaken somehow? If not, could they both be valid models (i  
 presume) but each showing something else, in terms of connections  
 between this variables?

Yes, I believe you are mistaken, but I have only a rudimentary  
understanding of mixed effects modeling so I won't comment further  
except to say that I'm pretty sure both sources you listed above are  
talking about the same kinds of models.

I don't know SAS at all, but I've used both HLM and R to run mixed  
effects models.  Part of the confusion may be due to the different  
ways that R and HLM expect the input data to be formatted. To run a 2- 
level model in HLM you need two separate files corresponding to the  
two levels. When you input the data into HLM you specify the grouping  
factor that links the two files together. In R you can use a single  
data file and specify the grouping factor in the model syntax.

I posted a similar question several months ago--see 
http://tolstoy.newcastle.edu.au/R/e4/help/08/02/3600.html
I also received an off-list reply pointing me to 
http://www.ats.ucla.edu/stat/examples/alda.htm
which gives both HLM and R syntax for the same models. One of the  
examples on the website is as follows:

HLM
Level 1 Model
COG = beta_0 + beta_1(TIME) + r
Level 2 model:
beta_0 = gamma_00 + gamma_01(PROGRAM) + mu_0
beta_1 = gamma_10 + gamma_11(PROGRAM) + mu_1
#

Notice that no grouping factor is specified because this is done  
separately, in the data input stage. The corresponding R syntax is
#
R syntax:
model1- lmer(cog~time*program + (time | id), data=dataframe)
#

Notice that there is an extra term in the R syntax (id) that is not in  
the HLM syntax. Again, this is because you have to tell HLM what the  
grouping variable is when you input the data, while in R you specify  
the grouping variable in the model. I'm guessing this may be the  
source of some of your confusion.

 Thank you and sorry for taking up your time.

No problem, HTH.
-Ista
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Re: [R] contrasts using adonis function

2008-06-16 Thread Gavin Simpson

On Mon, 2008-06-16 at 14:38 +0200, Luis San José García wrote:
 Hi,
 
 Somebody knows how to make contrasts if i'm using the function adonis?

What is wrong with the defaults? From ?adonis:

Usage:

 adonis(formula, data, permutations = 5, method = bray,
strata = NULL, contr.unordered = contr.sum,
contr.ordered = contr.poly, ...)

So look at ?contr.sum and ?contr.poly for what constrasts adonis uses by
default. If you want something different, then supply contr.helmert or
contr.treatment, contr.sum, or contr.poly to the relevant argument
in your adonis call.

If you want your own special contrasts, then I presume they make sense
in the context of the analysis? But for that you'll need to write your
own function to calculate the constrasts you want, along the lines of
contr.sum() or one of the other contrast functions.

HTH

G

-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Dr. Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
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[R] Generating Reports from R.

2008-06-16 Thread David Keegan


Hi,

I have a non-interactive R script that currently produces
various graphs in png or pdf format. I need to program the
script to combine the graphs with various pages of textual
information, including some in tabular format, into an
output report in pdf or html format.

What is the recommended way of doing this?

Regards,
David Keegan.
--

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Re: [R] Generating Reports from R.

Try using Sweave

On Mon, Jun 16, 2008 at 10:37 AM, David Keegan [EMAIL PROTECTED]
wrote:


 Hi,

 I have a non-interactive R script that currently produces
 various graphs in png or pdf format. I need to program the
 script to combine the graphs with various pages of textual
 information, including some in tabular format, into an
 output report in pdf or html format.

 What is the recommended way of doing this?

 Regards,
 David Keegan.
 --

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Re: [R] Delete Block of Rows

2008-06-16 Thread nmarti


Thanks for all the replies.
My data frame is actually 4000 rows and 10 columns.

This line,
df - subset( df, !( rownames(df) %in% 1:25 ) )  Did not work for me.  Is
there a certain library I need to load for this to work (I'm assuming there
isn't)?  I didn't get an error message, warning, or anything; it just didn't
work.

These two lines did work,
df = df[ -c(1:25), ]Or,
df = df[ 26:200, ]  I'm not sure why they didn't work the first time I
tried them.

Once again, thanks for the help!



milton ruser wrote:
 
 Hi there,
 
 I don´t know if you are trying to solve the delete job or to test how
 functions work.
 
 If you really want to delete lines from a data.frame, try something like
 this.
 
 
 rowcount-1:100
 x-runif(100)
 y-runif(100)
 
 df-data.frame(cbind(rowcount,x,y))
 
 df.subset-subset(df, !(rownames(df) %in% 1:25))
 
  #! (a condition) is the negation of the condition
 
  #   so the output will be those lines that are not on interval 1:25
 
 
 
 By the way, it is not a good idea to use data as a input argument on
 a function because data is a pre-defined function.
 Kind regards,
 
 miltinho
 
 
 
 On 6/15/08, nmarti [EMAIL PROTECTED] wrote:


 I am trying to delete a section of rows from a data frame (based on no
 condition).  Lets say my data frame has 200 rows and I want to delete
 rows
 1
 through 25.  How would I do this?

 I know x[ -1, ] deletes the first row (or any desired row).  I tried
 using
 different variations of this, like x[ -c(1:25), ] but that didn't work. 
 I
 also tried writting a few functions, for example:

 deleteRows - function( data, s, e, ) {
 for( i in s:e )
 data[ -i, ]
 }
 deleteRows( ds, 1, 25 )

 But that didn't work either, it only deleted row 25 (i'm new to writing
 functions).
 Any thoughts on how to solve my problem would be appreciated.
 --
 View this message in context:
 http://www.nabble.com/Delete-Block-of-Rows-tp17849775p17849775.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] using spec.pgram

2008-06-16 Thread Anthony Mathelier

Perhaps I'm applying spec.pgram wrong as you said. I will explain what I
want, so you can tell me why I'm wrong and perhaps what I have to do to do
it well.
I have some points in a 1-D space and I want to know if they are spaced at a
certain periodic distance. So, I computed all the distances between points
in my space. Then, I would like to know if a certain distance (period), or
multiples of a certain distance, is preferred to space my data. I made a
histogram of the distances and apply the spec.pgram function to know the
frequence (so the period) which is the most important to space the original
data.
But, when I have to sets of data (without necessarily the same number of
observation in each set), I want to compare the importance of the period
given by spec.pgram between the sets. Could I normalize the amplitude of the
peaks given by spec.pgram?
So, am I wrong to apply this methodology to exhibit a periodic distance
between my data? If, true, what could you recommend me to do this?
Thanks in advance for your answers.
Best regards,

Anthony

On Tue, Jun 10, 2008 at 6:13 PM, stephen sefick [EMAIL PROTECTED] wrote:

 I from a first thought I would say that you are apply this wrong!  The
 fourier transform convolves a function (cos(x)+isin(x)  (this may not be the
 exact formula but I don't have my books near)) to the data and then
 integrates over -1/2 to 1/2 takes the modulus and plots this- the
 periodogram.  The reason you preform a fourier transform is to look at
 recurring frequencies in the data, which are in the time domain.  The
 fourier transform converts the time series into the frequency domain and
 viola you have a peak into the hidden/recurring parts of your signal.  From
 your explaination your are applying this technique wrong-  look at schumway,
 MASS4, et al. books to get a handle on how this technique is used.  If you
 are to apply a time series analysis please use it on a time series.  Maybe
 your logic is not flawed but I don't see how a histogram with its associated
 binning is a better candidate for time series analysis than the original
 time series if at all.
 good luck

 Stephen

 On Tue, Jun 10, 2008 at 8:49 AM, Matthieu Stigler 
 [EMAIL PROTECTED] wrote:

 Hello

 I don't know exactly what you want to do but:

 -why do you use in your example h$counts and not h? Furthermore helpl file
 says it should be a time series, why then rather not your time series?

 -usually na.action will make the default action, which you can see by
 getOptions(na.action)

 -here in this function it is provided in the function values na.action =
 na.fail so it will just remove the NA in the time series

 -if you want to study a function, I advise you to copy it entirely, rename
 it and then just insert print(curiousobject...) in the function, this will
 allow you to let the function run and grasp the interessting objects, like:

 study-function (x, spans = NULL, kernel = NULL, taper = 0.1, pad = 0,
 fast = TRUE, demean = FALSE, detrend = TRUE, plot = TRUE,
   na.action = na.fail, ...)
 {
   series - deparse(substitute(x))
   x - na.action(as.ts(x))
   print(x)
   xfreq - frequency(x)
 ...}
 study(sunspots)

 -when you provide an example, instead of giving an external reference for
 the data, try to search a convenient internal data (accessed by data() ), so
 one will be able to reproduce your problems. Here you could use sunspots

 -to obtain the commented code... I don't know it...

 -good luck

 Matthieu





  Hi everyone,

 first of all, I would like to say that I am a newbie in R, so I apologize
 in
 advance if my questions seem to be too easy for you.

 Well, I'm looking for periodicity in histograms. I have histograms of
 certain phenomenons and I'm asking whether a periodicity exists in these
 data. So, I make a periodogram with the function spec.pgram. For
 instance,
 if I have a histogram h, I call spec.pgram by spec.pgram (h, log=no,
 taper=0.5). So, I have some peaks that appear and I would like to
 interpret
 them but I do not know how they are computed and so what a peak with a
 value
 of 1 represents in comparison with a peak of value 600 with another
 histogram.
 I looked at the source code of the function spec.pgram to better
 understand
 what is behind. But, when I apply the source code line by line, I've got
 a
 problem. For instance, I make:


 data = scan (file.txt)
 h = hist (data, breaks=max(data)/5000)


 #then I apply the first two lines of the spec.pgram function


 series - deparse(substitute(h$counts))
 x - na.action(as.ts(h$counts))
 x


 NULL
 I do not understand why when I apply the first two lines of the function
 I
 have x which is equal to NULL (which make a mistake in the following
 lines
 of the code) but if I apply the function directly with h$counts it gives
 me
 a result.
 So, if someone can explain to me what is the problem and/or how
 spec.pgram
 exactly computes the periodogram and how to interpret it with my data, I
 would be so grateful.
 And subsidiary

[R] Error in maximum likelihood estimation.

2008-06-16 Thread Dong-hyun Oh


Dear UseRs,

I wrote the following function to use MLE.

-
mlog - function(theta, nx = 1, nz = 1, dt){
  beta - matrix(theta[1:(nx+1)], ncol = 1)
  delta - matrix(theta[(nx+2):(nx+nz+1)], ncol = 1)
  sigma2 - theta[nx+nz+2]
  gamma - theta[nx+nz+3]
  y - as.matrix(dt[, 1], ncol = 1)
  x - as.matrix(data.frame(1, as.matrix(dt[, 2:(nx+1)], ncol = 2)))
  z - as.matrix(dt[, (nx+2):(nx+nz+1)], ncol = nz)

  d - z %*% delta / (gamma * sigma2)^.5
  mustar - (1-gamma) * z %*% delta - gamma * ( y - x %*% beta)
  sigmastar - (gamma * (1-gamma) * sigma2)^.5
  dstar - mustar / sigmastar

  loglik - (-0.5 * nrow(x) *(log(2*pi) + log(sigma2))
 -0.5 * sum(( y - x %*% beta + z %*% delta)^2/sigma2)
 -sum(log(pnorm(d))) + sum(log(pnorm(dstar
  return(-loglik)
}
---

Loglikelihood function is from page 21of Battese and Coelli (1993).  
(You can download this article at http://www.une.edu.au/economics/publications/econometrics/emwp69.PDF 
 )


To test the above function with an artificial data set, I created the  
following data.frame.


---
x1 - abs(rnorm(100))*100
x2 - abs(rnorm(100))*10
z1 - abs(rnorm(100))*5
z2 - abs(rnorm(100))*7
y - abs(0.3 + 0.3* log(x1) + 0.7* log(x2))
dat - data.frame(log(y), log(x1), log(x2), z1, z2)


The starting value I set up is as follows:
---
theta.start - c(0.09, 0.008, 0.008, 0.023, 0.0008, 0.008, 0.008)
--

Applying nlm() function to the above function with the starting values  
gives the following error message.



out - nlm(mlog, theta.start, nx = 2 , nz = 2, dt = dat, print.level =  
2, hessian\

 = T, iterlim = 500)

iteration = 0
Step:
[1] 0 0 0 0 0 0 0
Parameter:
[1] 0.0900 0.0080 0.0080 0.0230 0.0008 0.0080 0.0080
Function Value
[1] 6116.2
Gradient:
[1]  -10704.8  -46465.7  -22536.4   47168.2   54542.9 -776512.5  
579.9


Error in nlm(mlog, theta.start, nx = 2, nz = 2, dt = dat, print.level  
= 2,  :

  (converted from warning) NA/Inf replaced by maximum positive value



My questions are
1. Is my loglikelihood function set up appropriately?
2. How to set up starting values efficiently? I read a thread shown at https://stat.ethz.ch/pipermail/r-help/2005-September/079617.html 
 , but I cannot figure out how to set up starting values with  
expand.grid() of parameters I used (beta, delta, gamma, sigma2).


Thank you in advance.

Sincerely,
Dong-hyun Oh

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[R] Linear Regression ?

2008-06-16 Thread kayj


Hi All,

I have a data with 15 columns. the first 10 column are dependent(y’s)
variables and the following 5  an columns are the independent variables(x’s)
( MY DATA HAS A HEADER WITH THE NAME OF THE VAIABLES). I need to apply
linear regression y=a+bx where I need x to be one of the independent
variable and y to be one of the dependent variable, but for each x I have to
repeat the model for all 5 independent variable  one at a time .

Also each x is a categorical variable with 4 categories so I have to
introduce  three variables for each x 

New Variable 1  New Variable 2  New Variable 3
Category 1  0   0   0
Category 2  1   0   0
Category 3  0   1   0
Category 4  0   0   1


I want R to print out the summary and the analysis of variance for each
model.

How can this done in R? I want to use some loop that loops over the
independent variables for each x ?Also , I am not sure how to deal with the
problem of each x being a categorical variable and I have to create these
new variables for each x? 

I hope that someone can help. I really Appreciate it.

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Re: [R] ggplot2: How to remove legend component for geom_errorbar() ?

2008-06-16 Thread hadley wickham

Hi Carsten,

In my eagerness to get every component of the legend working
automatically, I've completed neglected a way to turn bits off of you
don't want them (the philosophy is that every geom that uses an
aesthetic should appear in the legend in some way).  The best thing I
can suggest at the moment is:

GeomErrorbar$guide_geom - function(.) point

to use points for the errorbar legend, which you won't see because
they'll overlap with the points from the points geom.  In the next
version, you'll be able to do

GeomLine$guide_geom - function(.) blank

Or I might come up with a better way of controlling the appearance.

Regards,

Hadley

On Mon, Jun 16, 2008 at 2:55 AM, Carsten Jaeger [EMAIL PROTECTED] wrote:
 Dear list (probably Hadley),

 I'm trying to do a plot like the following, composed of bars and error
 bars:

 df - data.frame(factor1=gl(2,5), factor2=gl(5,1), y=rnorm(10),
 err=0.1)
 ggplot(df, aes(x=factor1, y=y, fill=factor2)) +
geom_bar(position=dodge, stat=identity) +
geom_errorbar(aes(min=y-err, max=y+err), position=dodge,
 width=0.1)

 The plot obtained has a legend named factor2, giving as expected the
 colours used for the five factor levels of factor2. So far so good.
 However, the legend also contains solid horizontal lines, obviously
 representing the error bars. While I do not immediately see the sense of
 a legend symbol for error bars, is there any easy way to turn these off?

 Thanks,
 Carsten

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Re: [R] Delete Block of Rows

2008-06-16 Thread milton ruser

Hi there,

The subset function is available on your Base R. So you need not to load any
package.
May be it is not working because I suppose that your rownames is a
sequential one (1...4000). Case it is not true, the command will not work.

Anyway, I think that those line that are working find is simple than you use
%in%... etc.

Good luck,

miltinho
Brazil


On 6/16/08, nmarti [EMAIL PROTECTED] wrote:


 Thanks for all the replies.
 My data frame is actually 4000 rows and 10 columns.

 This line,
 df - subset( df, !( rownames(df) %in% 1:25 ) )  Did not work for me.  Is
 there a certain library I need to load for this to work (I'm assuming there
 isn't)?  I didn't get an error message, warning, or anything; it just
 didn't
 work.

 These two lines did work,
 df = df[ -c(1:25), ]Or,
 df = df[ 26:200, ]  I'm not sure why they didn't work the first time I
 tried them.

 Once again, thanks for the help!



 milton ruser wrote:
 
  Hi there,
 
  I don´t know if you are trying to solve the delete job or to test how
  functions work.
 
  If you really want to delete lines from a data.frame, try something like
  this.
 
 
  rowcount-1:100
  x-runif(100)
  y-runif(100)
 
  df-data.frame(cbind(rowcount,x,y))
 
  df.subset-subset(df, !(rownames(df) %in% 1:25))
 
   #! (a condition) is the negation of the condition
 
   #   so the output will be those lines that are not on interval 1:25
 
 
 
  By the way, it is not a good idea to use data as a input argument on
  a function because data is a pre-defined function.
  Kind regards,
 
  miltinho
 
 
 
  On 6/15/08, nmarti [EMAIL PROTECTED] wrote:
 
 
  I am trying to delete a section of rows from a data frame (based on no
  condition).  Lets say my data frame has 200 rows and I want to delete
  rows
  1
  through 25.  How would I do this?
 
  I know x[ -1, ] deletes the first row (or any desired row).  I tried
  using
  different variations of this, like x[ -c(1:25), ] but that didn't work.
  I
  also tried writting a few functions, for example:
 
  deleteRows - function( data, s, e, ) {
  for( i in s:e )
  data[ -i, ]
  }
  deleteRows( ds, 1, 25 )
 
  But that didn't work either, it only deleted row 25 (i'm new to writing
  functions).
  Any thoughts on how to solve my problem would be appreciated.
  --
  View this message in context:
  http://www.nabble.com/Delete-Block-of-Rows-tp17849775p17849775.html
  Sent from the R help mailing list archive at Nabble.com.
 
  __
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  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 
[[alternative HTML version deleted]]
 
 
  __
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  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 

 --
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Re: [R] sizing non-vector point shapes in ggplot2

2008-06-16 Thread hadley wickham

On Mon, Jun 16, 2008 at 7:52 AM, mfrumin [EMAIL PROTECTED] wrote:

 Dear all,

 With normal plotting, one can size a set of points in a plot using a vector
 argument to cex in the points() function.  This works whether you are using
 one of the standard R symbols (i.e. 19+) or some ascii symbol, such as '/'

 eg:
 plot(1:10, 1:10, type='n');
 points(1:10, 1:10, cex = 1:10, pch = '/')

 Trying to make the transition to ggplot2, I find that the aesthetic size
 mapping does not apply if i do geom_point(..., shape='/') -- the points show
 up looking like '/' but they are not sized.  Is there anything to do about
 this?

 eg:
 ggplot(data = data.frame(x = 1:10, y = 1:10, size = 1:10), aes(x = x, y =
 y)) + geom_point(mapping = aes(size = size), shape = '/')

 the plot that I'm making really needs a vertically oriented mark, not a
 round-ish point/square/triangle that takes up a lot of area. am I totally
 out of luck?

Hmmm, I'd never noticed this feature of grid before.  To size the
points, I'm using the size argument of grid.points, which doesn't seem
to affect the size of character based plotting symbols (because the
are using the fontsize graphical parameter).  I've cc'd Paul on this
email so he can confirm whether this is a bug or by design.

Regardless, you can make ggplot size the symbols correctly by running
the following code:

GeomPoint$draw - function(., data, scales, coordinates, ...) {
with(coordinates$transform(data),
  ggname(.$my_name(), pointsGrob(x, y, size=unit(size, mm), pch=shape,
  gp=gpar(col=colour, fill = fill, fontsize = size * .pt)))
)
  }

Hadley

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Re: [R] R on an ASUS eee PC, continued - installing packages


On Mon, 16 Jun 2008, Peter Dalgaard wrote:


Prof Brian Ripley wrote:

On Mon, 16 Jun 2008, Peter Dalgaard wrote:


Millo Giovanni wrote:


The only thing I wish I were able to do now is to have the graphics
windows defaults changed to a size fitting the small 9'' screen, as now
I have to reduce it and move it to the right by hand every time to
reproduce the results of 'windowstile in Windows. If anybody can
help...


That'll be something like X11.options(width=4,height=4,pointsize=8) in
your ~/.Rprofile


You need grDevices::X11.options, and it is better to use a load hook like

setHook(packageEvent(grDevices, onLoad),
function(...)
grDevices::X11.options(width=4,height=4,pointsize=8)
)


Thanks, Brian.

Obviously, I haven't used this for a while and overlooked the example in
the Examples section of ?Startup.

I do wonder if there are ways to make this less easily overlooked,
though. Pointers in any or all of ?x11, ?X11.options, and ?Devices could
help, but maybe we need it in an introductory document like R-intro,
along with the discussion of Rprofile files in Customizing the
environment.


I've added examples in ?X11.options, ?quartz.options and ?windows.options 
for now.  (?X11 and ?X11.options are the same help page.)


R-intro is slightly odd in that it and not R-admin has startup info, and 
it is scattered across 3 sections -- but at least one of those references 
?Startup.



There is another way to set the geometry, to set X11 resources as
described in ?X11.  I have in my ~/.Xresources

R_x11*geometry: 700x700-0+0

(for a 22 90dpi screen, not a 9 one)


That doesn't help with the pointsize, does it? (Quite crucial on low-res
screens, even at 1024x768, 12.3 I find that the default pointsize=12 is
a bit much.)


No -- I was really pointing it out for completeness and 'the geometry' was 
inserted to try to make this clear.


One thing that is a bit odd is that we have ways (two each) to set the 
initial position of the default device on X11 and Windows, but AFAICS none 
on Mac OS X.


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

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Re: [R] stretching text vertically

2008-06-16 Thread Greg Snow

You can embed plots using the subplot function in the TeachingDemos package.

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



 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Alex Reynolds
 Sent: Friday, June 13, 2008 4:56 PM
 To: r-help@r-project.org
 Subject: [R] stretching text vertically

 I'd like to stretch a plotted character vertically, to create
 a sequence logo.

 Is there a parameter to allow stretching text() output
 vertically or squeeze horizontally?

 I know about Oliver Bembom's seqLogo library, but this
 generates a sequence logo plot using a separate bitmap
 device. I want to recreate the sequence logo *inside* an
 existing plot.

 Alternatively, is there a way to embed one plot inside another?

 I could use imagemagick outside R to 'montage' separate
 bitmaps, but then the sequence logo is going to be very
 difficult to align (base for
 base) with the plot I'm trying to join it to.

 Thanks for any tips,
 Alex

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[R] aggregate() function, strange behavior for augmented data

2008-06-16 Thread David Afshartous



All,

I'm re-running some analysis that has been augmented with additional data.
When I use the exact same code for the augmented data, the behavior of the
aggregate function is very strange, viz., one of the resulting variables is
now coded as a factor while it was coded as numeric for the original data.
Unfortunately, I cannot provide a reproducible code example since it only
seems to occur with this data.  I've checked and re-checked the of both the
original and augmented data but nothing appears inconsistent.  Any
suggestions much appreciated.  See below for specifics.

Cheers,
David









# original data
 dim(junk1)
[1] 96  3
 junk1[1,]
  Hour Drug Aldo
10P9
 junk1$Hour
 [1] 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
5 0 3
[39] 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0
3 5 0
[77] 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5   ### Not coded as a factor
 junk1.mean.time.drug = aggregate(junk1[3], junk1[c(1,2)], mean)
 junk1.mean.time.drug$Hour
[1] 0 3 5 0 3 5  ### not coded as a factor

# augmented data
 dim(junk1)
[1] 108   3
 junk1[1,]
  Hour Drug Aldo
10P9
 junk1$Hour
  [1] 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
 [51] 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0
3 5 0 3 5 0 3 5 0 3 5 0 3 5 0
[101] 3 5 0 3 5 0 3 5### not coded as a factor
 junk1.mean.time.drug = aggregate(junk1[3], junk1[c(1,2)], mean)
 junk1.mean.time.drug$Hour
[1] 0 3 5 0 3 5
Levels: 0 3 5## coded as a factor now!

## of course, I get recode it again but I'm curious as to why this is
## changing here

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Re: [R] Importing data with different delimters

2008-06-16 Thread Greg Snow

It looks like your original data may be tab seperated, if that is the case then 
just use read.delim or use sep='\t' in read.table or scan.

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



 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of David Arnold
 Sent: Friday, June 13, 2008 5:15 PM
 To: r-help@r-project.org
 Subject: [R] Importing data with different delimters

 All,

 I have a data file with 56 entries that looks like this:

 City State  JanTemp Lat Long
 Mobile, AL  44  31.288.5
 Montgomery, AL  38  32.986.8
 Phoenix, AZ 35  33.6112.5
 Little Rock, AR 31  35.492.8
 Los Angeles, CA 47  34.3118.7
 San Francisco, CA   42  38.4123.0

 I would like to read this data into a dataframe. Is it
 possible to do without editing the datafile?

 D.

 __
 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
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Re: [R] using spec.pgram

2008-06-16 Thread stephen sefick

To get some sort of frequency which in your case seem to be cycles per
distance?  Is a valid use of a fourier transform as long as it is a distance
that is measured in a way that would be analogous to a time series-  In
other words if the distance proceeds from an origin in one direction-
geophysicists do this often with the realization of an earthquake picked up
by sensors that are a distance away from the origin of the epicenter, but
they are looking for coherencies in the signal from one place to the next in
the frequency domain seperated by distance- this is called beam forming-
They use the raw signal-  by binning (making a histogram) the data you are
loosing the signal-  you are looking at frequency of occurance of certain
values not for the underlying periodicities of the data (in time or
space).   You are fitting cos and isin function to you data to see if there
is periodicity-  the power is the integration of the convolution of this sin
and cosine function with your data-  It seems to me meaningless to preform
this convolution agianst something that is not a signal (the histogram).  If
you want to use a frequency domain technique you have to have a frequency to
investigate-  a histogram does not have this-  I is a frequency of occurance
by bin size which is NOT what you want (your would have cycles/binlength
that doesn't make any sense to me) to do this analysis on-  You want a
signal-  dissolved oxygen curve, sunspot record, etc. through time, or
distance as stated above- you are looking for the frequency of a waveform-
Anyway,  I may be misunderstanding- supply some code and explain the data
otherwise this line of though- in my limited expertise- is a dead end,  but
agian I still don't know what it is that you are, exactly, trying to do- and
what your dataset constits.  I hope these ruminations help

I recommend doing this analysis on the raw data- It doesn't matter that you
don't have the same amount of data points- as long as both sets of data have
circa ten times the length of (cycles/distance) what you want to detect-  If
things in your case are spaced by one meter then the lowest cycle
perdistance that you can reliably detect if 0.5 meters,  this is all
speculation because you don't have a problem with reproducible code, and we
have no idea what you are measuring or what your data looks like-  without
this information there is no way that I can say one way or the other that
you approach (suggested non-histogram) would be right or wrong.

Stephen

On Mon, Jun 16, 2008 at 9:33 AM, Anthony Mathelier 
[EMAIL PROTECTED] wrote:

 Perhaps I'm applying spec.pgram wrong as you said. I will explain what I
 want, so you can tell me why I'm wrong and perhaps what I have to do to do
 it well.
 I have some points in a 1-D space and I want to know if they are spaced at
 a certain periodic distance. So, I computed all the distances between points
 in my space. Then, I would like to know if a certain distance (period), or
 multiples of a certain distance, is preferred to space my data. I made a
 histogram of the distances and apply the spec.pgram function to know the
 frequence (so the period) which is the most important to space the original
 data.
 But, when I have to sets of data (without necessarily the same number of
 observation in each set), I want to compare the importance of the period
 given by spec.pgram between the sets. Could I normalize the amplitude of the
 peaks given by spec.pgram?
 So, am I wrong to apply this methodology to exhibit a periodic distance
 between my data? If, true, what could you recommend me to do this?
 Thanks in advance for your answers.
 Best regards,

 Anthony

 On Tue, Jun 10, 2008 at 6:13 PM, stephen sefick [EMAIL PROTECTED] wrote:

 I from a first thought I would say that you are apply this wrong!  The
 fourier transform convolves a function (cos(x)+isin(x)  (this may not be the
 exact formula but I don't have my books near)) to the data and then
 integrates over -1/2 to 1/2 takes the modulus and plots this- the
 periodogram.  The reason you preform a fourier transform is to look at
 recurring frequencies in the data, which are in the time domain.  The
 fourier transform converts the time series into the frequency domain and
 viola you have a peak into the hidden/recurring parts of your signal.  From
 your explaination your are applying this technique wrong-  look at schumway,
 MASS4, et al. books to get a handle on how this technique is used.  If you
 are to apply a time series analysis please use it on a time series.  Maybe
 your logic is not flawed but I don't see how a histogram with its associated
 binning is a better candidate for time series analysis than the original
 time series if at all.
 good luck

 Stephen

 On Tue, Jun 10, 2008 at 8:49 AM, Matthieu Stigler 
 [EMAIL PROTECTED] wrote:

 Hello

 I don't know exactly what you want to do but:

 -why do you use in your example h$counts and not h? Furthermore helpl
 file says it should be a time series, why

Re: [R] Importing data with different delimters

2008-06-16 Thread milton ruser

Hi David,

If the delimier is tab try this.

my.df-read.table(my_file.txt, head=T, sep=\t)

Cheers,
Miltinho
Brazil





  -Original Message-
  From: [EMAIL PROTECTED]
  [mailto:[EMAIL PROTECTED] On Behalf Of David Arnold
  Sent: Friday, June 13, 2008 5:15 PM
  To: r-help@r-project.org
  Subject: [R] Importing data with different delimters
 
  All,
 
  I have a data file with 56 entries that looks like this:
 
  City State  JanTemp Lat Long
  Mobile, AL  44  31.288.5
  Montgomery, AL  38  32.986.8
  Phoenix, AZ 35  33.6112.5
  Little Rock, AR 31  35.492.8
  Los Angeles, CA 47  34.3118.7
  San Francisco, CA   42  38.4123.0
 
  I would like to read this data into a dataframe. Is it
  possible to do without editing the datafile?
 
  D.
 
  __
  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
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Re: [R] Importing data with different delimters

Greg Snow wrote:
 It looks like your original data may be tab seperated, if that is the case 
 then just use read.delim or use sep='\t' in read.table or scan.
   
I think that was only half the problem. If you do that, you end up with
one column containing both City and State, comma-separated. Presumably,
the path of least resistance is just to do this and then use string
processing with sub()  or strsplit() to split the City, State strings.

-pd
   
 I have a data file with 56 entries that looks like this:

 City State  JanTemp Lat Long
 Mobile, AL  44  31.288.5
 Montgomery, AL  38  32.986.8
 Phoenix, AZ 35  33.6112.5
 Little Rock, AR 31  35.492.8
 Los Angeles, CA 47  34.3118.7
 San Francisco, CA   42  38.4123.0

 I would like to read this data into a dataframe. Is it
 possible to do without editing the datafile?

 D.
 

-- 
   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] aggregate() function, strange behavior for augmented data

2008-06-16 Thread David Afshartous


Everything was read in the same way, and str(junk1) confirms that they are
the same structure.  This is very strange.

## original data:
 str(junk1)
'data.frame':   96 obs. of  3 variables:
 $ Hour: int  0 3 5 0 3 5 0 3 5 0 ...
 $ Drug: Factor w/ 2 levels D,P: 2 2 2 1 1 1 2 2 2 1 ...
 $ Aldo: int  9 15 4 8 13 3 5 11 5 7 ...

## augmented data:
 str(junk1)
'data.frame':108 obs. of  3 variables:
 $ Hour: int  0 3 5 0 3 5 0 3 5 0 ...
 $ Drug: Factor w/ 2 levels D,P: 2 2 2 1 1 1 2 2 2 1 ...
 $ Aldo: int  9 15 4 8 13 3 5 11 5 7 ...






On 6/16/08 11:37 AM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:

 
 hi: do str(junk1) and it will tell you what  the components of junk1
 are.
 
 the only thing i can think of is that you used stringsAsFactors=FALSE
 when you ( probably ) used read.table to read in junk but you didn't use
 that
 options when you used read.table  to read in junk1 ?
 
 
 On Mon, Jun 16, 2008 at 11:30 AM, David Afshartous wrote:
 
 All,
 
 I'm re-running some analysis that has been augmented with additional
 data.
 When I use the exact same code for the augmented data, the behavior of
 the
 aggregate function is very strange, viz., one of the resulting
 variables is
 now coded as a factor while it was coded as numeric for the original
 data.
 Unfortunately, I cannot provide a reproducible code example since it
 only
 seems to occur with this data.  I've checked and re-checked the of
 both the
 original and augmented data but nothing appears inconsistent.  Any
 suggestions much appreciated.  See below for specifics.
 
 Cheers,
 David
 
 
 
 
 
 
 
 
 
 # original data
 dim(junk1)
 [1] 96  3
 junk1[1,]
   Hour Drug Aldo
 10P9
 junk1$Hour
  [1] 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5
 0 3
 5 0 3
 [39] 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
 5 0
 3 5 0
 [77] 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5   ### Not coded as a
 factor
 junk1.mean.time.drug = aggregate(junk1[3], junk1[c(1,2)], mean)
 junk1.mean.time.drug$Hour
 [1] 0 3 5 0 3 5  ### not coded as a factor
 
 # augmented data
  dim(junk1)
 [1] 108   3
 junk1[1,]
   Hour Drug Aldo
 10P9
 junk1$Hour
   [1] 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
 5 0 3
 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3
  [51] 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0
 3 5 0
 3 5 0 3 5 0 3 5 0 3 5 0 3 5 0
 [101] 3 5 0 3 5 0 3 5### not coded as a factor
 junk1.mean.time.drug = aggregate(junk1[3], junk1[c(1,2)], mean)
 junk1.mean.time.drug$Hour
 [1] 0 3 5 0 3 5
 Levels: 0 3 5## coded as a factor now!
 
 ## of course, I get recode it again but I'm curious as to why this is
 ## changing here
 
 __
 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] sizing non-vector point shapes in ggplot2

2008-06-16 Thread Michael Frumin

On Mon, Jun 16, 2008 at 3:55 PM, hadley wickham [EMAIL PROTECTED] wrote:

 On Mon, Jun 16, 2008 at 7:52 AM, mfrumin [EMAIL PROTECTED] wrote:
 
  Dear all,
 
  With normal plotting, one can size a set of points in a plot using a
 vector
  argument to cex in the points() function.  This works whether you are
 using
  one of the standard R symbols (i.e. 19+) or some ascii symbol, such as
 '/'
 
  eg:
  plot(1:10, 1:10, type='n');
  points(1:10, 1:10, cex = 1:10, pch = '/')
 
  Trying to make the transition to ggplot2, I find that the aesthetic size
  mapping does not apply if i do geom_point(..., shape='/') -- the points
 show
  up looking like '/' but they are not sized.  Is there anything to do
 about
  this?
 
  eg:
  ggplot(data = data.frame(x = 1:10, y = 1:10, size = 1:10), aes(x = x, y =
  y)) + geom_point(mapping = aes(size = size), shape = '/')
 
  the plot that I'm making really needs a vertically oriented mark, not a
  round-ish point/square/triangle that takes up a lot of area. am I totally
  out of luck?

 Hmmm, I'd never noticed this feature of grid before.  To size the
 points, I'm using the size argument of grid.points, which doesn't seem
 to affect the size of character based plotting symbols (because the
 are using the fontsize graphical parameter).  I've cc'd Paul on this
 email so he can confirm whether this is a bug or by design.

 Regardless, you can make ggplot size the symbols correctly by running
 the following code:

 GeomPoint$draw - function(., data, scales, coordinates, ...) {
with(coordinates$transform(data),
  ggname(.$my_name(), pointsGrob(x, y, size=unit(size, mm), pch=shape,
  gp=gpar(col=colour, fill = fill, fontsize = size * .pt)))
)
  }

 Hadley

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


Thanks for getting back to me so quickly.  This mostly works, in that:

- When I ran it just as you sent it, I get this error:
Error in inherits(x, factor) : object fill not found

- but when I took out fill = fill from your suggestion above, I get the
desired effect.

- but this does not change the shape shown in the scale for the sized
points.  they are still big circles.  but come to think of it, even setting
shape = 20 or some other vector point-type, this doesn't effect the shape
shown in the scale/legend thingy.  so I guess that's a separate question, I
will post it under separate cover.

thanks!

mike

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Re: [R] round(1.5) = round(2.5) = 2?

2008-06-16 Thread Greg Snow

The logic behind the round to even rule is that we are trying to represent an 
underlying continuous value and if x comes from a truly continuous 
distribution, then the probability that x==2.5 is 0 and the 2.5 was probably 
already rounded once from any values between 2.45 and 2.54..., if 
we use the round up on 0.5 rule that we learned in grade school, then the 
double rounding means that values between 2.45 and 2.50 will all round to 3 
(having been rounded first to 2.5).  This will tend to bias estimates upwards.  
To remove the bias we need to either go back to before the rounding to 2.5 
(which is often impossible to impractical), or just round up half the time and 
round down half the time (or better would be to round proportional to how 
likely we are to see values below or above 2.5 rounded to 2.5, but that will be 
close to 50/50 for most underlying distributions).  The stochastic approach 
would be to have the round function randomly choose which way to round, but 
deterministic types  are not comforatable with that, so round to even was 
chosen (round to odd should work about the same) as a consistent rule that 
rounds up and down about 50/50.

If you are dealing with data where 2.5 is likely to represent an exact value 
(money for example), then you may do better by multiplying all values by 10 or 
100 and working in integers, then converting back only for the final printing.  
Note that 2.5001 rounds to 3, so if you keep more digits of accuracy until 
the final printing, then rounding will go in the expected direction, or you can 
add 0.1 (or other small number) to your values just before rounding, 
but that can bias your estimates upwards.

Hope this helps,

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



 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Markus Didion
 Sent: Sunday, June 15, 2008 3:26 AM
 To: r-help@r-project.org
 Subject: [R] round(1.5) = round(2.5) = 2?

 Dear R-users

 with a bit of grief I had to repeat an extensive analysis
 because I did not suspect (and therefore did not read the
 documentation) that round was implemented as for rounding
 off a 5, the IEC 60559 standard is expected to be used, 'go
 to the even digit', resulting in
 round(1.5) = 2
 round (2.5) = 2.

 As a non-mathematician I am both puzzled and intrigued by
 this rule as it is against what I have learned in my math
 courses, i.e.
 round(1.5) = 2
 round (2.5) = 3.

 I would like to understand the reason behind this rule.

 Thanks for your comments.

 Markus

 --

 Markus Didion

 WaldökologieForest Ecology
 Inst. f. Terrestrische Oekosysteme  Inst. of
 Terrestrial Ecosystems
 Departement UmweltwissenschaftenDept. of
 Environmental Sciences
 Eidg. Technische Hochschule Swiss Fed.
 Inst. of Technology
 ETH-Zentrum CHN G78 ETH-Zentrum CHN G78
 Universitätstr. 22
   Universitaetstr. 22
 CH-8092 Zürich  CH-8092 Zurich
 Schweiz
   Switzerland

 Tel +41 (0)44 632 5629  Fax +41 (0)44 632 1358
 Email [EMAIL PROTECTED]
 homepage: http://www.fe.ethz.ch/people/didionm
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[R] optim() and starting values.

2008-06-16 Thread Dong-hyun Oh


Dear UseRs,

I wrote the following function to estimate parameters using MLE.

mlog - function(theta, nx = 1, nz = 1, dt){
 beta - matrix(theta[1:(nx+1)], ncol = 1)
 delta - matrix(theta[(nx+2):(nx+nz+1)], ncol = 1)
 sigma2 - theta[nx+nz+2]
 gamma - theta[nx+nz+3]
 y - as.matrix(dt[, 1], ncol = 1)
 x - as.matrix(data.frame(1, as.matrix(dt[, 2:(nx+1)], ncol = 2)))
 z - as.matrix(dt[, (nx+2):(nx+nz+1)], ncol = nz)

 d - z %*% delta / (gamma * sigma2)^.5
 mustar - (1-gamma) * z %*% delta - gamma * ( y - x %*% beta)
 sigmastar - (gamma * (1-gamma) * sigma2)^.5
 dstar - mustar / sigmastar

 loglik - (-0.5 * nrow(x) *(log(2*pi) + log(sigma2))
-0.5 * sum(( y - x %*% beta + z %*% delta)^2/sigma2)
-sum(log(pnorm(d))) + sum(log(pnorm(dstar
 return(-loglik)
}
--

To test my function, I created an artificial data set as follows:
--
x1 - abs(rnorm(100))*100
x2 - abs(rnorm(100))*10
z1 - abs(rnorm(100))*5
z2 - abs(rnorm(100))*7
y - abs(0.3 + 0.3* log(x1) + 0.7* log(x2))
dat - data.frame(log(y), log(x1), log(x2), z1, z2)
--

The following optimization results provides different estimates.

--
theta.start1 - c(1.06, 0.08, 0.04, 0.097, 0.008, 0.08, 0.008)
theta.start2 - c(1.06, 0.08, 0.04, 0.097, 0.8, 0.0008, 0.0008)
out.optim - optim(theta.start1, mlog, nx = 2, nz = 2, dt = dat,  
hessian = T)

par.theta1 - out.optim$par
out.optim - optim(theta.start2, mlog, nx = 2, nz = 2, dt = dat,  
hessian = T)

par.theta2 - out.optim$par
---

How can I set up concrete starting values?

Any advices will be appreciated.

Looking forward to hearing from you.

Sincerely,
Dong-hyun Oh

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] R as chart engine in web-service

2008-06-16 Thread jim holtman

I think that you have to define some type of criteria on what you
expect the performance to be.  There is a cost/benefit tradeoff in
having multiple servers with a load balancer and this all depends on
how much you want to spend on a solution.  You might want to prototype
it in R to at least get an understanding of the data transformations
that are required and then after you have some performance data and a
target for usage, you can then make a better decision.

On Mon, Jun 16, 2008 at 8:26 AM, Daniel Cegielka [EMAIL PROTECTED] wrote:
 Can you put some criteria on 'better'/'faster'.  What throughput are
 you expecting?  How many requests per second will you have to handle?


 It's very difficult to tell how many request per second I will have to
 handle. If you have normal site it will be max. 2-5 tps (on chart website)
 but if you have real time quotes it will be about 300 tps. In RT quotes I
 intend use java applet to make RT charts with independent connection.

 I thing about service like BigChart.com - put some symbol (GOOG, DJIA etc).

 http://bigcharts.marketwatch.com/quickchart/quickchart.asp?symb=djiasid=164
 3

 and the chart like:

 http://www.marketwatch.com/tools/quotes/intchart.asp?submitted=trueintflavo
 r=advancedsymb=DJIAorigurl=%2Ftools%2Fquotes%2Fintchart.asptime=8freq=1
 startdate=enddate=hiddenTrue=comp=Enter+Symbol(s)%3Acompidx=a~0comp
 ind=a~0uf=7168ma=1maval=50lf=1lf2=4lf3=0type=2size=2optstyle=10
 13

 In this kind of web services its important how faster the chart engine is in
 general. If there will be to many request per second I can add more machines
 and use load balancer (haproxy or pound).

 What size/type machine will you be running on?

 Probably 1-2 Quad Core Intel Xeon 2x6MB Cache, 2.0GHz per machine with 8GB
 (or 16GB) RAM. Zope 3 as web server on Gentoo Linux (x86). Memory will be
 mounted as a local disk (tmpfs) and python and R binaries will by copy into
 RAM. Python and R are compiled with ICC compiler.


 Is data being accessed from a database which
 may influence the timing?

 R will connect with Tree Data Server (database):
 http://tree.sourceforge.net/
 Data can be keep in memcached: http://www.danga.com/memcached/

 So I think there is very small influence on the timing.



 And I think about quantmod as a R chart package:
 http://www.quantmod.com/examples/charting/


 The main problem with R is that, R uses x11 server to render images and this
 way is very slow and I thing that faster would be the Cairo (or GDD) R
 package but I haven't tested this yet.

 R engine is very universal and this is the strong of R choice in this kind
 of web service... then begin write new (faster?) c/c++ plugin. What you
 think?

 daniel cegielka



 -Original Message-
 From: jim holtman [mailto:[EMAIL PROTECTED]
 Sent: Monday, June 16, 2008 1:55 AM
 To: Daniel Cegielka
 Cc: r-help@r-project.org
 Subject: Re: [R] R as chart engine in web-service

 Can you put some criteria on 'better'/'faster'.  What throughput are
 you expecting?  How many requests per second will you have to handle?
 What is the timing of your current method?  What size/type machine
 will you be running on?  Is data being accessed from a database which
 may influence the timing?  So you need to give some specificity of
 what your requirements are.

 You can probably get faster (but maybe not better) with a plug-in, but
 I would assume that it would also be more effort.

 On Sun, Jun 15, 2008 at 6:39 PM, Daniel Cegielka [EMAIL PROTECTED] wrote:
 Hi R-users

 I think about some web service with stock charts and I plan use python as
 web framework with R as mathematical and chart engine. I use rpy to
 connect
 R with python. It works good.

 Is it good idea to use R as chart engine? Maybe it's better (faster) to
 use
 c/c++ plugin?

 Daniel

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

 What is the problem you are trying to solve?





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

What is the problem you are trying to solve?

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Re: [R] ggplot2: How to remove legend component for geom_errorbar() ?

2008-06-16 Thread hadley wickham

 thanks for your tip! It worked fine (though I had to use tile instead
 of point). I see your point to handle legends the way you do, which is
 a very convenient feature of ggplot. It's great for points, lines, bars
 etc., but in the particular case of error bars, I think it would be
 reasonable to omit them by default and only turn them on if someone
 really wants them (by using something like legend=TRUE). Just my two
 cents, for now, I'm happy with your workaround.

Maybe this wouldn't have been such a problem if the legend for error
bars actually looked like errorbars!  I'll add that to my to do and
think about better ways to turn legends off when desired.

Hadley

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

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[R] prediction intervals for random effects in nlme

2008-06-16 Thread Marc Bernard

Dear all, 
nbsp;
This is a relist of my previous question. I noticed that some 
charactersnbsp;were nbsp;not displayed in the previous version.
nbsp;
nbsp;
Is there a function to calculate thenbsp; prediction intervals for random 
effects in non-linear mixed models? I found a way to do it for linear mixed 
models but not for non-linearnbsp;mixed one.
nbsp;
Many thanks
nbsp;
Bernard


  
_ 

o.fr
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Re: [R] candisc() error message

2008-06-16 Thread Michael Friendly

John is correct  the error message you got was clear:  the error SSP 
matrix is of deficient rank.

Possible solutions:
- Regroup the factor levels to fewer and/or
- Combine some of the species to broader classes so you have fewer than 
17 response variables

- Collect more samples

hope this helps,
-Michael

John Fox wrote:

Dear Tanya,

If I follow correctly what you're trying to do, you're in effect doing a
one-way MANOVA of 17 response variables on a six-level factor, with only 20
observations. That's not enough data. I'm copying to Michael Friendly in
case he hasn't seen your posting and would like to comment.

Regards,
 John

--
John Fox, Professor
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web: socserv.mcmaster.ca/jfox


  

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]


On
  

Behalf Of Tanya Yatsunenko
Sent: June-15-08 8:23 PM
To: r-help@r-project.org
Subject: [R] candisc() error message

Hi,
I am doing canonical discriminant analysis using candisc function from
the candisc package.
My input is a table of species distribution (columns = abundance of each
species in each sample)  in samples that are split by categories (rows),
and I want to know whether each category is associated with a particular
set of species and their abundances.
I have 20 rows (samples) split into 6 categories, and 17 columns


(species).
  

I am getting the following error message, which I don't understand:

  can-candisc(mod, data=canIN)
Error in linear.hypothesis.mlm(mod, hyp.matrix.2, SSPE = SSPE, V = V,  :
  The error SSP matrix is apparently of deficient rank = 16  17

Does anyone has any experience in candisc?

--
Tanya.


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--
Michael Friendly Email: [EMAIL PROTECTED] 
Professor, Psychology Dept.

York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele Streethttp://www.math.yorku.ca/SCS/friendly.html
Toronto, ONT  M3J 1P3 CANADA

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Re: [R] Linear Regression ?

2008-06-16 Thread Roland Rau


Hi,

did you have a look at the manual An Introduction to R? 
(http://cran.r-project.org/doc/manuals/R-intro.html).
Chapter Statistical Models in R should answer (most of) your questions 
concerning linear regression.


kayj wrote:


Also each x is a categorical variable with 4 categories so I have to
introduce  three variables for each x 



If I understood correctly what you mean (- coding of dummy variables), 
this is not necessary in R. If you variable is a factor, R does it for 
you automatically. If you are unhappy with the choice of reference 
category,  please check:

?relevel

I hope this helps,
Roland

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[R] Plotting 3d data?

2008-06-16 Thread Wesley Tansey

Hi all!

 

I'm very new to R, and I'm having trouble figuring out how to go from a file
of points that I have to a 3d surface plot of the data. I typically have
something like this:

 

X  Y  Z

0.0050.02334.45

0.0035  0.63  28.48

.

 

I've tried looking at the persp and wireframe packages, and the rgl package,
but I can't seem to figure out how to use any of them. I tried to take the
rgl.surface3d example and use it for myself, and this is what I have so far:

 

setwd(.)

 

data - read.csv(data.csv,header=T)

 

x - data$X

y - data$Y

z - data$Z

 

open3d()

surface3d(x, y, z) 

 

That gives me the following error:

 

Error in rgl.surface(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,  : 

  y length != x rows * z cols

 

I have no idea what it is trying to tell me there. Would anyone mind helping
me out?

 

 

Wesley

 


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Re: [R] Plotting 3d data?

Try scatterplot3d function in the package with same name.

On Mon, Jun 16, 2008 at 2:32 PM, Wesley Tansey [EMAIL PROTECTED] wrote:

 Hi all!



 I'm very new to R, and I'm having trouble figuring out how to go from a
 file
 of points that I have to a 3d surface plot of the data. I typically have
 something like this:



 X  Y  Z

 0.0050.02334.45

 0.0035  0.63  28.48

 .



 I've tried looking at the persp and wireframe packages, and the rgl
 package,
 but I can't seem to figure out how to use any of them. I tried to take the
 rgl.surface3d example and use it for myself, and this is what I have so
 far:



 setwd(.)



 data - read.csv(data.csv,header=T)



 x - data$X

 y - data$Y

 z - data$Z



 open3d()

 surface3d(x, y, z)



 That gives me the following error:



 Error in rgl.surface(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,  :

  y length != x rows * z cols



 I have no idea what it is trying to tell me there. Would anyone mind
 helping
 me out?





 Wesley




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

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[R] 回复: cch() and coxph() for case-coh ort

2008-06-16 Thread Jin Wang

I tried to compare if cch() and coxph() can generate same result for
same case cohort data

Use the standard data in cch(): nwtco

Since in cch contains the cohort size=4028, while ccoh.data size =1154
after selection, but coxph does not contain info of cohort size=4028.

The rough estimate between coxph() and cch() is same, but the lower
and upper CI and P-value are a little different.  Can we exactly use
coxph() to repeat cch() using with appropriate configuration in
coxph()?  Is SAS a better way(PHREG,CASECOH.SAS) to implement
time-dependent case-cohort?


 summary(fit2.ccP)
Call:
coxph(formula = Surv(edrel, rel) ~ stage + histol + age + offset(-100 *
(1 - subcohort)) + cluster(seqno), data = ccoh.data)

  n= 1154
   coef exp(coef) se(coef) robust sez   p
stageII  0.7363  2.09   0.12130.1699 4.33 1.5e-05
stageIII 0.5976  1.82   0.12330.1753 3.41 6.5e-04
stageIV  1.3921  4.02   0.13390.2081 6.69 2.2e-11
histolUH 1.5059  4.51   0.09110.1644 9.16 0.0e+00
age  0.0432  1.04   0.01460.0243 1.78 7.6e-02

 exp(coef) exp(-coef) lower .95 upper .95
stageII   2.09  0.479 1.497  2.91
stageIII  1.82  0.550 1.289  2.56
stageIV   4.02  0.249 2.676  6.05
histolUH  4.51  0.222 3.267  6.22
age   1.04  0.958 0.996  1.09

Rsquare= 0.273   (max possible= 1 )
Likelihood ratio test= 368  on 5 df,   p=0
Wald test= 134  on 5 df,   p=0
Score (logrank) test = 490  on 5 df,   p=0,   Robust = 165  p=0

  (Note: the likelihood ratio and score tests assume independence of
 observations within a cluster, the Wald and robust score tests do not).


 summary(fit.ccSP)
Case-cohort analysis,x$method, SelfPrentice
 with subcohort of 668 from cohort of 4028

Call: cch(formula = Surv(edrel, rel) ~ stage + histol + age, data = ccoh.data,
subcoh = ~subcohort, id = ~seqno, cohort.size = 4028, method = SelfPren)

Coefficients:
  CoefHR  (95%   CI) p
stageII  0.736 2.088 1.491 2.925 0.000
stageIII 0.597 1.818 1.285 2.571 0.001
stageIV  1.392 4.021 2.670 6.057 0.000
histolUH 1.506 4.507 3.274 6.203 0.000
age  0.043 1.044 0.996 1.095 0.069


2008/6/12, Terry Therneau [EMAIL PROTECTED]:
 Jin Wang had an error.  My original note specified a variable that was 1 for
 subjects NOT in the subcohort, so the correct coxph call is

   coxph(Surv(edrel, rel) ~ stage + histol + age +
offset(-100*(subcohort==0)) + cluster(seqno), data =ccoh.data)
   
 This gives the same coefficients as the cch example, along with the
 infinitesimal jackknife or robust variance estimate.

   Terry Therneau
   



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Re: [R] using spec.pgram

2008-06-16 Thread Anthony Mathelier

OK, it seems like I do not succeed in expressing what I do, or want to do.
So, I give you the example that bring me to this kind of analysis. I wrote
the paper Chromosomal periodicity of evolutionary conserved gene pairs
(which you can download at http://www.pnas.org/cgi/reprint/104/25/10559). In
figure 2, they have a histogram of distances between genes on a chromosome
and they make a discrete fourier transform analysis to exhibit a period of
117kb. They explain how they did in the first paragraph of Distributions of
distances and positions and fourier transform (last page). I thought that
this kind of analysis was made by spec.pgram with a histogram. But perhaps I
am wrong because I really do not understand what they mean by the histogram
was tranformed into a continuous probability density by using a Gaussian
smoothing window and normalizing the total density over the entire genome to
1. A discrete Fourier transform of the data were computed from 0 to 1,000kb
by using a Tukey window to taper the end (ratio of 0.5 for tapered to
untapered length..
I hope it explains better what I want to obtain from my distances.
Best regards,

Anthony

On Mon, Jun 16, 2008 at 5:25 PM, stephen sefick [EMAIL PROTECTED] wrote:

 To get some sort of frequency which in your case seem to be cycles per
 distance?  Is a valid use of a fourier transform as long as it is a distance
 that is measured in a way that would be analogous to a time series-  In
 other words if the distance proceeds from an origin in one direction-
 geophysicists do this often with the realization of an earthquake picked up
 by sensors that are a distance away from the origin of the epicenter, but
 they are looking for coherencies in the signal from one place to the next in
 the frequency domain seperated by distance- this is called beam forming-
 They use the raw signal-  by binning (making a histogram) the data you are
 loosing the signal-  you are looking at frequency of occurance of certain
 values not for the underlying periodicities of the data (in time or
 space).   You are fitting cos and isin function to you data to see if there
 is periodicity-  the power is the integration of the convolution of this sin
 and cosine function with your data-  It seems to me meaningless to preform
 this convolution agianst something that is not a signal (the histogram).  If
 you want to use a frequency domain technique you have to have a frequency to
 investigate-  a histogram does not have this-  I is a frequency of occurance
 by bin size which is NOT what you want (your would have cycles/binlength
 that doesn't make any sense to me) to do this analysis on-  You want a
 signal-  dissolved oxygen curve, sunspot record, etc. through time, or
 distance as stated above- you are looking for the frequency of a waveform-
 Anyway,  I may be misunderstanding- supply some code and explain the data
 otherwise this line of though- in my limited expertise- is a dead end,  but
 agian I still don't know what it is that you are, exactly, trying to do- and
 what your dataset constits.  I hope these ruminations help

 I recommend doing this analysis on the raw data- It doesn't matter that you
 don't have the same amount of data points- as long as both sets of data have
 circa ten times the length of (cycles/distance) what you want to detect-  If
 things in your case are spaced by one meter then the lowest cycle
 perdistance that you can reliably detect if 0.5 meters,  this is all
 speculation because you don't have a problem with reproducible code, and we
 have no idea what you are measuring or what your data looks like-  without
 this information there is no way that I can say one way or the other that
 you approach (suggested non-histogram) would be right or wrong.

 Stephen


 On Mon, Jun 16, 2008 at 9:33 AM, Anthony Mathelier 
 [EMAIL PROTECTED] wrote:

 Perhaps I'm applying spec.pgram wrong as you said. I will explain what I
 want, so you can tell me why I'm wrong and perhaps what I have to do to do
 it well.
 I have some points in a 1-D space and I want to know if they are spaced at
 a certain periodic distance. So, I computed all the distances between points
 in my space. Then, I would like to know if a certain distance (period), or
 multiples of a certain distance, is preferred to space my data. I made a
 histogram of the distances and apply the spec.pgram function to know the
 frequence (so the period) which is the most important to space the original
 data.
 But, when I have to sets of data (without necessarily the same number of
 observation in each set), I want to compare the importance of the period
 given by spec.pgram between the sets. Could I normalize the amplitude of the
 peaks given by spec.pgram?
 So, am I wrong to apply this methodology to exhibit a periodic distance
 between my data? If, true, what could you recommend me to do this?
 Thanks in advance for your answers.
 Best regards,

 Anthony

 On Tue, Jun 10, 2008 at 6:13 PM, stephen sefick [EMAIL

Re: [R] ggplot2: How to remove legend component for geom_errorbar() ?

2008-06-16 Thread Carsten Jaeger

Hi Hadley,

thanks for your tip! It worked fine (though I had to use tile instead
of point). I see your point to handle legends the way you do, which is
a very convenient feature of ggplot. It's great for points, lines, bars
etc., but in the particular case of error bars, I think it would be
reasonable to omit them by default and only turn them on if someone
really wants them (by using something like legend=TRUE). Just my two
cents, for now, I'm happy with your workaround.

Thanks again,
Carsten


On Mon, 2008-06-16 at 09:49 -0500, hadley wickham wrote:
 Hi Carsten,
 
 In my eagerness to get every component of the legend working
 automatically, I've completed neglected a way to turn bits off of you
 don't want them (the philosophy is that every geom that uses an
 aesthetic should appear in the legend in some way).  The best thing I
 can suggest at the moment is:
 
 GeomErrorbar$guide_geom - function(.) point
 
 to use points for the errorbar legend, which you won't see because
 they'll overlap with the points from the points geom.  In the next
 version, you'll be able to do
 
 GeomLine$guide_geom - function(.) blank
 
 Or I might come up with a better way of controlling the appearance.
 
 Regards,
 
 Hadley
 
 On Mon, Jun 16, 2008 at 2:55 AM, Carsten Jaeger [EMAIL PROTECTED] wrote:
  Dear list (probably Hadley),
 
  I'm trying to do a plot like the following, composed of bars and error
  bars:
 
  df - data.frame(factor1=gl(2,5), factor2=gl(5,1), y=rnorm(10),
  err=0.1)
  ggplot(df, aes(x=factor1, y=y, fill=factor2)) +
 geom_bar(position=dodge, stat=identity) +
 geom_errorbar(aes(min=y-err, max=y+err), position=dodge,
  width=0.1)
 
  The plot obtained has a legend named factor2, giving as expected the
  colours used for the five factor levels of factor2. So far so good.
  However, the legend also contains solid horizontal lines, obviously
  representing the error bars. While I do not immediately see the sense of
  a legend symbol for error bars, is there any easy way to turn these off?
 
  Thanks,
  Carsten
 
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[R] Lattice: Superpose bwplot on dotplot [Newbie Question]

2008-06-16 Thread Lord Yo


Hello everyone

I have dataset containing a monetary value (ABS) and two factors (Fct,
Group).  I am able to create useful using:

bwplot(ABS~Group|Fct)
and
dotplot(ABS~Group|Fct)

Question: What do I have to do to overlay the dotplot with the bwplot (same
data set)? 

I've found a couple of posts that hinted at the possibility of doing that,
and checked the panel.superpose() help, but the info was too complex for my
newbie brain.

Thanks for your help!
LY
-- 
View this message in context: 
http://www.nabble.com/Lattice%3A-Superpose-bwplot-on-dotplot--Newbie-Question--tp17867934p17867934.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Error in maximum likelihood estimation.

2008-06-16 Thread Dong-hyun Oh

Dear UseRs,

I wrote the following function to use MLE.

-
mlog - function(theta, nx = 1, nz = 1, dt){
  beta - matrix(theta[1:(nx+1)], ncol = 1)
  delta - matrix(theta[(nx+2):(nx+nz+1)], ncol = 1)
  sigma2 - theta[nx+nz+2]
  gamma - theta[nx+nz+3]
  y - as.matrix(dt[, 1], ncol = 1)
  x - as.matrix(data.frame(1, as.matrix(dt[, 2:(nx+1)], ncol = 2)))
  z - as.matrix(dt[, (nx+2):(nx+nz+1)], ncol = nz)

  d - z %*% delta / (gamma * sigma2)^.5
  mustar - (1-gamma) * z %*% delta - gamma * ( y - x %*% beta)
  sigmastar - (gamma * (1-gamma) * sigma2)^.5
  dstar - mustar / sigmastar

  loglik - (-0.5 * nrow(x) *(log(2*pi) + log(sigma2))
 -0.5 * sum(( y - x %*% beta + z %*% delta)^2/sigma2)
 -sum(log(pnorm(d))) + sum(log(pnorm(dstar
  return(-loglik)
}
---

Loglikelihood function is from page 21of Battese and Coelli (1993).  
(You can download this article 
athttp://www.une.edu.au/economics/publications/econometrics/emwp69.PDF 
  )

To test the above function with an artificial data set, I created the  
following data.frame.

---
x1 - abs(rnorm(100))*100
x2 - abs(rnorm(100))*10
z1 - abs(rnorm(100))*5
z2 - abs(rnorm(100))*7
y - abs(0.3 + 0.3* log(x1) + 0.7* log(x2))
dat - data.frame(log(y), log(x1), log(x2), z1, z2)


The starting value I set up is as follows:
---
theta.start - c(0.09, 0.008, 0.008, 0.023, 0.0008, 0.008, 0.008)
--

Applying nlm() function to the above function with the starting values  
gives the following error message.


out - nlm(mlog, theta.start, nx = 2 , nz = 2, dt = dat, print.level =  
2, hessian\
= T, iterlim = 500)

iteration = 0
Step:
[1] 0 0 0 0 0 0 0
Parameter:
[1] 0.0900 0.0080 0.0080 0.0230 0.0008 0.0080 0.0080
Function Value
[1] 6116.2
Gradient:
[1]  -10704.8  -46465.7  -22536.4   47168.2   54542.9 -776512.5  
579.9

Error in nlm(mlog, theta.start, nx = 2, nz = 2, dt = dat, print.level  
= 2,  :
  (converted from warning) NA/Inf replaced by maximum positive value



My questions are
1. Is my loglikelihood function set up appropriately?
2. How to set up starting values efficiently? I read a thread shown at 
https://stat.ethz.ch/pipermail/r-help/2005-September/079617.html 
  , but I cannot figure out how to set up starting values with  
expand.grid() of parameters I used (beta, delta, gamma, sigma2).

Thank you in advance.

Sincerely,
Dong-hyun Oh
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Re: [R] save workspace while running R on a cluster

2008-06-16 Thread Manisha Brahmachary

Thanks your suggestion works -:-)

From: milton ruser [mailto:[EMAIL PROTECTED] 
Sent: Thursday, June 12, 2008 2:42 PM
To: Manisha Brahmachary
Cc: r-help@r-project.org
Subject: Re: [R] save workspace while running R on a cluster

Hi Manisha,

How about you incluse something like this on your script.R: 

setwd(/your/full/working/directory)   # ?setwd

save.image()# or save.image(your_workspace.RDA).

By the way, I don´t know if you added the line below to run in background:

R --save  calculate.R  script.out

May be the rigth thing is

R --save  calculate.R  script.out 

Good luck.

miltinho

Brazil

On 6/12/08, Manisha Brahmachary [EMAIL PROTECTED] wrote: 

Hello,

I have a question about running R in a cluster environment. The shell script
I am running looks like this:

#!/bin/bash

cd /nfs/apollo/2/c2b2/users/mb0001/Data

/nfs/apollo/1/shares/software/core_facility/local/x86_64_rocks/R/current/bin/
R --save  calculate.R  script.out

I have used the -save command to save the R workspace (If, I understand it
correctly) . However, I am unable to find the workspace file.

Using find .RData did not help.

I would be obliged if someone can help me figure out where the workspace file
gets saved?

Thanks

manisha

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[R] Creating correlated multivariate dataset

2008-06-16 Thread sk

Hello list,
  I am trying to test a model but for the beginning I want to do this by using 
simulated dataset. The model is
  Y_t = X_t %*% beta + e 
  Where Y : (Nx1); X: (Nxp); beta: (0.6,0.3,0.1); e-uncorrelated normally 
distributed variates for each t. and later I want to use to use this dataset in 
a BUGS model to estimate the betas.
  Thank you for you consideration. 

   
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[R] pvclust distance matrix

2008-06-16 Thread Strobel, Bradley

Hello,
 
I am attempting to assign significance levels to a UPGMA cluster analysis as 
part my doctoral research.  The pvclust function works well but doesn't include 
the similarity index I need (morisita's) as an option for computing a distance 
matrix.  Morisita's is available in vegdist in the VEGAN library but I am 
having a hard time getting the vegdist  function to direct drop into pvclust 
as the vegdist documentation suggests.  I would appreciate any suggestions.
 
Thank you
 
Brad
 
Bradley N. Strobel
 
Department of Natural Resource Management
Texas Tech University
Lubbock, TX
(806)742-0841

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[R] minimizing title's police

2008-06-16 Thread hanen



in order to minimize the police of the title of my graph i tried to use:

title(nombre de fleurs données journellement par 6 cereus peruvianus du
1/07/02 au 31/09/02,font=0)

but this message appears:

Erreur dans title(nombre de fleurs données journellement par 6 cereus
peruvianus de 1/07/02 au 31/09/02,  : 
  valeur spécifiée pour le paramètre graphique font incorrect

what's the value that should i give to the parameter font ?
-- 
View this message in context: 
http://www.nabble.com/minimizing-title%27s-police-tp17870753p17870753.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Dual axis labeling of a single quantity

2008-06-16 Thread Thomas Adams

I have a problem where I need to label the vertical axes of a Boxplot 
with related, but different quantities (flow  height), which have a 
known relationship. Primarily I want to plot the variable as a flow on 
the left axis and on the opposing right axis, show the corresponding 
height. Is it possible to get the flow range (max  min) of the left 
axis and then supply the right axis with that range (corresponding 
max/min) which will be plotted automatically by R? It does not matter if 
the opposing tick marks exactly correspond to the tick mark values on 
the left side as long as the scale is correct (that is the correct 
corresponding range).


I'm not even sure where to start searching for this…

Regards,
Tom

--
Thomas E Adams
National Weather Service
Ohio River Forecast Center
1901 South State Route 134
Wilmington, OH 45177

EMAIL:  [EMAIL PROTECTED]

VOICE:  937-383-0528
FAX:937-383-0033

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Re: [R] left-aligned title?

Try with mtext:

plot(rnorm(100))
mtext(text = Left, side = 3, adj = 0)

On Mon, Jun 16, 2008 at 2:39 PM, Carlos Gershenson [EMAIL PROTECTED]
wrote:

 Hi,

 I am trying to insert a letter in a plot corner outside the plotting area.
 Thus, legend and text don't seem to work. title does the trick, but I
 cannot find a way of moving it from the center to the left corner... I
 already tried with a few parameters from par, but title does not take them.

 Would anyone have an idea on how to pull this one off?

 Thank you very much,

Carlos Gershenson

 http://homepages.vub.ac.be/~cgershen/http://homepages.vub.ac.be/%7Ecgershen/

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

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[R] in axis() suppressing axis line but keeping tick marks

2008-06-16 Thread Andrew Yee

I've been trying to figure out a parameter that will let you separately
adjust the parameters for the axis line from the tick mark.

In the following example, I would like to suppress the axis line, but keep
the tick marks.

Thanks,
Andrew


foo - data.frame(x=1:3, y=4:6)
plot(foo$x, foo$y, type=n, axes=F)
points(foo$x, foo$y)
axis(side=1, at=foo$x, lty=0) #would like to figure out way to keep tick
marks, but suppress axis line

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Re: [R] prediction intervals for random effects in nlme

2008-06-16 Thread Dieter Menne

Marc Bernard bernarduse1 at yahoo.fr writes:

 
 Is there a function to calculate thenbsp; prediction intervals for random
effects in non-linear mixed
 models? I found a way to do it for linear mixed models but not for
non-linearnbsp;mixed one.

Please do not send HTML mail to the list.

A discussion and a caveat by Douglas Bates for lmer/lme4 is in:

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/113978.html

It might work in a an analogous way for non-linear fits.

Dieter

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[R] barplot with color coding according to value

2008-06-16 Thread maria

Hello,

I want to use R to produce nice meteograms...

For this I would need to color-code my barplot according to the value plotted

say if value x  30 ... the bar is red ; if x between 20 and 10, the bar
is blue etc...

any ideas how to proceed ?

Thank You a lot,
maria

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Re: [R] cch() and coxph() for case-cohort

2008-06-16 Thread Terry Therneau

- begin included message -

 I tried to compare if cch() and coxph() can generate same result for
same case cohort data

Use the standard data in cch(): nwtco

Since in cch contains the cohort size=4028, while ccoh.data size =1154
after selection, but coxph does not contain info of cohort size=4028.

The rough estimate between coxph() and cch() is same, but the lower
and upper CI and P-value are a little different.  Can we exactly use
coxph() to repeat cch() using with appropriate configuration in
coxph()?  Is SAS a better way(PHREG,CASECOH.SAS) to implement
time-dependent case-cohort?

output deleted

--end included message 


I showed you how to do time-dependent case cohort analysis.  Stop complaining 
and use the advice.
 
 No, SAS will not be better.  It will be different due to defaults (different 
convergence criteria, different default ties method).  Not better, not worse, 
just different.  
 
  The code I sent results in one variance estimate type, the default cch code 
is 
another.  They are asymptotically equivalent, and in your sample are 
essentially 
identical: one CI was (1.497, 2.91) the other (1.491, 2.925).  Neither is 
'better' or 'correct', both are good.  If you feel that you must have the other 
estimate, then read the paper upon which cch is based (which I wrote); you can 
find it in the references for the cch help page.
  
  The cch output has some default information not in the coxph output and vice 
versa.  So what?  
  
Terry Therneau

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Re: [R] barplot with color coding according to value

Perhaps something like this:

x - sample(10:50, 10, rep=TRUE)
cols - as.character(cut(x,
   breaks = c(0, 10, 20, 30,max(x)),
   labels = c(green, blue, yellow,
red)))
barplot(x, col = cols)

On Mon, Jun 16, 2008 at 4:15 PM, [EMAIL PROTECTED] wrote:

 Hello,

 I want to use R to produce nice meteograms...

 For this I would need to color-code my barplot according to the value
 plotted

 say if value x  30 ... the bar is red ; if x between 20 and 10, the bar
 is blue etc...

 any ideas how to proceed ?

 Thank You a lot,
 maria

 __
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 http://www.R-project.org/posting-guide.html
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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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[R] error in runif

2008-06-16 Thread Yasin Hajizadeh

Hi all
  I would be grateful you can help me with my problem.
  I try to run an optimization code . in one line I have runif in order to 
sample the PDF. I get this error while i run it. 
   
  Error in runif(1, f$d[[n.of.u.vars + n.of.o.vars + j]][[2]][1], 
f$d[[n.of.u.vars +  : 
  invalid arguments
   
  Here is a part of that code:
   
   
  # initialize random numeber generator
  if (seed0) set.seed(seed)
# set parameters
  e.abs - e
  e.rel - e
  max.value - f$opt
  eval - 0
  last.impr - max.eval;
  nl - matrix(NA,k,k-1)
  iteration - 0
# separate unordered and ordered from continuous variables
  range.u - vector()
  range.o - vector()
  n.of.x.vars - 0
  n.of.o.vars - 0
  n.of.u.vars - 0
  for (i in f$d) {
if (i[[1]]==u) {
  range.u - c(range.u, length(i[[2]]))
  n.of.u.vars - n.of.u.vars + 1
}
if (i[[1]]==o) {
  range.o - c(range.o, length(i[[2]]))
  n.of.o.vars - n.of.o.vars + 1
}
if (i[[1]]==x) {
  n.of.x.vars - n.of.x.vars + 1
}
  }
   # initialize variables
  max.u - rep(NA,n.of.u.vars)
  max.o - rep(NA,n.of.o.vars)
  max.X - rep(NA,n.of.x.vars)
  max.y - -Inf
p.X - vector()
  p.u - vector()
  p.o - vector()
  p - data.frame(v=numeric(),sd=numeric(),gr=numeric());
# randomly choose the starting population 
  # (but based on the data given in the function definition)
  for (i in 1:k) {
if (n.of.u.vars0) {
  U - vector()
  for (j in 1:n.of.u.vars) {
U - c(U, sample(range.u[j],1))   
  }
  U - t(U)
}
else U - NULL
if (n.of.o.vars0) {
  O - vector()
  for (j in 1:n.of.o.vars) {
O - c(O, sample(range.o[j],1))   
  }
  O - t(O)
}
X - vector()
for (j in 1:n.of.x.vars) {
  X - c(X, runif(1,f$d[[n.of.u.vars+n.of.o.vars+j]][[2]][1],
f$d[[n.of.u.vars+n.of.o.vars+j]][[2]][2]))
}
X - t(X)

   
  Many Thanks
   
  Yasin


   
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Re: [R] minimizing title's police

2008-06-16 Thread Patrick Connolly

On Mon, 16-Jun-2008 at 11:35AM -0700, hanen wrote:

| 
| 
| in order to minimize the police of the title of my graph i tried to use:
| 
| title(nombre de fleurs données journellement par 6 cereus peruvianus du
| 1/07/02 au 31/09/02,font=0)
| 
| but this message appears:
| 
| Erreur dans title(nombre de fleurs données journellement par 6 cereus
| peruvianus de 1/07/02 au 31/09/02,  : 
|   valeur spécifiée pour le paramètre graphique font incorrect
| 
| what's the value that should i give to the parameter font ?

1 is probably what you want.  To change the size, try changing cex, or
perhaps one of the other cex.* par values.  Check out ?par and search
for cex.  While you're there, check out how to use font.



-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~} Great minds discuss ideas
 _( Y )_Middle minds discuss events 
(:_~*~_:)Small minds discuss people  
 (_)-(_)   . Anon
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] 回复: cch() and coxph() for case-c ohort


Jin Wang wrote:

I tried to compare if cch() and coxph() can generate same result for
same case cohort data

Use the standard data in cch(): nwtco

Since in cch contains the cohort size=4028, while ccoh.data size =1154
after selection, but coxph does not contain info of cohort size=4028.

The rough estimate between coxph() and cch() is same, but the lower
and upper CI and P-value are a little different.  Can we exactly use
coxph() to repeat cch() using with appropriate configuration in
coxph()?  Is SAS a better way(PHREG,CASECOH.SAS) to implement
time-dependent case-cohort?


  
I think you need to read the literature, in particular the paper by 
Therneau (!) and Li, which among other things details the implementation 
of the Self-Prentice estimator.  With that in mind, it should not be 
surprising that it is non-trivial how to get correct SE's out of coxph. 
What _is_ surprising (at least somewhat) is how close the robust SE are 
to those of the Self-Prentice method -- if I understand correctly, the 
connection is that Self-Prentice uses jackknifing for the contribution 
from subcohort sampling plus the standard Cox asymptotic variance and 
the robust method effectively uses jackknifing for both.


(I'm a bit puzzled about why cch() insists on having unique id's, 
though. Doesn't _look_ like it would be too hard to get rid of that 
restriction, at least for S-P, which admittedly is the only method I 
spent enough time studying. And that was a some years ago.)



summary(fit2.ccP)


Call:
coxph(formula = Surv(edrel, rel) ~ stage + histol + age + offset(-100 *
(1 - subcohort)) + cluster(seqno), data = ccoh.data)

  n= 1154
   coef exp(coef) se(coef) robust sez   p
stageII  0.7363  2.09   0.12130.1699 4.33 1.5e-05
stageIII 0.5976  1.82   0.12330.1753 3.41 6.5e-04
stageIV  1.3921  4.02   0.13390.2081 6.69 2.2e-11
histolUH 1.5059  4.51   0.09110.1644 9.16 0.0e+00
age  0.0432  1.04   0.01460.0243 1.78 7.6e-02

 exp(coef) exp(-coef) lower .95 upper .95
stageII   2.09  0.479 1.497  2.91
stageIII  1.82  0.550 1.289  2.56
stageIV   4.02  0.249 2.676  6.05
histolUH  4.51  0.222 3.267  6.22
age   1.04  0.958 0.996  1.09

Rsquare= 0.273   (max possible= 1 )
Likelihood ratio test= 368  on 5 df,   p=0
Wald test= 134  on 5 df,   p=0
Score (logrank) test = 490  on 5 df,   p=0,   Robust = 165  p=0

  (Note: the likelihood ratio and score tests assume independence of
 observations within a cluster, the Wald and robust score tests do not).


  

summary(fit.ccSP)


Case-cohort analysis,x$method, SelfPrentice
 with subcohort of 668 from cohort of 4028

Call: cch(formula = Surv(edrel, rel) ~ stage + histol + age, data = ccoh.data,
subcoh = ~subcohort, id = ~seqno, cohort.size = 4028, method = SelfPren)

Coefficients:
  CoefHR  (95%   CI) p
stageII  0.736 2.088 1.491 2.925 0.000
stageIII 0.597 1.818 1.285 2.571 0.001
stageIV  1.392 4.021 2.670 6.057 0.000
histolUH 1.506 4.507 3.274 6.203 0.000
age  0.043 1.044 0.996 1.095 0.069


2008/6/12, Terry Therneau [EMAIL PROTECTED]:
  

Jin Wang had an error.  My original note specified a variable that was 1 for
subjects NOT in the subcohort, so the correct coxph call is

  coxph(Surv(edrel, rel) ~ stage + histol + age +
 offset(-100*(subcohort==0)) + cluster(seqno), data =ccoh.data)

This gives the same coefficients as the cch example, along with the
infinitesimal jackknife or robust variance estimate.

Terry Therneau






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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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[R] Annoyance in as.numeric

2008-06-16 Thread Alberto Monteiro

Why does as.numeric convert matrices and arrays to vectors?

as.numeric(matrix(c(1, 2, 3, 4), 2, 2))
[1] 1 2 3 4

I could only figure out ugly ways to bypass this, like:

x - matrix(c(1, 2, 3, 4), 2, 2)
array(as.numeric(x), dim = dim(x), dimnames = dimnames(x))

Alberto Monteiro

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[R] weights in lmer

2008-06-16 Thread Brian S Cade

I originally sent this to Doug Bates but have received no reply yet so I 
thought I would expand to a wider source.

I've been trying to estimate linear mixed effect models in lmer() from the 
lme4 package using the weights option.  The help and code for lmer() 
suggest to me that this is implemented but I can't seem to get it to do 
anything with weights = , no error message reported it just seems to 
ignore the weights option as my weighted and unweigted models are 
identical in all summary statistics.  Below is an example of my model 
code.   The weights are lengths of observational bouts on wild horse 
behavior.   So is weights not really yet implemented or am I missing 
something? 


m2.wt- lmer(HD_RATE ~ NUM_PAIR + 
(NUM_PAIR|HMA),data=out2.5,weights=HOURS/max(HOURS), 
contrasts=list(NUM_PAIR=contr.treatment),family=gaussian)

Brian

Brian S. Cade, PhD

U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO  80526-8818

email:  [EMAIL PROTECTED]
tel:  970 226-9326
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[R] surface plotting

2008-06-16 Thread Floyd poole

I am new to R  and what to find out which package would be best to create a
surface plot, 2d or 3d if possible. I have a matrix (depth.dat) which has
over 15k depth measurements from an ultrasonic corrosion mapping
application. These depth measurements only have one axis (x) but I have used
Octave and gnuplot providing decent results, but R seems to have a web
interface with PERL which is what I am looking for.

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Re: [R] Plotting 3d data?

2008-06-16 Thread Duncan Murdoch


On 6/16/2008 1:32 PM, Wesley Tansey wrote:

Hi all!

 


I'm very new to R, and I'm having trouble figuring out how to go from a file
of points that I have to a 3d surface plot of the data. I typically have
something like this:

 


X  Y  Z

0.0050.02334.45

0.0035  0.63  28.48

.

 


I've tried looking at the persp and wireframe packages, and the rgl package,
but I can't seem to figure out how to use any of them. I tried to take the
rgl.surface3d example and use it for myself, and this is what I have so far:

 


setwd(.)

 


data - read.csv(data.csv,header=T)

 


x - data$X

y - data$Y

z - data$Z

 


open3d()

surface3d(x, y, z) 

 


That gives me the following error:

 

Error in rgl.surface(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,  : 


  y length != x rows * z cols

 


I have no idea what it is trying to tell me there. Would anyone mind helping
me out?


surface3d (and rgl.surface) expect you to have data that corresponds to 
a grid defining the surface.  So you need at least one of the parameters 
(z in surface3d, y in rgl.surface) to be a matrix with rows 
corresponding to x values and columns corresponding to the other variable.


If your data is really a bunch of points, not on a grid, then you need 
to use plot3d() to plot it as points, or convert it to a surface.  The 
interp() function in the akima package can do that.  (There are lots of 
other possibilities too.)


Duncan Murdoch

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Re: [R] Annoyance in as.numeric

2008-06-16 Thread markleeds


 hi: you can do below but i don't know if it's worth it ?

newx - data.matrix(data.frame(x))
print(newx)


On Mon, Jun 16, 2008 at  4:22 PM, Alberto Monteiro wrote:


Why does as.numeric convert matrices and arrays to vectors?

as.numeric(matrix(c(1, 2, 3, 4), 2, 2))
[1] 1 2 3 4

I could only figure out ugly ways to bypass this, like:

x - matrix(c(1, 2, 3, 4), 2, 2)
array(as.numeric(x), dim = dim(x), dimnames = dimnames(x))

Alberto Monteiro

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Re: [R] surface plotting

2008-06-16 Thread Duncan Murdoch


On 6/16/2008 4:40 PM, Duncan Murdoch wrote:

On 6/16/2008 4:25 PM, Floyd poole wrote:

I am new to R  and what to find out which package would be best to create a
surface plot, 2d or 3d if possible. I have a matrix (depth.dat) which has
over 15k depth measurements from an ultrasonic corrosion mapping
application. These depth measurements only have one axis (x) but I have used
Octave and gnuplot providing decent results, but R seems to have a web
interface with PERL which is what I am looking for.


For static plots, scatterplot3d (in the scatterplot3d package) is best. 
  For rotatable surfaces, persp3d in the rgl package.  (See 
example(persp3d) for some examples.)


Sorry, I changed my mind in the middle of writing that and didn't edit 
properly.


For surfaces:

persp() (in the base graphics package) for static plots
persp3d() (in rgl) for rotatable plots

For 3D scatterplots:

scatterplot3d() (in scatterplot3d) for static
plot3d() (in rgl) for rotatable.

Duncan Murdoch

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Re: [R] in axis() suppressing axis line but keeping tick marks