Re: [R] Copula fitting

2006-10-23 Thread Uwe Ligges


Puskás László wrote:
 Hi, 
 
 Is anybody using Copula package for fitting copulas to own
 datas? I always got this message: 
 
 Error in qnorm(p, mean, sd, lower.tail, log.p) : 
 Non-numeric argument to mathematical function


Please specify version of package and R as well as the code that 
produces the error message. Please read the posting guide which asks you 
to do so.
If this is not an user error but a bug, please report to the package 
maintainer.

Uwe Ligges


 thanks,
 Psl
 
 
 ___
 Ossza meg barátaival fényképen megörökített élményeit.
 http:\\fototar.t-online.hu
 
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[R] disaggregating table

2006-10-23 Thread Jeff Miller
Hi all,

 

This should be easy, but I can't seem to figure it out.

 

I have a table like this named newtable

 

a1  a2  a3  a4  Cnts  Score

1 100 4  3.28

1 011 2  2.63

 

I want the following:

 

a1  a2  a3  a4  Cnts  Score

1 100 4  3.28

1 100 4  3.28

1 100 4  3.28

1 100 4  3.28

1 011 2  2.63

1 011 2  2.63

 

Actually, the Cnts column could be removed, but it doesn't matter if it
stays.

 

Anyone know how to disaggregate with Cnts as the index?

 

Thanks,

Jeff

 


[[alternative HTML version deleted]]

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Re: [R] disaggregating table

2006-10-23 Thread Prof Brian Ripley
On Sun, 22 Oct 2006, Jeff Miller wrote:

 Hi all,

 This should be easy, but I can't seem to figure it out.

ind - rep(1:nrow(newtable), times=newtable$Cnts)
newtable[ind, -5]

gives

 a1 a2 a3 a4 Score
11  1  0  0  3.28
1.1  1  1  0  0  3.28
1.2  1  1  0  0  3.28
1.3  1  1  0  0  3.28
21  0  1  1  2.63
2.1  1  0  1  1  2.63

Note the row names: data frames must have unique row names.

 I have a table like this named newtable

 a1   a2   a3   a4  Cnts Score
 1 100 4  3.28
 1 011 2  2.63

 I want the following:

 a1   a2   a3   a4  Cnts Score
 1 100 4  3.28
 1 100 4  3.28
 1 100 4  3.28
 1 100 4  3.28
 1 011 2  2.63
 1 011 2  2.63

 Actually, the Cnts column could be removed, but it doesn't matter if it
 stays.

 Anyone know how to disaggregate with Cnts as the index?

   [[alternative HTML version deleted]]

Please use properly formatted plain text, as we do ask in the posting 
guide.  I've had to reformat the double-spaced text provided before 
pasting the table.

 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] Lmer, heteroscedasticity and permutation, need help please

2006-10-23 Thread Alan Juilland
Hi everybody,

I'm trying to analyse a set of data with a non-normal response, 2 fixed 
effects and 1 nested random effect with strong heteroscedasticity in the 
model.

I planned to use the function lmer : lmer(resp~var1*var2 + (1|rand)) and 
then use permutations based on the t-statistic given by lmer to get 
p-values.

1/ Is it a correct way to obtain p-values for my variables ? (see below)

2/ I read somewhere that lme is more adequate when heteroscedasticity is 
strong. Do I have to use lme instead of lmer ?

3/ It is possible to fit a glm in lmer using family= Is it 
possible to use it in spite of hard heteroscedasticity ?

4/ A last question concerning SAS. My model appears to not converge in 
SAS, indicating a structure in the variance. Is it implying something 
in lmer or lme ?

Many Thanks

Here is the function for the permutation :

permut-function(data,vardep,var1,var2,var.random,nboot=10){

# initialisation
ms.var1-numeric(length=nboot)
ms.var2-numeric(length=nboot)
ms.var3-numeric(length=nboot)
var1.permut-numeric(length=length(var1))
var2.permut-numeric(length=length(var2))
var3.permut-numeric(length=length(var2))
var1-factor(var1)
var2-factor(var2)

# model initial
model.ini-lmer(vardep~factor(var1)*factor(var2)+(1|var.random),data=data)
vc-vcov(model.ini)
b - fixef(model.ini)
se- sqrt(diag(vc))
z- b/se
p- 2*(1-pnorm(abs(z)))

# boucle de permutation
for(i in 1:nboot){

#boucles de randomisation des traitements
for (j in 1:nlevels(var1)){
var2.permut[var1==j]-sample(var2[var1==j])}
for (j in 1:nlevels(var2)){
var1.permut[var2==j]-sample(var1[var2==j])}
data2-data.frame(vardep,var1,var2,var.random,var1.permut,var2.permut)

#test var1:
m1-lmer(vardep~factor(var1.permut)*factor(var2)+(1|var.random),data=data2)
ms.var1[i]-as.numeric(fixef(m1)/sqrt(diag(vcov(m1[2]
#test var2:
m2-lmer(vardep~factor(var1)*factor(var2.permut)+(1|var.random),data=data2)
ms.var2[i]-as.numeric(fixef(m2)/sqrt(diag(vcov(m2[3]
#test interaction:
m3-lmer(vardep~factor(var1.permut)*factor(var2.permut)+(1|var.random),data=data2)
ms.var3[i]-as.numeric(fixef(m3)/sqrt(diag(vcov(m3[4]
}

max.boot-c(NA,max(ms.var1),max(ms.var2),max(ms.var3))
min.boot-c(NA,min(ms.var1),min(ms.var2),min(ms.var3))
pval-c(NA,length(ms.var1[ms.var1=z[2]])/nboot,length(ms.var2[ms.var2=z[3]])/nboot,length(ms.var3[ms.var3=z[4]])/nboot)

res-cbind(b,se,z,p,permut.min=min.boot,permut.max=max.boot,permut.pval=pval)
par(mfrow=c(2,2))
hist(ms.var1);abline(v=z[2],col=RED);hist(ms.var2);abline(v=z[3],col=RED);hist(ms.var3);abline(v=z[4],col=RED)

return(res)
}



-- 
Alan Juilland – PhD Student
Department of Ecology and Evolution
Biophore, University of Lausanne
1015 Dorigny
Switzerland
Tel : ++41 21 692 41 74
Fax :  +41 21 692 41 65

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Re: [R] R tcl/tk

2006-10-23 Thread Peter Dalgaard
Charles Annis, P.E. [EMAIL PROTECTED] writes:

 Friends:
 
 I am a long-time R user, learning tcl/tk, and am tying myself in knots over
 something that should be simple.
   
 I want to create a frame and put that frame inside the toplevel frame.
 
 This works (i.e. it places text in col 1, and the corresponding entry box in
 column 2) and later frame.2 resides where it should in the toplevel frame.
 
 tkgrid(tklabel(frame.2, text=FACTOR column ), tkentry(frame.2,
 textvariable=FACTOR.column.value, width=2),  sticky=e)
 
 
 This doesn't work because I don't know how to tell tcl that answer.button1
 belongs in frame2:
 
 tkconfigure(answer.button1, variable=buttonValue, value=TRUE)
 
 tkgrid(tklabel(frame.2, text=Question?yes ), answer.button1,
 sticky=e)
   ^^^
 
 I know this can not be as hard as I am making it.

Possibly, but there seems to be some context missing, so I can't quite
grasp what you want to do. It does look a bit suspicious that you
aren't saving the objects generated by tklabel and tkentry though.

In general, the hierarchy between widgets is defined at widget
creation time, so why is there no 

answer.button1 - tkbutton(frame.2,..whatever..) 

?
-- 
   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] sqlSave, fast=F option, bug?

2006-10-23 Thread Brandt, T. (Tobias)
I've also reproduced this behaviour on WinXP + MS SQL Server and would agree
that it's a bug with the fast=FALSE option. 

Have the package maintainers been contacted about this? 

-Original Message-
From: [EMAIL PROTECTED] 
[mailto:[EMAIL PROTECTED] On Behalf Of chao gai
Sent: 18 October 2006 08:10 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] sqlSave, fast=F option, bug?

The bug spot seems to be the fast option
sqlDrop(chan,'test')
df - data.frame(T=1, S=10)
sqlSave(chan, df, test, rownames=F)
df - data.frame(S=20, T=2)
sqlSave(chan, df, test, rownames=F, append=T,fast=T) df - 
data.frame(S=30, T=3) sqlSave(chan, df, test, rownames=F, 
append=T,fast=F)
sqlFetch(chan,'test')
   T  S
1  1 10
2  2 20
3 30  3
On Linux  MySQL  


On Wednesday 18 October 2006 17:06, Dieter Menne wrote:
 Paul MacManus paul.macmanus at gmail.com writes:
  Using the fast=F option, sqlSave saves without matching 
column names.
  It looks like a bug to me..

 

  In other words, sqlSave didn't check column names, it 
simply mapped 
  column 1 to column 1 and column 2 to column 2.

 This seems to be a problem with SQLServer. The following 
example with 
 Access works correctly for me, even if I close the 
connection in between.
 (Win2k, R 2.4.0)

 Dieter Menne

 -
 library(RODBC)
 channel - odbcConnectAccess(db1.mdb) df - data.frame(T=1, S=10) 
 sqlSave(channel, df, test, rownames=F)
 #odbcClose(channel)

 #channel - odbcConnectAccess(db1.mdb) df - data.frame(S=20, T=2) 
 sqlSave(channel, df, test, rownames=F, append=T,fast=T)
 odbcClose(channel)

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[R] calculate area of outer polygon

2006-10-23 Thread Tord Snäll
Dear all,
Does anyone know of a function that calculates the area of the outer 
polygon constructed from a data frame with co-ordinates? For example,
library(splancs)
test = rbind(c(0,0), c(10,0),c(10,10),c(0,20))
plot(test)
areapl(test)
[1] 150
This is OK, but I want the same area to be returned (150) when adding 
yet another co-ordinate in the middle of the polygon:
test = rbind(c(0,0), c(10,0),c(10,10),c(0,20),c(5,5))
plot(test)
areapl(test)
[1] 100

I have also tested area.owin() in library(spatstat) but the co-ordinates 
of the outer polygon is one of the arguments.


Thanks!

Tord

-- 

NOTE, new e-mail and contact information:

Tord Snäll
Department of Conservation Biology
Swedish University of Agricultural Sciences (SLU)
P.O. 7002, SE-750 07 Uppsala, Sweden
Office/Mobile/Fax
+46-18-672740/+46-730-891356/+46-18-673537
E-mail: [EMAIL PROTECTED]
www.nvb.slu.se/staff_tordsnall

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[R] Adjusted residuals

2006-10-23 Thread Dominik Heinzmann
Dear users,
Given a equation of the form y=a(1-exp(bt), where a,b parameters and t=age
and given prevalences for t=1,2,3,4,5
What is a nice way to plot a kind of adjusted residuals at those values 
for t?
How should one scale them to account for small counts at some values for t?
Thanks

-- 
Dominik Heinzmann
Institute of Mathematics
University of Zurich

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Re: [R] Key colour problem with lattice plot

2006-10-23 Thread Alex Brown
Hi Geoffrey,

I replicated your problem, and I came up with the same solution as  
Deepayan, so I won't reprint it here but:

Can anyone explain why the key is inverted versus the bar order?

I'd also like to thank you for showing me a different approach to  
lattice formulae.  I would have used the following formulation:

ex2 = stack(example)
ex3 = cbind(country=rep(rownames(example),2),ex2)
barchart(country ~ values, ex3,group=ind, auto.key=TRUE)

-Alex Brown


On 22 Oct 2006, at 12:21, Geoff Russell wrote:

 Hi,

 I seem to have a key colour problem with a lattice barchart. The  
 colours on my
 key rectangle don't match the colours on the barchart.

 Here is my data frame:

 LandFill Ruminants
 United States (USA) .21428 5528.16
 France 200.527083 1299.87
 Australia 185.878368 2448.17
 Russian Federation 1752.8334 2024.29
 Argentina 283.98732 2567.02
 Brazil 1048.42248 8839.61
 Colombia 265.125 1307.61
 Mexico 981.023351 1814.89
 Ethiopia 9.38020424 1237.49
 Sudan 16.0184936 1796.67
 India 553.425328 12360.3
 Pakistan 47.1593928 2346.71
 China 455.680191 8041.79

 example
LandFill Ruminants
 United States (USA) .214280   5528.16
 France   200.527083   1299.87
 Australia185.878368   2448.17
 Russian Federation  1752.833400   2024.29
 Argentina283.987320   2567.02
 Brazil  1048.422480   8839.61
 Colombia 265.125000   1307.61
 Mexico   981.023351   1814.89
 Ethiopia   9.380204   1237.49
 Sudan 16.018494   1796.67
 India553.425328  12360.30
 Pakistan  47.159393   2346.71
 China455.680191   8041.79


 And here is my plotting statement.

 barchart(rownames(example)~example$LandFill+example 
 $Ruminants,type=c(p,h),
 key=list(text=list(c(Landfill,Enteric Fermentation)),
 rectangles=Rows(trellis.par.get(superpose.symbol),c(1,2))



 Cheers,
 Geoff Russell

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Re: [R] calculate area of outer polygon

2006-10-23 Thread Barry Rowlingson
Tord Snäll wrote:
 Dear all,
 Does anyone know of a function that calculates the area of the outer 
 polygon constructed from a data frame with co-ordinates? For example,

  If by 'outer polygon' you mean 'convex hull', then look at help(chull).

Barry

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[R] How can I calculate adjusted means?

2006-10-23 Thread Gustaf Granath
Hi,
I have been looking for help about adjusted means but it seems like  
there is no specific information about it.

My problem is following:
In a 2 level ANOVA my model is - anova(Y~treat1+treat2+covarX)
I have 4 factors in treat1 and 2 factors in treat2.

Now I want the adjusted means for my measured variable. I.e. taking  
into account the covariate variable.

How can I do that?

Regards,
Gustaf Granath, phd student

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Re: [R] getting the by() function to work properly

2006-10-23 Thread Petr Pikal
Hi

maybe somebody can help you more but in meantime I would recommend to 
debug your function with

debug(f3)

and call by(ss3, sub, f3) to see where exactly comes the error from.

HTH
Petr
 
On 22 Oct 2006 at 22:49, Jancik, Jasper F. wrote:

Date sent:  Sun, 22 Oct 2006 22:49:23 -0400
From:   Jancik, Jasper F. [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] getting the by() function to work properly

 Hola everyone,
 
 Solving this issue would be a huge time saver for analyzing my thesis
 data, so I'll try to be as concise as possible and hope that someone
 can give me a hand.
 
 I am trying to run a function that I created, which works fine until I
 try to use it with the by function.  That function is listed at the
 bottom (f3) just in case that might be the issue.  Essentially I have
 compiled a large list of data coming from about 30 subjects with 88
 trials each and I am trying to get that function to apply to each
 subject without having to manually attach each subject's data.  The
 entire filename is 'ss2' so when i run ss2f3=f3(ss2) ss2f3
 everything is fine.  However when I try to use the by() function:
 by(ss3, sub, f3) #sub=subject header in the data I get the error
 message:  Error in var(x, na.rm = na.rm) : missing observations in
 cov/cor  There are no missing observations anywhere in the data so
 I'm really confused as to how this is happening.  My professor thought
 it might have something to do with the variance calculations in the
 function, so when we had it remove bad data the function would run,
 however it did not calculate everything and it erroneously gave the
 exact same calculations for each subject.
 
 I even get the message when I try to manually subset the data e.g.: 
 s2f3=f3(alldata[sub==2,]) Error in var(x, na.rm = na.rm) : missing
 observations in cov/cor
 
 I hope I've provided all the necessary info (the data set is attached
 just in case anyone has the time to make it work), but I'd be glad to
 give more if its needed.  I would really appreciate any suggestions at
 all.  Thanks in advance.
 
 
 
 f3=function(x) {
 + 
 +  twodied=x[died==2,]; 
 +  twosaved=x[saved==2,]; 
 +  hit2=x[saved==2  choice==+,]; 
 +  fa2=x [died==2  choice==+,];
 +  d2=(qnorm(((nrow(hit2))/(nrow(twosaved-qnorm((nrow(fa2))/(nrow(
 +  twodied; hitrate=((nrow(hit2))/(nrow(twosaved)));
 +  FArate=((nrow(fa2))/(nrow(twodied)));
 +  zhit2=qnorm(((nrow(hit2))/(nrow(twosaved;zfa2=qnorm((nrow(fa2))
 +  /(nrow(twodied)));B2= -.5*(zhit2+zfa2); tp2=mean(twosaved$resp);
 +  ta2=mean(twodied$resp); tpsd2=sd(twosaved$resp);
 +  tasd2=sd(twodied$resp); dmj2=((tp2-ta2)/((tasd2+tpsd2)/2));
 +  
 + L=list(d2=d2,B2=B2,dmj2=dmj2,tpsd2=tpsd2,tasd2=tasd2,tp2=tp2,ta2=ta2
 + ,hitrate=hitrate,FArate=FArate) }
 
 
 Jasper Jancik
 910.352.7546
 
 
 

Petr Pikal
[EMAIL PROTECTED]

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Re: [R] Book for Maximum Likelihood Methods in R

2006-10-23 Thread Ted Harding
On 23-Oct-06 Ben Bolker wrote:
 Milton Cezar Ribeiro milton_ruser at yahoo.com.br writes:
 
 
 Hi R-guys, 
 
 Is there a good book for maximum likelihood methods in R? 
 I´m looking mainly to Ecological approaches to apply ML.
 
 
 ** BLATANT PLUG warning **:
 
 I'm working on a book on statistical/ecological
 modeling in R, to be published by Princeton University
 Press sometime in (?) 2007, that focuses on constructing
 maximum likelihood models.  I have a draft up on
 my web site at http://www.zoo.ufl.edu/bolker/emdbook .
 
   I would be very grateful if other R aficionados
 wanted to have a look and tell me what they thought
 (keeping the target audience in mind --- ecology
 students and researchers who know a bit of classical
 statistics and want to be able to do more by
 rolling their own in R)

Ben, your plug is welcome. There is a hole to be filled
in this area, and I'm looking forward to checking your
plug for shape and size!

A question or two: I notice that one can get the whole book
in one shot from

  http://www.zoo.ufl.edu/bolker/emdbook/book.pdf

rather than piece by piece from the individual chapter
links (though the revision status may differ). But must
one also get the Exercises, R code, Data and Scripts
separately?

And/or can one use wget for it all?

Thanks, Ben, and best wishes for its future!
Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 23-Oct-06   Time: 10:31:29
-- XFMail --

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[R] Cube plots

2006-10-23 Thread Robert Kinley
Hi

does anyone know of any  R  functions or packages for the following :-
 . generating cube plots for displaying the results from a response 
surface experiment design 
 . generating ternary plots from a mixture design
 . tetrahedral plot or quaternary plot for displaying results from a four 
factor mixture experiment

thanks  Bob Kinley
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Re: [R] Cube plots

2006-10-23 Thread Duncan Murdoch
On 10/23/2006 7:17 AM, Robert Kinley wrote:
 Hi
 
 does anyone know of any  R  functions or packages for the following :-
  . generating cube plots for displaying the results from a response 
 surface experiment design 
  . generating ternary plots from a mixture design
  . tetrahedral plot or quaternary plot for displaying results from a four 
 factor mixture experiment

persp (in graphics) or persp3d (in rgl) can do surface plots; is that 
what a cube plot is?

I'm sure I've seen mixture plots before; try RSiteSearch or look in the 
R Graphics Gallery.

I don't think I've see tetrahedral plots, but the strategy above might 
turn them up.

Duncan Murdoch

 
 thanks  Bob Kinley
   [[alternative HTML version deleted]]
 
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Re: [R] R tcl/tk

2006-10-23 Thread Charles Annis, P.E.
Peter:

Thank you for your response.  I apologize for not sending a reproducible
example.  I have attached a reproducible example of the problem here.  This
is a much abbreviated version of my problem and even so it's rather long for
an example.  Run with 

tcl.tk.example(PROBLEM = TRUE)
tcl.tk.example(PROBLEM = FALSE)

to turn the problem on and off.

I have also attached a tcl.tk.example.txt file since these long inclusions
within an R-help note often have the line feeds missing.

Thank you for your insights.

Charles Annis, P.E.
###


tcl.tk.example - function(PROBLEM = TRUE) {
require(tcltk)
t2 - tktoplevel()

tktitle(t2) - Hit/Miss
group.button1 - tkradiobutton(t2)
group.button2 - tkradiobutton(t2)

repeated.measures.value - tclVar(3)
l.repeated.measures.value - tklabel(t2, text=   hit/miss column(s) )
e.repeated.measures.value - tkentry(t2,
textvariable=repeated.measures.value, width=5, borderwidth = 3,
relief=sunken)

tkgrid(tklabel(t2, text=Hit/Miss POD Setup), columnspan=2, sticky=n)

tkgrid.configure(l.repeated.measures.value, e.repeated.measures.value,
sticky=e)
tkgrid.configure(e.repeated.measures.value, sticky=w, padx=5)



frame.2 - tkframe(t2, borderwidth = 3, width=150, height=100,
relief=ridge)
GROUP.column.value - tclVar(4) 
l.GROUP.column.value - tklabel(frame.2, text=GROUP column )
e.GROUP.column.value - tkentry(frame.2, textvariable=GROUP.column.value,
width=2, borderwidth = 3, relief=sunken)
l.group.value.yes - tklabel(frame.2, text=group data?yes )
l.group.value.no - tklabel(frame.2, text=NO )
group.buttonValue - tclVar(GROUP.DATA)
tkconfigure(group.button1, variable=group.buttonValue, value=TRUE)
tkconfigure(group.button2, variable=group.buttonValue, value=FALSE)
if(PROBLEM){
### group.button1 not placed within frame.2 because I don't know how to
indicate frame.2
tkgrid(l.group.value.yes, group.button1, sticky=e)
tkgrid(group.button1, sticky=w)
tkgrid(l.group.value.no, group.button2, sticky=e)
tkgrid(group.button2, sticky=w) } else {
### second dummy lable is correctly placed because I can tell it that it
belongs in frame.2
tkgrid(l.group.value.yes, tklabel(frame.2, text=dummy), sticky=e)
tkgrid(l.group.value.no, tklabel(frame.2, text=dummy), sticky=e)
}
tkgrid(l.GROUP.column.value, e.GROUP.column.value,  sticky=e)
tkgrid(e.GROUP.column.value,  sticky=w, padx=5, pady=5)
tkgrid(frame.2, columnspan=2)


OnOK - function()
{
hit.miss.columns - eval(parse(text=paste(c(, substr(gsub(pattern =
-, replacement=:,
as.character(tclvalue(repeated.measures.value))),start=1,stop=1), 
n.columns - length(hit.miss.columns)
GROUP.column -
as.double(as.character(tclvalue(GROUP.column.value)))
repeated.measures.GROUP.names   -
as.character(tclvalue(repeated.measures.GROUP.value))

###read.hit.miss.input()
tkdestroy(t2)
}
OK.but - tkbutton(t2,text= OK , command=OnOK, borderwidth = 3,
relief=raised)
tkgrid(tklabel(t2,text=  Must click OK\nto register changes.  ),
(OK.but), sticky=e)
tkgrid(OK.but, sticky=w)
tkfocus(t2)
}
tcl.tk.example(PROBLEM = TRUE)
tcl.tk.example(PROBLEM = FALSE)

##



Charles Annis, P.E.

[EMAIL PROTECTED]
phone: 561-352-9699
eFax:  614-455-3265
http://www.StatisticalEngineering.com
 

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Peter Dalgaard
Sent: Monday, October 23, 2006 3:35 AM
To: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] R tcl/tk

Charles Annis, P.E. [EMAIL PROTECTED] writes:

 Friends:
 
 I am a long-time R user, learning tcl/tk, and am tying myself in knots
over
 something that should be simple.
   
 I want to create a frame and put that frame inside the toplevel frame.
 
 This works (i.e. it places text in col 1, and the corresponding entry box
in
 column 2) and later frame.2 resides where it should in the toplevel frame.
 
 tkgrid(tklabel(frame.2, text=FACTOR column ), tkentry(frame.2,
 textvariable=FACTOR.column.value, width=2),  sticky=e)
 
 
 This doesn't work because I don't know how to tell tcl that answer.button1
 belongs in frame2:
 
 tkconfigure(answer.button1, variable=buttonValue, value=TRUE)
 
 tkgrid(tklabel(frame.2, text=Question?yes ), answer.button1,
 sticky=e)
   ^^^
 
 I know this can not be as hard as I am making it.

Possibly, but there seems to be some context missing, so I can't quite
grasp what you want to do. It does look a bit suspicious that you
aren't saving the objects generated by tklabel and tkentry though.

In general, the hierarchy between widgets is defined at widget
creation time, so why is there no 

answer.button1 - tkbutton(frame.2,..whatever..) 

?
-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, 

[R] Recurrence plot in package tseriesChaos

2006-10-23 Thread Joydeep

Hi,
I'm a new user of the R package, and I need to generate recurrence plots for a 
set of 56 time series data. The package tseriesChaos has the function recurr 
that gives a plot. But what I really need are the recurrence plot parameters 
that are calculated from each plot (like %recur, %det, trend, etc...) and that 
can be stored dynamically in a output file for all the 56 datasets. 
It seems that the function recurr does not calculate these parameters. Is there 
another function in the package that does so? If not, is there another package 
that deals with recurrence quantification analysis? I would be very glad to 
have this problem solved in R itself.

Thanking in advance,
Joy

__
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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] Cube plots

2006-10-23 Thread hadley wickham
 I don't think I've see tetrahedral plots, but the strategy above might
 turn them up.

There was a userR presentation on them -
http://www.r-project.org/useR-2006/Abstracts/Matthews.pdf

With a little work, you could also do them in GGobi
(http://www.ggobi.org), which would generalise to more than 3d.  There
are some hints on how to do this (and why it might be useful) in an
old xgobi paper:
http://www.research.att.com/areas/stat/xgobi/papers/xgobi96.pdf

Regards,

Hadley

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Re: [R] Multilevel model (lme) question

2006-10-23 Thread Douglas Bates
On 10/22/06, Lukas Rode [EMAIL PROTECTED] wrote:
 Thanks Douglas for the detailed explanation; it was very helpful (I'll
 definitely give lme4 a try as well).
 For the record: I also found it helpful to play with model.matrix (e.g. p.
 14ff and p.28 of the book by Jose Pinheiro and Douglas) to see how the
 formulas for fixed and random effects are related to the LairdWare'82
 formulation for mixed-effects models.

 Now I have one follow-up question, about model selection:

 From your advice, I figure that the suggested procedure is to build a number
 of models (with differences in fixed and/or random effects) and then
 comparing them to select the most adequate one.
 When using anova(model1, model2) to compare two models that differ in both
 fixed and random effects, I get a warning message: Fitted objects with
 different fixed effects. REML comparisons are not meaningful. in:
 anova.lme(model1, model2).

You can compare models with different fixed-effects specifications if
you use method = ML.  The REML criterion does not behave like a
likelihood when you change the fixed-effects  specification.

Generally if you are using lme (and, as in your case, you have a
simple specification like (1|Subject) for the random effects) then you
can use anova applied to the model with the main effects and
interactions to test for the significance of the interaction term.  If
you decide to eliminate that term then you can test for the other
terms.

With lmer fits I recommend checking a Markov chain Monte Carlo sample
from the posterior distribtuion of the parameters to determine which
are signification (although this is not terribly well documented at
the present time).

 So apparently this means that anova only works for comparing models with
 same fixed-effects component but different random effects, right?
 So what is then the correct way to compare models with different fixed
 and/or random effects?
 For example, would it be correct to simply choose the model with the lowest
 AIC/BIC? (this seems to be a better choice than selecting by logLikelihood
 since AIC/BIC penalize higher #parameters, whereas likelihood does not).

Some people like to use AIC/BIC.  I find the definition of the cut-off
points to be rather arbitrary.  Also, one will frequently have the
model selection by AIC and by BIC disagree.

 Or is there another, better way?

 Many thanks so far,
   Lukas



 On 10/22/06, Douglas Bates [EMAIL PROTECTED] wrote:
  On 10/21/06, Lukas Rode [EMAIL PROTECTED] wrote:
   Dear list,
  
   I'm trying to fit a multilevel (mixed-effects) model using the lme
 function
   (package nlme) in R 2.4.0. As a mixed-effects newbie I'm neither sure
 about
   the modeling nor the correct R syntax.
 
  You may also want to consider using the lmer function from package
  lme4.  It's a later version of lme for R and is generally faster than
  lme.  I'll show the syntax for both.
 
   My data is structured as follows: For each subject, a quantity Y is
 measured
   at a number (= 2) of time points. Moreover, at time point 0
 (baseline), a
   quantity X is measured for each subject (I am interested to see whether
 X,
   or log(X), is a predictor of Y). I saw in some examples that
 time-invariant
   predictors should be repeated for all rows of a subject, which is why I
   copied the baseline value of X also to time points  0.
 
  That's correct.  For both lme and lmer the data need to be in the
  'long' form, also called the 'person-period' form.  You can reshape
  the data by hand, as you have done, or you can use the 'reshape'
  function or the more general capabilities in Hadley Wickham's reshape
  package.
 
   The resulting data frame looks like this:
 
   Grouped Data: Y ~ TIME | Subject
   Y TIMESubject X.Baseline
   9   0.0 11084
   7   3.7 11084
   11 0.0 27150
   8   9.2 27150
 
   Intra-subject trajectories of Y very close to linear. I'd like to check
   whether slope (and maybe also offset) of this line are (in part)
 predicted
   by X.baseline.
 
  I would say that this question would be addressed as part of the
  fixed-effects specification.  You would be comparing a model with main
  effects for TIME and X.Baseline and their interaction to various
  submodels.  In the S language formula notation these fixed effects
  specifications are
 
  Y ~ TIME * X.Baseline  # main effects for TIME and X.Baseline and
  their interaction
  Y ~ TIME + X.Baseline + TIME:X.Baseline # equivalent to first model
  Y ~ TIME + X.Baseline # main effects but no interaction.  That is
  X.Baseline does not affect slope
  Y ~ TIME + TIME: X.Baseline # X.Baseline affects slope (wrt TIME) but not
 offset
  Y ~ TIME # X.Baseline has no effect on Y
 
   Thus, I think the multilevel model specification should be as follows (i
 =
   subject, j=measurement):
   y_ij = \beta_i  + b_i * 

[R] likelihood question not so related to R but probably requires the use of R

2006-10-23 Thread Leeds, Mark \(IED\)
I have a question and it's only relation to R is that I probably need R
after I understand what to do.
 
Both models are delta y_t = Beta +  epslion 
and suppose I have a null hypothesis and alternative hypothesis
 
 
H_0 :   delta y_t = zero+ epsilonepsilon is normal ( 0,
sigmazero^2 )
 
 
H_1 delta y_t = beta   + epsilon  epsilon is normal (
sigmabeta^2 )
 
 


--
 
so, i calculate the MLE's under the null and the alternative as :
 
under H_0   beta hat = 0 and   sigmazero^2 hat = sum
over t ( delta y_t - zero )^2/ (n-1)
 
 
under H_1   beta hat =  ( sum of delta y_t ) /nand
sigmabeta^2 = sum over t ( delta y_t - beta hat )^2/(n-1)
 


---
 
what i have blanked out on is how i take the estimates above and test
which model is more likely given the data ?
I think I used to know this so I apologize if this is a stupid question.
I used to take my estimates and th use dnorm or pnorm or one of those
but
I can't remember what I did and I can't find my old code.  thanks.


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


[R] FW: Debug package question

2006-10-23 Thread Matthew Dowle

Dear list,

I received the response below from the package author of 'debug'.  I
post it to the list, with Mark's approval, in case it is useful to
others too.

Regards, Matthew


-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
Sent: 22 October 2006 23:53
To: Matthew Dowle
Cc: [EMAIL PROTECTED]
Subject: RE: Debug package question


Hi Matthew

That's an interesting point. If it's your own code, my recommendation
would be to predefine the function and then mtrace it, as you suggest.
But I can see that this isn't always easy, e.g. if you're trying to
debug a system function.

Because every little operation in R is actually a function call, it's
probably not very useful for 'mtrace' to have a step-into mode. [Anyway,
whether useful or not, it doesn't have one! ;)] 

The simplest way to get what you're after-- but it might not work--
would be to 'mtrace(apply)' and then call 'mtrace( FUN)' manually when
the debug window for 'apply' appears. The snag is that 'apply' is quite
likely to feature somewhere in the workings of 'debug' (I can't
remember!) so you might hang the thing altogether. Also, if your own
function happens to call 'apply' (directly or indirectly) you will get
stuck with a lot of debug windows.

A more robust, complicated, and devious way to get what you're after, is
to write a version of 'apply' that automatically calls 'mtrace' on its
function-argument and then invokes the real 'apply'. Something like
this:

mtrace.apply - function( X, MARGIN, FUN, ...) {
  FFF - FUN # don't know whether this is necessary
  mtrace( FFF) # naughty-- should really arrange for name to be unique
  mc - match.call( expand.dots=TRUE)
  mc$FUN - FFF
  mc[[1]] - quote( apply)
  eval( mc, envir=parent.frame())
}

Then call 'mtrace.apply' instead of 'apply'.

BTW: if your pass-in function isn't crashing the first time it's called,
you might find it useful to use the 'bp(1,F)' trick the first time it
appears in a debug window-- this means that it won't pop the window up
and wait for input until/unless there's a crash.

Hope this helps-- and glad you like the package

Mark

Mark Bravington
CSIRO Mathematical  Information Sciences
Marine Laboratory
Castray Esplanade
Hobart 7001
TAS

ph (+61) 3 6232 5118
fax (+61) 3 6232 5012
mob (+61) 438 315 623
 

 -Original Message-
 From: Matthew Dowle [mailto:[EMAIL PROTECTED]
 Sent: Friday, 20 October 2006 8:20 PM
 To: Bravington, Mark (CMIS, Hobart)
 Subject: Debug package question
 
 
  Hi Mark,
  
  I've been using your excellent debug for a while now.  Its truly
  excellent.
  
  The most common reason I find it useful is when R returns
 subscript
  out of bounds error.  R provides no context information in the
  message so I just mtrace() the function and your package 
 tells me the
  line straight away.  Perfect.
  
  I have one question ... when I use the apply() family to apply a
  function and the error occurs inside the function *defined* 
 inside the
  function,  e.g. :
  
  X = matrix(...)
  apply(X, 2, function(x) {
   lots of lines 
  ... M[m,] ...
   lots of lines 
  ... N[n,] ...
   lots of lines 
  ... O[o,] ...
   lots of lines 
  })
  
  Where it is the M[m,],  N[n,] or O[o,]  which generates the out of
  bounds on a particular iteration,  but you don't know which one.
  
  Mtrace() will skip over the apply() and say the same as R,
 that an out
  of bounds has occurred somewhere, but not where,  since
 mtrace has not
  been called on the function(x).
  
  One way to work around this, is to define the sub-function
 first with
  a name,   call mtace() on it,  then execute the apply.
 Is that the
  only way?   Is there a way to set mtrace() to automatically mtrace()
  any function it calls,  like step into and step over in other
  languages ?
  
  Regards,
  Matthew
  
  
   version
   _  
  platform i386-pc-mingw32
  arch i386   
  os   mingw32
  system   i386, mingw32  
  status  
  major2  
  minor1.1
  year 2005   
  month06 
  day  20 
  language R  
   
  
  and also
  
   version
   _   
  platform x86_64-unknown-linux-gnu
  arch x86_64  
  os   linux-gnu   
  system   x86_64, linux-gnu   
  status   
  major2   
  minor1.1 
  year 2005
  month06  
  day  20  
  language R   
   
  
  
 


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Re: [R] Recurrence plot in package tseriesChaos

2006-10-23 Thread Antonio, Fabio Di Narzo
Hi Joy.

2006/10/23, Joydeep [EMAIL PROTECTED]:

 Hi,
 I'm a new user of the R package, and I need to generate recurrence plots for 
 a set of 56 time series data. The package tseriesChaos has the function 
 recurr that gives a plot. But what I really need are the recurrence plot 
 parameters that are calculated from each plot (like %recur, %det, trend, 
 etc...) and that can be stored dynamically in a output file for all the 56 
 datasets.
 It seems that the function recurr does not calculate these parameters.

I'm sorry, but I don't ever know what these parameters are. The
'recurr' function only computes a matrix of distances in the state
space embedding-reconstructed from the given time series, and plots it
as an image plot.
I can only guess the parameters you're claiming can be computed from
the distance matrix.

Antonio.


 Is there another function in the package that does so? If not, is there 
 another package that deals with recurrence quantification analysis? I would 
 be very glad to have this problem solved in R itself.

 Thanking in advance,
 Joy

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


Re: [R] R tcl/tk

2006-10-23 Thread Peter Dalgaard
Charles Annis, P.E. [EMAIL PROTECTED] writes:

 Peter:
 
 Thank you for your response.  I apologize for not sending a reproducible
 example.  I have attached a reproducible example of the problem here.  This
 is a much abbreviated version of my problem and even so it's rather long for
 an example.  Run with 
 
 tcl.tk.example(PROBLEM = TRUE)
 tcl.tk.example(PROBLEM = FALSE)

...
 group.button1 - tkradiobutton(t2)

This would appear to be the problem. If you want button 1 to be a child
of frame.2, use  tkradiobutton(frame.2) (after creating the frame, of
course) 


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

__
R-help@stat.math.ethz.ch mailing list
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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 tcl/tk

2006-10-23 Thread Charles Annis, P.E.
D'oh!

You are right of course.  How very obvious these things are on the other
side of the solution.  I spent the entire weekend and completely missed it.

Thanks!

Charles Annis, P.E.

[EMAIL PROTECTED]
phone: 561-352-9699
eFax:  614-455-3265
http://www.StatisticalEngineering.com
 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Peter
Dalgaard
Sent: Monday, October 23, 2006 9:48 AM
To: [EMAIL PROTECTED]
Cc: 'Peter Dalgaard'; r-help@stat.math.ethz.ch
Subject: Re: [R] R tcl/tk

Charles Annis, P.E. [EMAIL PROTECTED] writes:

 Peter:
 
 Thank you for your response.  I apologize for not sending a reproducible
 example.  I have attached a reproducible example of the problem here.
This
 is a much abbreviated version of my problem and even so it's rather long
for
 an example.  Run with 
 
 tcl.tk.example(PROBLEM = TRUE)
 tcl.tk.example(PROBLEM = FALSE)

...
 group.button1 - tkradiobutton(t2)

This would appear to be the problem. If you want button 1 to be a child
of frame.2, use  tkradiobutton(frame.2) (after creating the frame, of
course) 


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

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


[R] R tcl/tk

2006-10-23 Thread Alberto Monteiro

This must be dumbest question ever asked, but...

When I ask help.search(tcltk), I get a reference to tcltk-package.

When I ask help(tcltk-package), I get rtfm(s) in 'R_HOME/Tcl/doc'.

But then when I ask help.search(R_HOME), I get nothing.

Is something missing here in those help pages?

Alberto Monteiro

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[R] Agnes Help

2006-10-23 Thread Vinicius Placco
Hi,

I'm trying to use the cluster package and I'm having some trouble... I
always get the message:

 myagnes - agnes(datafile.dat)
 Error: could not find function agnes

the package cluster is listed in the library() command, and I can reach the
help files from Agnes as well

I know that this can be some really easy thing to fix, but right now I have
no idea how to solve it.

Thanks.

Vinicius

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[R] Citation for tseries help pages for R

2006-10-23 Thread Jenny Vander Pluym
Good Morning/Afternoon!

I am editing a document and am not sure how to cite the Time series 
analysis and computational finance help pages that are accessible at 
this url: www.maths.lth.se/help/R/.R/library/tseries/html/00Index.html.  
Can anyone on this list help me with this?

Thank you for your time and have an excellent day!!

Jenny VP

-- 
Jenny Vander Pluym
NOAA, NOS, Center for Coastal Fisheries and Habitat Research
Applied Ecology and Restoration Research Branch
101 Pivers Island Rd.
Beaufort, NC 28516-9722 office: 252.728.8777 fax: 252.728.8784

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Re: [R] one is not one

2006-10-23 Thread Thomas Lumley
On Sat, 21 Oct 2006, Philipp Pagel wrote:

 On Sat, Oct 21, 2006 at 11:04:13AM +0200, Patrick Giraudoux wrote:
 Which means that actually sum(x) is NOT considered equal to 1...

 Any idea about what is going wrong?

 Others have already pointed out the problem and I would like to add a
 reference to a classical paper on this topic:

 David Goldberg
 What Every Computer Scientist Should Know About Floating-Point Arithmetic
 Computing Surveys, 1991

 Reprints can be found in many places online - e.g. here:


Yes, and one place is linked from the FAQ entry.


-thomas

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Re: [R] Citation for tseries help pages for R

2006-10-23 Thread Achim Zeileis
On Mon, 23 Oct 2006, Jenny Vander Pluym wrote:

 Good Morning/Afternoon!

 I am editing a document and am not sure how to cite the Time series
 analysis and computational finance help pages that are accessible at
 this url: www.maths.lth.se/help/R/.R/library/tseries/html/00Index.html.
 Can anyone on this list help me with this?

Cite the package itself using the citation provided by
  citation(tseries)

Best,
Z

 Thank you for your time and have an excellent day!!

 Jenny VP

 --
 Jenny Vander Pluym
 NOAA, NOS, Center for Coastal Fisheries and Habitat Research
 Applied Ecology and Restoration Research Branch
 101 Pivers Island Rd.
 Beaufort, NC 28516-9722 office: 252.728.8777 fax: 252.728.8784

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[R] problem of using list element in for cycle

2006-10-23 Thread Hu Chen
sorry, pressed sent by mistake.
for example
data - read.csv(data.txt)
data
V1  V2
1  YHR165C  CG8877
2  YJL130C CG18572
3  YDL171C  CG9674
4  YKR054C  CG7507
5  YDL140C  CG1554
6  YLR106C CG13185
7  YGL206C  CG9012
8  YNL262W  CG6768
9  YER172C  CG5931

typeof(data)
[1] list
for (i in 1:nrow(data)){
 cat(data[i,1]
   }

it'll not return things like YHR165C but number like 6,7,9..
is this a new feature of list? how to turn off it.
thanks

[[alternative HTML version deleted]]

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Re: [R] problem about using list element in for cycle

2006-10-23 Thread Hu Chen
sorry, pressed sent by mistake.
for example
data - read.csv(data.txt)
single
V1  V2
1  YHR165C  CG8877
2  YJL130C CG18572
3  YDL171C  CG9674
4  YKR054C  CG7507
5  YDL140C  CG1554
6  YLR106C CG13185
7  YGL206C  CG9012
8  YNL262W  CG6768
9  YER172C  CG5931

typeof(data)
[1] list
for (i in 1:nrow(data)){
 cat(data[i,1]
   }

it'll not return things like YHR165C but number like 6,7,9..
is this a new feature of list? how to turn off it.
thanks

On 10/23/06, Hu Chen [EMAIL PROTECTED] wrote:

 for example
 data - read.csv(data.txt)
 typeof(data)
 [1] list
 for (i in 1:nrow(data)){



[[alternative HTML version deleted]]

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[R] Agnes Help

2006-10-23 Thread Vinicius Placco
Hi

I'm trying to use the cluster package and I'm having some trouble... I
always get the message:

  myagnes - agnes(datafile.dat)
  Error: could not find function agnes

 the package cluster is listed in the library() command, and I can
reach the help files from Agnes as well

 I know that this can be some really easy thing to fix, but right now
I have no idea how to solve it.

 Thanks.

 Vinicius

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Re: [R] R tcl/tk

2006-10-23 Thread Duncan Murdoch
On 10/23/2006 10:11 AM, Alberto Monteiro wrote:
 This must be dumbest question ever asked, but...
 
 When I ask help.search(tcltk), I get a reference to tcltk-package.
 
 When I ask help(tcltk-package), I get rtfm(s) in 'R_HOME/Tcl/doc'.
 
 But then when I ask help.search(R_HOME), I get nothing.
 
 Is something missing here in those help pages?

R_HOME is a short form for the directory where you installed R. 
This is referring you to a directory on your system that holds 
documentation that's outside of R's help system.

Duncan Murdoch

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[R] problem about using list element in for cycle

2006-10-23 Thread Hu Chen
for example
data - read.csv(data.txt)
typeof(data)
[1] list
for (i in 1:nrow(data)){

[[alternative HTML version deleted]]

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Re: [R] Latex editor recommendations

2006-10-23 Thread Allen S. Rout
Tom Backer Johnsen [EMAIL PROTECTED] writes:

 This question is not oriented towards R, but is posted here because
 I have the impression that there are at least some Latex users among
 the contributors.  The question is: What editors for Latex are to be
 recommended?  I have located one:


EMACS.  

It's not just an editor, it's a religion.


- Allen S. Rout

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Re: [R] R tcl/tk

2006-10-23 Thread Gabor Grothendieck
On 10/23/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
 On 10/23/2006 10:11 AM, Alberto Monteiro wrote:
  This must be dumbest question ever asked, but...
 
  When I ask help.search(tcltk), I get a reference to tcltk-package.
 
  When I ask help(tcltk-package), I get rtfm(s) in 'R_HOME/Tcl/doc'.
 
  But then when I ask help.search(R_HOME), I get nothing.
 
  Is something missing here in those help pages?

 R_HOME is a short form for the directory where you installed R.
 This is referring you to a directory on your system that holds
 documentation that's outside of R's help system.


Note that R_HOME corresponds to the R.home() path
so you can do this on Windows (and the analogous operations
on other platforms):

 setwd(file.path(R.home(), Tcl, doc))
 dir()
[1] tcl84.cnt tcl84.GID TCL84.HLP tk84.cnt  TK84.HLP
 system(cmd /c start TCL84.HLP)
 system(cmd /c start TK84.HLP)

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Re: [R] R tcl/tk

2006-10-23 Thread Prof Brian Ripley
On Mon, 23 Oct 2006, Alberto Monteiro wrote:


 This must be dumbest question ever asked, but...

 When I ask help.search(tcltk), I get a reference to tcltk-package.

 When I ask help(tcltk-package), I get rtfm(s) in 'R_HOME/Tcl/doc'.

 But then when I ask help.search(R_HOME), I get nothing.

 Is something missing here in those help pages?

R_HOME is an environment variable: use Sys.getenv(R_HOME) to find its 
value (the same as R.home()), which will be the top-level directory of 
that particular R installation.

BTW, you have not been as careful as you require of others: your comments 
are only true if preceded by the missing 'library(tcltk)' and only if you 
are running on Windows.

-- 
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] character manipulation

2006-10-23 Thread Nicolas Degallier
Dear R'helpers,

I am reading lines in a .txt file
Each line is stocked into a n elements object, as this:



[958]  422 287 339  31 203 602 547 1026 500 366 346 227
[959]  410  67  11 220 110 451 562 598 732 163 163 220
[960]  179 513  95 186 102 595 333 1289 804 210 294 459
[961]  276 153 307 138 126 233 623 739 521 421 209  75
[962]   64 392  60  45 332 371 290 887 599 106  62 184
[963]  171 168 374  43  71 567 1110 799 960 189 339 279
[964]  151 353 648 150  85 422 883 962 923 296 228 273
[965]  213 805 403  43 271 445 463 846 1124 345 295  85
[966]  574 420 200  39 243 187 249 378 714 318 286 114
[967]   85  73 192  91 146 484 710 782 750 321 261  65
[968]   53 487  27  76 150 345 663 848 620 192 129 187
[969]  461 216  8 251 279 558 1012 838 1330 258 138 115
[970]   53  83  39 298 117 488 880 815 1087 390 266 355
[971]  225 184 173 121 167 416 411 667 605 172  60 157
[972]  318 138 552 192 100 461 450 852 190 361  17 192



Now, I would like to determine how many   (space) there are in each  
line ...

After to have wipes out the extra spaces (double spaces) with

  data1[i]-gsub(x=data1[i],pattern=  ,replacement=  
,fixed=FALSE,extended=FALSE)

I have attempted with regrexpr, sub, strsplit and many other  
character fonctions but I could not obtain the number of   in each  
line, in order to further do a test on it.

The answer may be trivial but at this time I gave up!!

Cheers

Nicolas Degallier

UMR 7159 / IRD UR182
Laboratoire d'Océanographie et du Climat, Expérimentation et  
Approches Numériques (LOCEAN)
Tour 45-55, 4e ét., case 100, 4 place Jussieu
75252  Paris Cedex 5  France
tél: (33) 01 44 27 51 57
fax: (33) 01 44 27 38 05

E-mail: [EMAIL PROTECTED]
pdf reprints (login=anonymous; password=[EMAIL PROTECTED]):  
ftp://ftp.lodyc.jussieu.fr/LOCEAN/ndelod

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Re: [R] character manipulation

2006-10-23 Thread jim holtman
Forgot the second part. To replace multiple spaces with one:

gsub( +,  , your.data)


On 10/23/06, Nicolas Degallier [EMAIL PROTECTED] wrote:

 Dear R'helpers,

 I am reading lines in a .txt file
 Each line is stocked into a n elements object, as this:



 [958]  422 287 339  31 203 602 547 1026 500 366 346 227
 [959]  410  67  11 220 110 451 562 598 732 163 163 220
 [960]  179 513  95 186 102 595 333 1289 804 210 294 459
 [961]  276 153 307 138 126 233 623 739 521 421 209  75
 [962]   64 392  60  45 332 371 290 887 599 106  62 184
 [963]  171 168 374  43  71 567 1110 799 960 189 339 279
 [964]  151 353 648 150  85 422 883 962 923 296 228 273
 [965]  213 805 403  43 271 445 463 846 1124 345 295  85
 [966]  574 420 200  39 243 187 249 378 714 318 286 114
 [967]   85  73 192  91 146 484 710 782 750 321 261  65
 [968]   53 487  27  76 150 345 663 848 620 192 129 187
 [969]  461 216  8 251 279 558 1012 838 1330 258 138 115
 [970]   53  83  39 298 117 488 880 815 1087 390 266 355
 [971]  225 184 173 121 167 416 411 667 605 172  60 157
 [972]  318 138 552 192 100 461 450 852 190 361  17 192



 Now, I would like to determine how many   (space) there are in each
 line ...

 After to have wipes out the extra spaces (double spaces) with

 data1[i]-gsub(x=data1[i],pattern=  ,replacement=
 ,fixed=FALSE,extended=FALSE)

 I have attempted with regrexpr, sub, strsplit and many other
 character fonctions but I could not obtain the number of   in each
 line, in order to further do a test on it.

 The answer may be trivial but at this time I gave up!!

 Cheers

 Nicolas Degallier

 UMR 7159 / IRD UR182
 Laboratoire d'Océanographie et du Climat, Expérimentation et
 Approches Numériques (LOCEAN)
 Tour 45-55, 4e ét., case 100, 4 place Jussieu
 75252  Paris Cedex 5  France
 tél: (33) 01 44 27 51 57
 fax: (33) 01 44 27 38 05

 E-mail: [EMAIL PROTECTED]
 pdf reprints (login=anonymous; password=[EMAIL PROTECTED]):
 ftp://ftp.lodyc.jussieu.fr/LOCEAN/ndelod

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




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

What is the problem you are trying to solve?

[[alternative HTML version deleted]]

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Re: [R] character manipulation

2006-10-23 Thread jim holtman
nchar(gsub([^ ], '', your.data))

This replaces all non-blanks with nulls leaving only blanks and then
determines number of characters.


On 10/23/06, Nicolas Degallier [EMAIL PROTECTED] wrote:

 Dear R'helpers,

 I am reading lines in a .txt file
 Each line is stocked into a n elements object, as this:



 [958]  422 287 339  31 203 602 547 1026 500 366 346 227
 [959]  410  67  11 220 110 451 562 598 732 163 163 220
 [960]  179 513  95 186 102 595 333 1289 804 210 294 459
 [961]  276 153 307 138 126 233 623 739 521 421 209  75
 [962]   64 392  60  45 332 371 290 887 599 106  62 184
 [963]  171 168 374  43  71 567 1110 799 960 189 339 279
 [964]  151 353 648 150  85 422 883 962 923 296 228 273
 [965]  213 805 403  43 271 445 463 846 1124 345 295  85
 [966]  574 420 200  39 243 187 249 378 714 318 286 114
 [967]   85  73 192  91 146 484 710 782 750 321 261  65
 [968]   53 487  27  76 150 345 663 848 620 192 129 187
 [969]  461 216  8 251 279 558 1012 838 1330 258 138 115
 [970]   53  83  39 298 117 488 880 815 1087 390 266 355
 [971]  225 184 173 121 167 416 411 667 605 172  60 157
 [972]  318 138 552 192 100 461 450 852 190 361  17 192



 Now, I would like to determine how many   (space) there are in each
 line ...

 After to have wipes out the extra spaces (double spaces) with

 data1[i]-gsub(x=data1[i],pattern=  ,replacement=
 ,fixed=FALSE,extended=FALSE)

 I have attempted with regrexpr, sub, strsplit and many other
 character fonctions but I could not obtain the number of   in each
 line, in order to further do a test on it.

 The answer may be trivial but at this time I gave up!!

 Cheers

 Nicolas Degallier

 UMR 7159 / IRD UR182
 Laboratoire d'Océanographie et du Climat, Expérimentation et
 Approches Numériques (LOCEAN)
 Tour 45-55, 4e ét., case 100, 4 place Jussieu
 75252  Paris Cedex 5  France
 tél: (33) 01 44 27 51 57
 fax: (33) 01 44 27 38 05

 E-mail: [EMAIL PROTECTED]
 pdf reprints (login=anonymous; password=[EMAIL PROTECTED]):
 ftp://ftp.lodyc.jussieu.fr/LOCEAN/ndelod

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




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

What is the problem you are trying to solve?

[[alternative HTML version deleted]]

__
R-help@stat.math.ethz.ch mailing list
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Re: [R] R tcl/tk

2006-10-23 Thread Alberto Monteiro
Gabor Grothendieck wrote:

 But then when I ask help.search(R_HOME), I get nothing.

 Is something missing here in those help pages?
 
 Note that R_HOME corresponds to the R.home() path

So I guess there _is_ something missing in those help pages, after
all! Instead of R_HOME, the help page should mention R.home()
evil grin

Alberto Monteiro

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[R] libz lib

2006-10-23 Thread Your EPEC ICT Team - Ricardo Rodríguez
Hi all,
 
Please, what libz must/could I use to build RMySQL in my Mac OS X 10.4.8 box? 
Thanks!
 
Best,
 
Ricardo
 
--
Ricardo Rodríguez
Your EPEC ICT Team

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Re: [R] problem about using list element in for cycle

2006-10-23 Thread Tony Plate
Your problem is that you are using cat() on a factor.  Use 
as.character() or format() to convert the factor to character data, 
which cat will then print in the way you want.

  x - data.frame(L=letters[1:3])
  x
   L
1 a
2 b
3 c
  x$L
[1] a b c
Levels: a b c
  cat(x$L, \n)
1 2 3
  cat(as.character(x$L), \n)
a b c
  cat(format(x$L), \n)
a b c
 


Hu Chen wrote:
 sorry, pressed sent by mistake.
 for example
 
data - read.csv(data.txt)
single
 
 V1  V2
 1  YHR165C  CG8877
 2  YJL130C CG18572
 3  YDL171C  CG9674
 4  YKR054C  CG7507
 5  YDL140C  CG1554
 6  YLR106C CG13185
 7  YGL206C  CG9012
 8  YNL262W  CG6768
 9  YER172C  CG5931
 
 
typeof(data)
 
 [1] list
 
for (i in 1:nrow(data)){
 
  cat(data[i,1]
}
 
 it'll not return things like YHR165C but number like 6,7,9..
 is this a new feature of list? how to turn off it.
 thanks
 
 On 10/23/06, Hu Chen [EMAIL PROTECTED] wrote:
 
for example
data - read.csv(data.txt)
typeof(data)
[1] list
for (i in 1:nrow(data)){


 
 
   [[alternative HTML version deleted]]
 
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 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Construction of Dataset for time varying COXPH analysis

2006-10-23 Thread yongchuan
Question: When survfit() function is used upon a coxph object, the 'n' returned 
is vastly smaller (n=6) than the number of distinct loans in the dataset used. 

I am trying to estimate a Cox proportional hazards model for a set of loans 
(over 6000) using using time varying covariates. For this 6000+ loans, I have 
some 62,000 different vectors representing the loans at different periods of 
time. I did the following:

resultsOpt - coxph(Surv(Start,Stop,PrepayDate)~ closingCoupon + loanPurposeId, 
data=latest)

which returned:

Call:
coxph(formula = Surv(Start, Stop, PrepayDate) ~ closingCoupon + 
loanPurposeId, data = latest)


   coef exp(coef) se(coef)z   p
closingCoupon 0.101  1.11   0.0271 3.73 1.9e-04
loanPurposeId 0.434  1.54   0.0624 6.96 3.3e-12

Likelihood ratio test=50.3  on 2 df, p=1.18e-11  n= 62297 


which seems fair.


However when I do:

 survfit(resultsOpt)
Call: survfit.coxph(object = resultsOpt)

  n  events  median 0.95LCL 0.95UCL 
  6 489 Inf Inf Inf 

the n = 6 when the number of distinct loans in the dataset is more like 6554.

My dataset looks like the following when I call it from within R:

 latest[1:5, 1:5]
  Start Stop PrepayDate modBalance closingCoupon
1 67  0   811.2769  8.35
2 78  0   811.2769  8.35
3 89  1   811.2769  8.35
4 45  0  2226.0825  8.70
5 56  0  2226.0825  8.70


where the first 3 rows present 1 loan, and the next 2 loans a new one. Am I 
putting the data in an incorrect format, and if so how should I correct it? 
Thanks much.

Pan

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Re: [R] Hi,about the png device

2006-10-23 Thread Alex Brown
Hi Li,

Sadly, the png and jpg devices are not really appropriate for your  
needs, since they rely on the X11 device for their operation.

In your case, the X11 device is not necessary, although if you did  
use it as you describe, it would let the png device work.

However, there are other ways to get png and jpg output from R which  
do not require the X11 device.

The pdf device gives beautiful pdf output, and the imagemagick tool  
can convert pdf into png.

The still-in-development GD device can also output png and it can do  
it faster than pdf + imagemagick, although it does not come with the  
standard R distribution - you will have to get it from CRAN, and I  
don't personally think the output is as nice, so far (although it has  
the potential to be better).

-Alex Brown

On 23 Oct 2006, at 14:29, lidaof wrote:

 Hi,
 Thank you for your reply:)

 i thought maybe that is because i use SSH to run R
 maybe i should run R in front of a linux server and under Genome or  
 KDE environment?

 i an a little confuse about that you said examine the pdf device  
 and the convert (imagemagick) tool for best quality png  
 output.,that seems i shoud install imagemagick fisr,then use pdf()  
 to generate png picture?

 and forgive my pool english:)

 Thanks again!

 Li

 On 10/23/06, Alex Brown [EMAIL PROTECTED] wrote:
 Hi, Li:

Your error message + the documention to png are sufficient  
 for you
 to work out what has happened here.

Here's a clue:

your error message:
'unable to open connection to X11 display '' '

?png ...
'They may not be usable unless the X11 display is available  
 to the
 owner of the R process.'

If you are using an interactive R session, you should  
 examine why
 your X11 session is not available.

If you are NOT using an interactive R session, I recommend you
 examine the pdf device and the convert (imagemagick) tool for best
 quality png output.

 -Alex Brown

 On 23 Oct 2006, at 03:45, lidaof wrote:

  Hi,
 
  i have a problem with png device
  that is my error log
   png(file=figure1.png);plot 
 (significant,type=scatter);dev.off()
  Error in X11(paste(png::, filename, sep = ), width, height,
  pointsize,
  :
  unable to start device PNG
  In addition: Warning message:
  unable to open connection to X11 display '' 
  i searched the mailing list,someone said because no libpng installed
  but i had installed the libpng and libpng-devel
  and i use Fedora Core4
  my R version is 2.4.0 and i use RPMs to install
 
  any solution?
 
  Thanks
 
  Li
  --
  DaoFeng,Li
  Box A1053 China Agriculture University,
  No.2 Yuanmingyuan Rd.West Haidian District
  Beijing,100094
  P.R.China
  Tel:+86-10-62814726
  Cell:+8613121987798
 
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 -- 
 DaoFeng,Li
 Box A1053 China Agriculture University,
 No.2 Yuanmingyuan Rd.West Haidian District
 Beijing,100094
 P.R.China
 Tel:+86-10-62814726
 Cell:+8613121987798


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Re: [R] getting the by() function to work properly

2006-10-23 Thread chao gai
Hi Jasper,

For me it seems that you are fooling around with attach() or something.
For one thing, it will not work like this,:
e.g. twodied=x[died==2,]is the first statement
died is not locally defined, hence it is gotten from somewhere else...
For the second thing, this will throw an error:
by(ss3, sub, f3)
sub is defind (unless attached from some data.frame), it is a statement.
Both these seem to suggest that somewhere there is a version of your dataframe 
attached from which values are being read. 
As a quick fix I made the following:
f3 - function(x) {
   attach(x)
   twodied - x[died==2,]
   twosaved=x[saved==2,]
   hit2=x[saved==2  choice==+,]
   fa2=x [died==2  choice==+,]
   
d2=(qnorm(((nrow(hit2))/(nrow(twosaved-qnorm((nrow(fa2))/(nrow(twodied
   hitrate=((nrow(hit2))/(nrow(twosaved)))
   FArate=((nrow(fa2))/(nrow(twodied)))
   zhit2=qnorm(((nrow(hit2))/(nrow(twosaved
   zfa2=qnorm((nrow(fa2))/(nrow(twodied)))
   B2= -.5*(zhit2+zfa2)
   tp2=mean(twosaved$resp)
   ta2=mean(twodied$resp)
   tpsd2=sd(twosaved$resp)
   tasd2=sd(twodied$resp)
   dmj2=((tp2-ta2)/((tasd2+tpsd2)/2))
   detach(x)
   
L=list(d2=d2,B2=B2,dmj2=dmj2,tpsd2=tpsd2,tasd2=tasd2,tp2=tp2,ta2=ta2,hitrate=hitrate,FArate=FArate)
   return(L)
}
f3(SS2)

by(SS2, SS2$sub, f3)

It did not throw an error for me.

As a note, could you not use table() or xtabs() for some of these 
calculations? All this nrow() stuff does not look very efficient.

Kees

On Monday 23 October 2006 04:49, Jancik, Jasper F. wrote:
 Hola everyone,

 Solving this issue would be a huge time saver for analyzing my thesis data,
 so I'll try to be as concise as possible and hope that someone can give me
 a hand.

 I am trying to run a function that I created, which works fine until I try
 to use it with the by function.  That function is listed at the bottom (f3)
 just in case that might be the issue.  Essentially I have compiled a large
 list of data coming from about 30 subjects with 88 trials each and I am
 trying to get that function to apply to each subject without having to
 manually attach each subject's data.  The entire filename is 'ss2' so when
 i run

 ss2f3=f3(ss2)
 ss2f3

 everything is fine.  However when I try to use the by() function:
 by(ss3, sub, f3)

 #sub=subject header in the data
 I get the error message:  Error in var(x, na.rm = na.rm) : missing
 observations in cov/cor  There are no missing observations anywhere in the
 data so I'm really confused as to how this is happening.  My professor
 thought it might have something to do with the variance calculations in the
 function, so when we had it remove bad data the function would run, however
 it did not calculate everything and it erroneously gave the exact same
 calculations for each subject.

 I even get the message when I try to manually subset the data e.g.:
  s2f3=f3(alldata[sub==2,])

 Error in var(x, na.rm = na.rm) : missing observations in cov/cor

 I hope I've provided all the necessary info (the data set is attached just
 in case anyone has the time to make it work), but I'd be glad to give more
 if its needed.  I would really appreciate any suggestions at all.  Thanks
 in advance.



 f3=function(x) {
 +
 +  twodied=x[died==2,];
 +  twosaved=x[saved==2,];
 +  hit2=x[saved==2  choice==+,];
 +  fa2=x [died==2  choice==+,];
 d2=(qnorm(((nrow(hit2))/(nrow(twosaved-qnorm((nrow(fa2))/(nrow(twodied)
))); +  hitrate=((nrow(hit2))/(nrow(twosaved)));
 FArate=((nrow(fa2))/(nrow(twodied))); + 
 zhit2=qnorm(((nrow(hit2))/(nrow(twosaved;zfa2=qnorm((nrow(fa2))/(nrow(t
wodied)));B2= -.5*(zhit2+zfa2); +  tp2=mean(twosaved$resp);
 +  ta2=mean(twodied$resp);
 +  tpsd2=sd(twosaved$resp);
 +  tasd2=sd(twodied$resp);
 +  dmj2=((tp2-ta2)/((tasd2+tpsd2)/2));
 +
 +
 L=list(d2=d2,B2=B2,dmj2=dmj2,tpsd2=tpsd2,tasd2=tasd2,tp2=tp2,ta2=ta2,hitrat
e=hitrate,FArate=FArate) }


 Jasper Jancik
 910.352.7546

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[R] Worm distribution :-)

2006-10-23 Thread Alberto Monteiro
I don't know if anyone has heard this tale, but it runs
more or like this way:

  A biologist was studing a (semi-spherical) cave where bats
  lives. He fell asleep in the cave, and he woke up in the
  middle of the night. Half-dreaming, he thought that he was
  outside, because glow-worms were living in the walls, and
  they looked like stars. However, he noticed that, unlike a
  real sky, these stars had no _pattern_: there were no
  recognized images like The Cross, a Scorpion, The Hunter, etc.

  When he woke up, he conjectured that the reason we _can_
  see patterns in the real sky is that the stars are randomly
  distributed, while the glow-worms tried to keep a distance
  to each other.

My question: what is the best way to generate a glow-worm-like
distribution? I imagine that using a Latin Hypercube would
leave too many holes in the (x,y) plane.

Alberto Monteiro

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Re: [R] Worm distribution :-)

2006-10-23 Thread rolf
You are talking about random point patterns, since the glow-worms
appear as ``stars'' (= points).   See the package ``spatial'' (which
comes with R) and try simulating a pattern using Strauss().

Or install the package ``spatstat'' from CRAN --- in this package
there is a variety of ways to simulate ``regular'' random point
patterns --- rMaternI, rMaternII, rSSI, and rmh (which simulates
several point pattern models depending on the specified Papangelou
conditional intensity function).

cheers,

Rolf Turner
[EMAIL PROTECTED]

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Re: [R] Worm distribution :-)

2006-10-23 Thread Ana Nelson
Have you heard of Boids? (google it if not) Depending upon what you  
want this for a simulation approach might be appropriate.

Regards,
Ana Nelson



On 23 Oct 2006, at 18:18, Alberto Monteiro wrote:

 I don't know if anyone has heard this tale, but it runs
 more or like this way:

   A biologist was studing a (semi-spherical) cave where bats
   lives. He fell asleep in the cave, and he woke up in the
   middle of the night. Half-dreaming, he thought that he was
   outside, because glow-worms were living in the walls, and
   they looked like stars. However, he noticed that, unlike a
   real sky, these stars had no _pattern_: there were no
   recognized images like The Cross, a Scorpion, The Hunter, etc.

   When he woke up, he conjectured that the reason we _can_
   see patterns in the real sky is that the stars are randomly
   distributed, while the glow-worms tried to keep a distance
   to each other.

 My question: what is the best way to generate a glow-worm-like
 distribution? I imagine that using a Latin Hypercube would
 leave too many holes in the (x,y) plane.

 Alberto Monteiro

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 guide.html
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[R] Methods of addressing multicollinearity in multiple linear regression with R

2006-10-23 Thread Ben Fairbank
In searching the R help archives I find a number of postings in April of
2005, but nothing since then.  If readers are aware of more recent
contributions addressing the problems arising from multicollinearity
(such as with the bootstrap, jackknife, or other techniques) I would
appreciate a reference.

 

Thank you,

 

Ben Fairbank

 


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Re: [R] Key colour problem with lattice plot

2006-10-23 Thread Deepayan Sarkar
On 10/23/06, Alex Brown [EMAIL PROTECTED] wrote:
 Hi Geoffrey,

 I replicated your problem, and I came up with the same solution as
 Deepayan, so I won't reprint it here but:

 Can anyone explain why the key is inverted versus the bar order?

Because the key goes from top to bottom, while cartesian coordinates
go from bottom to top. Neither can really be any other way by default,
if you think about it. I did promise someone that I would add an
option to reverse the order in key, but looks like I haven't done that
yet.

 I'd also like to thank you for showing me a different approach to
 lattice formulae.  I would have used the following formulation:

 ex2 = stack(example)
 ex3 = cbind(country=rep(rownames(example),2),ex2)
 barchart(country ~ values, ex3,group=ind, auto.key=TRUE)

Since we are talking about alternatives, here's another:

barchart(as.matrix(example), stack = FALSE, auto.key = TRUE)

-Deepayan


 -Alex Brown

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[R] Bayesian Information Criterion

2006-10-23 Thread Alexander Geisler
Hello!

I calculate a Bayesian logistic regression with the function MCMClogit 
from the package MCMCpack.

After the calculation I use the package CODA to test the model in 
respect to convergence.

Is there a compatible package to calculate the BIC (bayesian information 
criterion)? So that it is possible to write
BIC(mcmc object) with an mcmc object from the function MCMClogit.

I found the package nlme, but the function BIC in this package does not 
work with the mcmc object from the function MCMClogit.

Regards
Alex

-- 
Alexander Geisler * Moserhofgasse 36/1 * A-8010 Graz
StV Technische Mathematik | FakV TMTP
email: [EMAIL PROTECTED] | [EMAIL PROTECTED]
phone: +43 650 / 811 61 90

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[R] toggleDoc: clutter reduction mainly for S4 documentation files

2006-10-23 Thread Dieter Menne
The following perl-script is available from

http://www.menne-biomed.de/download/toggleDoc.zip

After it is tested on other systems (currently only Windows),
I plan to upload it to

http://cran.mirroring.de/other-software.html


toggleDoc.pl: Clutter reduction for S4-R documentation files.

Version 1.0, 23.10.2006


[EMAIL PROTECTED]

Perl script to optionally reduce clutter in the index file of S4 html and in
S3
files with many repetitive functions documentation. For an example using the
documentation of the coin package, see
http://www.menne-biomed.de/download/00Index.html .

Install and Convert

Put the file toggleDoc.pl and the javascript file toggleDoc.js into the
Rxxx/library
directory (you should  find the R.css file there).
For the protection of your valuable assets,  when you run this script
unmodified
new files 00IndexNew.html will be generated in each directory.

If you want to replace your original 00Index.html with the modified ones,
re-run
this script with the two commented lines a the bottom of the Perl script
commented out.

* Warning: this script has been tested under Windows only.
Please report required changes for Linux to [EMAIL PROTECTED]

What it does:

Only 00Index.html files are changed by adding a javascript reference and a
checkbox to switch between the Show All mode and the short mode; the
latter
is made the default on startup.
All references that have a , are hidden in the short mode, e.g.

dperm,IndependenceTest-method

In addition, repetitive similar entries with a common target file are
collapsed
to the first entry. For example, in R2HTML the following list only shows the
first HTML reference in short mode.

HTML  Outputs an object to a HTML file
HTML.anova.loglm  Outputs an object to a HTML file
HTML.aovOutputs an object to a HTML file

The program does not touch files with the string toggleDoc.js in the
header, so if you want to protect single files from being processed, simply
add this string in a comment to the html file (but not that this will be
overridden anyway on the next update.)


The function has been tested with IE and Firefox 1.5.

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[R] Plotting Text on a graph

2006-10-23 Thread Dan Chan
Hi,

 

I plotted 12 graphs on a page and output to a png file.  I wanted to
have an overall title for all 12 graphs.  What command can I use to do
this?  Below is the code that plotted the 12 graphs in one page.  

 

# FM10 by Month/ Export the plot to Wash2005FM10.png

png(file=Wash2005FM10.png,bg=white)

par(mfrow = c(3,4))

# Plot 12 Month of OFM10, FFM10 

for(i in 1:12) {

Temp - subset (Wash2005, MM == i)

plot (Temp$FM10.1,
Temp$FM10,main=month.name[i],xlim=c(5,25),ylim=c(5,25))

}

text(10,10, Overall Title)  'This will plot Overall Title at
coordinate (10,10) on the last graph, and so it is not what I want.  

dev.off()

 

Thank you. 

 

Daniel Chan

Meteorologist

Georgia Forestry Commission

P O Box 819

Macon, GA 

31202

Tel: 478-751-3508

Fax: 478-751-3465

 


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Re: [R] Plotting Text on a graph

2006-10-23 Thread David Barron
Have you thought of using split.screen instead?

Something like might work for you:

split.screen(c(1,1))
split.screen(c(3,4), 1)
for (i in 1:12) {
   screen(i+1)
   par(mar=c(2,2,4,2))
   plot(1:20,rnorm(20),xlab=,ylab=)
}
screen(1)
title(Overall title)


On 23/10/06, Dan Chan [EMAIL PROTECTED] wrote:
 Hi,



 I plotted 12 graphs on a page and output to a png file.  I wanted to
 have an overall title for all 12 graphs.  What command can I use to do
 this?  Below is the code that plotted the 12 graphs in one page.



 # FM10 by Month/ Export the plot to Wash2005FM10.png

 png(file=Wash2005FM10.png,bg=white)

 par(mfrow = c(3,4))

 # Plot 12 Month of OFM10, FFM10

 for(i in 1:12) {

 Temp - subset (Wash2005, MM == i)

 plot (Temp$FM10.1,
 Temp$FM10,main=month.name[i],xlim=c(5,25),ylim=c(5,25))

 }

 text(10,10, Overall Title)  'This will plot Overall Title at
 coordinate (10,10) on the last graph, and so it is not what I want.

 dev.off()



 Thank you.



 Daniel Chan

 Meteorologist

 Georgia Forestry Commission

 P O Box 819

 Macon, GA

 31202

 Tel: 478-751-3508

 Fax: 478-751-3465




 [[alternative HTML version deleted]]

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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[R] Plotting text on a graph

2006-10-23 Thread Erin Hodgess
Hi Daniel:

Here is a sample code that I have used:


z1 - rexp(100)
z2 - rexp(100)
z3 - rexp(100)
par(mfrow=c(2,2),oma = c(0, 0, 3, 0))
curve(dexp,from=0,to=5)
hist(z1,main=Size 5)
hist(z2,main=Size 15)
hist(z3,main=Size 30)
mtext(Densities, outer = TRUE, cex = 1.5)

And this will do the trick.

Hope this helps!

Sincerely,
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: [EMAIL PROTECTED]



Hi,

 

I plotted 12 graphs on a page and output to a png file.  I wanted to
have an overall title for all 12 graphs.  What command can I use to do
this?  Below is the code that plotted the 12 graphs in one page.  

 

# FM10 by Month/ Export the plot to Wash2005FM10.png

png(file=Wash2005FM10.png,bg=white)

par(mfrow = c(3,4))

# Plot 12 Month of OFM10, FFM10 

for(i in 1:12) {

Temp - subset (Wash2005, MM == i)

plot (Temp$FM10.1,
Temp$FM10,main=month.name[i],xlim=c(5,25),ylim=c(5,25))

}

text(10,10, Overall Title)  'This will plot Overall Title at
coordinate (10,10) on the last graph, and so it is not what I want.  

dev.off()

 

Thank you. 

 

Daniel Chan

Meteorologist

Georgia Forestry Commission

P O Box 819

Macon, GA 

31202

Tel: 478-751-3508

Fax: 478-751-3465

 


[[alternative HTML version deleted]]

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[R] Color eps/ps output from specialized plots?

2006-10-23 Thread Turgut Durduran
Hello,

First a disclaimer :) I am very new to using R.

I am generating some plots and eventhough I can get colored output in the 
encapsulated postscript files in the simplest of commands (e.g. plot(1:10,1:10, 
type=l, col=red) ), it does not work for the particular plots I want. It 
works on the screen.

Here is an example taken out from Mixed-Effects Models in S and S-Plus by 
Pinheiro and Bates;
   
library(nlme)
plot(CO2)
postscript(tmp.eps)
plot(CO2) 
dev.off()


The first plot has few colors on the screen, but postscript output is 
monochrome. I tried a more complex one too:
 postscript(test1.eps,horizontal = FALSE, onefile = FALSE, paper = special).

BTW, loading NLME chances the built-in (is it really?) data-set CO2  to a 
groupeddata with a formula 
uptake ~ conc | Plant , which could be recreated as:
 CO2 - groupedData( uptake ~ conc | Plant,data=CO2)


I could not dig out any pointers on figuring this issue out from the internet, 
specially since the simple plot command works.

Turgut

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[R] Extended CR model

2006-10-23 Thread Cougar Lawrence
I was wondering if someone could point me to the R package that implements
the partial proportional odds (PO) model discussed by Peterson and Harrell
(1990).  In Harrell (2001), the model is explained in Chapt. 13.

Respectfully,

Frank R. Lawrence

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[R] Help with recode and factor functions

2006-10-23 Thread Chris Linton
I have a data set with seven inputs.  Four of which are categorical.  For my
midterm, my professor wants us to scale all the inputs.  This means, I
pressume, that I have to use 'recode' or 'factor' to transform the
categorical data in numerical.  For example, one input variable is
'race=(b,w,h,o)'.  I just want to assign a numerical value to all
'b,w,h,o'.  I thought 'recode' should do this, but it doesn't work.  Here's
the code I'm using for recode:

recode(race, b='1';w='2';h='3';o='4')

this is the error I get:
Error in eval(expr, envir, enclos) : object o not found


It's not that there's no o.  If I change the order or combination of the
variables, it always can't find one of them.

I could also use 'factor', from what I hear.  But, I looked at the help
section on this function and I ended up more confused.


How do I code it so these variables take on numerial values?  I need to be
able to use:

race.centered = race - mean(race)


This scaling code doesn't really make sense if the values of 'race' are
non-numerical.  I might end up dividing by 2 SD's as well.  But, I don't
know if I need to.  I'll have to do some more reading.


Thank you for your help!

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[R] The Censummary Function...

2006-10-23 Thread nand kumar
Hi Forum,

I am new and learning this research tool.
was trying to run some tests on a censored dataset using the NADA library. 
Everything is hunky dory except 'censummary' throws an error and I am trying to 
figure out a reason for this.
Please help!
below you will see the error statement...
thnx
--nk


 library(NADA)
Loading required package: survival
Loading required package: splines

 AsExample -structure(list(AsLow = as.integer(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 
 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), AsHigh = as.integer(c(1, 1, 1, 
 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)), AsCen = 
 as.integer(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
 1, 1, 1))), .Names = c(AsLow, AsHigh, AsCen), class = data.frame, 
 row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24))
 AsExample
   AsLow AsHigh AsCen
1  0  1 1
2  0  1 1
3  0  1 1
4  0  1 1
5  0  1 1
6  0  1 1
7  0  1 1
8  0  1 1
9  0  1 1
10 0  1 1
11 0  1 1
12 1  3 1
13 1  3 1
14 1  3 1
15 1  3 1
16 1  3 1
17 1  3 1
18 1  3 1
19 1  3 1
20 1  3 1
21 1  3 1
22 1  3 1
23 1  3 1
24 1  3 1
 names(AsExample)
[1] AsLow  AsHigh AsCen 
 censummary(AsLow, AsHigh, AsCen)
Error: object AsLow not found
Error in censummary(AsLow, AsHigh, AsCen) : 
unable to find the argument 'obs' in selecting a method for function 
'censummary'


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Re: [R] how to create a vector with different categories in a simple way?

2006-10-23 Thread Jenny persson
Thank you so much, Marc and Phil. Unfortunenately, I misunderstood the problem 
myself and wasn't clear how i wanted the variables to be. I will describe the 
issue again and hope you can help me out.
   
  Here is part of data called layout
   
  Id Name block col row
  1 a1   11
  2 b1   21
  3 c131
  4 a14   1
  5 b112
  6 c122
  7 b132
  8 c14   2 
  9 d   1  1 3
  10 e   1 2  3
  11  a  1 3  3
  12  d  1 4  3 
  13  e  1 1  4
  14  a 1  2 4
  15  d  1 3  4
  16 c   1 4  4
  17  d  2 1  1
  18  c  2 2  1
  19  e  2 3  1
  20  d  2 4  1
  21  b  2 1  2
  22 e   2 2  2
  23 f2 3  2
  24 d   2 4  2
   
  32 a  2 4 4  and so on
  ..   .  .  .
  ..   .  .  .
  ..  .   .  . 
  768 f  48   44
   
  As you can see for each row there are 4 columns. the total observations in 
each block is 16. My real data contains 48 blocks which give  totally 768 
observations. The blocks ranged from 1-48 are displayed four and four after 
each other like below.
  Note there are 4 rows and 4 columns in each block. 
   1 2 3 4
 5 6 7 8
 9 10 11 12
 13 14 15 16
 17 18 19 20
 21 22 23 24
 25 26 27 28 
 29 30 31 32
 33 34 35 36
 37 38 39 40
 41 42 43 44 
 45 46 47 48
   
  What I want is to create two variables called blockrow respective blockcol in 
such a way that blockrow
 will have value 1 for block 1,2 3 and 4, blockrow=2 for blocks 5,6,7
 and 8 and so on. Similarly, blockcol = 1 for blocks
 1,5 ,9,13,17,21,25,29,33,37,41 and 44 and so on. As you can see there
 are 12 blockrows and 4 blockcols. The data should look like
   
  Id Name block col row blockrow blockcol
  1 a1   11  11
  2 b1   21  11
  3 c131 11
  4 a14   1  11
  5 b112 11
  6 c122 1   1
  7 b132 1   1
  8 c142  1   1
  9 d1   1  3 1   1
  10 e   1 2  3  1  1
  11  a  1 3  3  1  1
  12  d  1 4  3  1  1
  13  e  1 1  4  1  1
  14  a 1  2 4  1   1
  15  d  1 3  4  1 1
  16 c   1 4  4  1 1
  17  d  2 1  1  1 2 
  18  c  2 2  1  1 2
  19  e  2 3  1  1 2
  20  d  2 4  1  1 2
  21  b  2 1  2  1 2
  22 e   2 2  2  1 2
  23 f2 3  2  1 2
  24 d   2 4  2  1 2
   
  32 a  2 4 4  and so on
  ..   .  .  .
  ..   .  .  .
  ..  .   .  . 
  768 f  48   44 12  4
   
   
  I have an algorithm 
   
  blockrow -1
  if(layout$block = 4)  
  blockrow -1
  if(5=layout$block = 8)  
  blockrow -2
   
  if(9=layout$block = 12)  
  blockrow -3 and so on
   
  Can I do a for loop like :
   
  #-- Append some more columns to matrix layout--

blockrow-rep(0,nrow(layout))
blockcol-rep(0,nrow(layout))


   
  for (a in 1:12){
  if(4*a+1=layout$block=(a+1)*4)
  blockrow-(a+1)
  }
   
  Similarly,
 
  blockcol-1

if(layout$block = 5,9,13,17,21,25,29,33,37,41,45)  
  blockcol-1

  if(layout$block = 2,6,10,14,18,22,26,30,34,38,42,46)  
  blockcol-2 and so on
  which give the for loop


   
  blockcol-1
for (a in 1:12){
  if(layout$block==(4*a+1))
  blockrow-1
  }

   
  or how can i do it in R so I get blockrow and blockcol as i want ?
   
  Thanks again for your help,
  Best regards,
  Yen
   
   
   
   
   
   
  
 

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Re: [R] Color eps/ps output from specialized plots?

2006-10-23 Thread Deepayan Sarkar
On 10/23/06, Turgut Durduran [EMAIL PROTECTED] wrote:
 Hello,

 First a disclaimer :) I am very new to using R.

 I am generating some plots and eventhough I can get colored output in the 
 encapsulated postscript files in the simplest of commands (e.g. 
 plot(1:10,1:10, type=l, col=red) ), it does not work for the particular 
 plots I want. It works on the screen.

 Here is an example taken out from Mixed-Effects Models in S and S-Plus by 
 Pinheiro and Bates;

 library(nlme)
 plot(CO2)
 postscript(tmp.eps)
 plot(CO2)
 dev.off()

You are seeing the results of what was, in hindsight, an unfortunate
design decision in nlme. Since the plot function in R is generic,
not all plot() calls are guaranteed to behave the same way. In
particular, the (several) plot() functions in nlme don't actually plot
anything, instead they return an object of class trellis, which
themselves need to be 'plot'ted or 'print'ed to see any output.

Now, this is not normally a problem, but it is when you run your code
through source (which I presume you are, although you have neglected
to mention it). The solution is to replace the

plot(CO2)

command by either

print(plot(CO2))

or the even odder

plot(plot(CO2))

See

http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f

for a relevant discussion.

-Deepayan

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


[R] unexpected read.csv behavior

2006-10-23 Thread Robert Baer
Example data in a recent post was:
   LandFill Ruminants
United States (USA) .214280   5528.16
France   200.527083   1299.87
Australia185.878368   2448.17
Russian Federation  1752.833400   2024.29
Argentina283.987320   2567.02
Brazil  1048.422480   8839.61
Colombia 265.125000   1307.61
Mexico   981.023351   1814.89
Ethiopia   9.380204   1237.49
Sudan 16.018494   1796.67
India553.425328  12360.30
Pakistan  47.159393   2346.71
China455.680191   8041.79

In trying to play with this posting, I saved the data as a .csv file where the 
first row had only two entries and read:
 , LandFill, Reminants  

I tried reading the data in with 
read.csv(ex.csv, header=TRUE) 

To my surprise this created a 3 column dataframe with the first column labeled 
as X.  X was a factor.  According to the help file:

If there is a header and the first row contains one fewer field than the 
number of columns, the first column in the input is used for the row names. 
Otherwise if row.names is missing, the rows are numbered. 

I got the expected default behavior of having the unlabeled column become row 
labels when I tried,
read.csv(ex.csv, header=TRUE, row.names=1)

Is this unexpected behavior somehow related to how I designed a .csv file or is 
there something I'm misinterpreting in the documentation?

Thanks,
Rob Baer

 version
   _   
platform   i386-pc-mingw32 
arch   i386
os mingw32 
system i386, mingw32   
status 
major  2   
minor  4.0 
year   2006
month  10  
day03  
svn rev39566   
language   R   
version.string R version 2.4.0 (2006-10-03)


 

[[alternative HTML version deleted]]

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


[R] nested if/else very slow, more efficient ways?

2006-10-23 Thread Kim Milferstedt
Hello,

in the data.frame resultsfuzzy I would like to replace the 
characters in the second column (5a, 5b, ... 5e) with numbers 
from 1 to 5. The data.frame has 39150 entries. I seems to work on 
samples that are  nrow(resultsfuzzy) but it takes suspicously long.

Do you have any suggestions how to make the character replacing more efficient?

Code:

for (i in 1:nrow(resultsfuzzy))
{
if (resultsfuzzy[i,2] == 5a){resultsfuzzy[i,2] - 1} else
 if (resultsfuzzy[i,2] == 5b){resultsfuzzy[i,2] - 2} else
 if (resultsfuzzy[i,2] == 5c){resultsfuzzy[i,2] - 3} else
 if (resultsfuzzy[i,2] == 5d){resultsfuzzy[i,2] - 4} else
 resultsfuzzy[i,2] - 5
}

Thanks,

Kim

version

platform i386-pc-mingw32
arch i386
os   mingw32
system   i386, mingw32
status
major2
minor2.1
year 2005
month12
day  20
svn rev  36812
language R

__

Kim Milferstedt
University of Illinois at Urbana-Champaign
Department of Civil and Environmental Engineering
4125 Newmark Civil Engineering Laboratory
205 North Mathews Avenue MC-250
Urbana, IL 61801
USA
phone: (001) 217 333-9663
fax: (001) 217 333-6968
email: [EMAIL PROTECTED]
http://cee.uiuc.edu/research/morgenroth

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


Re: [R] The Censummary Function...

2006-10-23 Thread Jeffrey Robert Spies
Try:

censummary(AsExample$AsLow, AsExample$AsHigh, AsExample$AsCen)

 From the Introduction to R (http://cran.r-project.org/doc/manuals/R- 
intro.html#Lists-and-data-frames):
Components of lists may also be named, and in this case the  
component may be referred to either by giving the component name as a  
character string in place of the number in double square brackets,  
or, more conveniently, by giving an expression of the form

   name$component_name

The same goes for data.frames.

Jeff.

On Oct 23, 2006, at 5:38 PM, nand kumar wrote:

 Hi Forum,

 I am new and learning this research tool.
 was trying to run some tests on a censored dataset using the NADA  
 library. Everything is hunky dory except 'censummary' throws an  
 error and I am trying to figure out a reason for this.
 Please help!
 below you will see the error statement...
 thnx
 --nk


 library(NADA)
 Loading required package: survival
 Loading required package: splines

 AsExample -structure(list(AsLow = as.integer(c(0, 0, 0, 0, 0,  
 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), AsHigh  
 = as.integer(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3,  
 3, 3, 3, 3, 3, 3, 3)), AsCen = as.integer(c(1, 1, 1, 1, 1, 1, 1,  
 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))), .Names = c 
 (AsLow, AsHigh, AsCen), class = data.frame, row.names = c 
 (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,  
 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,  
 24))
 AsExample
AsLow AsHigh AsCen
 1  0  1 1
 2  0  1 1
 3  0  1 1
 4  0  1 1
 5  0  1 1
 6  0  1 1
 7  0  1 1
 8  0  1 1
 9  0  1 1
 10 0  1 1
 11 0  1 1
 12 1  3 1
 13 1  3 1
 14 1  3 1
 15 1  3 1
 16 1  3 1
 17 1  3 1
 18 1  3 1
 19 1  3 1
 20 1  3 1
 21 1  3 1
 22 1  3 1
 23 1  3 1
 24 1  3 1
 names(AsExample)
 [1] AsLow  AsHigh AsCen
 censummary(AsLow, AsHigh, AsCen)
 Error: object AsLow not found
 Error in censummary(AsLow, AsHigh, AsCen) :
 unable to find the argument 'obs' in selecting a method for  
 function 'censummary'


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


Re: [R] unexpected read.csv behavior

2006-10-23 Thread Duncan Murdoch
On 10/23/2006 5:54 PM, Robert Baer wrote:
 Example data in a recent post was:
LandFill Ruminants
 United States (USA) .214280   5528.16
 France   200.527083   1299.87
 Australia185.878368   2448.17
 Russian Federation  1752.833400   2024.29
 Argentina283.987320   2567.02
 Brazil  1048.422480   8839.61
 Colombia 265.125000   1307.61
 Mexico   981.023351   1814.89
 Ethiopia   9.380204   1237.49
 Sudan 16.018494   1796.67
 India553.425328  12360.30
 Pakistan  47.159393   2346.71
 China455.680191   8041.79
 
 In trying to play with this posting, I saved the data as a .csv file where 
 the first row had only two entries and read:
  , LandFill, Reminants  
 
 I tried reading the data in with 
 read.csv(ex.csv, header=TRUE) 
 
 To my surprise this created a 3 column dataframe with the first column 
 labeled as X.  X was a factor.  According to the help file:
 
 If there is a header and the first row contains one fewer field than the 
 number of columns, the first column in the input is used for the row names. 
 Otherwise if row.names is missing, the rows are numbered. 

Your first row contains 3 columns, the first one blank.  You don't want 
a comma before LandFill.

Duncan Murdoch

 
 I got the expected default behavior of having the unlabeled column become row 
 labels when I tried,
 read.csv(ex.csv, header=TRUE, row.names=1)
 
 Is this unexpected behavior somehow related to how I designed a .csv file or 
 is there something I'm misinterpreting in the documentation?
 
 Thanks,
 Rob Baer
 
 version
_   
 platform   i386-pc-mingw32 
 arch   i386
 os mingw32 
 system i386, mingw32   
 status 
 major  2   
 minor  4.0 
 year   2006
 month  10  
 day03  
 svn rev39566   
 language   R   
 version.string R version 2.4.0 (2006-10-03)
 
 
  
 
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Re: [R] Key colour problem with lattice plot

2006-10-23 Thread Alex Brown

 Can anyone explain why the key is inverted versus the bar  
 order?

 Because the key goes from top to bottom, while cartesian coordinates
 go from bottom to top. Neither can really be any other way by default,
 if you think about it. I did promise someone that I would add an
 option to reverse the order in key, but looks like I haven't done that
 yet.


In the case of barchart, what would be the justification for not  
ordering the bars (in a group) as for the key, or vice versa?  Please  
forgive my ignorance - I am curious to understand the design  
principles.  I am certainly very pleased to have the power of  
auto.key in lattice - it's the main reason why I stopped using the  
default plot function.

-Alex Brown

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Re: [R] nested if/else very slow, more efficient ways?

2006-10-23 Thread Mike Nielsen
One way that might do what you want is to change the character column
to a factor, and then apply as.numeric.

resultsfuzzy$x-as.numeric(factor(resultsfuzzy$x,levels=c(5a,5b,5c,5d,5e)))

This assumes, of course, that you know that the levels are going to be
in the set {5a,5b,5c,5d,5e}.

However, it may be better to just leave it as a factor, depending upon
what you intend to do with it later.

Hope this helps.

Regards,

Mike

On 10/23/06, Kim Milferstedt [EMAIL PROTECTED] wrote:
 Hello,

 in the data.frame resultsfuzzy I would like to replace the
 characters in the second column (5a, 5b, ... 5e) with numbers
 from 1 to 5. The data.frame has 39150 entries. I seems to work on
 samples that are  nrow(resultsfuzzy) but it takes suspicously long.

 Do you have any suggestions how to make the character replacing more 
 efficient?

 Code:

 for (i in 1:nrow(resultsfuzzy))
 {
 if (resultsfuzzy[i,2] == 5a){resultsfuzzy[i,2] - 1} else
  if (resultsfuzzy[i,2] == 5b){resultsfuzzy[i,2] - 2} else
  if (resultsfuzzy[i,2] == 5c){resultsfuzzy[i,2] - 3} else
  if (resultsfuzzy[i,2] == 5d){resultsfuzzy[i,2] - 4} else
  resultsfuzzy[i,2] - 5
 }

 Thanks,

 Kim

 version

 platform i386-pc-mingw32
 arch i386
 os   mingw32
 system   i386, mingw32
 status
 major2
 minor2.1
 year 2005
 month12
 day  20
 svn rev  36812
 language R

 __

 Kim Milferstedt
 University of Illinois at Urbana-Champaign
 Department of Civil and Environmental Engineering
 4125 Newmark Civil Engineering Laboratory
 205 North Mathews Avenue MC-250
 Urbana, IL 61801
 USA
 phone: (001) 217 333-9663
 fax: (001) 217 333-6968
 email: [EMAIL PROTECTED]
 http://cee.uiuc.edu/research/morgenroth

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



-- 
Regards,

Mike Nielsen

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Re: [R] nested if/else very slow, more efficient ways?

2006-10-23 Thread Alex Brown
There are a number of ways this might be approached.

Can you please give a sample of your data, and your desired output?

Are 5a ... 5e the only values that appear in that column, or are  
there other values, 4e for instance, that should stay the same  
during your conversion?

Do you wish to use the numbers 1 to 5 in the processed column in  
arithmetic processing, or are they just an enumeration of possible  
values?

While you think about it, I direct your attention to the functions:

sub
factor

-Alex

On 23 Oct 2006, at 23:03, Kim Milferstedt wrote:

 Hello,

 in the data.frame resultsfuzzy I would like to replace the
 characters in the second column (5a, 5b, ... 5e) with numbers
 from 1 to 5. The data.frame has 39150 entries. I seems to work on
 samples that are  nrow(resultsfuzzy) but it takes suspicously long.

 Do you have any suggestions how to make the character replacing  
 more efficient?

 Code:

 for (i in 1:nrow(resultsfuzzy))
 {
 if (resultsfuzzy[i,2] == 5a){resultsfuzzy[i,2] - 1} else
  if (resultsfuzzy[i,2] == 5b){resultsfuzzy[i,2] - 2} else
  if (resultsfuzzy[i,2] == 5c){resultsfuzzy[i,2] - 3} else
  if (resultsfuzzy[i,2] == 5d){resultsfuzzy[i,2] - 4}  
 else
  resultsfuzzy[i,2] - 5
 }

 Thanks,

 Kim

 version

 platform i386-pc-mingw32
 arch i386
 os   mingw32
 system   i386, mingw32
 status
 major2
 minor2.1
 year 2005
 month12
 day  20
 svn rev  36812
 language R

 __

 Kim Milferstedt
 University of Illinois at Urbana-Champaign
 Department of Civil and Environmental Engineering
 4125 Newmark Civil Engineering Laboratory
 205 North Mathews Avenue MC-250
 Urbana, IL 61801
 USA
 phone: (001) 217 333-9663
 fax: (001) 217 333-6968
 email: [EMAIL PROTECTED]
 http://cee.uiuc.edu/research/morgenroth

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

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


Re: [R] nested if/else very slow, more efficient ways?

2006-10-23 Thread Duncan Murdoch
On 10/23/2006 6:03 PM, Kim Milferstedt wrote:
 Hello,
 
 in the data.frame resultsfuzzy I would like to replace the 
 characters in the second column (5a, 5b, ... 5e) with numbers 
 from 1 to 5. The data.frame has 39150 entries. I seems to work on 
 samples that are  nrow(resultsfuzzy) but it takes suspicously long.
 
 Do you have any suggestions how to make the character replacing more 
 efficient?

I don't know if nested if/else is inefficient, but indexing of 
data.frames certainly is. You would do a lot better with ifelse() on the 
whole column.

For example, if column 2 is named B, this would be much faster:

resultsfuzzy$B - ifelse(resultsfuzzy$B == 5a, 1,
   ifelse(resultsfuzzy$B == 5b, 2, ... ))

(where you'll have to fill in the ... yourself in the obvious way.

But an even faster way for your particular problem would be

resultsfuzzy$B - match(resultsfuzzy$B, c(5a, 5b, 5c, 5d, 5e))

Duncan Murdoch
 
 Code:
 
 for (i in 1:nrow(resultsfuzzy))
 {
 if (resultsfuzzy[i,2] == 5a){resultsfuzzy[i,2] - 1} else
  if (resultsfuzzy[i,2] == 5b){resultsfuzzy[i,2] - 2} else
  if (resultsfuzzy[i,2] == 5c){resultsfuzzy[i,2] - 3} else
  if (resultsfuzzy[i,2] == 5d){resultsfuzzy[i,2] - 4} else
  resultsfuzzy[i,2] - 5
 }
 
 Thanks,
 
 Kim
 
 version
 
 platform i386-pc-mingw32
 arch i386
 os   mingw32
 system   i386, mingw32
 status
 major2
 minor2.1
 year 2005
 month12
 day  20
 svn rev  36812
 language R
 
 __
 
 Kim Milferstedt
 University of Illinois at Urbana-Champaign
 Department of Civil and Environmental Engineering
 4125 Newmark Civil Engineering Laboratory
 205 North Mathews Avenue MC-250
 Urbana, IL 61801
 USA
 phone: (001) 217 333-9663
 fax: (001) 217 333-6968
 email: [EMAIL PROTECTED]
 http://cee.uiuc.edu/research/morgenroth
 
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Re: [R] pdf/ convert/ font problem

2006-10-23 Thread Paul Murrell
Hi


Sam McClatchie wrote:
 Background:
 OS: Linux Ubuntu Dapper
 R : Version 2.3.1 (2006-06-01)
 Emacs 21.4.1
 ESS
 
 
 Colleagues
 
 This is neither a bug report, nor a complaint, so I think this is the 
 appropriate list.
 
 I am getting an odd font problem with a pdf generated using
  pdf(file = 
 ../figures/PCNM.models.larvae.fluor.zoo.stability.2001.and.2002.pdf,
   title=, family=serif, pointsize=10, width=6.75, height=8.625)
 
 The figure produced from my code is here:
 http://www.members.iinet.net.au/~s.mcclatchie/aslo/PCNM.models.larvae.fluor.zoo.stability.2001.and.2002.pdf
 (I can't really send all of the code as there are several subroutines 
 for the analysis that are not relevant here anyway).
 
 
 The pdf loads into acroread or xpdf fine, so is not corrupted, but 
 ghostscript gives an error:


Acroread (Windows) works for me, xpdf (Linux CentOS) thrashes my machine
for ever (or until I gave up), and ghostscript gives me ...

/undefinedresult in --currentpoint--

... but I can't see anything obvious in the raw PDF.


 ERROR: /invalidfont in --.type1execchar--ESP Ghostscript 815.02: 
 Unrecoverable error, exit code 1
 
 Operand stack:
--dict:4/4(L)--   --dict:16/17(ro)(G)--   108   a71   -string-
 Execution stack:
%interp_exit   .runexec2   --nostringval--   --nostringval--   
 --nostringval--   2   %stopped_push   --nostringval--   
 --nostringval--   --nostringval--   false   1   %stopped_push   1   3   
 %oparray_pop   1   3   %oparray_pop   1   3   %oparray_pop   1   3   
 %oparray_pop   .runexec2   --nostringval--   --nostringval--   
 --nostringval--   2   %stopped_push   --nostringval--   
 --nostringval--   --nostringval--   --nostringval--   --nostringval--   
 %array_continue   --nostringval--   false   1   %stopped_push   
 --nostringval--   %loop_continue   --nostringval--   --nostringval--   
 %finish_show   --nostringval--   --nostringval--   7   9   1   
 --nostringval--   (gs_show_enum)   %op_show_continue
 Dictionary stack:
--dict:1124/1686(ro)(G)--   --dict:0/20(G)--   --dict:85/200(L)--   
 --dict:104/127(ro)(G)--   --dict:241/347(ro)(G)--   --dict:20/24(L)--   
 --dict:4/6(L)--   --dict:20/20(L)--   --dict:12/13(L)--
 Current allocation mode is local
 --
 Also Imagemagick cannot display or convert the file to .png:


I think that's the same problem:  ImageMagick is just calling ghostscript.

Sorry, no solution to suggest yet.

Paul


 [EMAIL 
 PROTECTED]:~/dynamic_data/projects2005/sardine_egg_distribution/journal$ 
 convert PCNM.models.larvae.fluor.zoo.stability.2001.and.2002.pdf 
 PCNM.models.larvae.fluor.zoo.stability.2001.and.2002.png
 ERROR: /invalidfont in --.type1execchar--
 Operand stack:
--dict:4/4(L)--   --dict:15/16(ro)(G)--   108   a71   -string-
 Execution stack:
%interp_exit   .runexec2   --nostringval--   --nostringval--   
 --nostringval--   2   %stopped_push   --nostringval--   
 --nostringval--   --nostringval--   false   1   %stopped_push   1   3   
 %oparray_pop   1   3   %oparray_pop   1   3   %oparray_pop   
 --nostringval--   --nostringval--   2   1   1   --nostringval--   
 %for_pos_int_continue   --nostringval--   --nostringval--   
 --nostringval--   --nostringval--   %array_continue   --nostringval--   
 false   1   %stopped_push   --nostringval--   %loop_continue   
 --nostringval--   --nostringval--   %finish_show   --nostringval--  
 --nostringval--   9   10   1   --nostringval--   (gs_show_enum)   
 %op_show_continue
 Dictionary stack:
--dict:1125/1686(ro)(G)--   --dict:0/20(G)--   --dict:85/200(L)--   
 --dict:85/200(L)--   --dict:104/127(ro)(G)--   --dict:241/347(ro)(G)--   
 --dict:20/24(L)--   --dict:4/6(L)--   --dict:20/20(L)--   --dict:12/13(L)--
 Current allocation mode is local
 ESP Ghostscript 815.02: Unrecoverable error, exit code 1
 convert: Postscript delegate failed 
 `PCNM.models.larvae.fluor.zoo.stability.2001.and.2002.pdf'.
 
 
 I have all of the true type fonts installed.
 
 Probably the solution is just to use png() rather than pdf(), but I 
 thought this problem was curious enough to report.
 
 Best fishes
 
 Sam
 

-- 
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/

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Re: [R] Key colour problem with lattice plot

2006-10-23 Thread Deepayan Sarkar
On 10/23/06, Alex Brown [EMAIL PROTECTED] wrote:
 
  Can anyone explain why the key is inverted versus the bar
  order?
 
  Because the key goes from top to bottom, while cartesian coordinates
  go from bottom to top. Neither can really be any other way by default,
  if you think about it. I did promise someone that I would add an
  option to reverse the order in key, but looks like I haven't done that
  yet.
 

 In the case of barchart, what would be the justification for not
 ordering the bars (in a group) as for the key, or vice versa?  Please
 forgive my ignorance - I am curious to understand the design
 principles.  I am certainly very pleased to have the power of
 auto.key in lattice - it's the main reason why I stopped using the
 default plot function.

No particular reason, it's just (I think) one of the weaknesses in the
Trellis design. There is no actual communication between legends and
panel functions. Basically, the flexibility of Trellis comes from
almost completely separating the various elements of a graphic, and
you get an illusion of consistency if all the elements are allowed to
derive their defaults from the same settings. This is the reason there
will never be (without wrappers or other kinds of hacks) any hope of
'auto.key' ``detecting'' when you give a

col=c('green', 'red')

argument to xyplot (or panel.xyplot) and changing colors automatically
in response.

If you think about it, panel.barchart does the only reasonable thing,
namely order levels in the right direction. If you want things to be
ordered from top to bottom, you can reverse the axes:

barchart(as.matrix(example), stack = FALSE, ylim = c(14, 0))

which will lose the nice labels, but that's again a whole different
element of the graph. Changing the order for bars within y-levels is
one option, but would you really want that when the bars are vertical?

Similarly, the legend also does the only reasonable thing. It should
have an option to reverse the order, and it will in the next version
of lattice (which I plan to upload within a couple of days), where you
should be able to say:

barchart(as.matrix(example), stack = FALSE,
 auto.key = list(reverse.rows = TRUE))

-Deepayan

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[R] Replacing for loop with apply type function.

2006-10-23 Thread Ritwik Sinha
Hi Everyone,

I want to replace the following for loop with an apply type function
to (possibly) speed up the process.

# The for loop I want to replace

x - numeric(10)
for(i in 2:10)
x[i] - x[i-1]+1

x

To give you some more background, I am running a Metropolis Hastings
MCMC chain with k parameters. So, if x represents the sample from
the chain, the ith row (representing the ith iteration) depends on the
(i-1)th row.

Thanks in advance,
Ritwik Sinha
Graduate Student
Epidemiology and Biostatistics
Case Western Reserve University
[EMAIL PROTECTED] | +12163682366 | http://darwin.cwru.edu/~rsinha

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Re: [R] Extended CR model

2006-10-23 Thread Charles C. Berry


The posting guide (see below) suggests using something like

RSiteSearch(continuation ratio)

before posting to the list.

There are many relevant hits including several to packages on CRAN.


On Mon, 23 Oct 2006, Cougar Lawrence wrote:

 I was wondering if someone could point me to the R package that implements
 the partial proportional odds (PO) model discussed by Peterson and Harrell
 (1990).  In Harrell (2001), the model is explained in Chapt. 13.

 Respectfully,

 Frank R. Lawrence

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 R-help@stat.math.ethz.ch mailing list
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Charles C. Berry(858) 534-2098
  Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]   UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717

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Re: [R] Extended CR model

2006-10-23 Thread Charles C. Berry

Or better still

RSiteSearch(partial proportional odds)

and see the second hit.


On Mon, 23 Oct 2006, Charles C. Berry wrote:



 The posting guide (see below) suggests using something like

   RSiteSearch(continuation ratio)

 before posting to the list.

 There are many relevant hits including several to packages on CRAN.


 On Mon, 23 Oct 2006, Cougar Lawrence wrote:

 I was wondering if someone could point me to the R package that implements
 the partial proportional odds (PO) model discussed by Peterson and Harrell
 (1990).  In Harrell (2001), the model is explained in Chapt. 13.

 Respectfully,

 Frank R. Lawrence

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


 Charles C. Berry(858) 534-2098
  Dept of Family/Preventive Medicine
 E mailto:[EMAIL PROTECTED] UC San Diego
 http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717

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 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Charles C. Berry(858) 534-2098
  Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED]   UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717

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[R] why it returns list level number instead of its content?

2006-10-23 Thread Hu Chen
for example:
I have a data frame.
data$V4 returns:
.
[6936] P05796   P11096   P76174   P04475   P18775
[6941] P33225   P76387   P76388   P76388   P09375
[6946] P15300   P15723
1375 Levels:  O50190 O65938 O69415 P00274 P00363 P00364 P00370 P00373 ...
Q9AJ15
data$V5 returns something like data$V4
I want to cbind this two columns, so I use
new - cbind(data$V4,data$V5)
I expect it to return something like:
[1] P05761 P11986
[2] ...
however it returns
[1] 215 434
[2] 134 213
.
it uses level number instead of  its content like P05761. What's wrong
with  it?  how can I get its content instead of level number? I can use some
dirty ways to do that but I didn't understand why.

[[alternative HTML version deleted]]

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Re: [R] linear mixed effects models with breakpoints

2006-10-23 Thread Spencer Graves
  Are you familiar with how to include include the desired serial 
correlation structure in 'lme', ignoring the breakpoint question?  If 
no, please consult Pinheiro and Bates (2000) Mixed-Effects Models in S 
and S-Plus (Springer)?  If yes, it should be a relatively easy matter to 
add that to the 'lmeChgPt' function. 

  If you'd like further help from this listserve, please provide 
commented, minimal, self-contained, reproducible code, as suggested in 
the posting guide www.R-project.org/posting-guide.html. 

  Hope this helps. 
  Spencer Graves

Ewan Wakefield wrote:
 Hi folks

 I have some data to which I've been fitting linear mixed effects
 models. I am currently using a lme model in the nlme package, with terms
 for random effects due to repeated measures on individuals and the
 corCAR1 serial correlation structure. However, there is some suggestion
 in the data (and from theory) that a breakpoint (change point) model may
 be more appropriate. Scott, Norman and Berger's lmeChgPt model seems to
 go some way to fitting the requirements for such a model but does not
 incorporate, as I understand it, a serial correlation structure.

 Is anyone aware of a model fitting procedure that incorporates the
 correlation structures of the lme models in a breakpoint model?

 Kind regards,

 Ewan Wakefield



 British Antarctic Survey,
 High Cross,
 Madingley Road,
 Cambridge CB3 OET

 Tel. +44 (0) 1223 221215
 E-mail: [EMAIL PROTECTED]
 Website: antarctica.ac.uk

 --  
 This message (and any attachments) is for the recipient only...{{dropped}}

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Re: [R] why it returns list level number instead of its content?

2006-10-23 Thread Richard M. Heiberger
The class factor is defined for vectors, not matrices.
The attempt to use a factor in a matrix setting coerces it to numeric.
See the documentation for factor where it says
In particular, as.numeric applied to a factor is meaningless,
and may happen by implicit coercion.

 tmp - matrix(1:8, 4)
 tmp
 [,1] [,2]
[1,]15
[2,]26
[3,]37
[4,]48
 factor(tmp)
[1] 1 2 3 4 5 6 7 8
Levels: 1 2 3 4 5 6 7 8
 


## A simplified example based on yours:

 tmp - data.frame(a=factor(letters[1:4]), b=factor(letters[5:8]))
 tmp
  a b
1 a e
2 b f
3 c g
4 d h

 cbind(tmp$a, tmp$b)  ## coerced to numeric, both factors went to 1:4
 [,1] [,2]
[1,]11
[2,]22
[3,]33
[4,]44

 tmp[, c(a,b)]  ## in your example, subscripting is probably the right 
 method
  a b
1 a e
2 b f
3 c g
4 d h

 cbind.data.frame(tmp$a, tmp$b)  ## explicit use of cbind.data.frame works.
  tmp$a tmp$b
1 a e
2 b f
3 c g
4 d h

Rich

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[R] set.seed() and .Random.number

2006-10-23 Thread Taka Matzmoto
Hi R-users
I have two conditions. For each condition, 100 sets of 10 random numbers 
from N(0,1) need to be generated.  Here is my question.

At the begining I specify a seed number. I want to make the 100th set of the 
first condition and 1st set of the second conditon the same. What do I need 
to do ?

After generating 99th set of 10 random numbers and then saving .Random.seed 
then using .Random.seed for generating 100th set of the first condition and 
1st set of the second condtion. Is this right?

.Random.seed is a vector with 626 numbers, but set.seed() only accepts a 
integer number.
What do I need to do for saving .Random.seed and using the saved 
.Random.seed for setting set.seed() ?

Thanks

Taka

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