date:20081020


On Mon, 20 Oct 2008, [EMAIL PROTECTED] wrote:


I have a matrix x with repeating row names.



zz-matrix(0,4,4)

rownames(zz)=c(a,a,b,b)

data.frame(zz) (?)


The row names on a data frame should be unique.  You can try
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame.  If
you need the row name information, add it as a column in the data frame,
e.g. mydataframe$rnames - rownames(zz).  (Note to R-Core: the
documentation for as.data.frame doesn't mention the usage of
row.names=FALSE to ignore row names, but it seems to work consistently.
Does the help page for as.data.frame need updating?)


No.  row.names=FALSE is not intended to work, and did you check every 
single as.data.frame() method?


It just so happens that for the matrix method invalid input for 
'row.names' results in setting default row names.  Other methods may 
differ.





lm(as.formula(paste(final_dat[,5]~,paste(colnames(x),collapse=+))),x
)

this gives me a error



Error in model.frame.default(formula =
as.formula(paste(final_dat[,5]~,  :

  'data' must be a data.frame, not a matrix or an array


I suspect that if you try class(x), it will be a matrix, not the requisite
data frame.

Regards,
Richie.

Mathematical Sciences Unit
HSL



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[R] Bootstrap problem

2008-10-20 Thread Tom Willems

Dear R-users,
 
I'm having a small problem while bootstraping data.
What i would like to do, is resmple the data and calulate a function on 
this, so i can estimate the measure of reproducability for this data.
 
The function i wrote works fine, even while bootstraping.
The only problem is that bootstraping. 
 
The dataset existes of 10 trials, each divided in to 3 groups of high(3) 
medium(2) and low(1).
A bootstrap samlpe (trial) should always exist of 5 obs. taken from each 
group population, so to be representative.
 
example:
original data:
trial 1 : group(1) = (0,0,1,0,0);group(2) = (0,1,1,0,1);group(3) = 
(1,1,1,1,1)
...
bootstraped data:
trial 1 : group(1) = (0,0,0,0,1);group(2) = (1,1,0,0,1);group(3) = 
(1,0,1,1,1)
 
NOT
bootstraped data:
trial 1 : group(1) = (0,0,0,0,1,1,0);group(2) = (1,0,1);group(3) = 
(1,0,1,1,1,1,1,0,1,1)
 
Now I am familiar how to use function bootstrap (pkg bootstrap), but i 
read about a function called boot (pkg boot), however i can't seem to 
master this.
The explanation (help('boot') ) isn't making me any smarter.
I know I can always split the data up (wich is what i am doing) but i was 
wondering whether this would have an effect on the bootstrap, maby it is 
beter to keep all the groups together? 
 
 
here is a (this time WORKING) code example of what i did.
## proc
## generate data
datas - 
data.frame(protection=c(rep(c(0,1,0,1,0,0,1,0,1,1,1,0,1,1,1),2),c(0,0,0,0,1,0,1,0,1,1,1,0,1,0,1),rep(c(0,1,1,1,0,0,1,1,0,1,1,1,1,0,1),2),c(0,1,0,0,1,0,1,1,1,1,1,1,1,1,1),rep(c(0,1,0,1,0,0,1,1,1,1,1,0,1,1,1),2),c(0,1,0,0,0,0,1,1,0,1,1,1,1,1,1),c(0,0,1,1,0,0,1,1,1,0,0,1,1,0,1))
 
,group=rep(1:3,50),trial=c(rep(1,15),rep(2,15),rep(3,15),rep(4,15),rep(5,15),rep(6,15),rep(7,15),rep(8,15),rep(9,15),rep(10,15)))
## describe Function
Vacc.Vcon -function (dataset1 , trialdata , groupdata ) {

groups - unique (groupdata)
trials - unique (trialdata)
Tr - length(trials) 
G - length(groups)
Gl - length(dataset1)/(G*Tr) 
Tl - length(dataset1)/(Tr) 
iterg -data.frame(1:G,as.vector(groups))
trials - unique (trialdata)
Tr - length(trials) 
itert -data.frame(1:Tr,as.vector(trials))
triallist - c()
grouplist - c() 
for (x in 1:G){ 
ifelse(x==1,y-x,y- y+Tr)
grouplist[c(y:(y+Tr-1))] -rep(iterg[x,2],Tr)} 
iter -data.frame(1:(Tr),rep(trials,G),grouplist)
VACC - data.frame()
VACC.sub - function (dataset1,trialn,groupn){
p0 -sum(   ifelse(dataset1==1  trialdata==trialn  
groupdata==groupn, 1,0)  )
p1 -sum(   ifelse(dataset1==0  trialdata==trialn  
groupdata==groupn, 1,0)  ) 
p - p0+p1
VACC.group - 
list('Trial'=trialn,'Group'=groupn,'Vacc'=sum((p0/p)^2 
,(p1/p)^2),p0=(p0/p)  , p1=(p1/p) 
,n0=as.numeric(p0),n1=as.numeric(p1),'n'=as.numeric(p))
VACC.group}

for (i in 1:(G*Tr) ) {
  VACC[i,1] - VACC.sub (dataset1,iter[i,2],iter[i,3])[1] 
  VACC[i,2] - VACC.sub (dataset1,iter[i,2],iter[i,3])[2] 
  VACC[i,3] - VACC.sub (dataset1,iter[i,2],iter[i,3])[3]
  VACC[i,4] - VACC.sub (dataset1,iter[i,2],iter[i,3])[4] 
  VACC[i,5] - VACC.sub (dataset1,iter[i,2],iter[i,3])[5]
  VACC[i,6] - VACC.sub (dataset1,iter[i,2],iter[i,3])[6]
  VACC[i,7] - VACC.sub (dataset1,iter[i,2],iter[i,3])[7]
  VACC[i,8] - VACC.sub (dataset1,iter[i,2],iter[i,3])[8]
  VACC} 
  rownames(VACC) - NULL
  rownames(VACC) - paste(iter[,2],iter[,3],sep='_')
Pcalc - function(x) { 
out-(1/(Tr)) * sum(x)
out} 
P0 - tapply( VACC$p0,VACC$Group,Pcalc)
P1 -  tapply( VACC$p1,VACC$Group,Pcalc)
Vcon - mean(cbind(P0^2 + P1^2))
 
  Vacc.total - mean (tapply( VACC$Vacc,VACC$Group,mean))
  out - 
list(all=VACC,N=G,P0=P0,P1=P1,Vcon=Vcon*100,Vacc.total=Vacc.total*100)
  out   }
## end describe Function 
Vacc.Vcon (datas[,1] , datas[,3], datas[,2]) # example of how fun works

## data needs to be in matrix form for bootstrap function
xdata -matrix( 
cbind(datas$protection,datas$group,datas$trial),ncol=3,byrow=F) 
## function for bootstrap
  vacc.boot - function(x,xdata){ 
Vacc.Vcon(xdata[x,1],xdata[x,3],xdata[x,2]) }
bootk - 10 
results - bootstrap(1:150,bootk,vacc.boot,xdata) 
 
taccs - list() ;Vaccs - vector();Vcons - vector()
  boot.amp.vac2- for(i in 1:bootk) {
 m.i - results$thetastar[[i]] 
 taccs[i] - list(m.i )
 G.Vacc - round( 
tapply(taccs[[i]]$all$Vacc,rownames(taccs[[i]]$all),mean)*100 ,digits=3)
 Vaccs - round( mean(taccs[[i]]$Vacc.total),digits=3)
 Vcons-round( mean(taccs[[i]]$Vcon ),digits=3)
 tacc - list( 
data=taccs,Booted.means=list(Vacc.grouped=G.Vacc 
,Vacc.Total=Vaccs,Vcon.Total=Vcons)) 
 tacc} 
 
 
Rep.table - tacc$Booted.mean 
Rep.table 
## problem area = n should always be 5 in each group as in the original 
data
#calcues based on original data   last colon : n = 5
Vacc.Vcon (datas[,1] ,datas[,3], datas[,2])$all [1:5,]
#calcues based on Booted datan is not 5 !

Re: [R] Extracting variables with a given prefix in the parent environment

2008-10-20 Thread Duncan Murdoch


On 20/10/2008 2:06 AM, erwann rogard wrote:

hi,

get.vars.name.prefix-function(prefix){
 result-list()
 len-nchar(prefix)
 var.names-ls(name=1,pattern=prefix) #name=1 probably wrong option
 print(var.names)
 for(i in 1:length(var.names)){
  name-var.names[i]
  field-substr(name,len+1,nchar(name))
  result[[field]]-get(name)
 }
 result
}

for example
x.1-1
x.2-2
get.vars.name.prefix(x.) should return 1,2

this does not work if i call if from within another function:

g-function(){
  x.1-1
  x.2-2
  get.vars.name.prefix(x.) # should return 1,2, nothing else
}

i would like the environment specified within ls(...) to be that of the body
of the function that calls get.vars.name.prefix. apparently name=1 is
probably not the right option.


Use ls(name=parent.frame(), pattern=...)

And be careful about the pattern:  it's a regular expression, so you 
would need to start with a caret ^ to guarantee that it occurs at the 
start of the name.


Duncan Murdoch

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Re: [R] convert matrix to dataframe with repeating row names

2008-10-20 Thread Richard . Cotton

  The row names on a data frame should be unique.  You can try
  as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame. 
If
  you need the row name information, add it as a column in the data 
frame,
  e.g. mydataframe$rnames - rownames(zz).  (Note to R-Core: the
  documentation for as.data.frame doesn't mention the usage of
  row.names=FALSE to ignore row names, but it seems to work 
consistently.
  Does the help page for as.data.frame need updating?)
 
 No.  row.names=FALSE is not intended to work, and did you check every 
 single as.data.frame() method?
 
 It just so happens that for the matrix method invalid input for 
 'row.names' results in setting default row names.  Other methods may 
 differ.

row.names=FALSE seems a natural way of supressing existing row names to 
me, since it corresponds nicely to using row.names=FALSE in write.csv. 
Currently it seems that if a matrix has duplicate row names, then 
converting it to be a data frame requires

rnames - rownames(mymatrix)
rownames(mymatrix) - NULL
as.data.frame(mymatrix)
rownames(mymatrix) - rnames 

Ideally, three of these lines of code shouldn't really need to be there.

If you disagree that allowing row.names=FALSE is a good idea, or you don't 
want to change the function interface, then perhaps having as.data.frame 
check for duplicates and throwing a warning (rather than an error) would 
be preferable behaviour.  I do realise that there are dozens of 
as.data.frame methods, and the documentation does state that Few of the 
methods check for duplicated row names, but it would be beneficial from a 
user standpoint.

Regards,
Richie.

Mathematical Sciences Unit
HSL



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[R] Error reporting in R

2008-10-20 Thread A.Noufaily


Hello,

I am hoping someone can help me with the following:

I am applying the function mle on a single data set X n times, each
time using a different set of initial values v[i] (i=1,...,n).
The initial values are all finite. Two cases arise:
A- For some sets of initial values mle is giving parameter estimates.
B- For some other initial values, mle is reporting the following error: 

Error in optim(start,f,method=method,hessian=TRUE,...):
Initial value in vmin is not finite

My aim is to tell R to return the estimates whenever case A, and to
return the word Error whenever case B (using an if statement inside
a loop).

How to tell R to return the word Error whenever mle reports the error
given above?
Is there a function similar to is.nan which tests the existence of an
error?

Any assistance would be much appreciated,

Regards,

Amy

-
The Open University is incorporated by Royal Charter (RC 000391), an exempt 
charity in England  Wales and a charity registered in Scotland (SC 038302).

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Re: [R] pairs plots in R

2008-10-20 Thread Antony Unwin

If you want to do efficient exploratory data analysis on this kind of  
dataset, then interactive graphics with parallel coordinate plots  
(ipcp in iplots) should help.  Of course, it depends what you mean by  
large.  It might be worth looking at the book Graphics of Large  
Datasets for some ideas.

Antony Unwin
Professor of Computer-Oriented Statistics and Data Analysis,
Mathematics Institute,
University of Augsburg,
86135 Augsburg, Germany
Tel: + 49 821 5982218



 From: Sharma, Dhruv [EMAIL PROTECTED]
 Date: 19 October 2008 10:58:53 pm GMT+02:00
 To: r-help@r-project.org
 Subject: [R] pairs plots in R


 Hi,
  is there a way to take a data frame with 100+ columns and large  
 data set to do efficient exploratory analysis in R with pairs?

 I find using pairs on the whole matrix is slow and the resulting  
 matrix is tiny.

 Also the variable of interest for me is a binary var Y or N .

 Is there an efficient way to graphically view many variable  
 relationships that does not look teeny ?

 I could do pairs 10 at a time but this seems too brute force.

 thanks
 Dhruv

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Re: [R] Extracting variables with a given prefix in the parent environment

2008-10-20 Thread Henrique Dallazuanna

Try this:

Using 'envir' argument, in ls and get function:

get.vars.name.prefix-function(prefix, envir = parent.frame()){
   result-list()
   len-nchar(prefix)
   var.names-ls(envir = envir,pattern=prefix)
   print(var.names)
   for(i in 1:length(var.names)){
  name-var.names[i]
  field-substr(name,len+1,nchar(name))
  result[[field]]-get(name, envir = envir)
  }
 result
}

On Mon, Oct 20, 2008 at 4:06 AM, erwann rogard [EMAIL PROTECTED]wrote:

 hi,

 get.vars.name.prefix-function(prefix){
  result-list()
  len-nchar(prefix)
  var.names-ls(name=1,pattern=prefix) #name=1 probably wrong option
  print(var.names)
  for(i in 1:length(var.names)){
  name-var.names[i]
  field-substr(name,len+1,nchar(name))
  result[[field]]-get(name)
  }
  result
 }

 for example
 x.1-1
 x.2-2
 get.vars.name.prefix(x.) should return 1,2

 this does not work if i call if from within another function:

 g-function(){
  x.1-1
  x.2-2
  get.vars.name.prefix(x.) # should return 1,2, nothing else
 }

 i would like the environment specified within ls(...) to be that of the
 body
 of the function that calls get.vars.name.prefix. apparently name=1 is
 probably not the right option.

 thanks.

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

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[R] problem with multiple comparison

2008-10-20 Thread Patrice KOUAKOU



Dear all,

I'm a new user of R.
How to manage R to have homogeneous means groups with Duncan test? I've tried 
the Duncan test and the results is the difference between means taken in pairs.

Patrice


__


ble contre les messages non sollicitÃ©s 

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[R] Trying to pass arrays as arguments to a function

2008-10-20 Thread Kåre Edvardsen

I'd like to avoid looping through an array in order to change values in
the array as it takes too long.
I red from an earlier post it can be done by do.call but never got it
to work. The Idea is to change the value of y according to values in
x. Wherever x holds the value 3, the corresponding value in y
should be set to 1. 

So I tried the following giving an error message:

#
x - c(1,2,3,2,2,3,1,1,3,3)
y - c(0,0,1,1,0,0,1,0,0,1)

Change_y - function() {

if (x == 3) {y - 1}
  
}

do.call(Change_y, as.list(x,y))

Error in Change_y(1, 2, 3, 2, 2, 3, 1, 1, 3, 3) :
unused argument(s) ( ...)

##

How should it be done?

Cheers,
Kare

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Re: [R] Trying to pass arrays as arguments to a function

2008-10-20 Thread Kaom Te


y - ifelse(x == 3, 1, y)

-Kaom

On Oct 20, 2008, at 6:07 AM, Kåre Edvardsen wrote:

I'd like to avoid looping through an array in order to change values  
in

the array as it takes too long.
I red from an earlier post it can be done by do.call but never got  
it

to work. The Idea is to change the value of y according to values in
x. Wherever x holds the value 3, the corresponding value in y
should be set to 1.

So I tried the following giving an error message:

#
x - c(1,2,3,2,2,3,1,1,3,3)
y - c(0,0,1,1,0,0,1,0,0,1)

Change_y - function() {

if (x == 3) {y - 1}

}

do.call(Change_y, as.list(x,y))

Error in Change_y(1, 2, 3, 2, 2, 3, 1, 1, 3, 3) :
   unused argument(s) ( ...)

##

How should it be done?

Cheers,
Kare

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Re: [R] Trying to pass arrays as arguments to a function

2008-10-20 Thread Philipp Pagel


On Mon, Oct 20, 2008 at 03:07:45PM +0200, Kåre Edvardsen wrote:
 I'd like to avoid looping through an array in order to change values in
 the array as it takes too long.
 I red from an earlier post it can be done by do.call but never got it
 to work. The Idea is to change the value of y according to values in
 x. Wherever x holds the value 3, the corresponding value in y
 should be set to 1. 

This should do what you want - no looping required:

y[x==3] - 1

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://mips.gsf.de/staff/pagel

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Re: [R] Trying to pass arrays as arguments to a function

2008-10-20 Thread Claudia Beleites

 I'd like to avoid looping through an array in order to change values in
 the array as it takes too long.
 I red from an earlier post it can be done by do.call but never got it
 to work. The Idea is to change the value of y according to values in
 x. Wherever x holds the value 3, the corresponding value in y
 should be set to 1.
y [x == 3] - 1


-- 
Claudia Beleites
Dipartimento dei Materiali e delle Risorse Naturali
Università degli Studi di Trieste
Via Alfonso Valerio 6/a
I-34127 Trieste

phone: +39 (0 40) 5 58-34 47
email: [EMAIL PROTECTED]

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Re: [R] Dep setting for install of packages... EOT (I think) and my thanks

2008-10-20 Thread Brian Lunergan


Gabor Grothendieck wrote:

leaps is not in Depends or Imports for car.  It is in Suggests
and those don't get automatically pulled in when dependencies = NA.

What you could do is to replace the builtin menuInstallPkgs with
your own by running this:

assignInNamespace(menuInstallPkgs, function (type = getOption(pkgType)) {
install.packages(NULL, .libPaths()[1], dependencies = TRUE, type = type)
}, ns = utils)

You will need to do that in every session that you want to use the install
menu or you could just add it to your Rprofile.site file which you
can find by running this:

file.path(R.home(), etc, Rprofile.site)


So, if I'm understanding this correctly, for a simple, one libpath() 
situation like mine dep = NA will catch depends and imports, while dep = 
TRUE will catch those two plus anything under suggests. From prior 
discussion and suggested reading I take it I would use dep = FALSE to catch 
the first two if I had multiple locations in libpath() while dep = TRUE 
behaviour would stay the same.


In putting the above code snippet into the Rprofile file, does it need to 
be all on one line or can it spread over multiple lines as it appears above?


My thanks for your time and patient assistance in finding the answers. It 
is very much appreciated and useful.


Regards...
--

Brian Lunergan
Nepean, Ontario
Canada


---

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Re: [R] Date classes in ggplot2

2008-10-20 Thread Thompson, David (MNR)

Is not melt part of reshape package, i.e., require(reshape)? 

DaveT.
-Original Message-
From: stephen sefick [mailto:[EMAIL PROTECTED] 
Sent: October 19, 2008 12:12 PM
To: hadley wickham
Cc: R Help
Subject: Re: [R] Date classes in ggplot2

well I must have either resolve it without knowing it, ot something
funny is going on...  sorry I didn't run it in a clean R session.
Should have.  works fine
thanks

Stephen Sefick

On Sun, Oct 19, 2008 at 12:01 PM, hadley wickham 
[EMAIL PROTECTED] wrote:
 On Sun, Oct 19, 2008 at 10:49 AM, stephen sefick 
[EMAIL PROTECTED] wrote:
 updn.gg - (structure(list(date = structure(c(11808, 11869, 
11961, 11992,
 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057,
 13149), class = Date), unrestored = c(1.13789418691602, 
0.704948049842955,
 0.276777348238899, 0.417586861554189, 0.504870337754768, 
0.673201771716216,
 0.560704221510771, 0.835737007551542, 1.10773858390693, 
0.197070828834836,
 0.942350681588179, 0.950447141061461, 0.246637790002705, 
0.324035567509960
 ), restored = c(1.39981554315924, 0.89196314359498, 
0.407816250252697,
 0.823496839063978, 1.14429021220358, 1.23971035967413, 
0.960868900583432,
 0.927685306209829, 1.22072345292821, 0.249842897450642, 
1.00879641624694,
 0.925372139878243, 0.317259909172362, 0.382677149697482)), 
.Names = c(date,
 unrestored, restored), row.names = c(NA, -14L), class = 
data.frame))

 #I would like to do this in ggplot
 xyplot(unrestored+restored~date, data=updn.gg, type=c(l), 
auto.key=TRUE)
 #this is what I have tried and get an error message
 melt.updn - melt(updn.gg, id.var=date)

 What error message?  It works for me.

 Hadley

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

-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

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Re: [R] Date classes in ggplot2

yes it is,  I meant without the clutter of my previous analyses -
reshape is required by ggplot2 so I had that covered... If I am
understanding you correctly.
thanks for all of the help,

Stephen

On Mon, Oct 20, 2008 at 9:39 AM, Thompson, David (MNR)
[EMAIL PROTECTED] wrote:
 Is not melt part of reshape package, i.e., require(reshape)?

 DaveT.
-Original Message-
From: stephen sefick [mailto:[EMAIL PROTECTED]
Sent: October 19, 2008 12:12 PM
To: hadley wickham
Cc: R Help
Subject: Re: [R] Date classes in ggplot2

well I must have either resolve it without knowing it, ot something
funny is going on...  sorry I didn't run it in a clean R session.
Should have.  works fine
thanks

Stephen Sefick

On Sun, Oct 19, 2008 at 12:01 PM, hadley wickham
[EMAIL PROTECTED] wrote:
 On Sun, Oct 19, 2008 at 10:49 AM, stephen sefick
[EMAIL PROTECTED] wrote:
 updn.gg - (structure(list(date = structure(c(11808, 11869,
11961, 11992,
 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057,
 13149), class = Date), unrestored = c(1.13789418691602,
0.704948049842955,
 0.276777348238899, 0.417586861554189, 0.504870337754768,
0.673201771716216,
 0.560704221510771, 0.835737007551542, 1.10773858390693,
0.197070828834836,
 0.942350681588179, 0.950447141061461, 0.246637790002705,
0.324035567509960
 ), restored = c(1.39981554315924, 0.89196314359498,
0.407816250252697,
 0.823496839063978, 1.14429021220358, 1.23971035967413,
0.960868900583432,
 0.927685306209829, 1.22072345292821, 0.249842897450642,
1.00879641624694,
 0.925372139878243, 0.317259909172362, 0.382677149697482)),
.Names = c(date,
 unrestored, restored), row.names = c(NA, -14L), class =
data.frame))

 #I would like to do this in ggplot
 xyplot(unrestored+restored~date, data=updn.gg, type=c(l),
auto.key=TRUE)
 #this is what I have tried and get an error message
 melt.updn - melt(updn.gg, id.var=date)

 What error message?  It works for me.

 Hadley

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




--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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


-K. Mullis







-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] [R-pkgs] New verion 0.3-7 of gsubfn package

Version 0.3-7 of the gsubfn package is available on CRAN.
Changes to the package are:

- all known bugs have been fixed.

- in gsubfn and strapply the replacement object can be a
  list (or a function, formula, character string or proto
  object, as before).  In the case of a list, regexp matches
  are looked up in the list names and the corresponding list
  component used.

  # Example 1 - at string end replace M with e6  G with e9
  gsubfn(.$, list(M = e6, G = e9), c(19G, 32.5M))

- the backref= argument in gsubfn now defaults to the
  negative of its prior default.  Thus by default if there
  are back references only they and not the entire match is
  passed to the user function.  (Although the changed
  default introduces an incompatibility with prior versions
  this incompatability is small because it only affects
  situations where back references are present in the
  regular expression _and_ backref= was not used. Since the
  previous default for backref= was not useful in that case
  there would be very few, if any, such instances.)  On the
  other hand, it does mean that the backref= argument can be
  omitted most cases now.

For more info, see home page and brief description below.
http://gsubfn.googlecode.com

The gsubfn package - summary


A seemingly small enhancement to R's gsub function, allowing
the replacement string to be a function, formula, list or
proto object gives surprising additional power allowing
parsing of strings by content rather than delimiters.  e.g.
extract numbers:

# Example 2. Replace each number with its increment
gsubfn([0-9]+, ~ as.numeric(x) + 1, 90 and 17)

# Example 3. extract numbers
strapply(Over 90 stones in 17 places, [0-9]+)

The optional function passed to gsubfn and strapply may be
specified using usual R function notation or by a formula
whose left hand side represents the arguments (defaulting to
variables appearing in right hand side) and whose right hand
side represents the body.  In order to extend this
functionality beyond gsubfn and strapply to the rest of R
any R function may be prefaced with fn$ like this:

# Example 4. Integrate x^2
fn$integrate(~ x^2, 0, 1)

It also supports quasi-perl style string interpolation:

# Example 5. Quasi-perl style string interpolation
fn$cat(pi = $pi and e = `exp(1)`\n)

match.funfn is an alternative to match.fun which allows
developers to add this functionality to their own functions
by simply replacing match.fun with match.funfn -- a one line
change.  In that case even the fn$ prefix is not needed.

___
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Re: [R] pairs plots in R

2008-10-20 Thread Sharma, Dhruv

Thanks Felix.

Regards,
Dhruv 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Felix Andrews
Sent: Sunday, October 19, 2008 11:37 PM
To: Sharma, Dhruv
Cc: r-help@r-project.org
Subject: Re: [R] pairs plots in R

One idea:

if the primary variable of interest is a categorical (binary), I would rather 
look at univariate plots for each of your 100 variables, grouped by the primary 
one.

e.g.

library(latticeExtra)

marginal.plot(~ myBigDat, data = myBigData,
   groups = myBinaryVar, auto.key = TRUE,
   layout = c(4, 4))

(This is a convenient interface to lattice::densityplot and lattice::dotplot)

If you view 16 such densityplots per page, that still gives you 7 pages. You 
could use playwith() (from playwith package) to scroll through the pages.

-Felix

2008/10/20 Sharma, Dhruv [EMAIL PROTECTED]:
 Hi,
  is there a way to take a data frame with 100+ columns and large data set to 
 do efficient exploratory analysis in R with pairs?

 I find using pairs on the whole matrix is slow and the resulting matrix is 
 tiny.

 Also the variable of interest for me is a binary var Y or N .

 Is there an efficient way to graphically view many variable relationships 
 that does not look teeny ?

 I could do pairs 10 at a time but this seems too brute force.

 thanks
 Dhruv

[[alternative HTML version deleted]]

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

--
Felix Andrews / 安福立
http://www.neurofractal.org/felix/
3358 543D AAC6 22C2 D336  80D9 360B 72DD 3E4C F5D8
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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] Dep setting for install of packages... EOT (I think) and my thanks

You can put the code on several lines as shown or on one line.  Does
not matter.

Read the info on the dependencies= arg in ?install.packages
for the definitive explanation of how dependencies= works.

On Mon, Oct 20, 2008 at 9:32 AM, Brian Lunergan [EMAIL PROTECTED] wrote:
 Gabor Grothendieck wrote:

 leaps is not in Depends or Imports for car.  It is in Suggests
 and those don't get automatically pulled in when dependencies = NA.

 What you could do is to replace the builtin menuInstallPkgs with
 your own by running this:

 assignInNamespace(menuInstallPkgs, function (type =
 getOption(pkgType)) {
install.packages(NULL, .libPaths()[1], dependencies = TRUE, type =
 type)
 }, ns = utils)

 You will need to do that in every session that you want to use the install
 menu or you could just add it to your Rprofile.site file which you
 can find by running this:

 file.path(R.home(), etc, Rprofile.site)

 So, if I'm understanding this correctly, for a simple, one libpath()
 situation like mine dep = NA will catch depends and imports, while dep =
 TRUE will catch those two plus anything under suggests. From prior
 discussion and suggested reading I take it I would use dep = FALSE to catch
 the first two if I had multiple locations in libpath() while dep = TRUE
 behaviour would stay the same.

 In putting the above code snippet into the Rprofile file, does it need to be
 all on one line or can it spread over multiple lines as it appears above?

 My thanks for your time and patient assistance in finding the answers. It is
 very much appreciated and useful.

 Regards...
 --

 Brian Lunergan
 Nepean, Ontario
 Canada


 ---
 avast! Antivirus: Outbound message clean.
 Virus Database (VPS): 081018-0, 2008-10-18
 Tested on: 2008-10-20 09:32:23
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] pairs plots in R

2008-10-20 Thread Sharma, Dhruv

thanks Antony.

regards,
Dhruv

From: Antony Unwin [mailto:[EMAIL PROTECTED] 
Sent: Monday, October 20, 2008 7:00 AM
To: r-help@r-project.org
Cc: Sharma, Dhruv
Subject: Re: [R] pairs plots in R

If you want to do efficient exploratory data analysis on this kind of
dataset, then interactive graphics with parallel coordinate plots (ipcp
in iplots) should help.  Of course, it depends what you mean by large.
It might be worth looking at the book Graphics of Large Datasets for
some ideas.

Antony Unwin
Professor of Computer-Oriented Statistics and Data Analysis,
Mathematics Institute,
University of Augsburg, 
86135 Augsburg, Germany
Tel: + 49 821 5982218

From: Sharma, Dhruv [EMAIL PROTECTED]

Date: 19 October 2008 10:58:53 pm GMT+02:00

To: r-help@r-project.org

Subject: [R] pairs plots in R

Hi,
 is there a way to take a data frame with 100+ columns and large
data set to do efficient exploratory analysis in R with pairs?

I find using pairs on the whole matrix is slow and the resulting
matrix is tiny.

Also the variable of interest for me is a binary var Y or N .

Is there an efficient way to graphically view many variable
relationships that does not look teeny ?

I could do pairs 10 at a time but this seems too brute force.

thanks
Dhruv

[[alternative HTML version deleted]]

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[R] Package Deal

2008-10-20 Thread Leandro Marino

Hi,

I was trying to adjust an bayesian network with Deal package.

I know how to adjust when I give the structure and when I don't give the
struture. Now, I want to adjust when I know some variables relationships and
others I don't know.


Anybody knows how can I do that?

Regards,



Atenciosamente,
Leandro Lins Marino
Centro de Avaliação
Fundação CESGRANRIO
Rua Santa Alexandrina, 1011 - 2º andar
Rio de Janeiro, RJ - CEP: 20261-903
( (21) 2103-9600 R.:236
( (21) 8777-7907
* [EMAIL PROTECTED]
Aquele que suporta o peso da sociedade
é precisamente aquele que obtém
 as menores vantagens. (SMITH, Adam)

P  Antes de imprimir pense em sua responsabilidade e compromisso com o MEIO
AMBIENTE

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Re: [R] Error reporting in R

2008-10-20 Thread Duncan Murdoch


On 20/10/2008 5:39 AM, A.Noufaily wrote:

Hello,

I am hoping someone can help me with the following:

I am applying the function mle on a single data set X n times, each
time using a different set of initial values v[i] (i=1,...,n).
The initial values are all finite. Two cases arise:
A- For some sets of initial values mle is giving parameter estimates.
B- For some other initial values, mle is reporting the following error: 


Error in optim(start,f,method=method,hessian=TRUE,...):
Initial value in vmin is not finite

My aim is to tell R to return the estimates whenever case A, and to
return the word Error whenever case B (using an if statement inside
a loop).

How to tell R to return the word Error whenever mle reports the error
given above?
Is there a function similar to is.nan which tests the existence of an
error?


Normally an error halts evaluation, but see ?try.  The basic pattern is

result - try(mle(...))
if (inherits(result, try-error)) return(Error)
else return(result)

Duncan Murdoch

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Re: [R] The evaluation of optional function arguments

2008-10-20 Thread Sietse Brouwer

Hullo all,

 [***recap of the question: passing bar=harry fails when bar is among
 the ... args.]
 callTimes - function(tom, harry) {

timesDefineInside(foo=tom, bar=harry
 }

timesDefineInside - function(foo, ...) {
foo * bar
 }
 callTimes(3, 4)
 # Error: object bar not found
 [***end recap]


 On 20/10/2008, hadley wickham [EMAIL PROTECTED] wrote:
  Why don't you want to do this?
 
   timesDefineInside - function(foo, bar...) {
 foo * bar
   }


You're right that's it the obvious solution — the thing is, I'm
 hacking on somebody else's function, and in the interim I want to
 maintain backward compatibility. Part of that is expecting the same
 set of formals, so any extra arguments (in this case, 'bar' is
 actually 'experimentalMode') have to come throught the dots.


 Kaom Te wrote:
  Try this for timesDefineInside:
  timesDefineInside - function(foo, ...) {
 extra.args -   list(...)
 bar - extra.args$bar
 foo * bar
  }


Ah, that works! Apparently, there's a difference between
 list(...) and
 match.call(expand.dots=FALSE)$...   # wrapped in eval() or not.
 That makes sense, though, as the first will have evaluated bar=harry
 at the moment of calling, within the calling function, while the
 second will evaluated bar=harry within the called function. I think.

 Thank you both very much; and thanks, too, to Martin Morgan, who
 replied off-list pointing me at
 bar - eval.parent(match.call()$bar)
 .

 Grateful regards,

 Sietse
 Sietse Brouwer


 --

Sietse Brouwer -- [EMAIL PROTECTED] -- +31 6 13456848
 Wildekamp 32 -- 6721 JD Bennekom -- the Netherlands
 MSN: [EMAIL PROTECTED] -- ICQ: 341232104

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[R] Combining all possible values of variables into a new...

2008-10-20 Thread stefan . petersson


I'm trying to create a new column in my data.frame where subjects are 
categorized depending on values on four other columns. In any other case I 
would just nest a few ifelse statements, however, in this case i have 
4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I wonder 
if there is a more efficient way of doing this. 

For illustrational purposes, let's say i have:

x-c(1,0,0,1,0,0,1,0,0,1)
y-c(1,3,2,3,2,1,2,3,2,3)
z-c(1,2,1,2,1,2,1,2,1,2)
d-as.data.frame(cbind(x,y,z))

and i do:

d$myvar - ifelse(d$x == 0  d$y==1  d$z==1 , d$myvar - 1,
ifelse(d$x == 0  d$y==1  d$z==2 , d$myvar - 2,
ifelse(d$x == 0  d$y==2  d$z==1 , d$myvar - 3,
ifelse(d$x == 0  d$y==2  d$z==2 , d$myvar - 4,
ifelse(d$x == 0  d$y==3  d$z==1 , d$myvar - 5,
ifelse(d$x == 0  d$y==3  d$z==2 , d$myvar - 6,
ifelse(d$x == 1  d$y==1  d$z==1 , d$myvar - 7,
ifelse(d$x == 1  d$y==1  d$z==2 , d$myvar - 8,
ifelse(d$x == 1  d$y==2  d$z==1 , d$myvar - 9,
ifelse(d$x == 1  d$y==2  d$z==2 , d$myvar - 10,
ifelse(d$x == 1  d$y==3  d$z==1 , d$myvar - 11,
ifelse(d$x == 1  d$y==3  d$z==2 , d$myvar - 12, NA

Suggestions?

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Re: [R] Combining all possible values of variables into a new...

2008-10-20 Thread Dimitris Rizopoulos


try this:

x - c(1,0,0,1,0,0,1,0,0,1)
y - c(1,3,2,3,2,1,2,3,2,3)
z - c(1,2,1,2,1,2,1,2,1,2)
d - data.frame(x, y, z)

ind - do.call(paste, c(expand.grid(1:2, 1:3, 0:1)[3:1], sep = \r))
trg - do.call(paste, c(d, sep = \r))
d$myvar - match(trg, ind)


I hope it helps.

Best,
Dimitris


[EMAIL PROTECTED] wrote:
I'm trying to create a new column in my data.frame where subjects are categorized depending on values on four other columns. In any other case I would just nest a few ifelse statements, however, in this case i have 4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I wonder if there is a more efficient way of doing this. 


For illustrational purposes, let's say i have:

x-c(1,0,0,1,0,0,1,0,0,1)
y-c(1,3,2,3,2,1,2,3,2,3)
z-c(1,2,1,2,1,2,1,2,1,2)
d-as.data.frame(cbind(x,y,z))

and i do:

d$myvar - ifelse(d$x == 0  d$y==1  d$z==1 , d$myvar - 1,
ifelse(d$x == 0  d$y==1  d$z==2 , d$myvar - 2,
ifelse(d$x == 0  d$y==2  d$z==1 , d$myvar - 3,
ifelse(d$x == 0  d$y==2  d$z==2 , d$myvar - 4,
ifelse(d$x == 0  d$y==3  d$z==1 , d$myvar - 5,
ifelse(d$x == 0  d$y==3  d$z==2 , d$myvar - 6,
ifelse(d$x == 1  d$y==1  d$z==1 , d$myvar - 7,
ifelse(d$x == 1  d$y==1  d$z==2 , d$myvar - 8,
ifelse(d$x == 1  d$y==2  d$z==1 , d$myvar - 9,
ifelse(d$x == 1  d$y==2  d$z==2 , d$myvar - 10,
ifelse(d$x == 1  d$y==3  d$z==1 , d$myvar - 11,
ifelse(d$x == 1  d$y==3  d$z==2 , d$myvar - 12, NA

Suggestions?

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

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014

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Re: [R] use of empty space in split graph device

2008-10-20 Thread Benoit Boulinguiez

That's it.

thanks 


Regards/Cordialement


Benoit Boulinguiez 


-Message d'origine-
De : Greg Snow [mailto:[EMAIL PROTECTED] 
Envoyé : vendredi 17 octobre 2008 19:19
À : Benoit Boulinguiez; r-help@r-project.org
Objet : RE: [R] use of empty space in split graph device

The plot.new function will start a new plot in the next frame, but not plot
anything.  By default the coordinates go from 0 to 1 in both the x and y
directions.

Try:

 par(mfrow=c(2,2))
 boxplot(rnorm(100))
 boxplot(runif(100))
 boxplot(rexp(100, 1/3))
 plot.new()
 legend( 0,1, pch=1:3, legend=c('a','b','c'), lty=1, 
 col=c('red','green','blue'))


Hope this helps,

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111


 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
 project.org] On Behalf Of Benoit Boulinguiez
 Sent: Friday, October 17, 2008 7:30 AM
 To: r-help@r-project.org
 Subject: [R] use of empty space in split graph device

 Hi all,

 I divide the graphical device into 4 ---mfrow=c(2,2)--- to plot 3 
 boxplots graph.
 Thus I get an empty space for the fourth graph. I d'like to place a 
 legend in this space.

 legend() creates a legend in the last graph (the 3rd) I'd like a 
 legend in the empty space.

 How may I do it?

 #graph parameter
 par( fin=c(6,6),   #dim graph
  mai=c(1,1,0.5,0.5) #marges graph
  ,mfrow=c(2,2)
  )
 #boxplots
 boxplot(x)
 title(main=expression(1- residuals distribution))
 abline(0,0,lwd=2,col=2)
 boxplot(x)
 title(main=expression(2- residuals distribution))
 abline(0,0,lwd=2,col=2)
 boxplot(x)
 title(main=expression(3- residuals distribution))
 abline(0,0,lwd=2,col=2)



 Regards/Cordialement

 -
 Benoit Boulinguiez
 Ph.D
 Ecole de Chimie de Rennes (ENSCR) Bureau 1.20 Equipe CIP UMR CNRS 6226 
 Sciences Chimiques de Rennes
 Campus de Beaulieu, 263 Avenue du Général Leclerc 35700 Rennes, France 
 Tel 33 (0)2 23 23 80 83 Fax 33 (0)2 23 23 81 20 http://www.ensc- 
 rennes.fr/



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Re: [R] convert matrix to dataframe with repeating row names

2008-10-20 Thread Richard . Cotton

 I have a matrix x with repeating row names. 

 zz-matrix(0,4,4)
 
 rownames(zz)=c(a,a,b,b)
 
 data.frame(zz) (?)

The row names on a data frame should be unique.  You can try 
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame.  If 
you need the row name information, add it as a column in the data frame, 
e.g. mydataframe$rnames - rownames(zz).  (Note to R-Core: the 
documentation for as.data.frame doesn't mention the usage of 
row.names=FALSE to ignore row names, but it seems to work consistently. 
Does the help page for as.data.frame need updating?)

 lm(as.formula(paste(final_dat[,5]~,paste(colnames(x),collapse=+))),x
 )
 
 this gives me a error

 Error in model.frame.default(formula =
 as.formula(paste(final_dat[,5]~,  : 
 
   'data' must be a data.frame, not a matrix or an array

I suspect that if you try class(x), it will be a matrix, not the requisite 
data frame.

Regards,
Richie.

Mathematical Sciences Unit
HSL



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Re: [R] Combining all possible values of variables into a new...

2008-10-20 Thread David Hajage

or if x, y and z are factors :

 dbis - as.data.frame(apply(d, 2, as.factor)

 dbis$x:dbis$y:dbis$z
 [1] 1:1:1 0:3:2 0:2:1 1:3:2 0:2:1 0:1:2 1:2:1 0:3:2 0:2:1 1:3:2
12 Levels: 0:1:1 0:1:2 0:2:1 0:2:2 0:3:1 0:3:2 1:1:1 1:1:2 1:2:1 ... 1:3:2

and for your results :
 as.numeric(dbis$x:dbis$y:dbis$z)
 [1]  7  6  3 12  3  2  9  6  3 12




2008/10/20 Dimitris Rizopoulos [EMAIL PROTECTED]

 try this:

 x - c(1,0,0,1,0,0,1,0,0,1)
 y - c(1,3,2,3,2,1,2,3,2,3)
 z - c(1,2,1,2,1,2,1,2,1,2)
 d - data.frame(x, y, z)

 ind - do.call(paste, c(expand.grid(1:2, 1:3, 0:1)[3:1], sep = \r))
 trg - do.call(paste, c(d, sep = \r))
 d$myvar - match(trg, ind)


 I hope it helps.

 Best,
 Dimitris



 [EMAIL PROTECTED] wrote:

 I'm trying to create a new column in my data.frame where subjects are
 categorized depending on values on four other columns. In any other case I
 would just nest a few ifelse statements, however, in this case i have
 4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I
 wonder if there is a more efficient way of doing this.
 For illustrational purposes, let's say i have:

 x-c(1,0,0,1,0,0,1,0,0,1)
 y-c(1,3,2,3,2,1,2,3,2,3)
 z-c(1,2,1,2,1,2,1,2,1,2)
 d-as.data.frame(cbind(x,y,z))

 and i do:

 d$myvar - ifelse(d$x == 0  d$y==1  d$z==1 , d$myvar - 1,
 ifelse(d$x == 0  d$y==1  d$z==2 , d$myvar - 2,
 ifelse(d$x == 0  d$y==2  d$z==1 , d$myvar - 3,
 ifelse(d$x == 0  d$y==2  d$z==2 , d$myvar - 4,
 ifelse(d$x == 0  d$y==3  d$z==1 , d$myvar - 5,
 ifelse(d$x == 0  d$y==3  d$z==2 , d$myvar - 6,
 ifelse(d$x == 1  d$y==1  d$z==1 , d$myvar - 7,
 ifelse(d$x == 1  d$y==1  d$z==2 , d$myvar - 8,
 ifelse(d$x == 1  d$y==2  d$z==1 , d$myvar - 9,
 ifelse(d$x == 1  d$y==2  d$z==2 , d$myvar - 10,
 ifelse(d$x == 1  d$y==3  d$z==1 , d$myvar - 11,
 ifelse(d$x == 1  d$y==3  d$z==2 , d$myvar - 12, NA

 Suggestions?

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

 Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
 Tel: +31/(0)10/7043478
 Fax: +31/(0)10/7043014


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Re: [R] Calculate SPE in PLS package

2008-10-20 Thread Bjørn-Helge Mevik

Stella Sim [EMAIL PROTECTED] writes:

 I want to calculate SPE (squared prediction error) in x-space, can
 someone help?

 Here are my codes:

 fit.pls-
 plsr(Y~X,data=DAT,ncomp=3,scale=T,method='oscorespls',validation=CV,x=
 T) 
 actual-fit.pls$model$X

(The x = TRUE is not needed as long as model = TRUE (default).  x=TRUE
returns the predictors as fit.pls$x, and is included for compatibility
with lm().)

 pred-fit.pls$scores %*% t(fit.pls$loadings)
 SPE.x-rowSums((actual-pred)^2)

 Am I missing something here? 

You are missing the mean X spectrum.  See
matplot(t(pred), type = l, lty = 1) vs. matplot(t(actual), type = l, lty = 
1)

The Xmeans compontent of fit.pls contains this, so

pred - sweep(fit.pls$scores %*% t(fit.pls$loadings), 2, fit.pls$Xmeans, +)

would give you what you want.

Note, however, that this will calculate the _fitted_ SPE, not the
cross-validated SPE.  The crossvalidation implemented in the pls package
does not save the cross-validated scores/loadings -- that would consume
too much memory.  (Calculation of SPE withing the cross-validation
routines could have been implemented, but was not.)

-- 
Regards,
Bjørn-Helge Mevik

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Re: [R] Combining all possible values of variables into a new...

2008-10-20 Thread Gustaf Rydevik

On Mon, Oct 20, 2008 at 4:10 PM,  [EMAIL PROTECTED] wrote:

 I'm trying to create a new column in my data.frame where subjects are 
 categorized depending on values on four other columns. In any other case I 
 would just nest a few ifelse statements, however, in this case i have 
 4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I 
 wonder if there is a more efficient way of doing this.

 For illustrational purposes, let's say i have:

 x-c(1,0,0,1,0,0,1,0,0,1)
 y-c(1,3,2,3,2,1,2,3,2,3)
 z-c(1,2,1,2,1,2,1,2,1,2)
 d-as.data.frame(cbind(x,y,z))

 and i do:

 d$myvar - ifelse(d$x == 0  d$y==1  d$z==1 , d$myvar - 1,
 ifelse(d$x == 0  d$y==1  d$z==2 , d$myvar - 2,
 ifelse(d$x == 0  d$y==2  d$z==1 , d$myvar - 3,
 ifelse(d$x == 0  d$y==2  d$z==2 , d$myvar - 4,
 ifelse(d$x == 0  d$y==3  d$z==1 , d$myvar - 5,
 ifelse(d$x == 0  d$y==3  d$z==2 , d$myvar - 6,
 ifelse(d$x == 1  d$y==1  d$z==1 , d$myvar - 7,
 ifelse(d$x == 1  d$y==1  d$z==2 , d$myvar - 8,
 ifelse(d$x == 1  d$y==2  d$z==1 , d$myvar - 9,
 ifelse(d$x == 1  d$y==2  d$z==2 , d$myvar - 10,
 ifelse(d$x == 1  d$y==3  d$z==1 , d$myvar - 11,
 ifelse(d$x == 1  d$y==3  d$z==2 , d$myvar - 12, NA

 Suggestions?

How about the following?

x-c(1,0,0,1,0,0,1,0,0,1)
y-c(1,3,2,3,2,1,2,3,2,3)
z-c(1,2,1,2,1,2,1,2,1,2)
d-as.data.frame(cbind(x,y,z))

xyz.comb-interaction(x,y,z,lex.order=T)
d$myvar-match(xyz.comb,levels(xyz.comb))


/Gustaf


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

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[R] Odp: Combining all possible values of variables into a new...

2008-10-20 Thread Petr PIKAL

Hi

maybe you can change combination of your columns to factor

new -factor(paste(x,y,z, sep=))

and then reorder this factor (see reorder) change it to numeric to get 
index

index - as.numeric(factor)

and then choose from set of 144 data 

data[index] 

or use your index directly if you need just numbers from 1 to the final to 
get your myvar.

Maybe you could also look to permutations function from gtools package

Regards
Petr


[EMAIL PROTECTED] napsal dne 20.10.2008 16:10:26:

 
 I'm trying to create a new column in my data.frame where subjects are 
 categorized depending on values on four other columns. In any other case 
I 
 would just nest a few ifelse statements, however, in this case i have 
 4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I 

 wonder if there is a more efficient way of doing this. 
 
 For illustrational purposes, let's say i have:
 
 x-c(1,0,0,1,0,0,1,0,0,1)
 y-c(1,3,2,3,2,1,2,3,2,3)
 z-c(1,2,1,2,1,2,1,2,1,2)
 d-as.data.frame(cbind(x,y,z))
 
 and i do:
 
 d$myvar - ifelse(d$x == 0  d$y==1  d$z==1 , d$myvar - 1,
 ifelse(d$x == 0  d$y==1  d$z==2 , d$myvar - 2,
 ifelse(d$x == 0  d$y==2  d$z==1 , d$myvar - 3,
 ifelse(d$x == 0  d$y==2  d$z==2 , d$myvar - 4,
 ifelse(d$x == 0  d$y==3  d$z==1 , d$myvar - 5,
 ifelse(d$x == 0  d$y==3  d$z==2 , d$myvar - 6,
 ifelse(d$x == 1  d$y==1  d$z==1 , d$myvar - 7,
 ifelse(d$x == 1  d$y==1  d$z==2 , d$myvar - 8,
 ifelse(d$x == 1  d$y==2  d$z==1 , d$myvar - 9,
 ifelse(d$x == 1  d$y==2  d$z==2 , d$myvar - 10,
 ifelse(d$x == 1  d$y==3  d$z==1 , d$myvar - 11,
 ifelse(d$x == 1  d$y==3  d$z==2 , d$myvar - 12, NA
 
 Suggestions?
 
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Re: [R] Extracting variables with a given prefix in the parent environment

2008-10-20 Thread erwann rogard

Thank you all for your answers.

On Mon, Oct 20, 2008 at 6:32 AM, Henrique Dallazuanna [EMAIL PROTECTED]wrote:

 Try this:

 Using 'envir' argument, in ls and get function:

 get.vars.name.prefix-function(prefix, envir = parent.frame()){
result-list()
len-nchar(prefix)
var.names-ls(envir = envir,pattern=prefix)
print(var.names)
for(i in 1:length(var.names)){
   name-var.names[i]
   field-substr(name,len+1,nchar(name))
   result[[field]]-get(name, envir = envir)
   }
  result
 }

 On Mon, Oct 20, 2008 at 4:06 AM, erwann rogard [EMAIL PROTECTED]wrote:

 hi,

 get.vars.name.prefix-function(prefix){
  result-list()
  len-nchar(prefix)
  var.names-ls(name=1,pattern=prefix) #name=1 probably wrong option
  print(var.names)
  for(i in 1:length(var.names)){
  name-var.names[i]
  field-substr(name,len+1,nchar(name))
  result[[field]]-get(name)
  }
  result
 }

 for example
 x.1-1
 x.2-2
 get.vars.name.prefix(x.) should return 1,2

 this does not work if i call if from within another function:

 g-function(){
  x.1-1
  x.2-2
  get.vars.name.prefix(x.) # should return 1,2, nothing else
 }

 i would like the environment specified within ls(...) to be that of the
 body
 of the function that calls get.vars.name.prefix. apparently name=1 is
 probably not the right option.

 thanks.

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 --
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[R] Row Sampling

2008-10-20 Thread Alex99



Hi everyone,
does any one knows how can I make sample from rows?
i.e. I have a data like this:

 s1 s2 s3 s4
1   0  0  0  1
2   1  0  1  0
3   0  0  0  0
4   0  0  0  0
5   0  1  0  1
6   1  0  0  0
7   0  0  0  0
8   0  0  0  0
9   0  0  0  0
10  0  0  0  1

I need to make 5 different sample with 5 different persons(rows) in each
sample from it keeping s1,s3,s3,s4 but changing rows. I cant get replicate
or sample functions to make samples by changing the rows.

Thanks

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[R] How to get estimate of confidence interval?

2008-10-20 Thread Ted Byers

I thought I was finished, having gotten everything to work as intended. This
is a model of risk, and the short term forecasts look very good, given the
data collected after the estimates are produced (this model is intended to
be executed daily, to give a continuing picture of our risk). But now there
is a new requirement.

I have weekly samples from a non-autonomous process (i.e. although well
modelled as a decay process, with an exponential distribution fitting the
decay times well, the rate estimates and their sd vary considerably from one
week to the next). The total number of events to be expected from a given
sample over the next week can be easily estimated from a simple integral.
And the total number of these events from all samples, is just the sum of
these estimates over all samples. So far, so good (imagine you have a
sample of a variety of species of radionuclides all emitting alpha particles
with the same energy - so you can't tell from the decay event which species
produced the alpha particles).

I guess there are two parts of my question. I get a fit of the exponential
distribution to each sample using fitdistr(x,exponential). I am finding
the expected values vary by as much as a factor of 4, and the corresponding
estimates of sd vary by as much as a factor of 100 (some samples are MUCH
larger than others). How do I go from the sd it gives to a 99% confidence
interval for the integral for that function from now through a week from now
(or to the end of time, or through the next month/quarter)? And how do I
move from these estimates to get the expected value and confidence intervals
for the totals over all the samples? I am a bit rusty on figuring out how
error propagates through model calculations (an online reference for this
would be handy, if you know of one).

Thanks

Ted
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[R] Zero mean correlation Matrix

2008-10-20 Thread AliR


Hi,,

I want to use the existing cor function in R but with a different way to
compute the correlation method.. basically zero mean correlation. 

The forumula I have is 

'D' - function(c1, c2)
  sum(c1*c2, na.rm=T)/(sqrt(sum(c1*c1, na.rm=T))*sqrt(sum(c2*c2, na.rm=T)))


I am not sure how i can modify the method cor computes its square roots and
covariance matrixes? I only need to add this to get the answer 
   


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[R] slow graphics download

2008-10-20 Thread Paul Murtaugh


Hi,

I often work from home, running R on my office computer and displaying 
interactive graphics on my home computer using X11.  This was always 
more sluggish than sitting at my office computer, obviously, but 
recently the graphics transfer has become painfully slow -- so slow, 
that I now divert the images to a postscript file and display that file 
remotely (which takes a second or two).


I don't know what changed between the old days and now -- I may have 
installed a new version of R and updated the o.s. (ubuntu) on both 
computers, but am unsure of the timing.  I realize this makes 
troubleshooting difficult, but I am hoping someone can offer a tip on 
where to begin.


Thanks
-Paul

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[R] folded normal distribution in R

2008-10-20 Thread Andreas Wittmann


Dear R useRs,

i wanted to ask if the folded normal destribution (Y = abs(X) with X 
normal distributed)
with density and random number generator is implemented in R or in any 
R-related package
so far? Maybe i can use the non-central chi-square distribution and 
rchisq(n, df=1, ncp0) here?


Thanks and best regards

Andreas

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Re: [R] slow graphics download


I suspect that you need to set the type= in X11(): see its help page.
E.g.

 'type = nbcairo' is the same device as 'type=cairo' without
 buffering: which is faster will depend on the X11 connection.
 Both will be slower than 'type = Xlib', especially on a slow X11
 connection as all the rendering is done on the machine running R
 rather than in the X server.

Basically, R's X11 device is now by default optimized for quality rather 
than speed, and your unusual usage needs the reverse.



On Mon, 20 Oct 2008, Paul Murtaugh wrote:


Hi,

I often work from home, running R on my office computer and displaying 
interactive graphics on my home computer using X11.  This was always more 
sluggish than sitting at my office computer, obviously, but recently the 
graphics transfer has become painfully slow -- so slow, that I now divert the 
images to a postscript file and display that file remotely (which takes a 
second or two).


I don't know what changed between the old days and now -- I may have 
installed a new version of R and updated the o.s. (ubuntu) on both computers, 
but am unsure of the timing.  I realize this makes troubleshooting difficult, 
but I am hoping someone can offer a tip on where to begin.


Thanks
-Paul

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

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Re: [R] folded normal distribution in R

2008-10-20 Thread Matthias Kohl


Dear Andreas,

try:
library(distr)
X - Norm() ## standard normal distribution
Y - abs(X)
plot(Y)

Best regards,
Matthias

Andreas Wittmann wrote:

Dear R useRs,

i wanted to ask if the folded normal destribution (Y = abs(X) with X 
normal distributed)
with density and random number generator is implemented in R or in any 
R-related package
so far? Maybe i can use the non-central chi-square distribution and 
rchisq(n, df=1, ncp0) here?


Thanks and best regards

Andreas

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--
Dr. Matthias Kohl
www.stamats.de

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Re: [R] folded normal distribution in R

It's trivial to do.  In particular abs(rnorm(...)) will give you the 
random number generator.


The term 'folded normal distribution' is normally only used for mean=0, 
when you just double the density, but the general result is


ifelse(x  0, 1, 0) * (dnorm(x, ...) + dnorm(-x, ...))


On Mon, 20 Oct 2008, Andreas Wittmann wrote:


Dear R useRs,

i wanted to ask if the folded normal destribution (Y = abs(X) with X normal 
distributed)
with density and random number generator is implemented in R or in any 
R-related package
so far? Maybe i can use the non-central chi-square distribution and rchisq(n, 
df=1, ncp0) here?


Thanks and best regards

Andreas


--
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] R Newbie Question

2008-10-20 Thread Johannes Bauer

Hello list,

I just started R today and tried something quite simple. I wanted to
create a colored plot and eventually after hours of fiddling around got
it working. However, my solution seems very suboptimal and I'd really
appreciate your hints on how to improve. I believe that R already offers
many functions I coded (e.g. distance between two vectors, vector
length, vector normalization and so on). I generally didn't even figure
out how to create a simple vector or how to extract a row from a matrix
so the result is a vector (to get scalars, I use the sum functions,
which is an incredibly ugly workaround).

To sum the problem up: one has a binary star system with a large star
(e.g. red giant) and a small star (e.g. white dwarf). Gravitation
between them is directly proportional to the mass and indirectly
proportional to the square of the distance. If correctly plotted, one
should be able to see the inner lagrange point L1 which is the point
where the gravitational potentials of the stars cancel out, e.g. an
object would not be attracted to any star. Well, enough background
information, here's my rookie code - please feel free to comment on
anything :-)

Kind regards,
Johannes



star1center = vector(numeric, 2)
star1center[1] = -0.5
star1center[2] = 0
star1mass = 30

star2center = vector(numeric, 2)
star2center[1] = 0.5
star2center[2] = 0
star2mass = 1

sqr = function(x) {
return(x * x)
}

distance = function(a, b) {
return(sqrt(sqr(a[1] - b[1]) + sqr(a[2] - b[2])))
}

len = function(x) {
return(sqrt(sqr(x[1]) + sqr(x[2])))
}

norm = function(x) {
return(x / len(x))
}

gravitation = function(invecx, invecy) {
invec = vector(numeric, 2)
invec[1] = invecx
invec[2] = invecy
vec1 = star1mass * norm(star1center - invec) / sqr(distance(invec,
star1center))
vec2 = star2mass * norm(star2center - invec) / sqr(distance(invec,
star2center))
return(len(vec1 + vec2))
}

vmin = -1
vmax = 1
step = 0.1
vals = ((vmax - vmin) / step) + 1

xvals = seq(vmin, vmax, step)
yvals = seq(vmin, vmax, step)

a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
b = matrix(seq(1, vals*vals), vals)

for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(sum(a[x,][1]), sum(a[y,][1]))
}
}
filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)

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[R] pass longer arguments to adj of text()

2008-10-20 Thread Thomas Steiner

Hi,

text()'s pos does handle more dimensional labels, while adj does not -
or how can I pass an array to text() and use the adj argument (skyblue
text)?

plot(-1:1,-1:1)
text(c(0,0),c(0,0),c(a,b),pos=c(1,2),col=red)
text(c(0,0),c(0,0),c(a,b),adj=cbind(c(1,2),c(1,1)),col=skyblue)

Thomas

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[R] error message when plotting survival curves

2008-10-20 Thread A Van Dyke


I am trying to plot survival curves using the following code as an example:

rs1799964.coxph-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method=efron))

plot(rs1799964.coxph,lyt=c(1,3),xlab=Survival in Months,ylab=Proportion
Surviving)

I am gettingthe following error message:


Error in xy.coords(x, y, xlabel, ylabel, log) : 
  'x' and 'y' lengths differ


Any input to debugging this matter would be greatly appreciated.  Thank you
in advance.

-Alison
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[R] distributions and glm

2008-10-20 Thread drbn


Hello,
I have seen that some papers do this:

1.) Group data by year (e.g. 35 years) 

2.) Estimate the mean of the key variable through the distribution that fits
better (some years is a normal distribution , others is a more skewed, gamma
distribution, etc.)

3.) With these estimated means of each year do a GLM.

I'd like to know if it is possible (to use these means in a GLM) or is a
wrong idea.

Thanks in advance

David
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[R] calculating mean for samples

2008-10-20 Thread Alex99


Hi everyone,
 does any one knows how can I calculate mean for different samples
 i.e. I have a data like this:

  s1 s2 s3 s4
 1   0  0  0  1
 2   1  0  1  0
 3   0  0  0  0
 4   0  0  0  0
 5   0  1  0  1
 6   1  0  0  0
 7   0  0  0  0
 8   0  0  0  0
 9   0  0  0  0
 10  0  0  0  1

 I need to make 5 different sample with 5 different persons(rows) in each
 sample from it keeping s1,s3,s3,s4 but changing rows. and then calculate
 the mean for each S in each sample. and finally calculate the grand
 mean,which is the mean of means for each sample. i.e. if I sample 5 time I
 get 5 different means for s1, s2, s3, s4. then I need to add all five
 means for and divide it by 5.(of course I have to do it for s1,s2,s3,s4
 separately) 
-- 
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[R] Staging area for data before read into R

I am wondering if there is a better alternative than Excel for data
storage that does not require database knowledge (I will eventually
have to learn this, but it is not on my immediate todo list).  I need
something that is not limited to 256 columns... I don't need any of
the built in functions in excel just a spreadsheet like program with
cells that hold data in a data.frame format for a staging area before
I get it into R.  Any help would be greatly appreciated.  This is not
a direct r question, but all of you folks have more experience than I
do and I am having a time finding what I need with google.
thanks in advance

-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] error message when plotting survival curves

I believe that you have to specify which variable is the x and also
the y. I am not familiar with what you are doing, but it looks as
though you need to specify both x and y to get it to plot.

Stephen

On Mon, Oct 20, 2008 at 2:22 PM, A Van Dyke [EMAIL PROTECTED] wrote:

I am trying to plot survival curves using the following code as an example:

rs1799964.coxph-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method=efron))

plot(rs1799964.coxph,lyt=c(1,3),xlab=Survival in Months,ylab=Proportion
Surviving)

I am gettingthe following error message:

Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ

Any input to debugging this matter would be greatly appreciated. Thank you
in advance.

-Alison
--
View this message in context:
http://www.nabble.com/error-message-when-plotting-survival-curves-tp20075771p20075771.html
Sent from the R help mailing list archive at Nabble.com.

--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

Re: [R] R Newbie Question

2008-10-20 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Johannes Bauer
 Sent: Monday, October 20, 2008 10:42 AM
 To: r-help@r-project.org
 Subject: [R] R Newbie Question

 Hello list,

 I just started R today and tried something quite simple. I wanted to
 create a colored plot and eventually after hours of fiddling 
 around got
 it working. However, my solution seems very suboptimal and I'd really
 appreciate your hints on how to improve. I believe that R 
 already offers
 many functions I coded (e.g. distance between two vectors, vector
 length, vector normalization and so on). I generally didn't 
 even figure
 out how to create a simple vector 

See ?c (i.e. type ?c without quotes at R prompt)

star1center = c(-0.5, 0)

or how to extract a row 
 from a matrix
 so the result is a vector (to get scalars, I use the sum functions,
 which is an incredibly ugly workaround).

In your code a[] is not a matrix, it is a data frame.  However, you should be 
able to index it just like you would a matrix.  For your code,

b[x, y] = gravitation(a[x,1], a[y,1])

Without the need for the sum() function.

 To sum the problem up: one has a binary star system with a large star
 (e.g. red giant) and a small star (e.g. white dwarf). Gravitation
 between them is directly proportional to the mass and indirectly
 proportional to the square of the distance. If correctly plotted, one
 should be able to see the inner lagrange point L1 which is the point
 where the gravitational potentials of the stars cancel out, e.g. an
 object would not be attracted to any star. Well, enough background
 information, here's my rookie code - please feel free to comment on
 anything :-)

 Kind regards,
 Johannes

snip

I have edited your code to simplify some of your functions and use vectorized 
computations.  I'm sure others could provide even more optimized code.  I will 
leave it to others to respond concerning already written functions for norm, 
len, etc.  You could try using the various search functions with key terms 
(like distance) to see what may be available.

star1center = c(-0.5, 0)
star1mass = 30

star2center = c(0.5, 0)
star2mass = 1

sqr = function(x) {
return(x * x)
}

distance = function(a, b) {
return(sum((a-b)^2)^0.5)
}

len = function(x) {
return(sum(x^2)^0.5)
}

norm = function(x) {
return(x / len(x))
}

gravitation = function(invecx, invecy) {
invec = c(invecx, invecy)
vec1 = star1mass * norm(star1center - invec) / 
sqr(distance(invec,star1center))
vec2 = star2mass * norm(star2center - invec) / 
sqr(distance(invec,star2center))
return(len(vec1 + vec2))
}

vmin = -1
vmax = 1
step = 0.1
vals = ((vmax - vmin) / step) + 1

xvals = seq(vmin, vmax, step)
yvals = seq(vmin, vmax, step)

a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
b = matrix(seq(1, vals*vals), vals)

for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(a[x,1], a[y,1])
}
}
filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)

Hope this is helpful,

Dan  

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA  98504-5204

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Re: [R] calculating mean for samples

assuming that A1 is the data provided in the first post... it would be
helpful if you used dput() to copy into an email, so that we could
just copy the code and data right into a R session and be off and
running.  I hope this helps.

#I used dput() on the object A1
A1 - (structure(list(s1 = c(0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L),
s2 = c(0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L), s3 = c(0L,
1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), s4 = c(1L, 0L, 0L, 0L,
1L, 0L, 0L, 0L, 0L, 1L)), .Names = c(s1, s2, s3, s4
), class = data.frame, row.names = c(NA, -10L)))

f - do.call(rbind , rep(A1[sample(nrow(A1), 5),], 5))
d - apply(f, MARGIN=1, FUN=mean)
mean(d[names(d)==s1])
mean(d[names(d)==s2])
mean(d[names(d)==s3])
mean(d[names(d)==s4])

On Mon, Oct 20, 2008 at 2:44 PM, al ex [EMAIL PROTECTED] wrote:
 I dont have a problem with sampling and calculating mean:

 for(i in 1:5){
  res=(A1[sample(nrow(A1), 5),])
  Avg=colMeans(res)
  STD=sd(res)
  print(res)
  print(Avg)
  print(STD)
  }

 my problem is how to save the mean for each S in each sample and calculate
 the grand mean



 --- On Mon, 10/20/08, stephen sefick [EMAIL PROTECTED] wrote:

 From: stephen sefick [EMAIL PROTECTED]
 Subject: Re: [R] calculating mean for samples
 To: Alex99 [EMAIL PROTECTED]
 Date: Monday, October 20, 2008, 11:34 AM

 look at
 ?sample
 ?lapply
 ?mean

 On Mon, Oct 20, 2008 at 1:53 PM, Alex99 [EMAIL PROTECTED] wrote:


  Hi everyone,
 does any one knows how can I calculate mean for different samples
 i.e. I have a data like this:

  s1 s2 s3 s4
 1   0  0  0  1
 2   1  0  1  0
 3   0  0  0  0
 4   0  0  0  0
 5   0  1  0  1
 6   1  0  0  0
 7   0  0  0  0
 8   0  0  0  0
 9   0  0  0  0
 10  0  0  0  1

 I need to make 5 different sample with 5 different persons(rows) in
 each
 sample from it keeping s1,s3,s3,s4 but changing rows. and then
 calculate
 the mean for each S in each sample. and finally calculate
 the grand
 mean,which is the mean of means for each sample. i.e. if I sample 5
 time I
 get 5 different means for s1, s2, s3, s4. then I need to add all five
 means for and divide it by 5.(of course I have to do it
  for
 s1,s2,s3,s4
 separately)
 --
 View this message in context:
 http://www.nabble.com/calculating-mean-for-samples-tp20075174p20075174.html
 Sent from the R help mailing list archive at Nabble.com.

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




 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

   -K.
  Mullis

 __
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-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

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

Re: [R] pass longer arguments to adj of text()

2008-10-20 Thread Thomas Steiner

...and the offset argument can't handle arrays either:

text(c(0,0),c(0,0),c(a,b),pos=c(1,2),offset=c(1,0.5),col=darkgreen)

The second argument does not have any effect.
Can I pass arrays to text and use diffent locations?
Thomas

PS: I use R version 2.5.0 (2007-04-23) under Win2000.

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Re: [R] Staging area for data before read into R

sorry excel 2003 with no immediate update in the future.

On Mon, Oct 20, 2008 at 3:12 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
 You didn't say which version of Excel you are using but Excel 2007
 allows 16,384 columns.

 On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick [EMAIL PROTECTED] wrote:
 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance

 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis

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





-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] error message when plotting survival curves

2008-10-20 Thread Thomas Lumley


On Mon, 20 Oct 2008, A Van Dyke wrote:



I am trying to plot survival curves using the following code as an example:


rs1799964.coxph-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method=efron))



plot(rs1799964.coxph,lyt=c(1,3),xlab=Survival in Months,ylab=Proportion

Surviving)



You can't plot a coxph object. You probably want 
plot(survfit(rs1799965.copxh))


-thomas



I am gettingthe following error message:



Error in xy.coords(x, y, xlabel, ylabel, log) :

 'x' and 'y' lengths differ


Any input to debugging this matter would be greatly appreciated.  Thank you
in advance.

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

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Re: [R] par(mfrow=c(2,4))

reproducible code please.

On Mon, Oct 20, 2008 at 3:22 PM, phoebe kong [EMAIL PROTECTED] wrote:
 Hi All,

 I'm going to draw 8 plots in one page. I want the plots to be arranged
 in two rows, each row has 4 plots. So, I set the it par(mfrow=c(2,4)).

 However, there might be too many plots in a page, all the 8 plots were
 drawn in triangular shape, which makes the x-y coordinate scale not in
 same length, the x-axis length is much shorter than the y-axis length.

 Could anybody let me know how to make the x-y coordinate length be the
 same, like the square box?


 Thanks in advance.
 Sityee

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] R 2.8.0 is released

2008-10-20 Thread Jeroen Ooms


I'm getting a unexpected compiling error when doing make:

gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre  -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H   -fpic  -g -O2 -c platform.c -o
platform.o
platform.c: In function 'do_capabilities':
platform.c:1661: error: expected ',' or ';' before 'Rf_checkArityCall'
make[3]: *** [platform.o] Error 1

R.2.7.2 compiled without any problems on the same machine.
Here is my config.log: http://jeroen.xslhosting.net/config.log.zip
 
Am I doing anything wrong? And will their be binary Debian packages for this
release (they never appeared for R-2.7.2)?


thank you!



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Re: [R] R 2.8.0 is released

2008-10-20 Thread Jeroen Ooms


I'm getting a unexpected compiling error when doing make: 

gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre  -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H   -fpic  -g -O2 -c platform.c -o
platform.o 
platform.c: In function 'do_capabilities': 
platform.c:1661: error: expected ',' or ';' before 'Rf_checkArityCall' 
make[3]: *** [platform.o] Error 1 

R.2.7.2 compiled without any problems on the same machine. 
Here is my config.log: http://jeroen.xlshosting.net/config.log.zip
  
Am I doing anything wrong? And will their be binary Debian packages for this
release (they never appeared for R-2.7.2)? 


thank you! 
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Re: [R] pass longer arguments to adj of text()

2008-10-20 Thread Thomas Steiner

Brian, thank you for your reply.

 Why do you expect this to be vectorized when the help page clearly says
 otherwise?

Sure, it's in line with the help, no problem there. I just asked if
this feature is somehow possible.
I expected it, because pos does take arrays and it overrides any
adj value given (see help). So if pos does, why not adj?

 What is wrong with two or more text() calls?

it's quicker and more beautiful.
See my code at the end (last loop) of
http://commons.wikimedia.org/wiki/Image:Sonnenstand.png

Thomas

 NB: library(fortunes); fortune(WTFM) applies.

This one I didn't get. They help to do it without a loop?

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Re: [R] R 2.8.0 is released

2008-10-20 Thread Duncan Murdoch


On 10/20/2008 4:34 PM, Jeroen Ooms wrote:

I'm getting a unexpected compiling error when doing make:

gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre  -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H   -fpic  -g -O2 -c platform.c -o
platform.o
platform.c: In function 'do_capabilities':
platform.c:1661: error: expected ',' or ';' before 'Rf_checkArityCall'
make[3]: *** [platform.o] Error 1

R.2.7.2 compiled without any problems on the same machine.
Here is my config.log: http://jeroen.xslhosting.net/config.log.zip
 
Am I doing anything wrong? And will their be binary Debian packages for this

release (they never appeared for R-2.7.2)?


If you look at the file, it appears someone forgot a semicolon at the 
end of line 1657, which is only compiled for people on Unix with no X11. 
 That's likely a configuration problem, e.g. you don't have the X11 
development libraries installed, or the configure script couldn't find them.


Duncan Murdoch

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Re: [R] Staging area for data before read into R

2008-10-20 Thread Ted Byers


Define better.

Really, it depends on what you need to do (are all your data appropriately
represented in a 2D array?) and what resources are available.  If all your
data can be represented using a 2D array, then Excel is probably your best
bet for th enear term.  If not, you might as well bite the bullit and learn
to use an RDBMS, as there are few other data management options that can
cope with relational or hierarchical or object oriented data.

I use a number of different RDBMS (ranging from MS SQL to PostgreSQL and
MySQL).  I also use Excel on occasion, and plain text editors (like Emacs),
to create CSV files.  Which I use depends on the details of the particular
problem I am facing.

While I have not yet explored them, I did notice that R includes a number of
facilities for editing data (and the list of options is all the longer when
I use help.search(edit).

It may be a bit quicker for you to study up on basic use of something like
PostgreSQL, combined with pl/r (something I wish MySQL had), than it would
be to diligently examine all the different options open to you using R.  (I
have a couple books I could recommend that would likely be sufficient for
you to figure out what you need to do with either PostgreSQL or MySQL in a
matter of a week or two).

HTH

Ted


stephen sefick wrote:
 
 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance
 
 -- 
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy
 
 Let's not spend our time and resources thinking about things that are
 so little or so large that all they really do for us is puff us up and
 make us feel like gods.  We are mammals, and have not exhausted the
 annoying little problems of being mammals.
 
   -K. Mullis
 
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View this message in context: 
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Sent from the R help mailing list archive at Nabble.com.

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Re: [R] About Torus Rndamiztion

2008-10-20 Thread Ben Bolker

Surendra champanerkar surendrachampanerkar at gmail.com writes:

 
 Hallo
 I am going through the paper on Torus randomization by Kyle E. Harms
 torus randomization is an alternative to Chi-sq test of Independence When
 the assumption of independence within the groups is violated
 
 i want to know that Is there any library regarding torus randomization
 Someone said me that its in Utility.r but i could not find out this
 
 please help me in this regard
 

  I know the work you're referring to, but I don't know
whether Harms wrote the original code in R (quite possibly)
or whether someone has translated it (also quite possible).
Someone told me there was a function in Utility.r (freely
translating from your post) isn't enough information
for us to go on ... why don't you go back to whoever
told you and ask for more information?  Or write
to the author and ask him?

  Here are a couple of the results of RSiteSearch(torus)
that are likely to be relevant to your question
(especially the second).

http://finzi.psych.upenn.edu/R/library/hdeco/html/shift.html
http://finzi.psych.upenn.edu/R/library/spatstat/html/rshift.ppp.html

  You might also try further queries on the r-sig-eco list
or the r-sig-spatial list, since this is a fairly specialized
question ...

  good luck
Ben Bolker

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Re: [R] pass longer arguments to adj of text()

2008-10-20 Thread Peter Dalgaard


Thomas Steiner wrote:


NB: library(fortunes); fortune(WTFM) applies.


This one I didn't get. They help to do it without a loop?


No.

 library(fortunes); fortune(WTFM)

This is all documented in TFM. Those who WTFM don't want to have to WTFM 
again

on the mailing list. RTFM.
   -- Barry Rowlingson
  R-help (October 2003)

(TFM == The Friggin' Manual (or thereabouts), W==Write, R==Read)

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

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[R] ? extended rep()

2008-10-20 Thread Ted Harding

Hi Folks,
I'm wondering if there's a compact way to achieve the
following. The dream is that, by analogy with

  rep(c(0,1),times=c(3,4))
# [1] 0 0 0 1 1 1 1

one could write

  rep(c(0,1),times=c(3,4,5,6))

which would produce

# [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

in effect recycling x through 'times'.

The objective is to produce a vector of alternating runs of
0s and 1s, with the lengths of the runs supplied as a vector.
Indeed, more generally, something like

  rep(c(0,1,2), times=c(1,2,3,2,3,4))
# [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2

Suggestions appreciated! With thanks,
Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 20-Oct-08   Time: 21:57:15
-- XFMail --

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Re: [R] ? extended rep()

2008-10-20 Thread jim holtman

will this do what you want:

 f.rep - function(x, times){
+ # make sure the 'x' is long enough
+ x - head(rep(x, length(times)), length(times))
+ rep(x, times)
+ }

 f.rep(c(0,1), c(3,4,5,6,7))
 [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0



On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
[EMAIL PROTECTED] wrote:
 Hi Folks,
 I'm wondering if there's a compact way to achieve the
 following. The dream is that, by analogy with

  rep(c(0,1),times=c(3,4))
 # [1] 0 0 0 1 1 1 1

 one could write

  rep(c(0,1),times=c(3,4,5,6))

 which would produce

 # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

 in effect recycling x through 'times'.

 The objective is to produce a vector of alternating runs of
 0s and 1s, with the lengths of the runs supplied as a vector.
 Indeed, more generally, something like

  rep(c(0,1,2), times=c(1,2,3,2,3,4))
 # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2

 Suggestions appreciated! With thanks,
 Ted.

 
 E-Mail: (Ted Harding) [EMAIL PROTECTED]
 Fax-to-email: +44 (0)870 094 0861
 Date: 20-Oct-08   Time: 21:57:15
 -- XFMail --

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

What is the problem that you are trying to solve?

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Re: [R] ? extended rep()

Try this:

with(data.frame(x = 0:1, times = 3:6), rep(x, times))

or even shorter:

do.call(rep, data.frame(x = 0:1, times = 3:6))


On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
[EMAIL PROTECTED] wrote:
 Hi Folks,
 I'm wondering if there's a compact way to achieve the
 following. The dream is that, by analogy with

  rep(c(0,1),times=c(3,4))
 # [1] 0 0 0 1 1 1 1

 one could write

  rep(c(0,1),times=c(3,4,5,6))

 which would produce

 # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

 in effect recycling x through 'times'.

 The objective is to produce a vector of alternating runs of
 0s and 1s, with the lengths of the runs supplied as a vector.
 Indeed, more generally, something like

  rep(c(0,1,2), times=c(1,2,3,2,3,4))
 # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2

 Suggestions appreciated! With thanks,
 Ted.

 
 E-Mail: (Ted Harding) [EMAIL PROTECTED]
 Fax-to-email: +44 (0)870 094 0861
 Date: 20-Oct-08   Time: 21:57:15
 -- XFMail --

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 https://stat.ethz.ch/mailman/listinfo/r-help
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Re: [R] ? extended rep()

2008-10-20 Thread Stefan Evert



On 20 Oct 2008, at 22:57, (Ted Harding) wrote:


I'm wondering if there's a compact way to achieve the
following. The dream is that one could write

 rep(c(0,1),times=c(3,4,5,6))

which would produce

# [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

in effect recycling x through 'times'.


rep2 - function (x, times) rep(rep(x, length.out=length(times)), times)

rep2(c(0,1),times=c(3,4,5,6))
 [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

Any prizes for shortest solution? ;-)

Best,
Stefan

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Re: [R] ? extended rep()

Here is one other solution:

x - 0:1
times - 3:6
rep(x + 0*times, times)

This solution also works if the length of times is not a whole
number of lengths of x but in that case it does give a warning
which seems reasonable since that is the way recycling in
R works elsewhere too.

On Mon, Oct 20, 2008 at 5:17 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
 Try this:

 with(data.frame(x = 0:1, times = 3:6), rep(x, times))

 or even shorter:

 do.call(rep, data.frame(x = 0:1, times = 3:6))


 On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
 [EMAIL PROTECTED] wrote:
 Hi Folks,
 I'm wondering if there's a compact way to achieve the
 following. The dream is that, by analogy with

  rep(c(0,1),times=c(3,4))
 # [1] 0 0 0 1 1 1 1

 one could write

  rep(c(0,1),times=c(3,4,5,6))

 which would produce

 # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

 in effect recycling x through 'times'.

 The objective is to produce a vector of alternating runs of
 0s and 1s, with the lengths of the runs supplied as a vector.
 Indeed, more generally, something like

  rep(c(0,1,2), times=c(1,2,3,2,3,4))
 # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2

 Suggestions appreciated! With thanks,
 Ted.

 
 E-Mail: (Ted Harding) [EMAIL PROTECTED]
 Fax-to-email: +44 (0)870 094 0861
 Date: 20-Oct-08   Time: 21:57:15
 -- XFMail --

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Re: [R] Staging area for data before read into R

2008-10-20 Thread Carl Witthoft

Why not just write your data to a CSV (comma-spaced-variable) or a 
tab-spaced variable   text file?
You didn't say what software and/or hardware was generating your data, 
but most gizmos these days let you dump data to CSV.


No need for Excel at all.
I forget :-( how many rows/columns OpenOffice.org or KOffice can handle.

Carl

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Re: [R] ? extended rep()

2008-10-20 Thread Ted Harding

On 20-Oct-08 21:19:21, Stefan Evert wrote:
 On 20 Oct 2008, at 22:57, (Ted Harding) wrote:
 I'm wondering if there's a compact way to achieve the
 following. The dream is that one could write

  rep(c(0,1),times=c(3,4,5,6))

 which would produce

 # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

 in effect recycling x through 'times'.
 
 rep2 - function (x, times) rep(rep(x, length.out=length(times)),
 times)
 
 rep2(c(0,1),times=c(3,4,5,6))
   [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1
 
 Any prizes for shortest solution? ;-)
 
 Best,
 Stefan

If ever we are both within reach of 'en øl', then yes.
But Gabor came up with a shorter one.

I tried to shorten Gabor's but failed.

However, all competitors are entitled to a consolation prize!
(And that includes me ... )
Ted.


E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 20-Oct-08   Time: 22:59:16
-- XFMail --

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Re: [R] ? extended rep()

2008-10-20 Thread Ted Harding

On 20-Oct-08 21:17:22, Gabor Grothendieck wrote:
 Try this:
 with(data.frame(x = 0:1, times = 3:6), rep(x, times))
 
 or even shorter:
 do.call(rep, data.frame(x = 0:1, times = 3:6))

That is sneaky!

  data.frame(x = 0:1, times = 3:6)
#   x times
# 1 0 3
# 2 1 4
# 3 0 5
# 4 1 6

(Which is why it won't work with list(x=0:1,times=3:6))
Ted.

 On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
 [EMAIL PROTECTED] wrote:
 Hi Folks,
 I'm wondering if there's a compact way to achieve the
 following. The dream is that, by analogy with

  rep(c(0,1),times=c(3,4))
 # [1] 0 0 0 1 1 1 1

 one could write

  rep(c(0,1),times=c(3,4,5,6))

 which would produce

 # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1

 in effect recycling x through 'times'.

 The objective is to produce a vector of alternating runs of
 0s and 1s, with the lengths of the runs supplied as a vector.
 Indeed, more generally, something like

  rep(c(0,1,2), times=c(1,2,3,2,3,4))
 # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2

 Suggestions appreciated! With thanks,
 Ted.

 
 E-Mail: (Ted Harding) [EMAIL PROTECTED]
 Fax-to-email: +44 (0)870 094 0861
 Date: 20-Oct-08   Time: 21:57:15
 -- XFMail --

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
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E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 20-Oct-08   Time: 23:02:00
-- XFMail --

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Re: [R] Categorical Response Query

2008-10-20 Thread andyer weng

Hi all,

I have a queston about Categorical response.

i have a data frame containing age, sex, class, success(1=success,
0=non sucess).
age, sex,class are the explantory variables, and sucess is the
response variable.  and i can get n (the nunber of times each age
occurs) and r (the number of sucess of that age).

when I try to creat the regression relationship for these variables, I
have seen many different cases, i just wonder which one fits me the
best for this situation.

1st case,
xxx.glm-glm(success~age*sex*class,family=binomial, data=xxx.data)

2nd case

xxx.glm-glm(r/n~age*sex*class,family=binomial, data=xxx.data)

3rd case

xxx.glm-glm(cbind(r,n-r)~age*sex*class,family=binomial, data=xxx.data)

what is difference between the above 3 cases? which one is the best to use?

if Ii don't group the data, can I use the 1st case. if i group the
data, can i use 2nd or 3rd case?

please advise.

Cheers.
Andyer

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[R] aggregating along bins and bin-quantiles

2008-10-20 Thread Ivan Alves

Dear all,

I would like to aggregate a data frame (consisting of 2 columns - one  
for the bins, say factors, and one for the values) along bins and  
quantiles within the bins.

I have tried

aggregate(data.frame$values, list(bin = data.frame 
$bin,Quantile=cut2(data.frame$bin,g=10)),sum)

but then the quantiles apply to the population as a whole and not the  
individual bins. Upon this realisation I have tried

aggregate(data.frame$values, list(bin = data.frame 
$bin,Quantile=tapply(data.frame$values,data.frame$bin,cut2,g=10)),sum)

which gives the following error:

Error in sort.list(unique.default(x), na.last = TRUE) :
   'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?

clearly I am doing something wrong, but cannot figure out what.  I  
believe the error stems either from a. the output of tapply being a  
list of a dimension equal to the number of bins, and not a list of  
equal dimension as the values, or b. that somehow aggregate does not  
like that the second list (of the quantiles within the bins are not  
sorted nicely)

1. Do you have a reference for doing the summation on both bins and  
quantiles within the bins?
2. If not, can you give me some guidance as to what I am doing wrong  
and how I can solve the sort/list issue?

Any help would be greatly appreciated

Kind regards,

Ivan Alves


[[alternative HTML version deleted]]

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Re: [R] R Newbie Question

tIs a bit neater to create list objects for each of
star1 and star2 and then, since the various functions
are not used outside of gravitation stick them right
in gravitation.  Also we can use distance for len
as well by giving it a default second arg of 0 and
sqr does not seem to do much so we can eliminate
that too.

Making the vec computation a function eliminates
some redundant code and using outer eliminates
the two loops.

star1 - list(mass = 30, center = c(-0.5, 0))
star2 - list(mass = 1, center = c(0.5, 0))

gravitation2 - function(v1, v2) {
distance - function(a, b = 0) sqrt(sum((a-b)^2))
norm - function(x) x/distance(x)
vec - function(x) with(x, mass * norm(center - v) / distance(v, 
center)^2)

v - c(v1, v2)
distance(vec(star1) + vec(star2))
}

xvals2 - yvals2 - seq(vmin, vmax, step)
b2 - outer(xvals, yvals, Vectorize(gravitation2))


On Mon, Oct 20, 2008 at 1:41 PM, Johannes Bauer [EMAIL PROTECTED] wrote:
 Hello list,

 I just started R today and tried something quite simple. I wanted to
 create a colored plot and eventually after hours of fiddling around got
 it working. However, my solution seems very suboptimal and I'd really
 appreciate your hints on how to improve. I believe that R already offers
 many functions I coded (e.g. distance between two vectors, vector
 length, vector normalization and so on). I generally didn't even figure
 out how to create a simple vector or how to extract a row from a matrix
 so the result is a vector (to get scalars, I use the sum functions,
 which is an incredibly ugly workaround).

 To sum the problem up: one has a binary star system with a large star
 (e.g. red giant) and a small star (e.g. white dwarf). Gravitation
 between them is directly proportional to the mass and indirectly
 proportional to the square of the distance. If correctly plotted, one
 should be able to see the inner lagrange point L1 which is the point
 where the gravitational potentials of the stars cancel out, e.g. an
 object would not be attracted to any star. Well, enough background
 information, here's my rookie code - please feel free to comment on
 anything :-)

 Kind regards,
 Johannes



 star1center = vector(numeric, 2)
 star1center[1] = -0.5
 star1center[2] = 0
 star1mass = 30

 star2center = vector(numeric, 2)
 star2center[1] = 0.5
 star2center[2] = 0
 star2mass = 1

 sqr = function(x) {
return(x * x)
 }

 distance = function(a, b) {
return(sqrt(sqr(a[1] - b[1]) + sqr(a[2] - b[2])))
 }

 len = function(x) {
return(sqrt(sqr(x[1]) + sqr(x[2])))
 }

 norm = function(x) {
return(x / len(x))
 }

 gravitation = function(invecx, invecy) {
invec = vector(numeric, 2)
invec[1] = invecx
invec[2] = invecy
vec1 = star1mass * norm(star1center - invec) / sqr(distance(invec,
 star1center))
vec2 = star2mass * norm(star2center - invec) / sqr(distance(invec,
 star2center))
return(len(vec1 + vec2))
 }

 vmin = -1
 vmax = 1
 step = 0.1
 vals = ((vmax - vmin) / step) + 1

 xvals = seq(vmin, vmax, step)
 yvals = seq(vmin, vmax, step)

 a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
 b = matrix(seq(1, vals*vals), vals)

 for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(sum(a[x,][1]), sum(a[y,][1]))
}
 }
 filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
 1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)

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Re: [R] aggregating along bins and bin-quantiles

2008-10-20 Thread Ivan Alves

Apologies, just a typo in the first instruction (when translating the  
names), the question is still valid


On 21 Oct 2008, at 00:38, Ivan Alves wrote:


Dear all,

I would like to aggregate a data frame (consisting of 2 columns - one
for the bins, say factors, and one for the values) along bins and
quantiles within the bins.

I have tried

aggregate(data.frame$values, list(bin = data.frame
$bin,Quantile=cut2(data.frame$values,g=10)),sum)

but then the quantiles apply to the population as a whole and not the
individual bins. Upon this realisation I have tried

aggregate(data.frame$values, list(bin = data.frame
$bin,Quantile=tapply(data.frame$values,data.frame$bin,cut2,g=10)),sum)

which gives the following error:

Error in sort.list(unique.default(x), na.last = TRUE) :
  'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?

clearly I am doing something wrong, but cannot figure out what.  I
believe the error stems either from a. the output of tapply being a
list of a dimension equal to the number of bins, and not a list of
equal dimension as the values, or b. that somehow aggregate does not
like that the second list (of the quantiles within the bins which do  
not  appear to be

sorted nicely)

1. Do you have a reference for doing the summation on both bins and
quantiles within the bins?
2. If not, can you give me some guidance as to what I am doing wrong
and how I can solve the sort/list issue?

Any help would be greatly appreciated

Kind regards,

Ivan Alves


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[R] Mclust problem with mclust1Dplot: Error in to - from : non-numeric argument to binary operator

2008-10-20 Thread Emmanuel Levy

Dear list members,

I am using Mclust in order to deconvolute a distribution that I
believe is a sum of two gaussians.

First I can make a model:
 my.data.model = Mclust(my.data, modelNames=c(E), warn=T, G=1:3)

But then, when I try to plot the result, I get the following error:

 mclust1Dplot(my.data.model, parameters = my.data.model$parameters, what = 
 density)
Error in to - from : non-numeric argument to binary operator

Also, I'd like to allow for each gaussian to have a different variance
(modelNmaes=c(V)) , but then I get another error message:

 my.data.model = Mclust(my.data, modelNames=c(V), warn=T, G=1:3)
Warning messages:
1: In meV(data = data, z = z, prior = prior, control = control, warn = warn) :
  sigma-squared falls below threshold
2: In meV(data = data, z = z, prior = prior, control = control, warn = warn) :
  sigma-squared falls below threshold
3: In summary.mclustBIC(Bic, data, G = G, modelNames = modelNames) :
  best model occurs at the min or max # of components considered
4: In Mclust(my.data, modelNames = c(V), warn = T, G = 1:3) :
  optimal number of clusters occurs at min choice

Many thanks in advance for your help,

Best wishes,

Emmanuel


If you would like to reproduce the above, the dataset is:

my.data=c( 0.,0.0052,0.,-0.2136,0.4625,0.6047,0.,0.7370,0.5059
,-0.8060,-1.0790,0.,-1.5397,-0.0720,-3.2180,-1.6980,0.,2.2845
,-1.0741,0.,0.1020,-0.6010,0.2210,-0.0120,1.0785,0.,-0.4536
,-0.1127,-0.2032,-0.0421,-1.6818,-0.9935,-2.2105,-0.7963,-0.1820,-2.0468
,0.6161,-1.7663,-0.6800,-2.1290,-0.0167,0.,0.,0.,0.5427
,-0.0170,0.,0.,-0.6576,0.9055,0.1409,-0.1409,0.,0.3730
,-0.1800,-1.3141,0.6786,-0.2480,-2.5110,-0.1340,0.3000,-1.7350,0.
,-0.5464,0.,-0.7513,-1.9056,-1.4823,-0.5376,-0.4516,-1.1391,0.
,-2.2560,1.3770,0.3390,-2.6023,-1.0880,-0.1444,0.,-0.1459,0.1740
,0.,0.3310,0.0749,1.0360,-0.8345,-0.6843,-3.5171,-1.9482,-0.4972
,-0.0130,-2.0290,-0.2812,0.,0.,-0.0164,0.,-1.9220,-1.5941
,-1.0840,0.,0.0459,-2.2121,-1.1485,-1.1485,0.,-0.4449,-0.5001
,0.3520,1.9980,-3.8385,1.7160,1.0020,-0.2250,-0.8265,-0.2032)

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Re: [R] Staging area for data before read into R

2008-10-20 Thread Jim Porzak

Hi Stephen,

You don't say what staging is - do you mean for data entry or loading a
data file for review, or ... ?

In general, I keep away from Excel for data transfer purposes. It tends to
make intelligent decisions on data types leading to strange  bizarre
results (unless you explicitly type each column - which most users don't
do). Integers are interpreted as dates, high order zeros are stripped off of
ZIP codes, and the like.

HTH,
Jim Porzak
TGN.com
San Francisco, CA
http://www.linkedin.com/in/jimporzak
useR Group SF: http://ia.meetup.com/67/


On Mon, Oct 20, 2008 at 11:27 AM, stephen sefick [EMAIL PROTECTED] wrote:

 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance

 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis

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


[[alternative HTML version deleted]]

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

Re: [R] Staging area for data before read into R

There is a list of free spreadsheets with their row and column limits
at this link:
http://en.wikipedia.org/wiki/OpenOffice.org_Calc

On Mon, Oct 20, 2008 at 3:13 PM, stephen sefick [EMAIL PROTECTED] wrote:
 sorry excel 2003 with no immediate update in the future.

 On Mon, Oct 20, 2008 at 3:12 PM, Gabor Grothendieck
 [EMAIL PROTECTED] wrote:
 You didn't say which version of Excel you are using but Excel 2007
 allows 16,384 columns.

 On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick [EMAIL PROTECTED] wrote:
 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance

 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis

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





 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis


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

Re: [R] Staging area for data before read into R

Well, I am going to type in ever value because the data sheets are of
counts of insects that I identified, so I should be okay with
accuracy...  I really just need something that allows for more than
256 columns as I have encounter over 256 species of insects in even
small streams.  I think calc with it's 1000ish columns will do the
trick... thanks everbody for your help.

On Mon, Oct 20, 2008 at 8:25 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
 There is a list of free spreadsheets with their row and column limits
 at this link:
 http://en.wikipedia.org/wiki/OpenOffice.org_Calc

 On Mon, Oct 20, 2008 at 3:13 PM, stephen sefick [EMAIL PROTECTED] wrote:
 sorry excel 2003 with no immediate update in the future.

 On Mon, Oct 20, 2008 at 3:12 PM, Gabor Grothendieck
 [EMAIL PROTECTED] wrote:
 You didn't say which version of Excel you are using but Excel 2007
 allows 16,384 columns.

 On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick [EMAIL PROTECTED] wrote:
 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance

 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis

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





 --
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy

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

-K. Mullis





-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

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

-K. Mullis

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

Re: [R] Variable shortlisting for the logistic regression

2008-10-20 Thread Greg Snow

I nominate the below 2 paragraphs (or a possible shortening of them) as a new 
fortune.  While not as entertaining as many of the current fortunes, the wisdom 
gained and sentiment expressed deserves preservation and easy reference for 
future posters who think that Frank is only trying to be funny.

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111

 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
 project.org] On Behalf Of Darin Brooks
 Sent: Sunday, October 19, 2008 9:11 AM
 To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
 Cc: r-help@r-project.org
 Subject: Re: [R] Variable shortlisting for the logistic regression

 Frank's remark was made in response to my posting.  As funny as it
 was -
 it was the best thing that could have happened to me.  It sparked an
 enlightening discussion between my committee and me (in particular, the
 pros
  cons of stepwise vs. information theoretic approach to model
 selection).
 Being new to the R help list, I had no idea who Frank was.  I googled
 him
 (and asked around) and found very quickly that he should be taken
 seriously.
 And so should his remark.

 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
 project.org] On
 Behalf Of Rolf Turner
 Sent: Thursday, October 16, 2008 1:34 PM
 To: useR
 Cc: r-help@r-project.org
 Subject: Re: [R] Variable shortlisting for the logistic regression



 On 17/10/2008, at 8:22 AM, useR wrote:

  Let's try to bring this discussion back again after Frank made
  very funny remark!

 Frank's remark was *serious*.  Take it seriously.

 cheers,

 Rolf Turner

 ##
 Attention:\ This e-mail message is privileged and
 confid...{{dropped:9}}

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 No virus found in this incoming message.
 Checked by AVG - http://www.avg.com

 8:02 PM

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 guide.html
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[R] New mailing list r-sig-hpc

2008-10-20 Thread Dirk Eddelbuettel


As interest in topics that are often grouped under the catch-all umbrella of
'high performance computing with R' has grown over the last few months, it has
been suggested to create a dedicated mailing list.  

Now, thanks to Martin Maechler who tirelessly looks after all R mailing
matters, such a list has been set up as 'r-sig-hpc'.  

Suitable topics for discussion may comprise

 - parallel computing approaches for R with, but not limited to any of
   Rmpi, rpvm, snow, nws, rsprng, ... 

 - use of batch / queue management system such as Slurm, Condor, Sun Grid
 - Engine, LSF, ...

 - profiling and debugging R, in particular for compiled 

 - large-scale automation and scripting 

or any other topic you deem suitable for 'high performance computing with R'.

List information, as well as subscription information, is available at

  https://stat.ethz.ch/mailman/listinfo/r-sig-hpc

Regards, Dirk

-- 
Three out of two people have difficulties with fractions.

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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] Staging area for data before read into R

2008-10-20 Thread Christopher W. Ryan

How about simply a text editor, typing in your data, separated by commas
or tabs or spaces?  One row for each case/subject/observation?  R can
read that in easily.

A good, open-source, free data entry program is EpiData.
www.epidata.dk.  It is simple to use but probably more than you need for
task.

--Chris

Christopher W. Ryan, MD
SUNY Upstate Medical University Clinical Campus at Binghamton
40 Arch Street, Johnson City, NY  13790
cryanatbinghamtondotedu
PGP public keys available at http://home.stny.rr.com/ryancw/

If you want to build a ship, don't drum up the men to gather wood,
divide the work and give orders. Instead, teach them to yearn for the
vast and endless sea.  [Antoine de St. Exupery]

stephen sefick wrote:
 I am wondering if there is a better alternative than Excel for data
 storage that does not require database knowledge (I will eventually
 have to learn this, but it is not on my immediate todo list).  I need
 something that is not limited to 256 columns... I don't need any of
 the built in functions in excel just a spreadsheet like program with
 cells that hold data in a data.frame format for a staging area before
 I get it into R.  Any help would be greatly appreciated.  This is not
 a direct r question, but all of you folks have more experience than I
 do and I am having a time finding what I need with google.
 thanks in advance


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

Re: [R] Staging area for data before read into R

2008-10-20 Thread Seeliger . Curt

 Well, I am going to type in ever value because the data sheets are of
 counts of insects that I identified, so I should be okay with
 accuracy...  I really just need something that allows for more than
 256 columns as I have encounter over 256 species of insects in even
 small streams. ...

Oh, ugh, that sounds difficult and prone to entry errors.  You might be 
better off organizing in the 'long' format with 4-5 columns:
siteID, subSiteID, species, count, comments

You can then reshape(), or use the reshape package, or even the 'pivot 
table' available in excel and other spreadcheats.

Glad you have an answer.  Enjoy your day.
cur
-- 
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
[EMAIL PROTECTED]
541/754-4638


[EMAIL PROTECTED] wrote on 10/20/2008 06:01:43 PM:

 Well, I am going to type in ever value because the data sheets are of
 counts of insects that I identified, so I should be okay with
 accuracy...  I really just need something that allows for more than
 256 columns as I have encounter over 256 species of insects in even
 small streams.  I think calc with it's 1000ish columns will do the
 trick... thanks everbody for your help.
 
 On Mon, Oct 20, 2008 at 8:25 PM, Gabor Grothendieck
 [EMAIL PROTECTED] wrote:
  There is a list of free spreadsheets with their row and column limits
  at this link:
  http://en.wikipedia.org/wiki/OpenOffice.org_Calc
 
  On Mon, Oct 20, 2008 at 3:13 PM, stephen sefick [EMAIL PROTECTED] 
wrote:
  sorry excel 2003 with no immediate update in the future.
 
  On Mon, Oct 20, 2008 at 3:12 PM, Gabor Grothendieck
  [EMAIL PROTECTED] wrote:
  You didn't say which version of Excel you are using but Excel 2007
  allows 16,384 columns.
 
  On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick [EMAIL PROTECTED] 
wrote:
  I am wondering if there is a better alternative than Excel for data
  storage that does not require database knowledge (I will eventually
  have to learn this, but it is not on my immediate todo list).  I 
need
  something that is not limited to 256 columns... I don't need any of
  the built in functions in excel just a spreadsheet like program 
with
  cells that hold data in a data.frame format for a staging area 
before
  I get it into R.  Any help would be greatly appreciated.  This is 
not
  a direct r question, but all of you folks have more experience than 
I
  do and I am having a time finding what I need with google.
  thanks in advance
 
  --
  Stephen Sefick
  Research Scientist
  Southeastern Natural Sciences Academy
 
  Let's not spend our time and resources thinking about things that 
are
  so little or so large that all they really do for us is puff us up 
and
  make us feel like gods.  We are mammals, and have not exhausted the
  annoying little problems of being mammals.
 
 -K. 
Mullis
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide http://www.R-project.
 org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 
 
 
 
  --
  Stephen Sefick
  Research Scientist
  Southeastern Natural Sciences Academy
 
  Let's not spend our time and resources thinking about things that are
  so little or so large that all they really do for us is puff us up 
and
  make us feel like gods.  We are mammals, and have not exhausted the
  annoying little problems of being mammals.
 
 -K. 
Mullis
 
 
 
 
 
 -- 
 Stephen Sefick
 Research Scientist
 Southeastern Natural Sciences Academy
 
 Let's not spend our time and resources thinking about things that are
 so little or so large that all they really do for us is puff us up and
 make us feel like gods.  We are mammals, and have not exhausted the
 annoying little problems of being mammals.
 
 -K. Mullis
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

[[alternative HTML version deleted]]

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Re: [R] converting a row of a data.frame to a vector

2008-10-20 Thread z3025589


Try:

as.vect = function(x)

{
t1 = as.matrix(x)
t2 = aperm(t1)
t3 = as.vector(t2)

return(t3)
}

This function converts entire dataframe into a vector.


Alexander Pui
PhD Candidate in Hydroclimatology, 
University of New South Wales
Email: [EMAIL PROTECTED]

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[R] how to plot the histogram and the curve in the same graph

2008-10-20 Thread leo_wa


i want to plot the histogram and the curve in the same graph.if i have a set
of data ,i plot the histogram and also want to see what distribution it
was.So i want to plot the curve to know what distribution it like.
-- 
View this message in context: 
http://www.nabble.com/how-to-plot-the-histogram-and-the-curve-in--the-same-graph-tp20082506p20082506.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Simulation of data

2008-10-20 Thread Marcioestat


Hi listers,
I am working on a program of statistical analysis of simulated data and I've
been searching the error at the program, but I didn't find it!
It is something about the WHILE procedure, the error says: Error in while
(ecart = d) { : missing value where TRUE/FALSE needed
Thanks in advance!
Márcio

k-100
d-0.00112
z-rnorm(100, 0, 1)
prop-rep(0,100)
for (i in 1:100){
if (z[i]1.75){
prop[i]-1}
else{prop[i]-0}}
proportion-mean(prop)
ecart-sqrt((proportion*(1-proportion))/k)
while(ecart=d){
prop_-0
w- rnorm(1, 0, 1)
z-c(z,w)
{if (w1.75){
prop_-1}
else{
prop_-0}
}
prop-c(prop, prop_)
proportion-proportion+(prop[k+1]-proportion/(k+1))
ecart-sqrt((proportion*(1-proportion))/(k+1))
k-k+1
}

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Re: [R] Using an image background with graphics

2008-10-20 Thread Waichler, Scott R

Greg,

 The rimage package has functions for reading in and plotting 
 jpeg files that you could use for displaying the photograph.  
 If you then can find 2 points in the image (not on the same 
 horizontal or vertical line) for which you know the 
 coordinates in the coordinate system that you want to plot 
 in, then you can use the updateusr function from the 
 TeachingDemos package to set the user coordinates, then use 
 points/lines or other functions that can add to the current 
 plot (e.g. contour with add=TRUE) to overlay the information 
 of interest to the plot.

I tried this, and your approach works great except for one thing.  I
would like to plot to a pdf device, and when I do so my grayscale image
takes on a gridded appearance, with thick horizontal and vertical lines
of lighter gray.  These don't appear of course if I use jpeg() instead
of pdf().  How can I get rid of those?

Thanks,
Scott Waichler
[EMAIL PROTECTED]

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

Re: [R] Using an image background with graphics