Re: [R] How to get correct proportions/bounding box for latex figure?

[Martin Sandiford]

Does this do what you want?

library(lattice)

rand1 - rnorm(50)
rand2 - rnorm(50)
theplot - xyplot(rand1 ~ rand2, xlab=x axis,
ylab=y axis)

thefile - plotproblem.eps
trellis.device(postscript, file=thefile, color=F,
horizontal=FALSE, width=12, height=4, paper=special)
print(theplot, split=c(1,1,3,1), more=T)
print(theplot, split=c(2,1,3,1), more=T)
print(theplot, split=c(3,1,3,1), more=F)
dev.off()

(Paper size and type specified in trellis.device call).


Martin

On 15/03/2006, at 2:44 PM, context grey wrote:



 --- Duncan Murdoch [EMAIL PROTECTED] wrote:
 The R graphics model is that the drawing surface is
 established first,
 then the things you draw are adjusted to fit in it.
 R won't change the
 shape of the display because you are drawing more
 things on it.

 Thanks, this comment clarifies things somewhat.
 Though I find it
 an odd design choice for R, since it seems to entail
 then that
 the user has figure out the aspect ratio of the
 resulting plot,
 something that R could easily keep track of as it is
 drawing.

 Here's example code, producing 3 scatterplots
 side-by-side
 (here reusing the same plot for simplicity).

 What's desired is that the individual scatterplots
 have the
 natural aspect, e.g. square, with the axis units being
 the
 same for X, Y.   And to do this while producing a
 correct
 bounding box in the .eps file.

 As it stands the example code produces a correct
 bounding box,
 but the scatterplots are too stretched to be usable.
 Inserting
 aspect=1/1 in the xyplot() seems to cause the bounding
 box
 to be  incorrect.

 

 library(lattice)

 rand1 - rnorm(50)
 rand2 - rnorm(50)
 theplot - xyplot(rand1 ~ rand2, xlab=x axis,
 ylab=y axis)

 thefile - plotproblem.eps
 trellis.device(postscript, file=thefile, color=F,
 horizontal=FALSE)
 print(theplot, split=c(1,1,3,1), more=T)
 print(theplot, split=c(2,1,3,1), more=T)
 print(theplot, split=c(3,1,3,1), more=F)
 dev.off()

 ---


 I don't think I understand exactly what you want to
 achieve; sample code
 that produces something close would be helpful (even
 if it comes out the
 wrong shape).

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[R] difftime arguments

[Fred J.]

  Hi
I just started using RGui.exe under widnows.
  I have a text file containing date arranged in columns and rows, each column 
has the same format, each row with different formats. 3 of the columns are 
something like this 1/12/2006 3:59:45 PM
  I need to calculate the different in seconds between 2 selected periods using 
their row’s index
   
  My solution:
  Read the file in a data frame and collect the columns in concern in a 
variable.
  data1 - read.table(C:\\path\\data.txt, header=TRUE)
  ts - paste(data1[[1]], data1[[3]], data1[[7]])   
   
  #ts now looks like this 
  #... 1/12/2006 3:59:45 PM  1/12/2006 3:59:57 PM ...
   
  #now convert between character representations and object of classes POSIXct
  x - strptime(ts, %m/%d/%y %I:%M:%S %p)
   
  this last code line is putting out NA, what did I do wrong?
   
  After this then I do
  z - as.POSIXct(x)
   
  Is my whole approach efficient?
  Then I can use difftime, because using it without the above preparations 
gives this
   
   difftime(c(1/12/2006 3:59:45 PM, 1/12/2006 3:59:57 PM), format 
=%m/%d/%y %I:%M:%S %p)
  Error in difftime(c(1/12/2006 3:59:45 PM, 1/12/2006 3:59:57 PM), format = 
%m/%d/%y %I:%M:%S %p) : 
  unused argument(s) (format ...)
   
  Thanks
  
__



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[R] rational and the roundoff error

[Fred J.]

  Hi


I have a list of 12000 rational numbers as inputs, running some of R functions 
will surly accumulate some round-off errors,  Is there a way to have R do its 
calculations using rational numbers as input to minimize the round-off error?

thanks



-


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Re: [R] Help on factanal.fit.mle

[Sumanta Basak]

In mva package, try ?prcomp.

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of souvik banerjee
Sent: Wednesday, March 15, 2006 12:44 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Help on factanal.fit.mle

Hi
Can anybody please suggest me about the documentation of
factanal.fit.mle()
(Not factanal()-- searching factanal.fit.mle() in R always leads to
factanal()).
Is there any function for doing principal component factor analysis in
R.
Regards
Souvik Bandyopadhyay
JRF,
Dept Of Statistics
Calcutta University

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Re: [R] Help on factanal.fit.mle

[Sumanta Basak]

Try this link also.

http://tolstoy.newcastle.edu.au/R/help/02b/0943.html


-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of souvik banerjee
Sent: Wednesday, March 15, 2006 12:44 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Help on factanal.fit.mle

Hi
Can anybody please suggest me about the documentation of
factanal.fit.mle()
(Not factanal()-- searching factanal.fit.mle() in R always leads to
factanal()).
Is there any function for doing principal component factor analysis in
R.
Regards
Souvik Bandyopadhyay
JRF,
Dept Of Statistics
Calcutta University

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[R] create a list of vectors

[Arnau Mir Torres]

Hello.

I want to create a list of vectors but each component of the list has a 
different length.

For example:

Example=list()
Example=list(Example,c(1,2,3))
Example=list(Example,c(11,12,13,14,15))

If I want the first component of the Example list, I have to write:

Example[[1]][[2]]. R responses
1 2 3

The second component:
Example[[2]]:
11 12 13 14 15

If I iterate the previous example, Example=list(Example,c(...)) and I 
want the components, I have to write:
Example[[1]][[1]]...i times...[[1]][[2]]  (first component)
Example[[1]]...(i-1)times[[1]][[2]] (second component)
...

Can I make it in some other way?


Thanks,

Arnau.

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Re: [R] GAM using R tutorials?

[TEMPL Matthias]

Have you looked at:

An Introduction to R: Software for StatisticalModelling  Computing by
Petra Kuhnert and Bill Venables

which is available at http://cran.r-project.org/other-docs.html

Hope this helps.

Best,
Matthias


 
 Hi all,
 
 I am trying to use GAM to work on some data... Are there any 
 resources providing hands-on tutorial/guide on how to do GAM 
 on data in R? Specifically, I am not sure about which model 
 to choose, and smooth models with which effective 
 degree-of-freedom shall I use...
 
 I knew there is a book titled: GAM: an introduction using R. 
 Unfornately our local library does not have it... so that's 
 not an option given time constraint.
 
 Thanks a lot for your pointers!
 
 Michael.
 
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Re: [R] setMethod confusion

[Martin Maechler]

 Stephen == Stephen Henderson [EMAIL PROTECTED]
 on Tue, 14 Mar 2006 16:32:56 - writes:

Stephen Hello I've checked through previous postings but
Stephen don't see a fully equivalent problem-just a few
Stephen hints.  I have been trying to set a new method for
Stephen the existing function table or
Stephen as.data.frame.table for my class tfSites.
Stephen Taking out all the useful code and just returning
Stephen the input class I get the error

 setMethod(table, tfSites, function(.Object) .Object)

Stephen Error in conformMethod(signature, mnames, fnames,
Stephen f) : In method for function table: formal
Stephen arguments omitted in the method definition cannot
Stephen be in the signature (exclude = tfSites)

 
 setMethod(as.data.frame.table, tfSites,
 function(.Object) .Object )

Stephen Error in conformMethod(signature, mnames, fnames,
Stephen f) : In method for function as.data.frame.table:
Stephen formal arguments omitted in the method definition
Stephen cannot be in the signature (x = tfSites)

Stephen What does this mean? Is there something peculiar
Stephen about the table function? Is it because it takes
Stephen arguments beginning table(..., etc)

Yes.  Since table's  argument list starts with ...
you cannot directly write S4 methods for it.

One could consider changing table's argument list to become 
  (x, ..., exclude = c(NA, NaN), dnn = list.names(...), deparse.level = 1) 
but that's not entirely trivial to do back compatibly, since
table() produces *named* dimnames from its arguments in ...
and we'd want to make sure that this continues to work as now
even when the first argument is formally named 'x'.  E.g.,

  X - rpois(20, 1)
  table(X)
 X
  0  1  2  3 
  7 10  2  1 
  

should continue to contain X as  names(dimnames(.)).

Of course this has now become a topic for R-devel rather
than R-help.

Martin Maechler,
ETH Zurich

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[R] click on graph and select data points?

[Michael]

Hi all,

I am doing some clustering and the clustered results are presented in a
pairs plot showing 4 clusters...

I made the data points belong to 4 different clusters displaying different
colors.

Now I want to select the best clustered class, how can I click on the data
point, and the program returns the index of that cluster(its class number,
or color number)?

Also, if I would like to display only data points belong to one cluster, how
can I make data points belonging to other clusters disappear in the pairs
plot...
(I can do it manually, using a filtering scheme, but is there any most
convinient method for this?)

thanks a lot!

Michael.

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Re: [R] create a list of vectors

[François MICHONNEAU]

Hello,

I don't know if it is the most efficient way to do but my solution is:

x - vector(list, 10) #creates a list with a length = 10

then in my loop (where i is iterated) :

x[[i]] - my.vector

I hope this could help you

François Michonneau




Hello.

I want to create a list of vectors but each component of the list has a
different length.

For example:

Example=list()
Example=list(Example,c(1,2,3))
Example=list(Example,c(11,12,13,14,15))

If I want the first component of the Example list, I have to write:

Example[[1]][[2]]. R responses
1 2 3

The second component:
Example[[2]]:
11 12 13 14 15

If I iterate the previous example, Example=list(Example,c(...)) and I
want the components, I have to write:
Example[[1]][[1]]...i times...[[1]][[2]]  (first component)
Example[[1]]...(i-1)times[[1]][[2]] (second component)
...

Can I make it in some other way?


Thanks,

Arnau.

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Re: [R] difftime arguments

[Christoph Buser]

Dear Fred

You should change your code from
x - strptime(ts, %m/%d/%y %I:%M:%S %p)

to
x - strptime(ts, %m/%d/%Y %I:%M:%S %p)

Y instead of y, since your year includes the century 
(2006 and not 06)

Then it should work.

Regards,

Christoph

--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C13
ETH (Federal Inst. Technology)  8092 Zurich  SWITZERLAND
phone: x-41-44-632-4673 fax: 632-1228
http://stat.ethz.ch/~buser/
--



Fred J. writes:
Hi
  I just started using RGui.exe under widnows.
I have a text file containing date arranged in columns and rows, each 
  column has the same format, each row with different formats. 3 of the 
  columns are something like this 1/12/2006 3:59:45 PM
I need to calculate the different in seconds between 2 selected periods 
  using their row?s index
 
My solution:
Read the file in a data frame and collect the columns in concern in a 
  variable.
data1 - read.table(C:\\path\\data.txt, header=TRUE)
ts - paste(data1[[1]], data1[[3]], data1[[7]])   
 
#ts now looks like this 
#... 1/12/2006 3:59:45 PM  1/12/2006 3:59:57 PM ...
 
#now convert between character representations and object of classes 
  POSIXct
x - strptime(ts, %m/%d/%y %I:%M:%S %p)
 
this last code line is putting out NA, what did I do wrong?
 
After this then I do
z - as.POSIXct(x)
 
Is my whole approach efficient?
Then I can use difftime, because using it without the above preparations 
  gives this
 
 difftime(c(1/12/2006 3:59:45 PM, 1/12/2006 3:59:57 PM), format 
  =%m/%d/%y %I:%M:%S %p)
Error in difftime(c(1/12/2006 3:59:45 PM, 1/12/2006 3:59:57 PM), 
  format = %m/%d/%y %I:%M:%S %p) : 
unused argument(s) (format ...)
 
Thanks

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[R] Log Cholesky parametrization in lme

[Pryseley Assam]

Dear R-Users
   
  I used the nlme library to fit a linear mixed model (lme). The random effect 
standard errors and correlation reported are based on a Log-Cholesky 
parametrization. Can anyone tell me how to get the Covariance matrix of the 
random effects, given the above mentioned parameters based on the Log-Cholesky 
parametrization??
   
  Thanks in advance
   
  Pryseley


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[R] X11 fonts problem with ubuntu breezy

[Martijn van Iersel]

Hello

I have big trouble getting R to work correctly with X11 fonts on Ubuntu 
Breezy 5.10. I was hoping somebody could help me with this issue.

The first part of the problem is that I get the error could not find 
any X11 fonts for any command with graphical ouput, for example 
demo(graphics):


---
R : Copyright 2005, The R Foundation for Statistical Computing
Version 2.2.1  (2005-12-20 r36812)
ISBN 3-900051-07-0

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

  Natural language support but running in an English locale

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

[Previously saved workspace restored]

  demo (graphics);


demo(graphics)
 

Type  Return   to start :

  require(graphics)
[1] TRUE

  require(datasets)
[1] TRUE

  if (dev.cur() = 1) get(getOption(device))()
Error in get(getOption(device))() : could not find any X11 fonts
Check that the Font Path is correct.
---


After some googling, I found out that apparently this can be fixed by 
clearing the LANG environment variable:
LANG=
export LANG

(By default on my machine, the environment contains LANG=en_US.UTF-8 
and LANGUAGE=en_NL:en)

However, when I try again, I get a different error, like X11 font at 
size 14 could not be loaded


---
  demo(graphics)


demo(graphics)
 

Type  Return   to start :

  require(graphics)
[1] TRUE

  require(datasets)
[1] TRUE

  if (dev.cur() = 1) get(getOption(device))()

  opar - par(ask = interactive()  (.Device %in% c(X11,
GTK, gnome, windows, quartz)))

  x - rnorm(50)

  opar - c(opar, par(bg = white))

  plot(x, ann = FALSE, type = n)
Hit Return to see next plot:

  abline(h = 0, col = gray(0.9))

  lines(x, col = green4, lty = dotted)

  points(x, bg = limegreen, pch = 21)

  title(main = Simple Use of Color In a Plot, xlab = Just a Whisper 
of a Label,
col.main = blue, col.lab = gray(0.8), cex.main = 1.2, cex.lab = 1,
font.main = 4, font.lab = 3)
Error in title(main = Simple Use of Color In a Plot, xlab = Just a 
Whisper of a Label,  :
X11 font at size 14 could not be loaded
 
---


Any idea what the problem might be? I run in this problem both with the 
R 2.1.1. binary package from ubuntu, and when I build 2.2.1 myself from 
source.

regards,

Martijn van Iersel

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Re: [R] click on graph and select data points?

[Philipp Pagel]

 Now I want to select the best clustered class, how can I click on the data
 point, and the program returns the index of that cluster(its class number,
 or color number)?

Have a look at identify() 

cu
Philipp

-- 
Dr. Philipp PagelTel.  +49-8161-71 2131
Dept. of Genome Oriented Bioinformatics  Fax.  +49-8161-71 2186
Technical University of Munich
Science Center Weihenstephan
85350 Freising, Germany

 and

Institute for Bioinformatics / MIPS  Tel.  +49-89-3187 3675
GSF - National Research Center   Fax.  +49-89-3187 3585
  for Environment and Health
Ingolstädter Landstrasse 1
85764 Neuherberg, Germany
http://mips.gsf.de/staff/pagel

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[R] how to use the result of hclust?

[Michael]

Hi all,

Does hclust provide concrete clustered results? I could not see how to use
it to make 6 clusters... and it does not give the 6 cluster labels...

How to use the result of hclust?

thanks a lot,

Michael.

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[R] variance from correlated observations

[Antonio, Fabio Di Narzo]

Hi all.
A statistical question. I have to estimate the variance of the sum:

X(1) + X(2) + ...+ X(n)

from an observed sample, where X(i) are *correlated* and not necessarly
identically distributed. Someone can suggest a simple strategy
(I hope by exploiting some already present R package) for obtaining such
estimate from an observed vector X[1:n]?

Tnx all,
Antonio, Fabio Di Narzo.

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Re: [R] how to use the result of hclust?

[Jacques VESLOT]

?cutree
?plot.hclust  ?identify.hclust
   
hc- hclust(dist(tab, manhattan), ward)
plot(hc, hang=-1)
(x - identify(hc))

cutree(hc, 2)



Michael a écrit :

Hi all,

Does hclust provide concrete clustered results? I could not see how to use
it to make 6 clusters... and it does not give the 6 cluster labels...

How to use the result of hclust?

thanks a lot,

Michael.

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Re: [R] how to use the result of hclust?

[TEMPL Matthias]

?cutree

 
 Hi all,
 
 Does hclust provide concrete clustered results? I could not 
 see how to use it to make 6 clusters... and it does not give 
 the 6 cluster labels...
 
 How to use the result of hclust?
 
 thanks a lot,
 
 Michael.
 
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Re: [R] setMethod confusion

[Henrik Bengtsson]

On 3/15/06, Martin Maechler [EMAIL PROTECTED] wrote:
  Stephen == Stephen Henderson [EMAIL PROTECTED]
  on Tue, 14 Mar 2006 16:32:56 - writes:

 Stephen Hello I've checked through previous postings but
 Stephen don't see a fully equivalent problem-just a few
 Stephen hints.  I have been trying to set a new method for
 Stephen the existing function table or
 Stephen as.data.frame.table for my class tfSites.
 Stephen Taking out all the useful code and just returning
 Stephen the input class I get the error

  setMethod(table, tfSites, function(.Object) .Object)

 Stephen Error in conformMethod(signature, mnames, fnames,
 Stephen f) : In method for function table: formal
 Stephen arguments omitted in the method definition cannot
 Stephen be in the signature (exclude = tfSites)


  setMethod(as.data.frame.table, tfSites,
  function(.Object) .Object )

 Stephen Error in conformMethod(signature, mnames, fnames,
 Stephen f) : In method for function as.data.frame.table:
 Stephen formal arguments omitted in the method definition
 Stephen cannot be in the signature (x = tfSites)

 Stephen What does this mean? Is there something peculiar
 Stephen about the table function? Is it because it takes
 Stephen arguments beginning table(..., etc)

 Yes.  Since table's  argument list starts with ...
 you cannot directly write S4 methods for it.

Although not fully tested, but a workaround could be to i) define an
S4 method tableS4(), then ii) rename table() to table.default() and
iii) make table() an S3 generic function, and finally iv) define
table.tfSites() to call tableS4(). Would this work?  If so, step
(ii)-(iv) can be done in one step using the R.oo package. Example:

library(R.oo)
setClass(fSites, representation(x=numeric, y=numeric))
setGeneric(tableS4, function(.Object, ...) standardGeneric(tableS4))
setMethod(tableS4, fSites, function(.Object) .Object)
setMethodS3(table, fSites, function(.Object, ...) tableS4(.Object, ...))

Test;
 x - new(fSites)
 table(x)
An object of class fSites
Slot x:
numeric(0)

Slot y:
numeric(0)

 X - rpois(20, 1)
 table(X)
X
 0  1  2  3
10  7  2  1

But, what's wrong with S3 in the first place? ;)

/Henrik




 One could consider changing table's argument list to become
   (x, ..., exclude = c(NA, NaN), dnn = list.names(...), deparse.level = 1)
 but that's not entirely trivial to do back compatibly, since
 table() produces *named* dimnames from its arguments in ...
 and we'd want to make sure that this continues to work as now
 even when the first argument is formally named 'x'.  E.g.,

   X - rpois(20, 1)
   table(X)
  X
   0  1  2  3
   7 10  2  1
  

 should continue to contain X as  names(dimnames(.)).

 Of course this has now become a topic for R-devel rather
 than R-help.

 Martin Maechler,
 ETH Zurich

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--
Henrik Bengtsson
Mobile: +46 708 909208 (+1h UTC)

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Re: [R] R-help Digest, Vol 37, Issue 15

[isaac . martin]

Mi nueva dirección de correo es: [EMAIL PROTECTED]

New e-mail address: [EMAIL PROTECTED]

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Re: [R] variance from correlated observations

[Liaw, Andy]

You need to know the covariance of the Xs.  The sum is just a linear
function of the Xs, so its variance is a function of a quadratic form
involving the covariance matrix of the Xs.

Andy

From: Antonio, Fabio Di Narzo
 
 Hi all.
 A statistical question. I have to estimate the variance of the sum:
 
 X(1) + X(2) + ...+ X(n)
 
 from an observed sample, where X(i) are *correlated* and not 
 necessarly identically distributed. Someone can suggest a 
 simple strategy (I hope by exploiting some already present R 
 package) for obtaining such estimate from an observed vector X[1:n]?
 
 Tnx all,
 Antonio, Fabio Di Narzo.
 
   [[alternative HTML version deleted]]
 
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Re: [R] rational and the roundoff error

[Duncan Murdoch]

On 3/15/2006 3:21 AM, Fred J. wrote:
   Hi
 
 
 I have a list of 12000 rational numbers as inputs, running some of R 
 functions will surly accumulate some round-off errors,  Is there a way to 
 have R do its calculations using rational numbers as input to minimize the 
 round-off error?

R has no internal support for rational numbers, but several packages 
support calculations using rationals.  Try RSiteSearch('rational 
numbers').

Duncan Murdoch

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Re: [R] setMethod confusion -best reference for S4 programming

[Stephen Henderson]

Thanks I think you have both answered my question (reckon Ill go S3 on
that). As an adjunct to this do you know what might be the best
reference to the S4 methods current implementation.

I have ordered the Chambers book Programming with Data, and I have a
short tutorial-- S4 Classes in 15 pages, more or less.

Have I missed any other useful resources?

Stephen Henderson
Wolfson Inst. for Biomedical Research
Cruciform Bldg., Gower Street
University College London
United Kingdom, WC1E 6BT
+44 (0)207 679 6827


-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Henrik Bengtsson
Sent: 15 March 2006 10:59
To: Martin Maechler
Cc: Stephen Henderson; r-help@stat.math.ethz.ch
Subject: Re: [R] setMethod confusion

On 3/15/06, Martin Maechler [EMAIL PROTECTED] wrote:
  Stephen == Stephen Henderson [EMAIL PROTECTED]
  on Tue, 14 Mar 2006 16:32:56 - writes:

 Stephen Hello I've checked through previous postings but
 Stephen don't see a fully equivalent problem-just a few
 Stephen hints.  I have been trying to set a new method for
 Stephen the existing function table or
 Stephen as.data.frame.table for my class tfSites.
 Stephen Taking out all the useful code and just returning
 Stephen the input class I get the error

  setMethod(table, tfSites, function(.Object) .Object)

 Stephen Error in conformMethod(signature, mnames, fnames,
 Stephen f) : In method for function table: formal
 Stephen arguments omitted in the method definition cannot
 Stephen be in the signature (exclude = tfSites)


  setMethod(as.data.frame.table, tfSites,
  function(.Object) .Object )

 Stephen Error in conformMethod(signature, mnames, fnames,
 Stephen f) : In method for function as.data.frame.table:
 Stephen formal arguments omitted in the method definition
 Stephen cannot be in the signature (x = tfSites)

 Stephen What does this mean? Is there something peculiar
 Stephen about the table function? Is it because it takes
 Stephen arguments beginning table(..., etc)

 Yes.  Since table's  argument list starts with ...
 you cannot directly write S4 methods for it.

Although not fully tested, but a workaround could be to i) define an
S4 method tableS4(), then ii) rename table() to table.default() and
iii) make table() an S3 generic function, and finally iv) define
table.tfSites() to call tableS4(). Would this work?  If so, step
(ii)-(iv) can be done in one step using the R.oo package. Example:

library(R.oo)
setClass(fSites, representation(x=numeric, y=numeric))
setGeneric(tableS4, function(.Object, ...) standardGeneric(tableS4))
setMethod(tableS4, fSites, function(.Object) .Object)
setMethodS3(table, fSites, function(.Object, ...) tableS4(.Object,
...))

Test;
 x - new(fSites)
 table(x)
An object of class fSites
Slot x:
numeric(0)

Slot y:
numeric(0)

 X - rpois(20, 1)
 table(X)
X
 0  1  2  3
10  7  2  1

But, what's wrong with S3 in the first place? ;)

/Henrik




 One could consider changing table's argument list to become
   (x, ..., exclude = c(NA, NaN), dnn = list.names(...), deparse.level
= 1)
 but that's not entirely trivial to do back compatibly, since
 table() produces *named* dimnames from its arguments in ...
 and we'd want to make sure that this continues to work as now
 even when the first argument is formally named 'x'.  E.g.,

   X - rpois(20, 1)
   table(X)
  X
   0  1  2  3
   7 10  2  1
  

 should continue to contain X as  names(dimnames(.)).

 Of course this has now become a topic for R-devel rather
 than R-help.

 Martin Maechler,
 ETH Zurich

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--
Henrik Bengtsson
Mobile: +46 708 909208 (+1h UTC)

**
This email and any files transmitted with it are confidentia...{{dropped}}

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Re: [R] GAM using R tutorials?

[Gavin Simpson]

On Tue, 2006-03-14 at 23:52 -0800, Michael wrote:
 Hi all,
 
 I am trying to use GAM to work on some data... Are there any resources
 providing hands-on tutorial/guide on how to do GAM on data in R?
 Specifically, I am not sure about which model to choose, and smooth models
 with which effective degree-of-freedom shall I use...
 
 I knew there is a book titled: GAM: an introduction using R. Unfornately our
 local library does not have it... so that's not an option given time
 constraint.
 
 Thanks a lot for your pointers!
 
 Michael.

Michael,

Please learn to use the search tools provided for you! You have posted
numerous emails to the list recently, many of which you could have
solved for yourself if only you'd heeded peoples' advice and searched
for yourself.

For this problem;

1) I'd suggest to the local library that they might consider buying the
book, but in the meantime...

2) ...in R, do RSiteSearch(GAM) and look at the list shown in your
browser. The first hit is the help page for package mgcv. Look at the
references included on that help page - most are technical/statistical
papers, but a starting point might be the RNews article Simon Wood
wrote.

That should get yourself started. But if you'd done the search yourself,
you wouldn't have had to wait for someone on the list to do it for you.

Finally - Please read the posting guide - it is there for a reason.

HTH

G
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin Simpson [T] +44 (0)20 7679 5522
ENSIS Research Fellow [F] +44 (0)20 7679 7565
ENSIS Ltd.  ECRC [E] gavin.simpsonATNOSPAMucl.ac.uk
UCL Department of Geography   [W] http://www.ucl.ac.uk/~ucfagls/cv/
26 Bedford Way[W] http://www.ucl.ac.uk/~ucfagls/
London.  WC1H 0AP.
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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[R] manipulating weeks dates

[Ronaldo Reis-Jr.]

Hi,

I have these vectors:

 WEEK - rep(c(1:52),2)
 YEAR - rep(c(2000,1999),c(52,52))

How to make a vector of Date with weeks in years? I try as.date from survival 
package, but it dont work with weeks, just only with days, months etcs.

Thanks
Ronaldo
-- 
Realmente minha cidade e muito facultativa

--Elivelton, ao repórter da Jovem Pan que falava das muitas
  faculdades que existiam em sua cidade natal.
--
|   // | \\   [***]
|   ( õ   õ )  [Ronaldo Reis Júnior]
|  V  [UFV/DBA-Entomologia]
|/ \   [36570-000 Viçosa - MG  ]
|  /(.''`.)\  [Fone: 31-3899-4007 ]
|  /(: :'  :)\ [EMAIL PROTECTED]]
|/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
|( `-  )   [***]
|  _/   \_Powered by GNU/Debian Woody/Sarge

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[R] manipulating weeks dates

[Ronaldo Reis-Jr.]

Hi,

I have these vectors:

 WEEK - rep(c(1:52),2)
 YEAR - rep(c(2000,1999),c(52,52))

How to make a vector of Date with weeks in years? I try as.date from survival 
package, but it dont work with weeks, just only with days, months etcs.

Thanks
Ronaldo
-- 
Realmente minha cidade e muito facultativa

--Elivelton, ao repórter da Jovem Pan que falava das muitas
  faculdades que existiam em sua cidade natal.
--
|   // | \\   [***]
|   ( õ   õ )  [Ronaldo Reis Júnior]
|  V  [UFV/DBA-Entomologia]
|/ \   [36570-000 Viçosa - MG  ]
|  /(.''`.)\  [Fone: 31-3899-4007 ]
|  /(: :'  :)\ [EMAIL PROTECTED]]
|/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
|( `-  )   [***]
|  _/   \_Powered by GNU/Debian Woody/Sarge  

-- 
Cerveja? Serve já!
--
|   // | \\   [***]
|   ( õ   õ )  [Ronaldo Reis Júnior]
|  V  [UFV/DBA-Entomologia]
|/ \   [36570-000 Viçosa - MG  ]
|  /(.''`.)\  [Fone: 31-3899-4007 ]
|  /(: :'  :)\ [EMAIL PROTECTED]]
|/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
|( `-  )   [***]
|  _/   \_Powered by GNU/Debian Woody/Sarge

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Re: [R] click on graph and select data points?

[Katharine Mullen]

if you are interested in a solution using the tcltk package, then an
idea is to base a solution on the code for the demo tkcanvas.

after installing the tcltk package, then

require(tcltk)
demo(tkcanvas)


Katharine Mullen
Department of Physics and Astronomy
Faculty of Sciences
Vrije Universiteit Amsterdam
de Boelelaan 1081
1081 HV Amsterdam
The Netherlands
room: T.1.06
tel: +31 205987870
fax: +31 205987992
e-mail: [EMAIL PROTECTED]
http://www.nat.vu.nl/~kate/

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[R] errorbars with xyplot

[Ronny Hannemann]

Hi everyone,

I'm very new to R and I like to learn a lot... actually I have a little
problem concerning errorbars with xyplot.

My data look like
  run target   hemi x
1   1 Nichts  Links  0.0007743240
2   2 Nichts  Links -0.0008153365
3   1 Target  Links -0.0015825950
4   2 Target  Links  0.0088743785
5   1 Nichts Rechts  0.0015898995
6   2 Nichts Rechts -0.0011465190
7   1 Target Rechts -0.005320
8   2 Target Rechts  0.0039010500

these data I plotted with xyplot(data=dummyy,
x~hemi|target,groups=run,type=p)

the same I do have with standard errors of means...

How could I attach errorbars to the datapoints?  I found some R-code from
Dec 2002 but I dont work (and I did not understand it...)

Thanks for helping me, maybe others out there will find it usefull too.

Best
Ronny

[[alternative HTML version deleted]]

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Re: [R] manipulating weeks dates

[Gabor Grothendieck]

Try:

as.Date(paste(YEAR, WEEK, 0), %Y %U %w)

On 3/15/06, Ronaldo Reis-Jr. [EMAIL PROTECTED] wrote:
 Hi,

 I have these vectors:

  WEEK - rep(c(1:52),2)
  YEAR - rep(c(2000,1999),c(52,52))

 How to make a vector of Date with weeks in years? I try as.date from survival
 package, but it dont work with weeks, just only with days, months etcs.

 Thanks
 Ronaldo
 --
 Realmente minha cidade e muito facultativa

 --Elivelton, ao repórter da Jovem Pan que falava das muitas
  faculdades que existiam em sua cidade natal.
 --
 |   // | \\   [***]
 |   ( õ   õ )  [Ronaldo Reis Júnior]
 |  V  [UFV/DBA-Entomologia]
 |/ \   [36570-000 Viçosa - MG  ]
 |  /(.''`.)\  [Fone: 31-3899-4007 ]
 |  /(: :'  :)\ [EMAIL PROTECTED]]
 |/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
 |( `-  )   [***]
 |  _/   \_Powered by GNU/Debian Woody/Sarge

 --
 Cerveja? Serve já!
 --
 |   // | \\   [***]
 |   ( õ   õ )  [Ronaldo Reis Júnior]
 |  V  [UFV/DBA-Entomologia]
 |/ \   [36570-000 Viçosa - MG  ]
 |  /(.''`.)\  [Fone: 31-3899-4007 ]
 |  /(: :'  :)\ [EMAIL PROTECTED]]
 |/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
 |( `-  )   [***]
 |  _/   \_Powered by GNU/Debian Woody/Sarge

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[R] (no subject)

[Erez Shabo]

Hello all,

I'm trying to calculate the Maximum likelihood of individuals to get the
ancestry.
I mixd 3 populations 15 generations in proportion of 20% 20% 60% when each
population
sorce have diferent genome (0 1 and 2) with frequencies for each one.
So now i have individuals looks like 0 0 2 1 1 2 0 . and i don't now how
to calculate the
mle although i try to figure out from the package state4.
can somebody help me please?

Thanks

[[alternative HTML version deleted]]

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[R] (newbie) Weighted qqplot?

[Vivek Satsangi]

Folks,
Normally, in a data frame, one observation counts as one observation
of the distribution. Thus one can easily produce a CDF and (in Splus
atleast) use cdf.compare to compare the CDF (BTW: what is the R
equivalent of the SPlus cdf.compare() function, if any?)

However, if each point should not count equally, how can I weight the
points before comparing the distributions? I was thinking of somehow
creating multiple observations for each actual observation based on
weights and creating a new dataframe etc. -- but that seem excessive.
Surely there is a simpler way?

 x - rnorm(100)
 y - rnorm(10)
 xw - rnorm(100) * 1.73 # The weights. These won't add up to 1 or N or 
 anything because of missing values.
 yw - rnorm(10) * 6.23 # The weights. These won't add up to 1 or to the same 
 number as xw.
 # The question to answer is, how can I create a qq plot or cdf compare of x 
 vs. y, weighted by their weights, xw and yw (to eventually figure out if y 
 comes from the population x, similar to Kolmogorov-Smirnov GOF)?
 qqplot(x,y) # What now?

Thanks for any help,

--
-- Vivek Satsangi
Student, Rochester, NY USA

Life is short, the art long, opportunity fleeting, experiment
treacherous, judgement difficult.

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Re: [R] create a list of vectors

[Gabor Grothendieck]

Try this:

L - list(letters, head(LETTERS))
L[[2]][[4]] # D
sapply(L, length) # 26 6
L - c(L, 1:4)
L[[3]][[2]] # 2
L - c(L, list(1:5))


On 3/15/06, Arnau Mir Torres [EMAIL PROTECTED] wrote:
 Hello.

 I want to create a list of vectors but each component of the list has a
 different length.

 For example:

 Example=list()
 Example=list(Example,c(1,2,3))
 Example=list(Example,c(11,12,13,14,15))

 If I want the first component of the Example list, I have to write:

 Example[[1]][[2]]. R responses
 1 2 3

 The second component:
 Example[[2]]:
 11 12 13 14 15

 If I iterate the previous example, Example=list(Example,c(...)) and I
 want the components, I have to write:
 Example[[1]][[1]]...i times...[[1]][[2]]  (first component)
 Example[[1]]...(i-1)times[[1]][[2]] (second component)
 ...

 Can I make it in some other way?


 Thanks,

 Arnau.

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Re: [R] errorbars with xyplot

[Frank E Harrell Jr]

Ronny Hannemann wrote:
 Hi everyone,
 
 I'm very new to R and I like to learn a lot... actually I have a little
 problem concerning errorbars with xyplot.
 
 My data look like
   run target   hemi x
 1   1 Nichts  Links  0.0007743240
 2   2 Nichts  Links -0.0008153365
 3   1 Target  Links -0.0015825950
 4   2 Target  Links  0.0088743785
 5   1 Nichts Rechts  0.0015898995
 6   2 Nichts Rechts -0.0011465190
 7   1 Target Rechts -0.005320
 8   2 Target Rechts  0.0039010500
 
 these data I plotted with xyplot(data=dummyy,
 x~hemi|target,groups=run,type=p)
 
 the same I do have with standard errors of means...
 
 How could I attach errorbars to the datapoints?  I found some R-code from
 Dec 2002 but I dont work (and I did not understand it...)
 
 Thanks for helping me, maybe others out there will find it usefull too.
 
 Best
 Ronny

library(Hmisc)
?xYplot

Frank Harrell

-- 
Frank E Harrell Jr   Professor and Chair   School of Medicine
  Department of Biostatistics   Vanderbilt University

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Re: [R] (newbie) Weighted qqplot?

[Duncan Murdoch]

On 3/15/2006 8:31 AM, Vivek Satsangi wrote:
 Folks,
 Normally, in a data frame, one observation counts as one observation
 of the distribution. Thus one can easily produce a CDF and (in Splus
 atleast) use cdf.compare to compare the CDF (BTW: what is the R
 equivalent of the SPlus cdf.compare() function, if any?)
 
 However, if each point should not count equally, how can I weight the
 points before comparing the distributions? I was thinking of somehow
 creating multiple observations for each actual observation based on
 weights and creating a new dataframe etc. -- but that seem excessive.
 Surely there is a simpler way?
 
 x - rnorm(100)
 y - rnorm(10)
 xw - rnorm(100) * 1.73 # The weights. These won't add up to 1 or N or 
 anything because of missing values.
 yw - rnorm(10) * 6.23 # The weights. These won't add up to 1 or to the same 
 number as xw.
 # The question to answer is, how can I create a qq plot or cdf compare of x 
 vs. y, weighted by their weights, xw and yw (to eventually figure out if y 
 comes from the population x, similar to Kolmogorov-Smirnov GOF)?
 qqplot(x,y) # What now?

qqplot doesn't support weights, but it's a simple enough function that 
you could write a version that did.  Look at the cases where length(x) 
is not equal to length(y):  e.g. if length(y)  length(x), qqplot 
constructs a linear approximation to a function mapping 1:nx onto the 
sorted x values, then takes length(y) evenly spaced values from that 
function.  You want to do the same sort of thing, except that instead of 
even spacing, you want to look at the cumulative sums of the weights.

You might want to use some kind of graphical indicator of whether points 
are heavily weighted or not, but I don't know what to recommend for that.

By the way, your example above will give negative weights in xw and yw; 
you probably won't like the results if you do that.

Duncan Murdoch

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[R] par(mfrow, fin) incompatibility?

[Dan Bebber]

Hello,

I want a 2x1 multi-figure, with each plot 5 square.
Test code:

x-rnorm(10,0,1)
y-rnorm(10,0,1)
par(pty=s, mfrow=c(2,1), fin=c(5,5))
plot(x,y)
plot(y,x)

but this does not work (overplots the two figures). Substituting pin for fin
works, but is not what I want. Are mfrow and fin incompatible?
I am basing my code on Fig. 4.6 in MASS4.
Running R 2.2.1  WinXP.

Thanks
Dan Bebber

Department of Plant Sciences
University of Oxford

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[R] matrix indexing

[tom wright]

Can someone please give me a pointer here.
I have two matrices

matA
A   B   C
1   5   2   4
2   2   4   3
3   1   2   4

matB
A   B   C
1   TRUEFALSE   TRUE
2   FALSE   TRUETRUE
3   FALSE   FALSE   FALSE

how do I extract all the values from matA where the coresponding entry
in matB == TRUE (or FALSE), perferably in vector form.

Many thanks
tom

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[R] installation problem with Biobase

[Haleh Yasrebi]

Hello,
 I wanted to install MergeMaid package in v 2.2.1. I could install it but 
couldn't use without its dependant, Biobase. at biobase installation, I got the 
following error message
 
In method for function split: expanding the signature
to include omitted arguments in definition: drop = missing
Error in .MakeSignature(new(signature), def, signature) :
the names in signature for method (x, f, ) do not match function's 
arguments (x, f, drop)
Execution halted
ERROR: execution of package source for 'Biobase' failed
** Removing '/usr/lib/R/library/Biobase'
 Do you know any solution?
 
 Regards,
 
 haleh
 

-

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Re: [R] matrix indexing

[Gabor Grothendieck]

Try this:

matA[c(matB)]

In fact even this works for your example although in general it
couldbe problematic since a two column matrix index has
special meaning:

matA[matB]


On 3/15/06, tom wright [EMAIL PROTECTED] wrote:
 Can someone please give me a pointer here.
 I have two matrices

 matA
A   B   C
 1   5   2   4
 2   2   4   3
 3   1   2   4

 matB
A   B   C
 1   TRUEFALSE   TRUE
 2   FALSE   TRUETRUE
 3   FALSE   FALSE   FALSE

 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.

 Many thanks
 tom

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Re: [R] matrix indexing

[Marc Schwartz (via MN)]

On Wed, 2006-03-15 at 06:03 -0500, tom wright wrote:
 Can someone please give me a pointer here.
 I have two matrices
 
 matA
   A   B   C
 1 5   2   4
 2 2   4   3
 3 1   2   4
 
 matB
   A   B   C
 1 TRUEFALSE   TRUE
 2 FALSE   TRUETRUE
 3 FALSE   FALSE   FALSE
 
 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.
 
 Many thanks
 tom

The subsetting/indexing is premised on the index values being TRUE,
thus:

 matA[matB]
[1] 5 4 4 3

 matA[!matB]
[1] 2 1 2 2 4

HTH,

Marc Schwartz

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Re: [R] matrix indexing

[Andy Bunn]

 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] Behalf Of tom wright
 Sent: Wednesday, March 15, 2006 6:04 AM
 To: R-Stat Help
 Subject: [R] matrix indexing
 
 
 Can someone please give me a pointer here.
 I have two matrices
 
 matA
   A   B   C
 1 5   2   4
 2 2   4   3
 3 1   2   4
 
 matB
   A   B   C
 1 TRUEFALSE   TRUE
 2 FALSE   TRUETRUE
 3 FALSE   FALSE   FALSE
 
 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.

Lie this?
matA - matrix(c(5,2,1,2,4,2,4,3,4),3,3)
matB - matrix(c(T,F,F,F,T,F,T,T,F),3,3)
matA[matB]
matA[!matB]


 
 Many thanks
 tom
 
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Re: [R] matrix indexing

[Dimitris Rizopoulos]

matA[matB] or matA[!matB]

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

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


- Original Message - 
From: tom wright [EMAIL PROTECTED]
To: R-Stat Help R-help@stat.math.ethz.ch
Sent: Wednesday, March 15, 2006 12:03 PM
Subject: [R] matrix indexing


 Can someone please give me a pointer here.
 I have two matrices

 matA
 A B C
 1 5 2 4
 2 2 4 3
 3 1 2 4

 matB
 A B C
 1 TRUE FALSE TRUE
 2 FALSE TRUE TRUE
 3 FALSE FALSE FALSE

 how do I extract all the values from matA where the coresponding 
 entry
 in matB == TRUE (or FALSE), perferably in vector form.

 Many thanks
 tom

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


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

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Re: [R] matrix indexing

[Philippe Grosjean]

This is really elementary indexing in S language:

matA[matB]

Best,

Philippe Grosjean

tom wright wrote:
 Can someone please give me a pointer here.
 I have two matrices
 
 matA
   A   B   C
 1 5   2   4
 2 2   4   3
 3 1   2   4
 
 matB
   A   B   C
 1 TRUEFALSE   TRUE
 2 FALSE   TRUETRUE
 3 FALSE   FALSE   FALSE
 
 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.
 
 Many thanks
 tom
 
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 R-help@stat.math.ethz.ch mailing list
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Re: [R] matrix indexing

[vito muggeo]

Dear tom,
is the following what you are looking for?

  a=matrix(runif(9),3,3)
  a
   [,1]  [,2]  [,3]
[1,] 0.9484247 0.9765431 0.6169739
[2,] 0.8423545 0.3137295 0.4031847
[3,] 0.6724235 0.1076373 0.2356923
  b-matrix(sample(c(TRUE,FALSE),size=9,replace=TRUE),3,3)
  b
   [,1]  [,2]  [,3]
[1,] FALSE  TRUE  TRUE
[2,]  TRUE  TRUE  TRUE
[3,] FALSE FALSE FALSE
  a[b]
[1] 0.8423545 0.9765431 0.3137295 0.6169739 0.4031847
 

best,
vito

tom wright wrote:
 Can someone please give me a pointer here.
 I have two matrices
 
 matA
   A   B   C
 1 5   2   4
 2 2   4   3
 3 1   2   4
 
 matB
   A   B   C
 1 TRUEFALSE   TRUE
 2 FALSE   TRUETRUE
 3 FALSE   FALSE   FALSE
 
 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.
 
 Many thanks
 tom
 
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 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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-- 

Vito M.R. Muggeo
Dip.to Sc Statist e Matem `Vianelli'
Università di Palermo
viale delle Scienze, edificio 13
90128 Palermo - ITALY
tel: 091 6626240
fax: 091 485726/485612

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[R] Setting xlim in lattice plots

[Mike White]

I am having difficulty setting different xlim values in the lattice
histogram plot function.
An example is shown below.  I think I need to convert the limits data.frame
to a list of paired values but don't know how. Any help would be
appreciated.

library(lattice)
mat - as.data.frame(matrix(abs(c(rnorm(100),
10*rnorm(100),20*rnorm(100),30*rnorm(100))),ncol=4))
colnames(mat)-c(C1,C2,C3,C4)
mat2-stack(mat)

limits-cbind(rep(0, ncol(mat)),apply(mat,2,max))
histogram(~ values | ind, data=mat2, xlim=limits,
scales=list(x=list(relation=free)))

Thanks
Mike White

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[R] Surv object in data.frame

[Heinz Tuechler]

Dear All,

a Surv object I put in a data frame behaves somehow unexpected (see example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly it has to do with the class attribute, because if I change the
class attribute to let Surv appeare first, again it works.
Is this known? Should I have found information on it?
Any comments?

Thanks

Heinz Tüchler

## example data
starttime - rep(0,5)
stoptime  - 1:5
event - c(1,0,1,1,1)
group - c(1,1,1,2,2)
## Surv object
survobj   - Surv(starttime, stoptime, event)
## Cox-regression
coxph(survobj~group) # this works
## put Surv object in data.frame
df.test - data.frame(survobj=I(survobj), group)
## Cox-regression on data.frame
coxph(survobj~group, data=df.test) # this does not work
attr(df.test$survobj, 'class') # survobject has class AsIs, Surv
attr(df.test$survobj, 'class') - c('Surv', 'AsIs') # put Surv first
attr(df.test$survobj, 'class') # survobject has class Surv, AsIs
coxph(survobj~group, data=df.test) # now it works

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Re: [R] setMethod confusion -best reference for S4 programming

[Berton Gunter]

You might also wish to read the relevant chapter of VR's S PROGRAMMING.

-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
The business of the statistician is to catalyze the scientific learning
process.  - George E. P. Box
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of 
 Stephen Henderson
 Sent: Wednesday, March 15, 2006 4:28 AM
 To: r-help@stat.math.ethz.ch
 Cc: Henrik Bengtsson; Martin Maechler
 Subject: Re: [R] setMethod confusion -best reference for S4 
 programming
 
 Thanks I think you have both answered my question (reckon Ill go S3 on
 that). As an adjunct to this do you know what might be the best
 reference to the S4 methods current implementation.
 
 I have ordered the Chambers book Programming with Data, and I have a
 short tutorial-- S4 Classes in 15 pages, more or less.
 
 Have I missed any other useful resources?
 
 Stephen Henderson
 Wolfson Inst. for Biomedical Research
 Cruciform Bldg., Gower Street
 University College London
 United Kingdom, WC1E 6BT
 +44 (0)207 679 6827
 
 
 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On
 Behalf Of Henrik Bengtsson
 Sent: 15 March 2006 10:59
 To: Martin Maechler
 Cc: Stephen Henderson; r-help@stat.math.ethz.ch
 Subject: Re: [R] setMethod confusion
 
 On 3/15/06, Martin Maechler [EMAIL PROTECTED] wrote:
   Stephen == Stephen Henderson [EMAIL PROTECTED]
   on Tue, 14 Mar 2006 16:32:56 - writes:
 
  Stephen Hello I've checked through previous postings but
  Stephen don't see a fully equivalent problem-just a few
  Stephen hints.  I have been trying to set a new method for
  Stephen the existing function table or
  Stephen as.data.frame.table for my class tfSites.
  Stephen Taking out all the useful code and just returning
  Stephen the input class I get the error
 
   setMethod(table, tfSites, function(.Object) .Object)
 
  Stephen Error in conformMethod(signature, mnames, fnames,
  Stephen f) : In method for function table: formal
  Stephen arguments omitted in the method definition cannot
  Stephen be in the signature (exclude = tfSites)
 
 
   setMethod(as.data.frame.table, tfSites,
   function(.Object) .Object )
 
  Stephen Error in conformMethod(signature, mnames, fnames,
  Stephen f) : In method for function as.data.frame.table:
  Stephen formal arguments omitted in the method definition
  Stephen cannot be in the signature (x = tfSites)
 
  Stephen What does this mean? Is there something peculiar
  Stephen about the table function? Is it because it takes
  Stephen arguments beginning table(..., etc)
 
  Yes.  Since table's  argument list starts with ...
  you cannot directly write S4 methods for it.
 
 Although not fully tested, but a workaround could be to i) define an
 S4 method tableS4(), then ii) rename table() to table.default() and
 iii) make table() an S3 generic function, and finally iv) define
 table.tfSites() to call tableS4(). Would this work?  If so, step
 (ii)-(iv) can be done in one step using the R.oo package. Example:
 
 library(R.oo)
 setClass(fSites, representation(x=numeric, y=numeric))
 setGeneric(tableS4, function(.Object, ...) 
 standardGeneric(tableS4))
 setMethod(tableS4, fSites, function(.Object) .Object)
 setMethodS3(table, fSites, function(.Object, ...) tableS4(.Object,
 ...))
 
 Test;
  x - new(fSites)
  table(x)
 An object of class fSites
 Slot x:
 numeric(0)
 
 Slot y:
 numeric(0)
 
  X - rpois(20, 1)
  table(X)
 X
  0  1  2  3
 10  7  2  1
 
 But, what's wrong with S3 in the first place? ;)
 
 /Henrik
 
 
 
 
  One could consider changing table's argument list to become
(x, ..., exclude = c(NA, NaN), dnn = list.names(...), 
 deparse.level
 = 1)
  but that's not entirely trivial to do back compatibly, since
  table() produces *named* dimnames from its arguments in ...
  and we'd want to make sure that this continues to work as now
  even when the first argument is formally named 'x'.  E.g.,
 
X - rpois(20, 1)
table(X)
   X
0  1  2  3
7 10  2  1
   
 
  should continue to contain X as  names(dimnames(.)).
 
  Of course this has now become a topic for R-devel rather
  than R-help.
 
  Martin Maechler,
  ETH Zurich
 
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  PLEASE do read the posting guide!
 http://www.R-project.org/posting-guide.html
 
 
 
 
 --
 Henrik Bengtsson
 Mobile: +46 708 909208 (+1h UTC)
 
 **
 This email and any files transmitted with it are 
 confidentia...{{dropped}}
 
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[R] Question about 'lables' ect.

[jia ding]

Hi,

I have a file named:
test_R.txt
aaa  2
bbb  5
ccc  7
sss  3
xxx  8

I want to have a plot:
test-read.table(test_R.txt,col.name=c(Name,Score))
par(mfrow=c(1,2))
barplot(test$Score)
name-test$Name
axis(1,at=1:length(test$Name),labels=paste(name))

Q1, if you try the script above,you will get 5 bars, the axis only shows
aaa, ccc,xxx, but where are bbbsss?

Q2, pls have a look this x-axis again, you will find the middle of the bars
are not pointing to the x-axes.

Q3, how can i change the width of the bars? I feel they are too fat.

Thanks!
Nina

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Re: [R] manipulating weeks dates

[Ronaldo Reis-Jr.]

Em Quarta 15 Março 2006 10:26, Gabor Grothendieck escreveu:
 as.Date(paste(YEAR, WEEK, 0), %Y %U %w)

Hi,

it works, but it use a year with 53 weeks, I need to use with 52 weeks, how to 
change this?

Thanks
Ronaldo
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-- anagrama
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|   ( õ   õ )  [Ronaldo Reis Júnior]
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|/ \   [36570-000 Viçosa - MG  ]
|  /(.''`.)\  [Fone: 31-3899-4007 ]
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Re: [R] Setting xlim in lattice plots

[Sundar Dorai-Raj]

Mike White wrote:
 I am having difficulty setting different xlim values in the lattice
 histogram plot function.
 An example is shown below.  I think I need to convert the limits data.frame
 to a list of paired values but don't know how. Any help would be
 appreciated.
 
 library(lattice)
 mat - as.data.frame(matrix(abs(c(rnorm(100),
 10*rnorm(100),20*rnorm(100),30*rnorm(100))),ncol=4))
 colnames(mat)-c(C1,C2,C3,C4)
 mat2-stack(mat)
 
 limits-cbind(rep(0, ncol(mat)),apply(mat,2,max))
 histogram(~ values | ind, data=mat2, xlim=limits,
 scales=list(x=list(relation=free)))
 
 Thanks
 Mike White
 


I think you want a breaks=NULL in your call to histogram:

library(lattice)
set.seed(42)
mat - data.frame(C1 = abs(rnorm(100)),
   C2 = abs(10 * rnorm(100)),
   C3 = abs(20 * rnorm(100)),
   C4 = abs(30 * rnorm(100)))
mat2 - stack(mat)
histogram(~ values | ind, data = mat2, breaks = NULL,
   scales = list(x = list(relation = free)))

HTH,

--sundar

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Re: [R] installation problem with Biobase

[Seth Falcon]

Haleh,

This question would be better asked on the Bioconductor mailing list.
You haven't told us what version of R you are using.  I suspect you
have a version mismatch.

With R 2.2.x you should be able to do the following to get MergeMaid
installed:

From the R prompt do:

source(http://bioconductor.org/biocLite.R;)
biocLite(MergeMaid)

And if you have further questions, please review the posting guide__ and
then send a msg to the bioc list!

__ http://www.bioconductor.org/docs/postingGuide.html


Best,

+ seth

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Re: [R] Surv object in data.frame

[Thomas Lumley]

On Wed, 15 Mar 2006, Heinz Tuechler wrote:

 Dear All,

 a Surv object I put in a data frame behaves somehow unexpected (see example).
 If I do a Cox regression on the original Surv object it works. If I put it
 in a data.frame and do the regression on the data frame it does not work.
 Seemingly it has to do with the class attribute, because if I change the
 class attribute to let Surv appeare first, again it works.
 Is this known? Should I have found information on it?

Well, this is the sort of thing that happens when you use kludges like 
AsIs.

The problem is with [.AsIs
survobj[,1] is supposed to be a vector of times (that's what [.Surv 
returns), but [.AsIs sticks the original class attribute on to it.

 str(survobj[,1])
  num [1:5] 0 0 0 0 0
 str(I(survobj)[,1])
Classes 'AsIs', 'Surv'  num [1:5] 0 0 0 0 0

The solution is not to use I() -- there's no problem with putting survival 
objects in a data frame
 df.right-data.frame(survobj,group)
 df.right
   survobj group
1  (0,1 ] 1
2  (0,2+] 1
3  (0,3 ] 1
4  (0,4 ] 2
5  (0,5 ] 2


-thomas

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Re: [R] Question about 'lables' ect.

[Liaw, Andy]

Try something like:

xp - barplot(test$Score, space=.5)
axis(1, at=xp, labels=as.character(test$Name))

See ?barplot more more detail.

Andy

From: jia ding
 
 Hi,
 
 I have a file named:
 test_R.txt
 aaa  2
 bbb  5
 ccc  7
 sss  3
 xxx  8
 
 I want to have a plot:
 test-read.table(test_R.txt,col.name=c(Name,Score))
 par(mfrow=c(1,2))
 barplot(test$Score)
 name-test$Name
 axis(1,at=1:length(test$Name),labels=paste(name))
 
 Q1, if you try the script above,you will get 5 bars, the axis 
 only shows aaa, ccc,xxx, but where are bbbsss?
 
 Q2, pls have a look this x-axis again, you will find the 
 middle of the bars are not pointing to the x-axes.
 
 Q3, how can i change the width of the bars? I feel they are too fat.
 
 Thanks!
 Nina
 
   [[alternative HTML version deleted]]
 
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Re: [R] Question about 'lables' ect.

[Marc Schwartz (via MN)]

On Wed, 2006-03-15 at 17:54 +0100, jia ding wrote:
 Hi,
 
 I have a file named:
 test_R.txt
 aaa  2
 bbb  5
 ccc  7
 sss  3
 xxx  8
 
 I want to have a plot:
 test-read.table(test_R.txt,col.name=c(Name,Score))

 par(mfrow=c(1,2))

It's not clear what the purpose is here, at least in this example. Do
you plan on creating a second plot?

 barplot(test$Score)
 name-test$Name
 axis(1,at=1:length(test$Name),labels=paste(name))
 
 Q1, if you try the script above,you will get 5 bars, the axis only shows
 aaa, ccc,xxx, but where are bbbsss?

The easiest way to do this is to use the 'names.arg' argument in
barplot():

barplot(test$Score, names.arg = as.character(test$Name))

Note that the 'Name' column in the 'test' data frame will be a factor by
default, so you need to convert it to a character vector here.

 Q2, pls have a look this x-axis again, you will find the middle of the bars
 are not pointing to the x-axes.

Note that in the Value section of ?barplot, it indicates that barplot()
returns the bar midpoints, which are not at integer values along the x
axis.

You would need to do something like:

mp - barplot(test$Score)

axis(1, at = mp, labels = as.character(test$Name))


 Q3, how can i change the width of the bars? I feel they are too fat.

You can use the 'space' argument:

barplot(test$Score, names.arg = as.character(test$Name), space = 0.5)


See the descriptions of the 'width' and 'space' arguments in ?barplot
for some of the subtleties here.

See ?barplot for more information and further examples.

HTH,

Marc Schwartz

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Re: [R] matrix indexing

[tom wright]

Thanks everyone.
Obvious when you think about it, and you check that both the matrices
your trying it with are actually matrices... instead of one being a
list.

On Wed, 2006-15-03 at 06:03 -0500, tom wright wrote:
 Can someone please give me a pointer here.
 I have two matrices
 
 matA
   A   B   C
 1 5   2   4
 2 2   4   3
 3 1   2   4
 
 matB
   A   B   C
 1 TRUEFALSE   TRUE
 2 FALSE   TRUETRUE
 3 FALSE   FALSE   FALSE
 
 how do I extract all the values from matA where the coresponding entry
 in matB == TRUE (or FALSE), perferably in vector form.
 
 Many thanks
 tom
 
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Re: [R] Surv object in data.frame

[Robert Baer]

This does work:
coxph(survobj~group, data=df.test[[1]]) # this works like your original

To get insight compare:
str(survobj)
str(df.test)
str(df.test[[1]])

Then note the 2nd sentence of the  following from ?coxph
Arguments:

 formula: a formula object, with the response on the left of a '~'
  operator, and the terms on the right.  The response must be a
  survival object as returned by the 'Surv' function.



Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA


Dear All,

a Surv object I put in a data frame behaves somehow unexpected (see
example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly it has to do with the class attribute, because if I change the
class attribute to let Surv appeare first, again it works.
Is this known? Should I have found information on it?
Any comments?

Thanks

Heinz Tüchler

## example data
starttime - rep(0,5)
stoptime  - 1:5
event - c(1,0,1,1,1)
group - c(1,1,1,2,2)
## Surv object
survobj   - Surv(starttime, stoptime, event)
## Cox-regression
coxph(survobj~group) # this works
## put Surv object in data.frame
df.test - data.frame(survobj=I(survobj), group)
## Cox-regression on data.frame
coxph(survobj~group, data=df.test) # this does not work
attr(df.test$survobj, 'class') # survobject has class AsIs, Surv
attr(df.test$survobj, 'class') - c('Surv', 'AsIs') # put Surv first
attr(df.test$survobj, 'class') # survobject has class Surv, AsIs
coxph(survobj~group, data=df.test) # now it works

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Re: [R] Address matching (was R-help Digest, Vol 37, Issue 12)

[Roger Bivand]

On Sun, 12 Mar 2006, Ferran Carrascosa wrote:

 Hi r-users,
 
 I would like to know if R have any solution to the Address standardization.
 The problem is to classify a database of addresses with the real
 addresses of a streets of Spain. Ideally, I would like to assign
 Postal code, census data and other geographic information.

There are no such built-in databases in R, and commercial solutions are 
typically costly. I assume you have addresses, and need a spatial index or 
key to associate the addresses with polygon containers like postcodes or 
census tracts. If there are not too many and they are all close to each 
other, a GPS and a bike are very effective ... Watch out for false 
positives in commercial automatic address matching, they may be about 10% 
(mostly typing errors in input data, but can be honest errors in the 
software).

 
 If this is not possible I would like to know solutions in R about text
 mining, text classification, distance within text data,...
 

RSiteSearch(text mining) is quite productive.

 Any help will be appreciate
 
 Thanks in advance
 
 Ferran Carrascosa
 
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-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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[R] zeros to the left of the decimal point

[Dimitri Joe]

Hi,

This one is quick: how to ask R to print 0.1 as .1, i.e, what I want is

  0.1
.1

Many thanks,

Dimitri

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Re: [R] display postal codes for Germany on a map

[Roger Bivand]

On Tue, 14 Mar 2006, Stefan Pohl wrote:

 Hi,
 
 is there an R package with which is it possible to display postal codes
 for Germany on a map?

If you know the geographical or projected coordinates of the postcodes, 
then there are plenty of possibilities, but this kind of data is typically 
only available commercially and is not provided within R or its 
contributed packages. If you do have, have a look at the classes and 
methods in the sp package.

Since this is the second query from within the EU about postcode 
coordinates and open acces to geospatial data in a short time, I'm 
providing a link to:

http://okfn.org/geo/manifesto.php

and you may also refer to the EU INSPIRE initiative, but Europe is already 
so far behind the US that I'm not optimistic about sensible access to 
public geospatial data any time soon.

 
 Thanks,
 
 Stefan.
   [[alternative HTML version deleted]]
 
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-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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Re: [R] map question

[Roger Bivand]

On Tue, 14 Mar 2006, Dean Sonneborn wrote:

 Would anyone with experience with the map functions know how to divide 
 Czechoslovakia into the Czech Republic and Slovakia. They have been two 
 separate countries for some time now. I'm thinking about the worldhires 
 map database  in particular.
 

There are several borders in that database that are missing, although the 
country polygons are useful for people mapping cold-war data! I think most 
users tend to need the coastlines more than country polygons, so the data 
sets are very useful anyway.

Adding borders can be done manually by adding the longitude,latitude 
coordinates to the input data shipped with the source package, then 
rebuild the database.

If you don't need a large area, but a smaller set of country borders, 
there has been a discussion recently on the R-sig-geo list about this, and 
I'd suggest your following up there:

https://stat.ethz.ch/pipermail/r-sig-geo/2006-March/thread.html

at Adding some data as colors to world map

 

-- 
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Helleveien 30, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43
e-mail: [EMAIL PROTECTED]

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Re: [R] (newbie) Weighted qqplot?

[Vivek Satsangi]

Folks,
I am documenting what I finally did, for the next person who comes along...

Following Dr. Murdoch's suggestion, I looked at qqplot. The following
approach might be helpful to get to the same information as given by
qqplot.
To summarize the ask: given x, y, xw and yw, show (visually is okay)
whether a and b are from the same distribution. xw is the weight of
each x observation and yw is the weight of each y observation.

Put x and xw into a dataframe.
Sort by x.
Calculate cumulative x weights, normalized to total 1.

Put y and yw into a dataframe.
Sort by y
Calculate cumulative weights, normalized to total 1.

Plot x and y against cumulative normalized weights. The shapes of the
two lines should be similar (to the eye)-- or the distribution is
different.

Vivek

On 3/15/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
 On 3/15/2006 8:31 AM, Vivek Satsangi wrote:
  Folks,
  Normally, in a data frame, one observation counts as one observation
  of the distribution. Thus one can easily produce a CDF and (in Splus
  atleast) use cdf.compare to compare the CDF (BTW: what is the R
  equivalent of the SPlus cdf.compare() function, if any?)
 
  However, if each point should not count equally, how can I weight the
  points before comparing the distributions? I was thinking of somehow
  creating multiple observations for each actual observation based on
  weights and creating a new dataframe etc. -- but that seem excessive.
  Surely there is a simpler way?
 
  x - rnorm(100)
  y - rnorm(10)
  xw - rnorm(100) * 1.73 # The weights. These won't add up to 1 or N or 
  anything because of missing values.
  yw - rnorm(10) * 6.23 # The weights. These won't add up to 1 or to the 
  same number as xw.
  # The question to answer is, how can I create a qq plot or cdf compare of 
  x vs. y, weighted by their weights, xw and yw (to eventually figure out if 
  y comes from the population x, similar to Kolmogorov-Smirnov GOF)?
  qqplot(x,y) # What now?

 qqplot doesn't support weights, but it's a simple enough function that
 you could write a version that did.  Look at the cases where length(x)
 is not equal to length(y):  e.g. if length(y)  length(x), qqplot
 constructs a linear approximation to a function mapping 1:nx onto the
 sorted x values, then takes length(y) evenly spaced values from that
 function.  You want to do the same sort of thing, except that instead of
 even spacing, you want to look at the cumulative sums of the weights.

 You might want to use some kind of graphical indicator of whether points
 are heavily weighted or not, but I don't know what to recommend for that.

 By the way, your example above will give negative weights in xw and yw;
 you probably won't like the results if you do that.

 Duncan Murdoch



--
-- Vivek Satsangi
Student, Rochester, NY USA

Life is short, the art long, opportunity fleeting, experiment
treacherous, judgement difficult.

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[R] \r with RSQLite

[Mikkel Grum]

What am I doing wrong, or is the \r that I'm getting
in the example below a bug?

 a - (1:10)
 b - (LETTERS[1:10])
 df - as.data.frame(cbind(a, b))
 
 df
a b
1   1 A
2   2 B
3   3 C
4   4 D
5   5 E
6   6 F
7   7 G
8   8 H
9   9 I
10 10 J
 library(RSQLite)
 drv - dbDriver(SQLite)
 con - dbConnect(drv, dbname = Test)
 dbWriteTable(con, DF, df, row.names = FALSE,
overwrite = TRUE)
[1] TRUE
 df2 - dbGetQuery(con, SELECT DISTINCT * FROM
DF)
 dbDisconnect(con)
[1] TRUE
 df2
a   b
1   1 A\r
2   2 B\r
3   3 C\r
4   4 D\r
5   5 E\r
6   6 F\r
7   7 G\r
8   8 H\r
9   9 I\r
10 10 J\r

 sessionInfo()
R version 2.2.1, 2005-12-20, i386-pc-mingw32 

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

other attached packages:
 RSQLite  DBI 
 0.4-1 0.1-10 


Mikkel Grum
Genetic Diversity
International Plant Genetic Resources Institute

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Re: [R] (newbie) Weighted qqplot?

[Duncan Murdoch]

On 3/15/2006 1:38 PM, Vivek Satsangi wrote:
 Folks,
 I am documenting what I finally did, for the next person who comes along...
 
 Following Dr. Murdoch's suggestion, I looked at qqplot. The following
 approach might be helpful to get to the same information as given by
 qqplot.
 To summarize the ask: given x, y, xw and yw, show (visually is okay)
 whether a and b are from the same distribution. xw is the weight of
 each x observation and yw is the weight of each y observation.
 
 Put x and xw into a dataframe.
 Sort by x.
 Calculate cumulative x weights, normalized to total 1.
 
 Put y and yw into a dataframe.
 Sort by y
 Calculate cumulative weights, normalized to total 1.
 
 Plot x and y against cumulative normalized weights. The shapes of the
 two lines should be similar (to the eye)-- or the distribution is
 different.

One variation that would make the result more like a qqplot:  you could 
work out a vector of weights w (perhaps the cumulative weights from x or 
from y or perhaps something else) and plot y(w) versus x(w), where y(w) 
and x(w) are the linear interpolation values that approx gives you.

Duncan Murdoch
 
 Vivek
 
 On 3/15/06, Duncan Murdoch [EMAIL PROTECTED] wrote:
 On 3/15/2006 8:31 AM, Vivek Satsangi wrote:
  Folks,
  Normally, in a data frame, one observation counts as one observation
  of the distribution. Thus one can easily produce a CDF and (in Splus
  atleast) use cdf.compare to compare the CDF (BTW: what is the R
  equivalent of the SPlus cdf.compare() function, if any?)
 
  However, if each point should not count equally, how can I weight the
  points before comparing the distributions? I was thinking of somehow
  creating multiple observations for each actual observation based on
  weights and creating a new dataframe etc. -- but that seem excessive.
  Surely there is a simpler way?
 
  x - rnorm(100)
  y - rnorm(10)
  xw - rnorm(100) * 1.73 # The weights. These won't add up to 1 or N or 
  anything because of missing values.
  yw - rnorm(10) * 6.23 # The weights. These won't add up to 1 or to the 
  same number as xw.
  # The question to answer is, how can I create a qq plot or cdf compare of 
  x vs. y, weighted by their weights, xw and yw (to eventually figure out 
  if y comes from the population x, similar to Kolmogorov-Smirnov GOF)?
  qqplot(x,y) # What now?

 qqplot doesn't support weights, but it's a simple enough function that
 you could write a version that did.  Look at the cases where length(x)
 is not equal to length(y):  e.g. if length(y)  length(x), qqplot
 constructs a linear approximation to a function mapping 1:nx onto the
 sorted x values, then takes length(y) evenly spaced values from that
 function.  You want to do the same sort of thing, except that instead of
 even spacing, you want to look at the cumulative sums of the weights.

 You might want to use some kind of graphical indicator of whether points
 are heavily weighted or not, but I don't know what to recommend for that.

 By the way, your example above will give negative weights in xw and yw;
 you probably won't like the results if you do that.

 Duncan Murdoch

 
 
 --
 -- Vivek Satsangi
 Student, Rochester, NY USA
 
 Life is short, the art long, opportunity fleeting, experiment
 treacherous, judgement difficult.
 
 __
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Re: [R] manipulating weeks dates

[Gabor Grothendieck]

Given that there are more than 7 * 52 days in a week you may
need to think about what it is that you want.

This will give you Jan 1 for Week 1, Jan 8 for Week 2, etc.

as.Date(paste(YEAR,1,1,sep=-)) + 7 * (WEEK - 1)

That or some variation of that might be suitable.  See the help
desk article in R News 4/1 for more info on dates.


On 3/15/06, Ronaldo Reis-Jr. [EMAIL PROTECTED] wrote:
 Em Quarta 15 Março 2006 10:26, Gabor Grothendieck escreveu:
  as.Date(paste(YEAR, WEEK, 0), %Y %U %w)

 Hi,

 it works, but it use a year with 53 weeks, I need to use with 52 weeks, how to
 change this?

 Thanks
 Ronaldo
 --
 Errigal Mountains -- Tailoring manure
-- anagrama
 --
 |   // | \\   [***]
 |   ( õ   õ )  [Ronaldo Reis Júnior]
 |  V  [UFV/DBA-Entomologia]
 |/ \   [36570-000 Viçosa - MG  ]
 |  /(.''`.)\  [Fone: 31-3899-4007 ]
 |  /(: :'  :)\ [EMAIL PROTECTED]]
 |/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
 |( `-  )   [***]
 |  _/   \_Powered by GNU/Debian Woody/Sarge

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Re: [R] manipulating weeks dates

[Gabor Grothendieck]

I meant there are more than 7 * 52 days in a year.

On 3/15/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
 Given that there are more than 7 * 52 days in a week you may
 need to think about what it is that you want.

 This will give you Jan 1 for Week 1, Jan 8 for Week 2, etc.

 as.Date(paste(YEAR,1,1,sep=-)) + 7 * (WEEK - 1)

 That or some variation of that might be suitable.  See the help
 desk article in R News 4/1 for more info on dates.


 On 3/15/06, Ronaldo Reis-Jr. [EMAIL PROTECTED] wrote:
  Em Quarta 15 Março 2006 10:26, Gabor Grothendieck escreveu:
   as.Date(paste(YEAR, WEEK, 0), %Y %U %w)
 
  Hi,
 
  it works, but it use a year with 53 weeks, I need to use with 52 weeks, how 
  to
  change this?
 
  Thanks
  Ronaldo
  --
  Errigal Mountains -- Tailoring manure
 -- anagrama
  --
  |   // | \\   [***]
  |   ( õ   õ )  [Ronaldo Reis Júnior]
  |  V  [UFV/DBA-Entomologia]
  |/ \   [36570-000 Viçosa - MG  ]
  |  /(.''`.)\  [Fone: 31-3899-4007 ]
  |  /(: :'  :)\ [EMAIL PROTECTED]]
  |/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ]
  |( `-  )   [***]
  |  _/   \_Powered by GNU/Debian Woody/Sarge
 
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Re: [R] click on graph and select data points?

[justin bem]

I think you ca try 
  identify()


-

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[R] which.minimums not which.min

[Fred J.]

  Hi
   
  Is there a function which determines the location, i.e., index of the all 
minimums or maximums of a numeric vector.
  Which.min(x) only finds the (first) of such.
   
   x - c(1:4,0:5, 4, 11)
   x
   [1]  1  2  3  4  0  1  2  3  4  5 4 11
   which.min(x)
  [1] 5
   which.max(x)
  [1] 11
  
   
  but I need 
  which.min(x)  to be 5 11
  which.max(x) to be 4 10
   
  thanks
   
  
__



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[R] R icon image file

[Erin Hodgess]

Dear R People:

I would like to include a link to the R home page
on a web page for students.

I would like to have the R icon as part of the link.

Where is the image file please? (for the icon)

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

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Re: [R] \r with RSQLite

[bogdan romocea]

\r is a carriage return character which some editors may use as a line
terminator when writing files.  My guess is that RSQLite writes your
data frame to a temp file using \r as a line terminator and then runs
a script to have SQLite import the data (together with \r - this would
be the problem), but I have no idea if that's really the case. Check
the documentation or ask the maintainer.


 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Mikkel Grum
 Sent: Wednesday, March 15, 2006 1:46 PM
 To: r-help@stat.math.ethz.ch
 Cc: [EMAIL PROTECTED]
 Subject: [R] \r with RSQLite

 What am I doing wrong, or is the \r that I'm getting
 in the example below a bug?

  a - (1:10)
  b - (LETTERS[1:10])
  df - as.data.frame(cbind(a, b))
 
  df
 a b
 1   1 A
 2   2 B
 3   3 C
 4   4 D
 5   5 E
 6   6 F
 7   7 G
 8   8 H
 9   9 I
 10 10 J
  library(RSQLite)
  drv - dbDriver(SQLite)
  con - dbConnect(drv, dbname = Test)
  dbWriteTable(con, DF, df, row.names = FALSE,
 overwrite = TRUE)
 [1] TRUE
  df2 - dbGetQuery(con, SELECT DISTINCT * FROM
 DF)
  dbDisconnect(con)
 [1] TRUE
  df2
 a   b
 1   1 A\r
 2   2 B\r
 3   3 C\r
 4   4 D\r
 5   5 E\r
 6   6 F\r
 7   7 G\r
 8   8 H\r
 9   9 I\r
 10 10 J\r

  sessionInfo()
 R version 2.2.1, 2005-12-20, i386-pc-mingw32

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

 other attached packages:
  RSQLite  DBI
  0.4-1 0.1-10


 Mikkel Grum
 Genetic Diversity
 International Plant Genetic Resources Institute

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 R-help@stat.math.ethz.ch mailing list
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Re: [R] which.minimums not which.min

[Christos Hatzis]

Try 

order(x, decreasing=TRUE/FALSE)
  

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Fred J.
Sent: Wednesday, March 15, 2006 2:32 PM
To: r-help@stat.math.ethz.ch
Subject: [R] which.minimums not which.min

  Hi
   
  Is there a function which determines the location, i.e., index of the all
minimums or maximums of a numeric vector.
  Which.min(x) only finds the (first) of such.
   
   x - c(1:4,0:5, 4, 11)
   x
   [1]  1  2  3  4  0  1  2  3  4  5 4 11
   which.min(x)
  [1] 5
   which.max(x)
  [1] 11
  
   
  but I need
  which.min(x)  to be 5 11
  which.max(x) to be 4 10
   
  thanks
   
  
__



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Re: [R] R icon image file

[Christos Hatzis]

http://www.r-project.org/Rlogo.jpg 

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erin Hodgess
Sent: Wednesday, March 15, 2006 2:39 PM
To: r-help@stat.math.ethz.ch
Subject: [R] R icon image file

Dear R People:

I would like to include a link to the R home page on a web page for
students.

I would like to have the R icon as part of the link.

Where is the image file please? (for the icon)

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

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Re: [R] which.minimums not which.min

[Marc Schwartz (via MN)]

On Wed, 2006-03-15 at 11:32 -0800, Fred J. wrote:
   Hi

   Is there a function which determines the location, i.e., index of
 the all minimums or maximums of a numeric vector.
   Which.min(x) only finds the (first) of such.

x - c(1:4,0:5, 4, 11)
x
[1]  1  2  3  4  0  1  2  3  4  5 4 11
which.min(x)
   [1] 5
which.max(x)
   [1] 11
   

   but I need 
   which.min(x)  to be 5 11
   which.max(x) to be 4 10

   thanks


There is something wrong with your example code versus data here, since:

 x
 [1]  1  2  3  4  0  1  2  3  4  5  4 11

 which.min(x)
[1] 5

 which.max(x)
[1] 12


There is one one minimum value of 0 in that vector and only one maximum
value of 11.

If you had a vector 'x':

 x - c(1:4, 0:5, 4, 0, 5)

 x
 [1] 1 2 3 4 0 1 2 3 4 5 4 0 5


You could then do the following to get the indices of the multiple
min/max values:

 which(x == min(x))
[1]  5 12

 which(x == max(x))
[1] 10 13


The only other thing that I can think you might be considering would be
local minima/maxima in the vector and if that is what you want using:

  RSiteSearch(local minima)

or

  RSiteSearch(peaks)


should lead you to some solutions that have been discussed previously.

HTH,

Marc Schwartz

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Re: [R] which.minimums not which.min

[Liaw, Andy]

What you want seems to be the valleys and peaks in the data.  If so, try:

  RSiteSearch(find peaks)

which points to a post by Philippe Grosjean, pointing to the pastesc
package:

 library(pastecs)
Loading required package: boot
 tp - turnpoints(x)
 which(tp$peaks)
[1]  4 10
 which(tp$pits)
[1]  5 11

Andy
 

From: Fred J.
 
   Hi

   Is there a function which determines the location, i.e., 
 index of the all minimums or maximums of a numeric vector.
   Which.min(x) only finds the (first) of such.

x - c(1:4,0:5, 4, 11)
x
[1]  1  2  3  4  0  1  2  3  4  5 4 11
which.min(x)
   [1] 5
which.max(x)
   [1] 11
   

   but I need 
   which.min(x)  to be 5 11
   which.max(x) to be 4 10

   thanks

   
 __
 
 
 
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Re: [R] which.minimums not which.min

[Philippe Grosjean]

What Fred is looking for is local minima/maxima, also known as turning 
points, or pits/peaks in a series.  You can look at ?turnpoints in 
pastecs library.

  x - c(1:4,0:5, 4, 11)
  x
  [1]  1  2  3  4  0  1  2  3  4  5  4 11
  tp - turnpoints(x)
  summary(tp)
Turning points for: x

nbr observations  : 12
nbr ex-aequos : 0
nbr turning points: 4 (first point is a peak)
E(p) = 6.67 Var(p) = 1.81 (theoretical)

   point type   proba  info
1 4 peak 0.1 3.3219281
2 5  pit 0.002380952 8.7142455
310 peak 0.005952381 7.3923174
411  pit 0.7 0.5849625
  plot(tp) # Only useful for a longer and more complex series!
  # Get the position of peaks
  (1:length(x))[extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)]
[1]  4 10
Warning message:
arguments after the first two are ignored in: UseMethod(extract, e, n, 
...)
  (1:length(x))[extract(tp, no.tp = FALSE, peak = FALSE, pit = TRUE)]
[1]  5 11
Warning message:
arguments after the first two are ignored in: UseMethod(extract, e, n, 
...)
  # By the way, there are warnings although it works well (I ask on R-Help)

Now, you can easily code your which.minima() function using turnpoints:

x - c(1:4,0:5, 4, 11)
x
tp - turnpoints(x)
summary(tp)
plot(tp) # Only useful for a longer and more complex series!
# Get the position of peaks
(1:length(x))[extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)]
(1:length(x))[extract(tp, no.tp = FALSE, peak = FALSE, pit = TRUE)]
# By the way, there are warnings although it works well (I ask on R-Help)

which.minima - function(x) {
if (!require(pastecs)) stop(pastecs library is required!)
x - as.vector(x)
(1:length(x))[extract(turnpoints(x), no.tp = FALSE, peak = FALSE, pit = 
TRUE)]
}

which.minima(x)

Of course, you could optimize this code. This is just a rough solution!
Best,

Philippe Grosjean

..°}))
  ) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
  ) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
  ) ) ) ) )   Mons-Hainaut University, Pentagone (3D08)
( ( ( ( (Academie Universitaire Wallonie-Bruxelles
  ) ) ) ) )   8, av du Champ de Mars, 7000 Mons, Belgium
( ( ( ( (
  ) ) ) ) )   phone: + 32.65.37.34.97, fax: + 32.65.37.30.54
( ( ( ( (email: [EMAIL PROTECTED]
  ) ) ) ) )
( ( ( ( (web:   http://www.umh.ac.be/~econum
  ) ) ) ) )  http://www.sciviews.org
( ( ( ( (
..

Marc Schwartz (via MN) wrote:
 On Wed, 2006-03-15 at 11:32 -0800, Fred J. wrote:
 
  Hi
   
  Is there a function which determines the location, i.e., index of
the all minimums or maximums of a numeric vector.
  Which.min(x) only finds the (first) of such.
   
   x - c(1:4,0:5, 4, 11)
   x
   [1]  1  2  3  4  0  1  2  3  4  5 4 11
   which.min(x)
  [1] 5
   which.max(x)
  [1] 11
  
   
  but I need 
  which.min(x)  to be 5 11
  which.max(x) to be 4 10
   
  thanks
   
 
 
 There is something wrong with your example code versus data here, since:
 
 
x
 
  [1]  1  2  3  4  0  1  2  3  4  5  4 11
 
 
which.min(x)
 
 [1] 5
 
 
which.max(x)
 
 [1] 12
 
 
 There is one one minimum value of 0 in that vector and only one maximum
 value of 11.
 
 If you had a vector 'x':
 
 
x - c(1:4, 0:5, 4, 0, 5)
 
 
x
 
  [1] 1 2 3 4 0 1 2 3 4 5 4 0 5
 
 
 You could then do the following to get the indices of the multiple
 min/max values:
 
 
which(x == min(x))
 
 [1]  5 12
 
 
which(x == max(x))
 
 [1] 10 13
 
 
 The only other thing that I can think you might be considering would be
 local minima/maxima in the vector and if that is what you want using:
 
   RSiteSearch(local minima)
 
 or
 
   RSiteSearch(peaks)
 
 
 should lead you to some solutions that have been discussed previously.
 
 HTH,
 
 Marc Schwartz
 
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Re: [R] which.minimums not which.min

[Marc Schwartz (via MN)]

On Wed, 2006-03-15 at 21:45 +0100, Philippe Grosjean wrote:
 What Fred is looking for is local minima/maxima, also known as turning 
 points, or pits/peaks in a series.  You can look at ?turnpoints in 
 pastecs library.
 
   x - c(1:4,0:5, 4, 11)
   x
   [1]  1  2  3  4  0  1  2  3  4  5  4 11
   tp - turnpoints(x)
   summary(tp)
 Turning points for: x
 
 nbr observations  : 12
 nbr ex-aequos : 0
 nbr turning points: 4 (first point is a peak)
 E(p) = 6.67 Var(p) = 1.81 (theoretical)
 
point type   proba  info
 1 4 peak 0.1 3.3219281
 2 5  pit 0.002380952 8.7142455
 310 peak 0.005952381 7.3923174
 411  pit 0.7 0.5849625
   plot(tp) # Only useful for a longer and more complex series!
   # Get the position of peaks
   (1:length(x))[extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)]
 [1]  4 10
 Warning message:
 arguments after the first two are ignored in: UseMethod(extract, e, n, 
 ...)
   (1:length(x))[extract(tp, no.tp = FALSE, peak = FALSE, pit = TRUE)]
 [1]  5 11
 Warning message:
 arguments after the first two are ignored in: UseMethod(extract, e, n, 
 ...)
   # By the way, there are warnings although it works well (I ask on R-Help)
 
 Now, you can easily code your which.minima() function using turnpoints:
 
 x - c(1:4,0:5, 4, 11)
 x
 tp - turnpoints(x)
 summary(tp)
 plot(tp) # Only useful for a longer and more complex series!
 # Get the position of peaks
 (1:length(x))[extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)]
 (1:length(x))[extract(tp, no.tp = FALSE, peak = FALSE, pit = TRUE)]
 # By the way, there are warnings although it works well (I ask on R-Help)
 
 which.minima - function(x) {
   if (!require(pastecs)) stop(pastecs library is required!)
   x - as.vector(x)
   (1:length(x))[extract(turnpoints(x), no.tp = FALSE, peak = FALSE, pit = 
 TRUE)]
 }
 
 which.minima(x)
 
 Of course, you could optimize this code. This is just a rough solution!
 Best,
 
 Philippe Grosjean


Philippe,

Thanks for the clarification. As with Andy's reply, it seems that my
closing thoughts were correct.

I was confused since the actual result of which.max() in Fred's post did
not match the data provided.

Best regards,

Marc

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[R] Additional arguments in S3 method produces a warning

[Philippe Grosjean]

Hello,
I just notice this:
  x - c(1:4,0:5, 4, 11)
  library(pastecs)
Loading required package: boot
  tp - turnpoints(x)
  extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)
  [1] FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE 
FALSE
Warning message:
arguments after the first two are ignored in: UseMethod(extract, e, n, 
...)
  extract(tp)
  [1]  0  0  0  1 -1  0  0  0  0  1 -1  0
Warning message:
arguments after the first two are ignored in: UseMethod(extract, e, n, 
...)

My extract.turnpoints() function produces warnings. I can easily spot 
the origin of this warning:

  extract
function (e, n, ...)
UseMethod(extract, e, n, ...)
  extract.turnpoints
function (e, n, no.tp = 0, peak = 1, pit = -1, ...)
{
 if (missing(n))
 n - length(e)
 res - rep(no.tp, length.out = e$n)
 res[e$pos[e$peaks]] - peak
 res[e$pos[e$pits]] - pit
 if (n  length(res)  n  0)
 res - res[1:n]
 res
}

This is because my extract.turnpoints() method defines more arguments 
than 'e' and 'n' in the generic function. However,

1) I though that the '...' argument in S3 generic function was there to 
allow defining/passing additional arguments in/to S3 methods. Is this 
correct? If yes, why the warning?

2) Despite the warning says arguments after the first two are ignored, 
this appears not to be the case: in this example, 'no.tp', 'peak' and 
'pit' arguments are taken into account, as you can see (different 
behaviour if you give other values to them).

I am a little bit lost. Could someone help me, please.

Best,

Philippe Grosjean


-- 
..°}))
  ) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
  ) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
  ) ) ) ) )   Mons-Hainaut University, Pentagone (3D08)
( ( ( ( (Academie Universitaire Wallonie-Bruxelles
  ) ) ) ) )   8, av du Champ de Mars, 7000 Mons, Belgium
( ( ( ( (
  ) ) ) ) )   phone: + 32.65.37.34.97, fax: + 32.65.37.30.54
( ( ( ( (email: [EMAIL PROTECTED]
  ) ) ) ) )
( ( ( ( (web:   http://www.umh.ac.be/~econum
  ) ) ) ) )  http://www.sciviews.org
( ( ( ( (
..

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[R] generating RANDOM ROWS from matrix

[mark salsburg]

Dear group,

I would like to generate a 1000 random rows from a MATRIX with dimensions
12,000 by 20 (i.e. to generate a 1000 by 20 matrix of random rows)

Does the function sample() work for this???

thank you in advance

[[alternative HTML version deleted]]

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Re: [R] Additional arguments in S3 method produces a warning

[Gabor Grothendieck]

Define extract like this:

extract - function(e, n, ...) UseMethod(extract)

# test -- no warning
extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)

On 3/15/06, Philippe Grosjean [EMAIL PROTECTED] wrote:
 Hello,
 I just notice this:
   x - c(1:4,0:5, 4, 11)
   library(pastecs)
 Loading required package: boot
   tp - turnpoints(x)
   extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)
  [1] FALSE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE
 FALSE
 Warning message:
 arguments after the first two are ignored in: UseMethod(extract, e, n,
 ...)
   extract(tp)
  [1]  0  0  0  1 -1  0  0  0  0  1 -1  0
 Warning message:
 arguments after the first two are ignored in: UseMethod(extract, e, n,
 ...)

 My extract.turnpoints() function produces warnings. I can easily spot
 the origin of this warning:

   extract
 function (e, n, ...)
 UseMethod(extract, e, n, ...)
   extract.turnpoints
 function (e, n, no.tp = 0, peak = 1, pit = -1, ...)
 {
 if (missing(n))
 n - length(e)
 res - rep(no.tp, length.out = e$n)
 res[e$pos[e$peaks]] - peak
 res[e$pos[e$pits]] - pit
 if (n  length(res)  n  0)
 res - res[1:n]
 res
 }

 This is because my extract.turnpoints() method defines more arguments
 than 'e' and 'n' in the generic function. However,

 1) I though that the '...' argument in S3 generic function was there to
 allow defining/passing additional arguments in/to S3 methods. Is this
 correct? If yes, why the warning?

 2) Despite the warning says arguments after the first two are ignored,
 this appears not to be the case: in this example, 'no.tp', 'peak' and
 'pit' arguments are taken into account, as you can see (different
 behaviour if you give other values to them).

 I am a little bit lost. Could someone help me, please.

 Best,

 Philippe Grosjean


 --
 ..°}))
  ) ) ) ) )
 ( ( ( ( (Prof. Philippe Grosjean
  ) ) ) ) )
 ( ( ( ( (Numerical Ecology of Aquatic Systems
  ) ) ) ) )   Mons-Hainaut University, Pentagone (3D08)
 ( ( ( ( (Academie Universitaire Wallonie-Bruxelles
  ) ) ) ) )   8, av du Champ de Mars, 7000 Mons, Belgium
 ( ( ( ( (
  ) ) ) ) )   phone: + 32.65.37.34.97, fax: + 32.65.37.30.54
 ( ( ( ( (email: [EMAIL PROTECTED]
  ) ) ) ) )
 ( ( ( ( (web:   http://www.umh.ac.be/~econum
  ) ) ) ) )  http://www.sciviews.org
 ( ( ( ( (
 ..

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Re: [R] generating RANDOM ROWS from matrix

[Doran, Harold]

Is something like this what your looking for:

x   - matrix(c(rnorm(100)),ncol=10)
sub - sample(5, replace=TRUE) # For sampling with replacement
x[sub,]

Harold

 

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of mark salsburg
Sent: Wednesday, March 15, 2006 4:02 PM
To: R-help@stat.math.ethz.ch
Subject: [R] generating RANDOM ROWS from matrix

Dear group,

I would like to generate a 1000 random rows from a MATRIX with
dimensions 12,000 by 20 (i.e. to generate a 1000 by 20 matrix of random
rows)

Does the function sample() work for this???

thank you in advance

[[alternative HTML version deleted]]

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Re: [R] generating RANDOM ROWS from matrix

[Berton Gunter]

?sample

You must use replace=FALSE to guarantee 1000 different rows

mymatrix[sample(12000,1000,replace=FALSE),]

-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
The business of the statistician is to catalyze the scientific learning
process.  - George E. P. Box
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of mark salsburg
 Sent: Wednesday, March 15, 2006 1:02 PM
 To: R-help@stat.math.ethz.ch
 Subject: [R] generating RANDOM ROWS from matrix
 
 Dear group,
 
 I would like to generate a 1000 random rows from a MATRIX 
 with dimensions
 12,000 by 20 (i.e. to generate a 1000 by 20 matrix of random rows)
 
 Does the function sample() work for this???
 
 thank you in advance
 
   [[alternative HTML version deleted]]
 
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[R] lapply vs. for (was: Incrementing a counter in lapply)

[Gregor Gorjanc]

 From: Thomas Lumley
 
 On Tue, 14 Mar 2006, John McHenry wrote:
 
  Thanks, Gabor  Thomas.
 
  Apologies, but I used an example that obfuscated the question that I
  wanted to ask.
 
  I really wanted to know how to have extra arguments in 
 functions that
  would allow, per the example code, for something like a 
 counter to be 
  incremented. Thomas's suggestion of using mapply 
 (reproduced below with 
  corrections) is probably closest.
 
 It is probably worth pointing out here that the R 
 documentation does not 
 specify the order in which lapply() does the computation.
 
 If you could work out how to increment a counter (and you could, with 
 sufficient effort), it would not necessarily work, because the 'i'th 
 evaluation would not necessarily be of the 'i'th element.
 
 [lapply() does in fact start at the beginning, go on until it 
 gets to the 
 end, and then stop, but this isn't documented.   Suppose R became 
 multithreaded, for example]
 
 The corollary, it seems to me, is that sometimes it's better to leave the
 good old for loop alone.  It's not always profitable to turn for loops into
 some *apply construct.  The trick is learning to know when to do it and when
 not to.

Can someone share some of this tricks with me? Up to now I have always
done things with for loop. Just recently I started to pay attention to
*apply* constructs and I already wanted to start implementing them
instead of good old for, but then a stroke of lightning came from this
thread. Based on words from Thomas, lapply should not be used for tasks
where order is critical. Did I get this clear enough. Additionally, I
have read notes (I lost link, but was posted on R-help, I think) from
Thomas on R and he mentioned that it is commonly assumed that *apply* (I
do not remember which one of *apply*) is faster than loop, but that this
is not true. Any additional pointers to literature?

-- 
Lep pozdrav / With regards,
Gregor Gorjanc

--
University of Ljubljana PhD student
Biotechnical Faculty
Zootechnical Department URI: http://www.bfro.uni-lj.si/MR/ggorjan
Groblje 3   mail: gregor.gorjanc at bfro.uni-lj.si

SI-1230 Domzale tel: +386 (0)1 72 17 861
Slovenia, Europefax: +386 (0)1 72 17 888

--
One must learn by doing the thing; for though you think you know it,
 you have no certainty until you try. Sophocles ~ 450 B.C.

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Re: [R] lapply vs. for (was: Incrementing a counter in lapply)

[Philippe Grosjean]

the for() loop is very slow in S-PLUS. This is probably one of the 
motivation of developing the apply() family of functions (as well as the 
ugly For() loop) under this system.

Now, for() loops are much faster in R. Also, if you look at the R code 
in apply(), you will realize that there is a for() loop in it!

So, why would you prefer using apply() or the like?
1) If you write code to be run both in S-PLUS and R,

2) If you want more concise code (much housekeeping is done by apply() 
and co),

3) Because the apply() family is more in the phylosophy of vectorized 
calculation, that is, the favored approach in S language.

Take care, however, that the optimal approach is not just to replace 
for() loops with apply() and co, but to *rethink* completelly your 
algorithm in a vectorized way. This often ends up with a very different 
solution!
Best,

Philippe Grosjean

Gregor Gorjanc wrote:
From: Thomas Lumley

On Tue, 14 Mar 2006, John McHenry wrote:


Thanks, Gabor  Thomas.

Apologies, but I used an example that obfuscated the question that I
wanted to ask.

I really wanted to know how to have extra arguments in 

functions that

would allow, per the example code, for something like a 

counter to be 

incremented. Thomas's suggestion of using mapply 

(reproduced below with 

corrections) is probably closest.

It is probably worth pointing out here that the R 
documentation does not 
specify the order in which lapply() does the computation.

If you could work out how to increment a counter (and you could, with 
sufficient effort), it would not necessarily work, because the 'i'th 
evaluation would not necessarily be of the 'i'th element.

[lapply() does in fact start at the beginning, go on until it 
gets to the 
end, and then stop, but this isn't documented.   Suppose R became 
multithreaded, for example]

The corollary, it seems to me, is that sometimes it's better to leave the
good old for loop alone.  It's not always profitable to turn for loops into
some *apply construct.  The trick is learning to know when to do it and when
not to.
 
 
 Can someone share some of this tricks with me? Up to now I have always
 done things with for loop. Just recently I started to pay attention to
 *apply* constructs and I already wanted to start implementing them
 instead of good old for, but then a stroke of lightning came from this
 thread. Based on words from Thomas, lapply should not be used for tasks
 where order is critical. Did I get this clear enough. Additionally, I
 have read notes (I lost link, but was posted on R-help, I think) from
 Thomas on R and he mentioned that it is commonly assumed that *apply* (I
 do not remember which one of *apply*) is faster than loop, but that this
 is not true. Any additional pointers to literature?


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Re: [R] lapply vs. for (was: Incrementing a counter in lapply)

[Patrick Burns]

In my opinion the main issue between using 'for' and
an apply function is the simplicity of the code.  If it is
simpler and more understandable to use 'lapply' than
a 'for' loop in a situation, then use 'lapply'.  If in a
different situation it is the 'for' loop that is simpler, then
use the 'for' loop.

In modern day R whatever timing differences there may
be are likely to be slight, and virtually certain not to be
critical.

Where the confusion comes in is because in the olden
days of S-PLUS, the timing differences could be quite
substantial in some cases.  The hangover from that is
that apply functions are too often recommended in R.


Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)

Gregor Gorjanc wrote:

From: Thomas Lumley


On Tue, 14 Mar 2006, John McHenry wrote:

  

Thanks, Gabor  Thomas.

Apologies, but I used an example that obfuscated the question that I
wanted to ask.

I really wanted to know how to have extra arguments in 


functions that
  

would allow, per the example code, for something like a 


counter to be 
  

incremented. Thomas's suggestion of using mapply 


(reproduced below with 
  

corrections) is probably closest.


It is probably worth pointing out here that the R 
documentation does not 
specify the order in which lapply() does the computation.

If you could work out how to increment a counter (and you could, with 
sufficient effort), it would not necessarily work, because the 'i'th 
evaluation would not necessarily be of the 'i'th element.

[lapply() does in fact start at the beginning, go on until it 
gets to the 
end, and then stop, but this isn't documented.   Suppose R became 
multithreaded, for example]
  

The corollary, it seems to me, is that sometimes it's better to leave the
good old for loop alone.  It's not always profitable to turn for loops into
some *apply construct.  The trick is learning to know when to do it and when
not to.



Can someone share some of this tricks with me? Up to now I have always
done things with for loop. Just recently I started to pay attention to
*apply* constructs and I already wanted to start implementing them
instead of good old for, but then a stroke of lightning came from this
thread. Based on words from Thomas, lapply should not be used for tasks
where order is critical. Did I get this clear enough. Additionally, I
have read notes (I lost link, but was posted on R-help, I think) from
Thomas on R and he mentioned that it is commonly assumed that *apply* (I
do not remember which one of *apply*) is faster than loop, but that this
is not true. Any additional pointers to literature?

  


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Re: [R] lapply vs. for (was: Incrementing a counter in lapply)

[Thomas Lumley]

On Wed, 15 Mar 2006, Philippe Grosjean wrote:

 the for() loop is very slow in S-PLUS. This is probably one of the
 motivation of developing the apply() family of functions (as well as the
 ugly For() loop) under this system.

 Now, for() loops are much faster in R. Also, if you look at the R code
 in apply(), you will realize that there is a for() loop in it!


lapply(), on the other hand, can be faster than a loop. That's why it went 
from being a loop to being internal C code.

-thomas

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Re: [R] Surface plot

[Cashorali, Tanya]

Hi, thanks for the help but I'm still having issues.

Basically, I have two matrices of equal dimension, one should produce
something similar to a heatmap.. The 2nd matrix should be the heights
for each value of the heatmap - producing a sort of surface plot.

Viewing this seems like a problem too, as I need to scale the x axis so
that ~140 labels are visible. 

Maybe wireframe isn't the way to go?  I've tried persp() too.

Thank you!!
~Tanya

-Original Message-
From: Uwe Ligges [mailto:[EMAIL PROTECTED] 
Sent: Tuesday, March 14, 2006 3:22 AM
To: Cashorali, Tanya
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Wireframe axis labels

Cashorali, Tanya wrote:

 Hi,
 I'm trying to do a surface plot using the wireframe function.
 Everything is working beautifully except that I want to be able to 
 re-scale it a LOT so that I can fit ~145 labels on the x-axis or
y-axis.
 I've tried using zoom, scales, aspect, .. nothing seems to work.  The 
 help on wireframe in R says that you can input a list of labels for 
 any of the axes, but this has also failed.


H, you can do so, but I do not believe you really want 145
labels:


g - expand.grid(x = 1:145, y = 1:145, gr = 1:2) g$z - log((g$x^g$g +
g$y^2) * g$gr)

wireframe(z ~ x * y, data = g, groups = gr,
   scales = list(arrows = FALSE, at=1:145))

Uwe Ligges

 Thanks for any help!
 -Tanya
 
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[R] concatenating factor from list

[Sebastian Luque]

Hi,

I've run into a ridiculous problem I can't find any solutions for in the
archives or help pages:

data(barley)
cutYield - with(barley, by(yield, variety, cut, breaks = c(0, 30, 60, 90)))

As in this example, I'm using 'by' to return a factor for each level of
another factor.  The problem is that 'by' returns a list of the factors,
and I need all these factors concatenated.  No problem, I said:

unlist(cutYield)

which returns an 'integer' class object, so it's no longer a factor.  The
same happens with:

do.call(c, cutYield)

I could recreate the factor from the integer level codes returned above,
but this is not good because this means redoing the job already done by
the function called in 'by', making the code more prone to errors.

Thanks in advance for any pointers,

-- 
Sebastian

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Re: [R] How to get correct proportions/bounding box for latex figure?

[context grey]

That's approximately right, but the individual
scatterplots
are slightly stretched horizontally.   

Is there not any way to have the plots have true 1:1
aspect ratio
(given that the range of the data is the same on both
axes)
and still get a bounding box?.   And, without getting
out
a ruler and manually calculating width/height values?

(I'm guessing you arrived at width=12,height=4 just
by eyeballing it.?)

--- Martin Sandiford [EMAIL PROTECTED] wrote:

 Does this do what you want?
 
 library(lattice)
 
 rand1 - rnorm(50)
 rand2 - rnorm(50)
 theplot - xyplot(rand1 ~ rand2, xlab=x axis,
 ylab=y axis)
 
 thefile - plotproblem.eps
 trellis.device(postscript, file=thefile, color=F,
 horizontal=FALSE, width=12, height=4,
 paper=special)
 print(theplot, split=c(1,1,3,1), more=T)
 print(theplot, split=c(2,1,3,1), more=T)
 print(theplot, split=c(3,1,3,1), more=F)
 dev.off()
 
 (Paper size and type specified in trellis.device
 call).
 
 
 Martin
 
 On 15/03/2006, at 2:44 PM, context grey wrote:
 
 
 
  --- Duncan Murdoch [EMAIL PROTECTED] wrote:
  The R graphics model is that the drawing surface
 is
  established first,
  then the things you draw are adjusted to fit in
 it.
  R won't change the
  shape of the display because you are drawing more
  things on it.
 
  Thanks, this comment clarifies things somewhat.
  Though I find it
  an odd design choice for R, since it seems to
 entail
  then that
  the user has figure out the aspect ratio of the
  resulting plot,
  something that R could easily keep track of as it
 is
  drawing.
 
  Here's example code, producing 3 scatterplots
  side-by-side
  (here reusing the same plot for simplicity).
 
  What's desired is that the individual scatterplots
  have the
  natural aspect, e.g. square, with the axis units
 being
  the
  same for X, Y.   And to do this while producing a
  correct
  bounding box in the .eps file.
 
  As it stands the example code produces a correct
  bounding box,
  but the scatterplots are too stretched to be
 usable.
  Inserting
  aspect=1/1 in the xyplot() seems to cause the
 bounding
  box
  to be  incorrect.
 
  
 
  library(lattice)
 
  rand1 - rnorm(50)
  rand2 - rnorm(50)
  theplot - xyplot(rand1 ~ rand2, xlab=x axis,
  ylab=y axis)
 
  thefile - plotproblem.eps
  trellis.device(postscript, file=thefile, color=F,
  horizontal=FALSE)
  print(theplot, split=c(1,1,3,1), more=T)
  print(theplot, split=c(2,1,3,1), more=T)
  print(theplot, split=c(3,1,3,1), more=F)
  dev.off()
 
  ---
 
 
  I don't think I understand exactly what you want
 to
  achieve; sample code
  that produces something close would be helpful
 (even
  if it comes out the
  wrong shape).
 
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Re: [R] concatenating factor from list

[Gabor Grothendieck]

Is this ok or is it what you are trying to avoid:

  factor(unlist(lapply(cutYield, as.character)))


On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote:
 Hi,

 I've run into a ridiculous problem I can't find any solutions for in the
 archives or help pages:

 data(barley)
 cutYield - with(barley, by(yield, variety, cut, breaks = c(0, 30, 60, 90)))

 As in this example, I'm using 'by' to return a factor for each level of
 another factor.  The problem is that 'by' returns a list of the factors,
 and I need all these factors concatenated.  No problem, I said:

 unlist(cutYield)

 which returns an 'integer' class object, so it's no longer a factor.  The
 same happens with:

 do.call(c, cutYield)

 I could recreate the factor from the integer level codes returned above,
 but this is not good because this means redoing the job already done by
 the function called in 'by', making the code more prone to errors.

 Thanks in advance for any pointers,

 --
 Sebastian

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[R] (no subject)

[Linda Lei]

Hi there,

 

Can R use principal component analysis (PCA) to do the clustering? Or
does PCA only be used to pick up the important variables?

 

Thank you!


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Re: [R] concatenating factor from list

[Sebastian Luque]

Gabor Grothendieck [EMAIL PROTECTED] wrote:

 Is this ok or is it what you are trying to avoid:

 factor(unlist(lapply(cutYield, as.character)))

Thank you Gabor.  The problem with that is what if some levels do not
appear in any member of cutYield?  In that case, the factor created above
would contain fewer levels than those present in every member of cutYield.

Cheers,

-- 
Sebastian P. Luque

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Re: [R] concatenating factor from list

[Gabor Grothendieck]

Since all components of cutYield have the same levels, one
could do this to ensure that all levels are represented:

factor(unlist(lapply(cutYield, as.character)), levels = levels(cutYield[[1]]))


On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote:
 Gabor Grothendieck [EMAIL PROTECTED] wrote:

  Is this ok or is it what you are trying to avoid:

  factor(unlist(lapply(cutYield, as.character)))

 Thank you Gabor.  The problem with that is what if some levels do not
 appear in any member of cutYield?  In that case, the factor created above
 would contain fewer levels than those present in every member of cutYield.

 Cheers,

 --
 Sebastian P. Luque

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Re: [R] Surv object in data.frame

[Heinz Tuechler]

At 09:23 15.03.2006 -0800, Thomas Lumley wrote:
On Wed, 15 Mar 2006, Heinz Tuechler wrote:

 Dear All,

 a Surv object I put in a data frame behaves somehow unexpected (see
example).
 If I do a Cox regression on the original Surv object it works. If I put it
 in a data.frame and do the regression on the data frame it does not work.
 Seemingly it has to do with the class attribute, because if I change the
 class attribute to let Surv appeare first, again it works.
 Is this known? Should I have found information on it?

Well, this is the sort of thing that happens when you use kludges like 
AsIs.

The problem is with [.AsIs
survobj[,1] is supposed to be a vector of times (that's what [.Surv 
returns), but [.AsIs sticks the original class attribute on to it.

 str(survobj[,1])
  num [1:5] 0 0 0 0 0
 str(I(survobj)[,1])
Classes 'AsIs', 'Surv'  num [1:5] 0 0 0 0 0

The solution is not to use I() -- there's no problem with putting survival 
objects in a data frame
 df.right-data.frame(survobj,group)
 df.right
   survobj group
1  (0,1 ] 1
2  (0,2+] 1
3  (0,3 ] 1
4  (0,4 ] 2
5  (0,5 ] 2


   -thomas

Thank you, Thomas. You are right, it works, but why then I find on the help
page for Surv{survival} the following sentence:
To include a survival object inside a data frame, use the I() function.
Surv objects are implemented as a matrix of 2 or 3 columns.

Heinz

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Re: [R] concatenating factor from list

[Sebastian Luque]

Sebastian Luque [EMAIL PROTECTED] wrote:

Gabor Grothendieck [EMAIL PROTECTED] wrote:

 Is this ok or is it what you are trying to avoid:

 factor(unlist(lapply(cutYield, as.character)))

 Thank you Gabor.  The problem with that is what if some levels do not
 appear in any member of cutYield?

This addition to your code takes care of that, although it's a bit
expensive:

factor(unlist(lapply(cutYield, as.character)),
   levels = unique(unlist(lapply(test, levels


Thanks!

-- 
Sebastian P. Luque

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Re: [R] Surv object in data.frame

[Heinz Tuechler]

At 11:59 15.03.2006 -0600, Robert Baer wrote:
This does work:
coxph(survobj~group, data=df.test[[1]]) # this works like your original

To get insight compare:
str(survobj)
str(df.test)
str(df.test[[1]])

Thank you for your answer. It seems to me that your solution only works, as
long as the original objects are in the search path. If I do rm(survobj,
group) then coxph(survobj~group, data=df.test[[1]]) does not work, because
it is not found in data=df.test[[1]]. As long as the objects are present also
data=data.frame(NULL) works.

## example data
starttime - rep(0,5)
stoptime  - 1:5
event - c(1,0,1,1,1)
group - c(1,1,1,2,2)
## Surv object
survobj   - Surv(starttime, stoptime, event)
## put Surv object in data.frame
df.test - data.frame(survobj=I(survobj), group)

## following Robert Baer
coxph(survobj~group, data=df.test[[1]]) # this works like your original

coxph(survobj~group, data=data.frame(NULL)) # give an empty data frame

## remove objects to verify that df.test is used
rm(starttime, stoptime, event, group, survobj)
coxph(survobj~group, data=df.test[[1]]) # now it doesn't work



Then note the 2nd sentence of the  following from ?coxph
Arguments:

 formula: a formula object, with the response on the left of a '~'
  operator, and the terms on the right.  The response must be a
  survival object as returned by the 'Surv' function.
I know that the response must be a survival object as returned by the
'Surv' function. The following sentence on the help page for Surv{survival}:
To include a survival object inside a data frame, use the I() function.
Surv objects are implemented as a matrix of 2 or 3 columns. gave me the
impression that a survival object retains its class if it is included via
I() in a data frame. I was in error.

Heinz






Robert W. Baer, Ph.D.
Associate Professor
Department of Physiology
A. T. Still University of Health Science
800 W. Jefferson St.
Kirksville, MO 63501-1497 USA


Dear All,

a Surv object I put in a data frame behaves somehow unexpected (see
example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly it has to do with the class attribute, because if I change the
class attribute to let Surv appeare first, again it works.
Is this known? Should I have found information on it?
Any comments?

Thanks

Heinz Tüchler

## example data
starttime - rep(0,5)
stoptime  - 1:5
event - c(1,0,1,1,1)
group - c(1,1,1,2,2)
## Surv object
survobj   - Surv(starttime, stoptime, event)
## Cox-regression
coxph(survobj~group) # this works
## put Surv object in data.frame
df.test - data.frame(survobj=I(survobj), group)
## Cox-regression on data.frame
coxph(survobj~group, data=df.test) # this does not work
attr(df.test$survobj, 'class') # survobject has class AsIs, Surv
attr(df.test$survobj, 'class') - c('Surv', 'AsIs') # put Surv first
attr(df.test$survobj, 'class') # survobject has class Surv, AsIs
coxph(survobj~group, data=df.test) # now it works

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Re: [R] Surv object in data.frame

[Thomas Lumley]

On Thu, 16 Mar 2006, Heinz Tuechler wrote:


 Thank you, Thomas. You are right, it works, but why then I find on the help
 page for Surv{survival} the following sentence:
 To include a survival object inside a data frame, use the I() function.
 Surv objects are implemented as a matrix of 2 or 3 columns.


That was probably true when it was written, but that was a long time ago. 
I will change it.

-thomas

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[R] Problem compiling R-Patched

[Kevin E. Thorpe]

I downloaded R-Patched today (to see if another problem I want to
ask about is still present or if its just me - as per the posting
guide).  I ran tools/rsync-recommended successfully.  I then ran
configure --enable-R-shlib successfully.  Then make stops with the
following error.

gcc -shared -L/usr/local/lib -o tools.so text.o init.o Rmd5.o md5.o 
-L../../../../lib -lR
mkdir -p -- ../../../../library/tools/libs
make[5]: Leaving directory `/home/src/R-patched/src/library/tools/src'
make[4]: Leaving directory `/home/src/R-patched/src/library/tools/src'
Error in file(datafile, wb) : unable to open connection
In addition: Warning message:
cannot open file '/usr/local/lib/R/library/tools/R/tools.rdb', reason 
'Permission denied'
Execution halted
make[3]: *** [all] Error 1
make[3]: Leaving directory `/home/src/R-patched/src/library/tools'
make[2]: *** [R] Error 1
make[2]: Leaving directory `/home/src/R-patched/src/library'
make[1]: *** [R] Error 1
make[1]: Leaving directory `/home/src/R-patched/src'
make: *** [R] Error 1

The tools.rdb file is on my system from a previous build

-rw-r--r--  1 root root 90486 2005-08-04 18:54 /usr/local/.../tools.rdb

As you can see from the output above, I am not trying to build in 
/usr/local/... at all so I am puzzled as to why anything is being done
in /usr/local/ before the make install.  Can anyone suggest what I 
should try?  I am compiling on SuSE Linux version 9.2.

Thank you.

Kevin

-- 
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED]  Tel: 416.946.8081  Fax: 416.946.3297

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Re: [R] Problem compiling R-Patched

[Liaw, Andy]

Could it be that you have the environment variable R_HOME (or something like
that) defined somewhere?  Just a wild guess...

Andy

From: Kevin E. Thorpe
 
 I downloaded R-Patched today (to see if another problem I 
 want to ask about is still present or if its just me - as per 
 the posting guide).  I ran tools/rsync-recommended 
 successfully.  I then ran configure --enable-R-shlib 
 successfully.  Then make stops with the following error.
 
 gcc -shared -L/usr/local/lib -o tools.so text.o init.o Rmd5.o md5.o 
 -L../../../../lib -lR
 mkdir -p -- ../../../../library/tools/libs
 make[5]: Leaving directory `/home/src/R-patched/src/library/tools/src'
 make[4]: Leaving directory `/home/src/R-patched/src/library/tools/src'
 Error in file(datafile, wb) : unable to open connection
 In addition: Warning message:
 cannot open file '/usr/local/lib/R/library/tools/R/tools.rdb', reason 
 'Permission denied'
 Execution halted
 make[3]: *** [all] Error 1
 make[3]: Leaving directory `/home/src/R-patched/src/library/tools'
 make[2]: *** [R] Error 1
 make[2]: Leaving directory `/home/src/R-patched/src/library'
 make[1]: *** [R] Error 1
 make[1]: Leaving directory `/home/src/R-patched/src'
 make: *** [R] Error 1
 
 The tools.rdb file is on my system from a previous build
 
 -rw-r--r--  1 root root 90486 2005-08-04 18:54 
 /usr/local/.../tools.rdb
 
 As you can see from the output above, I am not trying to build in 
 /usr/local/... at all so I am puzzled as to why anything is 
 being done in /usr/local/ before the make install.  Can 
 anyone suggest what I 
 should try?  I am compiling on SuSE Linux version 9.2.
 
 Thank you.
 
 Kevin
 
 -- 
 Kevin E. Thorpe
 Biostatistician/Trialist, Knowledge Translation Program 
 Assistant Professor, Department of Public Health Sciences 
 Faculty of Medicine, University of Toronto
 email: [EMAIL PROTECTED]  Tel: 416.946.8081  Fax: 416.946.3297
 
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Re: [R] Problem compiling R-Patched

[Kevin E. Thorpe]

Liaw, Andy wrote:
 Could it be that you have the environment variable R_HOME (or something like
 that) defined somewhere?  Just a wild guess...

No, R_HOME is not defined in the shell I was compiling in, but R_LIBS
was!  Brilliant wild guess.

Thanks,

Kevin

 Andy
 
 From: Kevin E. Thorpe
 
I downloaded R-Patched today (to see if another problem I 
want to ask about is still present or if its just me - as per 
the posting guide).  I ran tools/rsync-recommended 
successfully.  I then ran configure --enable-R-shlib 
successfully.  Then make stops with the following error.

gcc -shared -L/usr/local/lib -o tools.so text.o init.o Rmd5.o md5.o 
-L../../../../lib -lR
mkdir -p -- ../../../../library/tools/libs
make[5]: Leaving directory `/home/src/R-patched/src/library/tools/src'
make[4]: Leaving directory `/home/src/R-patched/src/library/tools/src'
Error in file(datafile, wb) : unable to open connection
In addition: Warning message:
cannot open file '/usr/local/lib/R/library/tools/R/tools.rdb', reason 
'Permission denied'
Execution halted
make[3]: *** [all] Error 1
make[3]: Leaving directory `/home/src/R-patched/src/library/tools'
make[2]: *** [R] Error 1
make[2]: Leaving directory `/home/src/R-patched/src/library'
make[1]: *** [R] Error 1
make[1]: Leaving directory `/home/src/R-patched/src'
make: *** [R] Error 1

The tools.rdb file is on my system from a previous build

-rw-r--r--  1 root root 90486 2005-08-04 18:54 
/usr/local/.../tools.rdb

As you can see from the output above, I am not trying to build in 
/usr/local/... at all so I am puzzled as to why anything is 
being done in /usr/local/ before the make install.  Can 
anyone suggest what I 
should try?  I am compiling on SuSE Linux version 9.2.

Thank you.

Kevin


-- 
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED]  Tel: 416.946.8081  Fax: 416.946.3297

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Re: [R] concatenating factor from list

[Gabor Grothendieck]

On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote:
 Sebastian Luque [EMAIL PROTECTED] wrote:

 Gabor Grothendieck [EMAIL PROTECTED] wrote:

  Is this ok or is it what you are trying to avoid:

  factor(unlist(lapply(cutYield, as.character)))

  Thank you Gabor.  The problem with that is what if some levels do not
  appear in any member of cutYield?

 This addition to your code takes care of that, although it's a bit
 expensive:

 factor(unlist(lapply(cutYield, as.character)),
   levels = unique(unlist(lapply(test, levels



In thinking about this a bit more here is another solution which
does not disassemble and reassemble the factors and so may
be more along the lines you were looking for:

do.call(rbind, lapply(cutYield, function(x) data.frame(x = x)))$x

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Re: [R] comparing AIC values of models with transformed, untransformed, and weighted variables

[Ben Bolker]

Patrick Baker patrick.baker at sci.monash.edu.au writes:
What I'd like to get some advice or insight on is whether 
 there is an appropriate way to rescale the AIC values to permit  
 comparisons across these models. Any suggestions would be very welcome. 
 Cheers, Patrick Baker
 
 

  Not a complete solution, but you could take a look at
the likelihoods associated with Box-Cox transformations
(e.g. Venables and Ripley MASS pp. 170-172).

  Ben Bolker

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[R] Having trouble with plot.survfit and fun=cloglog

[Kevin E. Thorpe]

I'm having trouble getting fun=cloglog to work with plot on
a survfit object.  Here are the data I used for the commands
that follow.

days status
2 0
2 0
5 1
9 0
14 1
16 0
16 0
17 0
29 1
30 0
37 1
37 0
39 1
44 0
44 0
58 0
60 1
67 1
68 1
82 1
82 1
86 0
86 0
89 1
93 0
97 1
100 0
100 0
100 0


  library(survival)
Loading required package: splines
  eg1.km - survfit(Surv(days,status),data=eg1)
  plot(eg1.km,mark.time=FALSE,conf.int=FALSE)  # Works
  plot(eg1.km,mark.time=FALSE,conf.int=FALSE,fun=cumhaz)  # Works
  plot(eg1.km,mark.time=FALSE,conf.int=FALSE,fun=cloglog)  # Error
Error in rep.default(2, n2 - 1) : invalid number of copies in rep()
In addition: Warning message:
2 x values = 0 omitted from logarithmic plot in: xy.coords(x, y, 
xlabel, ylabel
, log)

The axes are set and drawn up but nothing else is plotted.

  plot(eg1.km,mark.time=FALSE,conf.int=FALSE,fun=cumhaz,log=xy) # OK
Warning messages:
1: 2 x values = 0 omitted from logarithmic plot in: xy.coords(x, y, 
xlabel, yla
bel, log)
2: 1 y value = 0 omitted from logarithmic plot in: xy.coords(x, y, 
xlabel, ylab
el, log)

This does display the right plot but I am confused about how the xaxis
is computed.  The first tick mark is labeled with 100, the next with
200 right up to 600, but the entire plot is drawn between 0 and 100.

Please elighten me as to what I'm not understanding about fun=cloglog
and log scales.

  R.version
  _
platform i686-pc-linux-gnu
arch i686
os   linux-gnu
system   i686, linux-gnu
status   Patched
major2
minor2.1
year 2006
month03
day  13
svn rev  37540
language R

The survival package is version 2.21

Thank you,

Kevin

-- 
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto
email: [EMAIL PROTECTED]  Tel: 416.946.8081  Fax: 416.946.3297

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