date:20090917


On 09/17/2009 07:30 AM, _ wrote:


Hi all,
when using .jinit() I get the message .jinit() : Cannot create Java
virtual machine (-1).


You probably also need to set JAVA_HOME to that location, and have the 
bin directory (the one that contains java.exe) in your PATH as well.


Usually, support for rJava goes in this mailing list :
http://mailman.rz.uni-augsburg.de/mailman/listinfo/stats-rosuda-devel

Romain


I set the classpath variable to jre and to the jdk but nothing works.
  Sys.getenv(CLASSPATH)
CLASSPATH
C:\\Programme\\Java\\jre6\\bin
  Sys.getenv(CLASSPATH)
CLASSPATH
C:\\Programme\\Java\\jdk1.6.0_13

Java is correct installed because other Java-Applications are running.

I would be nice if someone could help me.

Thanks !




--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/yw8E : New R package : sos
|- http://tr.im/y8y0 : search the graph gallery from R
`- http://tr.im/y8wY : new R package : ant

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[R] latex code in R - convert to pdf

2009-09-17 Thread Martin Batholdy


hi,


is it possible to convert latex code to pdf in R (like a latex-program  
would do it)?

Is there a package that comes with this capabilities?


My problem is that I want to generate tables automatically -
and I can't use a latex editor at that computer ...



Besides latex ... are there good ways to generate tables in R?




thanks for any suggestions!

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Re: [R] How to do rotation for polygon?

2009-09-17 Thread Hemavathi Ramulu

hi everyone,
Its really helped, it almost as what I wanted.

Thanks alot.

On Sat, Sep 12, 2009 at 4:30 AM, Greg Snow greg.s...@imail.org wrote:

  Does this do what you want?



 library(TeachingDemos)

 ms.pent - function(ang=0,...) {

 theta - seq(ang, length.out=6, by=2*pi/5)

 cbind( cumsum(cos(theta)/2), cumsum(sin(theta)/2) )

 }



 par(xpd=NA)

 my.symbols( rep(1,5), rep(1,5), ms.pent, ang=seq(0, by=2*pi/5,
 length.out=5),

 add=FALSE, col=2:6 )





 --

 Gregory (Greg) L. Snow Ph.D.

 Statistical Data Center

 Intermountain Healthcare

 greg.s...@imail.org

 801.408.8111



 *From:* Hemavathi Ramulu [mailto:hema.ram...@gmail.com]
 *Sent:* Thursday, September 10, 2009 2:44 AM
 *To:* Greg Snow
 *Cc:* r-help@r-project.org
 *Subject:* Re: [R] How to do rotation for polygon?



 Hi everyone,



 I still couldn't get the diagram as I mentioned before. I try Grey and
 Milton suggestion but

 it confusing.

 I hope anyone helped me.



 Thanks in advance.



 Regards,

 Hema.

 On Thu, Sep 3, 2009 at 11:39 PM, Greg Snow greg.s...@imail.org wrote:

 The my.symbols and ms.polygon functions in the TeachingDemos package may
 help.

 --
 Gregory (Greg) L. Snow Ph.D.
 Statistical Data Center
 Intermountain Healthcare
 greg.s...@imail.org
 801.408.8111



  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
  project.org] On Behalf Of Hemavathi Ramulu
  Sent: Wednesday, September 02, 2009 11:05 PM
  To: r-help@r-project.org
  Subject: [R] How to do rotation for polygon?
 
  Hi everyone,
  I have coding for repeating pentagon as below:
 
  plot(0:11,type=n)
  for (i in 1:10 )polygon(rep(c(4,5,7,8,6)), i*c(.5,.3,.3,.5,.7), bor=2)
 
  which are increasing vertically.
 
  Now, I want to know how to rotate the pentagon, so that I will get
  pattern
  like flower.
  Basicly, repeating pentagon in circle.
 
  Thanks alot for helping me to solve this problem.
  --
  Hemavathi
 

[[alternative HTML version deleted]]
 
  __
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  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide http://www.R-project.org/posting-
  guide.html
  and provide commented, minimal, self-contained, reproducible code.




 --
 Hemavathi Ramulu




-- 
Hemavathi Ramulu

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[R] R functions with array arguments

2009-09-17 Thread Maurizio Paolillo

Dear R users,
I'm trying to implement a self-defined function with multiple  
arguments, one of which is an array, but I find that the result is a  
single value instead of an array.
This is the example I'm working on:


# define integration limit vector
  Mabslim - c(-17.95, -16.65, -17.27, -17.62, -16.76, -17.07, -17.02)

# Define function
  schech-function(x,alpha,xstar) (10^(0.4*(alpha+1)*(xstar- 
x))*exp(-10^(0.4*(xstar-x

# Define new function integrating the previous one
  my_gamma-function(alpha,xstar,xlim,xmax)  
integrate(schech,xmax,xlim,alpha,xstar)$value

  my_gamma(-1,-21,Mabslim,-27)
[1] 2.487746

Note that 'Mabslim' is used as an upper integration limit within the  
function. I tried to use sapply() but it looks to me as if this  
can be used only if the array is the first argument of the function.
I'm a beginner to R, apologies if I'm just using the wrong approach.
Any suggestion on how to solve this?
Thanks,
Maurizio


[[alternative HTML version deleted]]

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Re: [R] latex code in R - convert to pdf

2009-09-17 Thread Philipp Pagel

 
 is it possible to convert latex code to pdf in R (like a
 latex-program would do it)?
 Is there a package that comes with this capabilities?
 
 
 My problem is that I want to generate tables automatically -
 and I can't use a latex editor at that computer ...
 
 
 Besides latex ... are there good ways to generate tables in R?

Have a look at Sweave and xtable - I think that's what you want.

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

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

[R] Turning points in a series

2009-09-17 Thread ogbos okike

Good morning once more. My problem of yesterday has been addressed. Having
learned a few tricks from that, I wish to ask another question in connection
with that. My data is a cosmic ray data consisting of dates and counts. When
I plot a graph of counts versus dates, the resultant signal shows a number
of maximum and minimum points. These minimum points (turning points) are of
interest to me. Reading these dates and counts off from the plot is
difficult as I am dealing with a large data. I have been looking at
turnpoints function in pastecs library but have not been able to figure out
the appropriate commands that one can use to find the minima/maxima (turning
points) or pits/peaks in a series. My data is of the form shown below where
y stands for year, m month, d day and finally count. Is there a way I could
find these minima together with the dates they occurred?
I would be indebted to those of you who will show me the way out of these
problem.
Thank you.
Best regards
Ogbos


y  m   d  count
93 02 07 3974.6
93 02 08 3976.7
93 02 09 3955.2
93 02 10 3955.0
93 02 11 3971.8
93 02 12 3972.8
93 02 13 3961.0
93 02 14 3972.8
93 02 15 4008.0
93 02 16 4004.2
93 02 17 3981.2
93 02 18 3996.8
93 02 19 4028.2
93 02 20 4029.5
93 02 21 3953.4
93 02 22 3857.3
93 02 23 3848.3
93 02 24 3869.8
93 02 25 3898.1
93 02 26 3920.5
93 02 27 3936.7
93 02 28 3931.9

[[alternative HTML version deleted]]

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

Re: [R] latex code in R - convert to pdf

2009-09-17 Thread cls59




Martin Batholdy wrote:
 
 is it possible to convert latex code to pdf in R (like a latex-program  
 would do it)?
 Is there a package that comes with this capabilities?
 

Unfortunately you're out of luck if you're seeking a direct path from LaTeX
code generated in R to pdf without passing through a LaTeX compiler.
Re-implementing the pdfTeX compiler in R would be a monumental undertaking
as the TeX macro-expansion language is pretty hairy-- low-level TeX can make
the worst Perl screen-vomit look tame by comparison.

And that's just TeX-- throw in a pile of LaTeX macro packages and the
difficulty shoots up another order of magnitude or two.


Martin Batholdy wrote:
 
 My problem is that I want to generate tables automatically -
 and I can't use a latex editor at that computer ...
 
 Besides latex ... are there good ways to generate tables in R?
 

There's the xtable package which can perform some automagical formatting of
R objects to LaTeX code. If you don't have access to LaTeX at a workstation,
you could try using xtable's HTML output mode-- then use Word or OpenOffice
to open the HTML file and copy the table.

There are also a few utilities out there that can perform a conversion from
TeX to HTML-- they might be worth Googling if xtable's HTML output isn't
working for you.

Good luck!

-Charlie

-
Charlie Sharpsteen
Undergraduate
Environmental Resources Engineering
Humboldt State University
-- 
View this message in context: 
http://www.nabble.com/latex-code-in-R--%3E-convert-to-pdf-tp25486430p25487067.html
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[R] geoR, variofit

2009-09-17 Thread Sascha Bellaire

Hello All!

I calculate a variogram using the function variog (package geoR)  
afterwards I use variofit to fit a spherical model (see code below).  
Now I just changed the units of the variable (in this case MPa to kPa  
just a factor of 1000). If I do so, I get a different fit and  
therefore different ranges etc. Why? The semi-variance is of course 6  
orders of magnitude higher but the values are the same. So from my  
point of view it should be the same fit. Maybe there is a reason for  
this I do not get.

Hopefully someone could reply to this question.
Thanks, Sascha Bellaire!

Parts of code, just multiplying wl1$PSI with 1000 changed it.

#Calculate the extent of the current sampling design

locations - cbind(wl1$X,wl1$Y)

extent - max(dist(locations))

max.dist - extent/2



data1 - cbind(wl1$X,wl1$Y,wl1$PSI*1000)

geodata1 - as.geodata(data1, coords.col = 1:2, data.col = 3)



# Calculate the sample variogram

vario_var1 - variog(geodata1,

  uvec = nlags,

  trend = trend,

  option = bin,

  estimator.type = modulus,

  max.dist = max.dist,

  pairs.min = 10.,

  direction = omnidirectional,

  messages = TRUE)



# Fit a model

variofit_var1 - variofit(vario_var1,

   cov.model = variotype,

   fix.nugget = FALSE,

   max.dist = vario_var1$max.dist,

   messages = TRUE)








___
Sascha Bellaire
WSL Institute for Snow and Avalanche Research SLF
Formation of Alpine Natural Hazards
Flüelastrasse 11
7260 Davos Dorf, Switzerland

Tel.: +41 81 4170 292   
Fax: +41 81 4170 110

www.slf.ch






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[R] Quadradic constraint in optimization of linear program

2009-09-17 Thread pragathichi


Can someone suggest which package is required for optimization of linear
program with quadradic constraint.

Eg max x1+x2+x3
subject to
x1+x2+x3=10
x3^2 = 0
-- 
View this message in context: 
http://www.nabble.com/Quadradic-constraint-in-optimization-of-linear-program-tp25487057p25487057.html
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Re: [R] latex code in R - convert to pdf

2009-09-17 Thread baptiste auguie

Hi,

for basic tables (e.g. display a data.frame without fancy formatting),
you could try the textplot() function from the gplots package, or this
rough function for Grid graphics,

source(http://gridextra.googlecode.com/svn/trunk/R/tableGrob.r;)
# install.packages(gridextra, repos=http://R-Forge.R-project.org;)

 tc  = textConnection(
      carat   VeryLongWordIndeed color clarity depth
 14513  1.35 Ideal     J     VS2  61.4
 28685  0.30  Good     G    VVS1  64.0
 50368  0.75 Ideal     F     SI2  59.2)
 d = read.table(tc,head=T)
 close(tc)
 grid.newpage()
 grid.table(d)


HTH,

baptiste



2009/9/17 Martin Batholdy batho...@googlemail.com

 hi,


 is it possible to convert latex code to pdf in R (like a latex-program would 
 do it)?
 Is there a package that comes with this capabilities?


 My problem is that I want to generate tables automatically -
 and I can't use a latex editor at that computer ...



 Besides latex ... are there good ways to generate tables in R?




 thanks for any suggestions!

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

__
R-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] latex code in R - convert to pdf

2009-09-17 Thread Philipp Pagel

On Thu, Sep 17, 2009 at 10:08:57AM +0200, Philipp Pagel wrote:
  
  is it possible to convert latex code to pdf in R (like a
  latex-program would do it)?
  Is there a package that comes with this capabilities?
  
  
  My problem is that I want to generate tables automatically -
  and I can't use a latex editor at that computer ...
  
  
  Besides latex ... are there good ways to generate tables in R?
 
 Have a look at Sweave and xtable - I think that's what you want.

Charlies post made me aware that by latex editor you may mean that
there is no LaTeX installation on your machine. In that case Sweave
and xtable will obviously be of little use. If you have Openoffice on
that computer package odfWeave may be the solution. If openoffice is
not available, either, maybe package HTMLUtils would be another option
(I haven't used it so far, so I may be wrong here).

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

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Re: [R] latex code in R - convert to pdf

2009-09-17 Thread David Hajage

Hello,

You could try :
 - odfWeave : to generate odf (open office...) report
 - R2HTML and hwriter : to generate html report
 - ascii : to generate asciidoc http://www.methods.co.nz/asciidoc/ report
and then convert it to html, xml, pdf and more.

2009/9/17 Philipp Pagel p.pa...@wzw.tum.de

 On Thu, Sep 17, 2009 at 10:08:57AM +0200, Philipp Pagel wrote:
  
   is it possible to convert latex code to pdf in R (like a
   latex-program would do it)?
   Is there a package that comes with this capabilities?
  
  
   My problem is that I want to generate tables automatically -
   and I can't use a latex editor at that computer ...
  
  
   Besides latex ... are there good ways to generate tables in R?
 
  Have a look at Sweave and xtable - I think that's what you want.

 Charlies post made me aware that by latex editor you may mean that
 there is no LaTeX installation on your machine. In that case Sweave
 and xtable will obviously be of little use. If you have Openoffice on
 that computer package odfWeave may be the solution. If openoffice is
 not available, either, maybe package HTMLUtils would be another option
 (I haven't used it so far, so I may be wrong here).

 cu
Philipp

 --
 Dr. Philipp Pagel
 Lehrstuhl für Genomorientierte Bioinformatik
 Technische Universität München
 Wissenschaftszentrum Weihenstephan
 85350 Freising, Germany
 http://webclu.bio.wzw.tum.de/~pagel/http://webclu.bio.wzw.tum.de/%7Epagel/

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 http://www.R-project.org/posting-guide.html
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Re: [R] R functions with array arguments

2009-09-17 Thread baptiste auguie

Try this,

sapply(Mabslim , my_gamma, alpha=-1, xstar = -21, xmax = -27)

or wrap it with ?Vectorize,

vmy_gamma = Vectorize(my_gamma, vectorize.args = xlim)
vmy_gamma(alpha=-1, xstar = -21, xlim= Mabslim, xmax = -27)


HTH,

baptiste


2009/9/17 Maurizio Paolillo paoli...@na.infn.it:
 Dear R users,
 I'm trying to implement a self-defined function with multiple
 arguments, one of which is an array, but I find that the result is a
 single value instead of an array.
 This is the example I'm working on:


 # define integration limit vector
   Mabslim - c(-17.95, -16.65, -17.27, -17.62, -16.76, -17.07, -17.02)

 # Define function
   schech-function(x,alpha,xstar) (10^(0.4*(alpha+1)*(xstar-
 x))*exp(-10^(0.4*(xstar-x

 # Define new function integrating the previous one
   my_gamma-function(alpha,xstar,xlim,xmax)
 integrate(schech,xmax,xlim,alpha,xstar)$value

   my_gamma(-1,-21,Mabslim,-27)
 [1] 2.487746

 Note that 'Mabslim' is used as an upper integration limit within the
 function. I tried to use sapply() but it looks to me as if this
 can be used only if the array is the first argument of the function.
 I'm a beginner to R, apologies if I'm just using the wrong approach.
 Any suggestion on how to solve this?
 Thanks,
 Maurizio


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Re: [R] Turning points in a series

2009-09-17 Thread Ted Harding

On 17-Sep-09 08:10:47, ogbos okike wrote:
 Good morning once more. My problem of yesterday has been addressed.
 Having learned a few tricks from that, I wish to ask another question
 in connection with that. My data is a cosmic ray data consisting of
 dates and counts.
 When I plot a graph of counts versus dates, the resultant signal
 shows a number of maximum and minimum points. These minimum points
 (turning points) are of interest to me. Reading these dates and counts
 off from the plot is difficult as I am dealing with a large data.
 I have been looking at turnpoints function in pastecs library but have
 not been able to figure out the appropriate commands that one can use
 to find the minima/maxima (turning points) or pits/peaks in a series.
 My data is of the form shown below where y stands for year, m month,
 d day and finally count. Is there a way I could find these minima
 together with the dates they occurred?
 I would be indebted to those of you who will show me the way out of
 these problem.
 Thank you.
 Best regards
 Ogbos
 
 
 y  m   d  count
 93 02 07 3974.6
 93 02 08 3976.7
 93 02 09 3955.2
 93 02 10 3955.0
 93 02 11 3971.8
 93 02 12 3972.8
 93 02 13 3961.0
 93 02 14 3972.8
 93 02 15 4008.0
 93 02 16 4004.2
 93 02 17 3981.2
 93 02 18 3996.8
 93 02 19 4028.2
 93 02 20 4029.5
 93 02 21 3953.4
 93 02 22 3857.3
 93 02 23 3848.3
 93 02 24 3869.8
 93 02 25 3898.1
 93 02 26 3920.5
 93 02 27 3936.7
 93 02 28 3931.9

The following simple function TP() (for Turning Point) locates
the positions i where x[i] is greater than both of its immediate
neighbours (local maximum) or less than both of its neighbours
(local minimum).

  TP - function(x){
L - length(x)
which( ((x[1:(L-2)]x[2:(N-1)])(x[2:(L-1)]x[3:L]))
  |((x[1:(L-2)]x[2:(N-1)])(x[2:(L-1)]x[3:L])) ) + 1
  }

Applied to your series count above:

  TP(count)
  # [1]  2  4  6  7  9 11 14 17 21

If you assign these values to an index:

  ix - TP(count)
  rbind(d[ix],count[ix])

  # [1,]8.0   10   12.0   13   15   17.0   20.0   23.0   27.0
  # [2,] 3976.7 3955 3972.8 3961 4008 3981.2 4029.5 3848.3 3936.7

Of course, this is only a very simplistic view of turning point,
and will pick out everything which is a local minimum or maximum.

The above function can be extended (in a fairly obvious way) to
identify each position i where x[i] is greater than its neighbours
out to 2 on either side, or less than these neighbours; or more
generally out to k on either side.

A lot depends on how you want to interpret turning point. With
your count series, it might be that you were only interested
in identifying the relatively extreme turning points, such as
i=4 (maybe), i=9 (maybe), i=14, i=17, i=21(maybe).

Hoping this helps,
Ted.


E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 17-Sep-09   Time: 09:47:53
-- XFMail --

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Re: [R] How to extract data.frame columns using regex?

I think this is what you want:
 df - data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)
 grep('^x',names(df))
[1] 1 2 3

The returned indexes refer to the column positions, so you could do:
names(df)[grep('^x',names(df))]
[1] x1 x2 x3
or
df[,grep('^x',names(df))]
   x1 x2 x3
1   1  2  3
2   2  3  4
3   3  4  5
4   4  5  6
5   5  6  7
6   6  7  8
7   7  8  9
8   8  9 10
9   9 10 11
10 10 11 12
11 11 12 13

HTH

Schalk Heunis


On Thu, Sep 17, 2009 at 5:03 AM, Peng Yu pengyu...@gmail.com wrote:

 Hi,

 data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)

 I want to extract all the columns that with the name 'x?'. Is there a
 general way to do this in R?

 Regards,
 Peng

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[R] package documentation of S4 methods

2009-09-17 Thread El Gorgonzola


Hi,

I'm new to this mailing list and to R-programming so if the question is stupid 
I apologize.

I have to create a package, which includes an S4-class called BList. For 
objects of this class I implemented a method show, which displays the first 
15 data-lines of the object. I further implemented a method showall, which 
displays the whole data in the object. For the showall-method i first defined 
a generic method:

setGeneric(showall,
function(object){
out- standardGeneric(showall)
})

and then defined the method using

setMethod(showall, signature(object=BList),
function(object){
...
}

My problem: I don't know how to document the showall-method for the package. 
If I just add the method in the documentation 'BList-class.Rd' as an alias 
using:

\alias{showall,BList-method}

and then add it to the list of methods:

\section{Methods}{
  \describe{
\item{show}{\code{signature(object = BList)}: Short output}
\item{showall}{\code{signature(object = BList)}: Long output}
}
}

I get the following warning (when I check the package for installation):

* checking for missing documentation entries ... WARNING
Undocumented code objects:
  showall
All user-level objects in a package should have documentation entries.


I then used promptMethods() to build a seperate documentation-file 
'showall-methods.Rd'. If I then run the check I get the warning:

* checking Rd files ... WARNING
Rd files with duplicated alias 'showall,BList-method':
  BList-class.Rd showall-methods.Rd

If I then remove the alias line '\alias{showall,BList-method}' from either 
'BList-class.Rd' or 'showall-methods.Rd' I get the first warning (Undocumented 
code objects...). 

My question: how do you document methods of S4-classes in a package? It works 
fine for the method show but it does not work for showall, for which I had 
to define the generic function first. What can I do to fix this, or is it 
better to just define showall as a function and not as a method?

Thanks a lot,

Elton G.




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[R] JGR install and run question

2009-09-17 Thread David Young

Hello all,

I tried to install the GUI interface JGR for R yesterday on my Windows
Vista machine running R 2.9.2.

It did install and run but whenever I ran the package manager it
crashed.  It also did not appear to display all the available
packages.  For instance Hmisc and bayesm were not in the list.

Since I am new to R, and more so to JGR, I'd just like to know if JGR
is still supported and should run or if I'm wasting my time with it.
It had the nice feature of color coding the programs so that errors
were easier to spot, like in SAS.

Is anyone else successfully running JGR on R 2.9.2 with Vista?

Thanks in advance for any comments.

-- 
Best regards,

David Young
Marketing and Statistical Consultant
Madrid, Spain
+34 913 540 381
http://www.linkedin.com/in/europedavidyoung

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[R] JGR install and run question

2009-09-17 Thread David Young

Hello all,

I tried to install the GUI interface JGR for R yesterday on my Windows
Vista machine running R 2.9.2.

It did install and run but whenever I ran the package manager it
crashed.  It also did not appear to display all the available
packages.  For instance Hmisc and bayesm were not in the list.

Since I am new to R, and more so to JGR, I'd just like to know if JGR
is still supported and should run or if I'm wasting my time with it.
It had the nice feature of color coding the programs so that errors
were easier to spot, like in SAS.

Is anyone else successfully running JGR on R 2.9.2 with Vista?

Thanks in advance for any comments.

-- 
Best regards,

David Young
Marketing and Statistical Consultant
Madrid, Spain
+34 913 540 381
http://www.linkedin.com/in/europedavidyoung

__
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Re: [R] How to extract data.frame columns using regex?

2009-09-17 Thread Martin Maechler

 SH == Schalk Heunis schalk.heu...@enerweb.co.za
 on Thu, 17 Sep 2009 11:15:16 +0200 writes:

SH I think this is what you want:
 df - data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)
 grep('^x',names(df))
SH [1] 1 2 3

SH The returned indexes refer to the column positions, so you could do:
 names(df)[grep('^x',names(df))]
SH [1] x1 x2 x3

yes, or slightly more elegant and efficient

   grep('^x',names(df), value = TRUE)
  [1] x1 x2 x3



SH or
 df[,grep('^x',names(df))]
SH x1 x2 x3
SH 1   1  2  3
SH 2   2  3  4
SH 3   3  4  5
SH 4   4  5  6
SH 5   5  6  7
SH 6   6  7  8
SH 7   7  8  9
SH 8   8  9 10
SH 9   9 10 11
SH 10 10 11 12
SH 11 11 12 13

SH HTH

SH Schalk Heunis


SH On Thu, Sep 17, 2009 at 5:03 AM, Peng Yu pengyu...@gmail.com wrote:

 Hi,
 
 data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)
 
 I want to extract all the columns that with the name 'x?'. Is there a
 general way to do this in R?
 
 Regards,
 Peng
 
 __
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SH [[alternative HTML version deleted]]

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Re: [R] Turning points in a series

2009-09-17 Thread Philippe Grosjean


Hello,

I don't see what's wrong with turnpoints() from pastecs. It is easy to 
use, and provides additional information for each turnpoints, i.e., 
probability of occurrence against the null hypothesis that the series is 
purely random, and the number of bits of information associated with the 
points according to Kendall's information theory. See ?turnpoints.


Rewriting the function is a nice exercise, but a small explanation on 
how to use turnpoints() is much easier. So, nobody is able to tell that 
simply using:


 library(pastecs)
 turnpoints(dat$count)

does the job? Am I the only one interested by the extra information 
provided by turnpoints()?


Here is a more extensive example that shows also how to get date or 
counts associated to pits/peaks:


txt - 
y  m   d  count
93 02 07 3974.6
93 02 08 3976.7
93 02 09 3955.2
93 02 10 3955.0
93 02 11 3971.8
93 02 12 3972.8
93 02 13 3961.0
93 02 14 3972.8
93 02 15 4008.0
93 02 16 4004.2
93 02 17 3981.2
93 02 18 3996.8
93 02 19 4028.2
93 02 20 4029.5
93 02 21 3953.4
93 02 22 3857.3
93 02 23 3848.3
93 02 24 3869.8
93 02 25 3898.1
93 02 26 3920.5
93 02 27 3936.7
93 02 28 3931.9

con - textConnection(txt)
dat - read.table(con, header = TRUE)
close(con)
dat$date - as.Date(paste(dat$y, dat$m, dat$d), format = %y %m %d)

library(pastecs)
tp - turnpoints(dat$count)
tp
summary(tp)
# Indicate which turnpoints are significant (see ?turnpoints)
plot(tp, level = 0.05)
# Another plot
plot(dat$count, type = l)
lines(tp)

# Get counts for all turnpoints
allcounts - - dat$count[extract(tp, no.tp = FALSE, peak = TRUE, pit = 
TRUE)]


# Get dates for all turnpoints
alldates - dat$date[extract(tp, no.tp = FALSE, peak = TRUE, pit = TRUE)]
alldates
# Get dates for informative turnpoints (5%) only (see ?turnpoints)
alldates[tp$proba  0.05]
# Get dates for peaks only
dat$date[extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)]
# Etc...

Best,

Philippe Grosjean

..°}))
 ) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
 ) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
 ) ) ) ) )   Mons-Hainaut University, Belgium
( ( ( ( (
..

(Ted Harding) wrote:

On 17-Sep-09 08:10:47, ogbos okike wrote:

Good morning once more. My problem of yesterday has been addressed.
Having learned a few tricks from that, I wish to ask another question
in connection with that. My data is a cosmic ray data consisting of
dates and counts.
When I plot a graph of counts versus dates, the resultant signal
shows a number of maximum and minimum points. These minimum points
(turning points) are of interest to me. Reading these dates and counts
off from the plot is difficult as I am dealing with a large data.
I have been looking at turnpoints function in pastecs library but have
not been able to figure out the appropriate commands that one can use
to find the minima/maxima (turning points) or pits/peaks in a series.
My data is of the form shown below where y stands for year, m month,
d day and finally count. Is there a way I could find these minima
together with the dates they occurred?
I would be indebted to those of you who will show me the way out of
these problem.
Thank you.
Best regards
Ogbos


y  m   d  count
93 02 07 3974.6
93 02 08 3976.7
93 02 09 3955.2
93 02 10 3955.0
93 02 11 3971.8
93 02 12 3972.8
93 02 13 3961.0
93 02 14 3972.8
93 02 15 4008.0
93 02 16 4004.2
93 02 17 3981.2
93 02 18 3996.8
93 02 19 4028.2
93 02 20 4029.5
93 02 21 3953.4
93 02 22 3857.3
93 02 23 3848.3
93 02 24 3869.8
93 02 25 3898.1
93 02 26 3920.5
93 02 27 3936.7
93 02 28 3931.9


The following simple function TP() (for Turning Point) locates
the positions i where x[i] is greater than both of its immediate
neighbours (local maximum) or less than both of its neighbours
(local minimum).

  TP - function(x){
L - length(x)
which( ((x[1:(L-2)]x[2:(N-1)])(x[2:(L-1)]x[3:L]))
  |((x[1:(L-2)]x[2:(N-1)])(x[2:(L-1)]x[3:L])) ) + 1
  }

Applied to your series count above:

  TP(count)
  # [1]  2  4  6  7  9 11 14 17 21

If you assign these values to an index:

  ix - TP(count)
  rbind(d[ix],count[ix])

  # [1,]8.0   10   12.0   13   15   17.0   20.0   23.0   27.0
  # [2,] 3976.7 3955 3972.8 3961 4008 3981.2 4029.5 3848.3 3936.7

Of course, this is only a very simplistic view of turning point,
and will pick out everything which is a local minimum or maximum.

The above function can be extended (in a fairly obvious way) to
identify each position i where x[i] is greater than its neighbours
out to 2 on either side, or less than these neighbours; or more
generally out to k on either side.

A lot depends on how you want to interpret turning point. With
your count series, it might be that you were only interested
in identifying the relatively extreme turning points, such as
i=4 (maybe), i=9 (maybe), i=14, i=17, i=21(maybe).

Hoping this helps,
Ted.

Re: [R] How to extract data.frame columns using regex?


On 09/17/2009 12:04 PM, Martin Maechler wrote:



SH == Schalk Heunisschalk.heu...@enerweb.co.za
 on Thu, 17 Sep 2009 11:15:16 +0200 writes:


 SH  I think this is what you want:
   df- data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)
   grep('^x',names(df))
 SH  [1] 1 2 3

 SH  The returned indexes refer to the column positions, so you could do:
   names(df)[grep('^x',names(df))]
 SH  [1] x1 x2 x3

yes, or slightly more elegant and efficient

 grep('^x',names(df), value = TRUE)
   [1] x1 x2 x3


or, if you have the operators package:

 require( operators )
 names(df) %~|% ^x
[1] x1 x2 x3

... note that is has been declared confusing in this mailing list 
previously. http://article.gmane.org/gmane.comp.lang.r.general/154749





 SH  or
   df[,grep('^x',names(df))]
 SH  x1 x2 x3
 SH  1   1  2  3
 SH  2   2  3  4
 SH  3   3  4  5
 SH  4   4  5  6
 SH  5   5  6  7
 SH  6   6  7  8
 SH  7   7  8  9
 SH  8   8  9 10
 SH  9   9 10 11
 SH  10 10 11 12
 SH  11 11 12 13

 SH  HTH

 SH  Schalk Heunis


 SH  On Thu, Sep 17, 2009 at 5:03 AM, Peng Yupengyu...@gmail.com  wrote:

   Hi,
 
   data.frame(x1=1:11,x2=2:12,x3=3:13,y=4:14)
 
   I want to extract all the columns that with the name 'x?'. Is there a
   general way to do this in R?
 
   Regards,
   Peng




--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/yw8E : New R package : sos
|- http://tr.im/y8y0 : search the graph gallery from R
`- http://tr.im/y8wY : new R package : ant

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[R] Problems with the commands FUNCTION and DERIV to build a polynomial

2009-09-17 Thread Noela Sánchez

Hi all,

I need to automate a process in order to prepare a  a big loop in the future
but I have a problem with the *command function*

First I fit a model with lm


model1-lm(data2[,2]~data2[,1]+I(data2[,1]^2)+I(data2[,1]^3)+I(data2[,1]^4))

I extract the coefficients to build the polynomial.

coef-as.matrix(model1$coefficients)

In the next step I need to define the polynomial to derive it. If I write
the coefficients manually (writing the numbers by hand) the deriv command
works fine!

bb-deriv(~2847.22015 -463.06063*x+ 25.43829*x^2 -0.17896*x^3, namevec=x,
function.arg=40)

I would like to automate this step by being able to extract the coefficients
from the linear model and adding them into the polynomial (and not write
them by hand)!

 But if I build the polynomial with the function(x) command calling the *
coef* values, the numeric values are not interpreted, the command function
does not read properly the coefficients from the linear model.

  fun-function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3

 fun

function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3

How can i avoid to write the values of the coefficients by hand??

I need to do this many many times, this is the reason i need to be able to
automate the process and then build a loop to repeat it many times with
different outputs of a linear model!

Somebody can help me?


-- 
Noela
Grupo de Recursos Marinos y Pesquerías
Universidad de A Coruña

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Re: [R] JGR install and run question

2009-09-17 Thread Gabor Grothendieck

I've had similar problems. See:
http://article.gmane.org/gmane.comp.lang.r.rosuda.devel/747

Although a bit out of date, there is a page on GUIs for R at
these two links:
http://www.sciviews.org/_rgui/
http://wiki.r-project.org/rwiki/doku.php?id=guis:guis

On Thu, Sep 17, 2009 at 5:58 AM, David Young dyo...@telefonica.net wrote:
 Hello all,

 I tried to install the GUI interface JGR for R yesterday on my Windows
 Vista machine running R 2.9.2.

 It did install and run but whenever I ran the package manager it
 crashed.  It also did not appear to display all the available
 packages.  For instance Hmisc and bayesm were not in the list.

 Since I am new to R, and more so to JGR, I'd just like to know if JGR
 is still supported and should run or if I'm wasting my time with it.
 It had the nice feature of color coding the programs so that errors
 were easier to spot, like in SAS.

 Is anyone else successfully running JGR on R 2.9.2 with Vista?

 Thanks in advance for any comments.

 --
 Best regards,

 David Young
 Marketing and Statistical Consultant
 Madrid, Spain
 +34 913 540 381
 http://www.linkedin.com/in/europedavidyoung

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


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[R] RCurl and Google Scholar's EndNote references

2009-09-17 Thread Jarno Tuimala

Hi!

I've performed a Google Scholar Search using a query, let's say Frank
Harrell, and parsed the links to the EndNote references from the resulting
HTML code. Now I'd like to download all the references automatically. For
this, I have tried to use RCurl, but I can't seem to get it working: I
always get error code 403 Forbidden from the web server.

Initially I tried to do this without using cookies:

library(RCurl)
getURL(
http://scholar.google.fi/scholar.enw?q=info:U6Gfb4QPVFMJ:scholar.google.com/output=citationhl=fioe=ASCIIct=citationcd=0
)

or

getURLContent(
http://scholar.google.fi/scholar.enw?q=info:U6Gfb4QPVFMJ:scholar.google.com/output=citationhl=fioe=ASCIIct=citationcd=0
)
Error: Forbidden
and then with cookies:

 getURL(
http://scholar.google.fi/scholar.enw?q=info:U6Gfb4QPVFMJ:scholar.google.com/output=citationhl=fioe=ASCIIct=citationcd=0;,
.opts=list(cookiejar=cookiejar.txt))

But they both consistently fail the same way. What am I doing wrong?

sessionInfo()
R version 2.9.0 (2009-04-17)
i386-pc-mingw32
locale:
LC_COLLATE=Finnish_Finland.1252;LC_CTYPE=Finnish_Finland.1252;LC_MONETARY=Finnish_Finland.1252;LC_NUMERIC=C;LC_TIME=Finnish_Finland.1252
attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base
other attached packages:
[1] RCurl_0.98-1   bitops_1.0-4.1

Thanks!
Jarno

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Re: [R] Filling Empty Column with String in read.table

2009-09-17 Thread Henrique Dallazuanna

Try this after read the data file:

dat[is.na(dat[,5]),5] - '-'

On Thu, Sep 17, 2009 at 3:33 AM, Gundala Viswanath gunda...@gmail.com wrote:
 I have a data file that looks like this.

 __DATA__
 D7KAR5Z02F447V  176 G   0.22
 D7KAR5Z02J3WLG  94  A   1.05    29
 D7KAR5Z02F4K6L  198 a   0.13
 D7KAR5Z02J4SYO  67  C   0.95    28
 D7KAR5Z02J4SYO  83  C   1.01    29
 D7KAR5Z02J4SYO  97  T   0.13
 D7KAR5Z02J4SYO  166 A   0.94    27


 I want the rows where the last column has no-entry to be filled
 with -.

 Is there a way to do it with read.table?
 I don't find any option to do it with read.table.

 This only fill the empty 5th column with blank.

 dat - read.table(myfile.txt,na.strings=-,header=FALSE,fill=T);

 G.V.

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

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Re: [R] Quadradic constraint in optimization of linear program

2009-09-17 Thread Gabor Grothendieck

The 2nd constraint holds trivially and the 1st constraint implies that the
maximum is 10.  The solution is not unique and any values of the
variables satisfying the 1st constraint are optimum.  No software
needed.

On Thu, Sep 17, 2009 at 4:19 AM, pragathichi pragathichitr...@tcs.com wrote:

 Can someone suggest which package is required for optimization of linear
 program with quadradic constraint.

 Eg max x1+x2+x3
 subject to
 x1+x2+x3=10
 x3^2 = 0
 --
 View this message in context: 
 http://www.nabble.com/Quadradic-constraint-in-optimization-of-linear-program-tp25487057p25487057.html
 Sent from the R help mailing list archive at Nabble.com.

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[R] What is the time complexity of 'match()'?

Hi,

Suppose 'x' is a vector of length n and 'y' is a vector of length m, I
am wondering what the time complexity of 'match(x,y)' is. Is it n
times m?

Regards,
Peng

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[R] heatmap.2() problems with re-ordering of rows and columns

2009-09-17 Thread bioinformatics_guy


I have a file of the following form

-11 -10 -9 -8
-10 -9   -8   NA
-9   -7NA NA
-8NA  NA NA

So basically a NxN matrix of log scores.  I want to get a heatmap of these
log scores but I'm having a problem. 

I'm using the following code

library(gplots)
data=read.table(filein.txt,header=FALSE)
mat=as.matrix(data)
heatmap.2(mat,dendrogram=c(none))

But on the picture, it rearranges all my row,columns.  I want it the y axis
to be labeled from [10,-10] and the x axis to be the same [-10,10] so that
the bottom left cell is -10,-10 and the top right cell is 10,10 -- which is
the way the matrix is laid out.  Why is it rearranging my cells?

-- 
View this message in context: 
http://www.nabble.com/heatmap.2%28%29-problems-with-re-ordering-of-rows-and-columns-tp25490249p25490249.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] poisson lognormal regression

2009-09-17 Thread Ruben Roa Ureta

 Dear all,
 I want to directly write the loglike function of poisson- lognormal
 regression and then estimate the parameters. the response (Y)is count
 and I have two explanatory variable (x1:nominal) and (x2:continiuouse).
 which package in R can be used ?.How can input the data and write the
 loglike function and then optimize.
 regards  

I suggest you define a function with the process model and the likelihood
and then use nlm to minimiza the negative log-likelihood. See the examples
at the end of the nlm help.
HTH
Ruben

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[R] Error message in Design library

2009-09-17 Thread Richardson, Patrick

This was working a few weeks ago, but perhaps the package has been updated 
since then.

model.1 - lrm(response ~ p_value, data=c_abl_oncogene_1_RTK)

When I run the following command . . . .
prediction.1 - predict(model.1, type=c(fitted))
I get the following error message. . . .
Error in predictDesign(object, ..., type = lp, se.fit = FALSE) :
  could not find function Varcov

It seems like a required function of predict may be missing in the Design 
package (although I doubt Professor Harrell would have overlooked this). 
Perhaps its my own stupidity with something. Any ideas what could be causing 
this?  I'm sorry I didn't post a reproducable example, but it a lot of code.

Any help would be appreciated.

Best regards,

Patrick


R version 2.9.2 (2009-08-24)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United 
States.1252;LC_MONETARY=English_United 
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

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

other attached packages:
 [1] Design_2.2-0ROCR_1.0-2  gplots_2.7.1caTools_1.9 
bitops_1.0-4.1  gdata_2.6.1 gtools_2.6.1svSocket_0.9-43 svMisc_0.9.48   
TinnR_1.0.3 R2HTML_1.59-1
[12] Hmisc_3.7-0 MASS_7.2-48 survival_2.35-7

loaded via a namespace (and not attached):
[1] cluster_1.12.0  lattice_0.17-25 tools_2.9.2

This email message, including any attachments, is for th...{{dropped:9}}

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Re: [R] latex code in R - convert to pdf



On Sep 17, 2009, at 4:31 AM, Philipp Pagel wrote:


On Thu, Sep 17, 2009 at 10:08:57AM +0200, Philipp Pagel wrote:


is it possible to convert latex code to pdf in R (like a
latex-program would do it)?
Is there a package that comes with this capabilities?


My problem is that I want to generate tables automatically -
and I can't use a latex editor at that computer ...


Besides latex ... are there good ways to generate tables in R?


Have a look at Sweave and xtable - I think that's what you want.


Charlies post made me aware that by latex editor you may mean that
there is no LaTeX installation on your machine. In that case Sweave
and xtable will obviously be of little use. If you have Openoffice on
that computer package odfWeave may be the solution. If openoffice is
not available, either, maybe package HTMLUtils would be another option
(I haven't used it so far, so I may be wrong here).


The Sweave help page has a set of examples  followed by a line that is  
commented out. If one runs the examples and then this line:

tools::texi2dvi(Sweave-test-1.tex, pdf=TRUE)

One gets a) an error message but also 4 files two of which are the  
expected pdf files and two if which are eps file type. I'm not sure  
what the error message is telling me but it is not correct to my eyes  
that it could be called failure:


Error in tools::texi2dvi(Sweave-test-1.tex, pdf = TRUE) :
  Running 'texi2dvi' on 'Sweave-test-1.tex' failed.



David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] How to colour the tip labels in a phylogenetic tree

2009-09-17 Thread Graham Etherington


Hi,
Using Ape, I have constructed an object of class phylo, using the 
method 'nj' (lets call the object 'tree_ja').
I also have a given subset of 'tree_ja' in a vector (lets call the 
vector 'subspecies').
What I want to do, is construct a nj tree - plot(tree_ja) - but have the 
species in vector 'subspecies' shown as red at the tips of the tree.


The closest I've come is this:
Given that 'tree_ja$tip.label' provides the following:
 [1] 1_T1  2_T1  3_T1  4_T1  5_T1  6_T1
 [7] 7_T1  8_T1  9_T1 10_T1 11_T1 12_T1

and that my 'subspecies' vector is:
subspecies - c(1_T1, 2_T1, 3_T1, 4_T1, 6_T1)
which can also be written as:
subspecies - c(tree_ja$tip.label[1:4], tree_ja$tip.label[5])

I can construct a method which gives me the following statement:

plot(tree_ja, tip.col = c('red', 'red', 'red', 'red', 'black', 'red', 
'black', 'black', 'black', 'black', 'black', 'black'))


But this doesn't work (at least not on my full dataset, which as 118 
tips - reduced to 12 here for brevity) and I'm SURE there must be a 
better way of doing it.


Could anyone help me with this?
Many thanks,
Graham

--
Dr. Graham Etherington
Post-doctoral Bioinformatician,
Department of Computational and Systems Biology
John Innes Centre
Norwich Research Park
Colney
Norwich
NR4 7UH
UK

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Re: [R] inline error message

2009-09-17 Thread Uwe Ligges




_ wrote:

Hi all,
I installed the library inline to my default R-enviroment 
(c:\Programme\R.. )
downloaded and installed RTools from 
http://www.murdoch-sutherland.com/Rtools/

to c:\Rtools.
Path variable is set right (with respect to order) but I get still the 
error message

from my R
Error in compileCode(f, code, language, verbose) :
 Compilation ERROR, function(s)/method(s) not created!

Where is my failure ?


If you tell us what library (I guess package?) you mean and how you 
traied to install it and what command gave the error message (including 
a complete logfile perhaps), we may be able to help.


Uwe Ligges



Thank you !

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Re: [R] Error message in Design library



On Sep 17, 2009, at 8:58 AM, Richardson, Patrick wrote:

This was working a few weeks ago, but perhaps the package has been  
updated since then.


model.1 - lrm(response ~ p_value, data=c_abl_oncogene_1_RTK)

When I run the following command . . . .
prediction.1 - predict(model.1, type=c(fitted))
I get the following error message. . . .
Error in predictDesign(object, ..., type = lp, se.fit = FALSE) :
 could not find function Varcov

It seems like a required function of predict may be missing in the  
Design package (although I doubt Professor Harrell would have  
overlooked this). Perhaps its my own stupidity with something. Any  
ideas what could be causing this?  I'm sorry I didn't post a  
reproducable example, but it a lot of code.




No, Prof Harrell did not overlook this. Instead he rewrote the whole  
package which is now rms. The missing Varcov function was made  
available in an earlier posting to r-help just a couple of days ago.


https://stat.ethz.ch/pipermail/r-help/2009-September/211306.html

--
David



Any help would be appreciated.

Best regards,

Patrick


R version 2.9.2 (2009-08-24)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States. 
1252;LC_MONETARY=English_United States. 
1252;LC_NUMERIC=C;LC_TIME=English_United States.1252


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


other attached packages:
[1] Design_2.2-0ROCR_1.0-2  gplots_2.7.1caTools_1.9  
bitops_1.0-4.1  gdata_2.6.1 gtools_2.6.1svSocket_0.9-43  
svMisc_0.9.48   TinnR_1.0.3 R2HTML_1.59-1

[12] Hmisc_3.7-0 MASS_7.2-48 survival_2.35-7

loaded via a namespace (and not attached):
[1] cluster_1.12.0  lattice_0.17-25 tools_2.9.2

This email message, including any attachments, is for th...{{dropped: 
9}}


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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] heatmap.2() problems with re-ordering of rows and columns

Hi bioinformatics_guy
I think you are looking for the image function:
image(mat)

The heatmap.2 function does hierarchical clustering on rows and columns and
then orders the rows and columns according to the results of the clustering.
Image simply plots the matrix.

HTH
Schalk Heunis

On Thu, Sep 17, 2009 at 2:23 PM, bioinformatics_guy wwwhite...@gmail.comwrote:

I have a file of the following form

-11 -10 -9 -8
-10 -9 -8 NA
-9 -7NA NA
-8NA NA NA

So basically a NxN matrix of log scores. I want to get a heatmap of these
log scores but I'm having a problem.

I'm using the following code

library(gplots)
data=read.table(filein.txt,header=FALSE)
mat=as.matrix(data)
heatmap.2(mat,dendrogram=c(none))

But on the picture, it rearranges all my row,columns. I want it the y axis
to be labeled from [10,-10] and the x axis to be the same [-10,10] so that
the bottom left cell is -10,-10 and the top right cell is 10,10 -- which is
the way the matrix is laid out. Why is it rearranging my cells?

--
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Re: [R] What is the time complexity of 'match()'?

2009-09-17 Thread Tal Galili

try running it on a grid within system.time()
And the answer will be revealed.

Tal





On Thu, Sep 17, 2009 at 3:15 PM, Peng Yu pengyu...@gmail.com wrote:

 Hi,

 Suppose 'x' is a vector of length n and 'y' is a vector of length m, I
 am wondering what the time complexity of 'match(x,y)' is. Is it n
 times m?

 Regards,
 Peng

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-- 
--


My contact information:
Tal Galili
Phone number: 972-50-3373767
FaceBook: Tal Galili
My Blogs:
http://www.r-statistics.com/
http://www.talgalili.com
http://www.biostatistics.co.il

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Re: [R] What is the time complexity of 'match()'?

2009-09-17 Thread Thomas Lumley


On Thu, 17 Sep 2009, Peng Yu wrote:


Hi,

Suppose 'x' is a vector of length n and 'y' is a vector of length m, I
am wondering what the time complexity of 'match(x,y)' is. Is it n
times m?


match() hashes the second argument then does hash lookups for the first 
argument (the source is in src/main/unique.c), so its time complexity will be 
close to m+n.

Even just sorting the second argument and doing binary search would allow 
(m+n)log(m) complexity -- it really would take a brute force and ignorance 
approach to give mn time complexity, and match() is sometimes used for quite 
large m and n.

 -thomas


Thomas Lumley   Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle

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Re: [R] heatmap.2() problems with re-ordering of rows and columns

2009-09-17 Thread bioinformatics_guy


Schalk, 

Thats a great function! The only question is, is it as flexible as
heatmap.2?

I figured out how to get it from rearranging the rows and columns but I
can't figure out how to label the rows and columns? What I like about the
heatmap.2 is that it gives a grid and histogram of the heatmap which is
nice.

I'm trying to 

library(gplots)
data=read.table(filein.txt,header=FALSE)
mat=as.matrix(data)
heatmap.2(mat,dendrogram=c(none),trace=c(none),Rowv=F,Colv=F)

which works but labels my columns V1 thru V21 and rows 1-21.  Id like them
to be different an under the man pages for heatmap.2 it states:

 # Row/Column Labeling
  margins = c(5, 5),
  ColSideColors,
  RowSideColors,
  cexRow = 0.2 + 1/log10(nr),
  cexCol = 0.2 + 1/log10(nc),
  labRow = NULL,
  labCol = NULL,

So I'm adding cexRow=30-3(nr) (as I want it to decrement by 3 for each row
but R spits back an error say nr is not recognized.  I was looking at other
help pages but couldn't find out how to lable the axis the way I wanted to. 

Schalk Heunis wrote:
 
 Hi bioinformatics_guy
 I think you are looking for the image function:
 image(mat)
 
 The heatmap.2 function does hierarchical clustering on rows and columns
 and
 then orders the rows and columns according to the results of the
 clustering.
  Image simply plots the matrix.
 
 HTH
 Schalk Heunis
 
 
 On Thu, Sep 17, 2009 at 2:23 PM, bioinformatics_guy
 wwwhite...@gmail.comwrote:
 

 I have a file of the following form

 -11 -10 -9 -8
 -10 -9   -8   NA
 -9   -7NA NA
 -8NA  NA NA

 So basically a NxN matrix of log scores.  I want to get a heatmap of
 these
 log scores but I'm having a problem.

 I'm using the following code

 library(gplots)
 data=read.table(filein.txt,header=FALSE)
 mat=as.matrix(data)
 heatmap.2(mat,dendrogram=c(none))

 But on the picture, it rearranges all my row,columns.  I want it the y
 axis
 to be labeled from [10,-10] and the x axis to be the same [-10,10] so
 that
 the bottom left cell is -10,-10 and the top right cell is 10,10 -- which
 is
 the way the matrix is laid out.  Why is it rearranging my cells?

 --
 View this message in context:
 http://www.nabble.com/heatmap.2%28%29-problems-with-re-ordering-of-rows-and-columns-tp25490249p25490249.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Problems with the commands FUNCTION and DERIV to build a polynomial

2009-09-17 Thread Thomas Lumley


On Thu, 17 Sep 2009, [ISO-8859-1] Noela Sánchez wrote:


Hi all,

I need to automate a process in order to prepare a  a big loop in the future
but I have a problem with the *command function*

First I fit a model with lm




model1-lm(data2[,2]~data2[,1]+I(data2[,1]^2)+I(data2[,1]^3)+I(data2[,1]^4))

I extract the coefficients to build the polynomial.

coef-as.matrix(model1$coefficients)

In the next step I need to define the polynomial to derive it. If I write
the coefficients manually (writing the numbers by hand) the deriv command
works fine!

bb-deriv(~2847.22015 -463.06063*x+ 25.43829*x^2 -0.17896*x^3, namevec=x,
function.arg=40)

I would like to automate this step by being able to extract the coefficients
from the linear model and adding them into the polynomial (and not write
them by hand)!

But if I build the polynomial with the function(x) command calling the *
coef* values, the numeric values are not interpreted, the command function
does not read properly the coefficients from the linear model.


 fun-function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3



fun


function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3

How can i avoid to write the values of the coefficients by hand??



As is often the case there are two parts to this answer: how to do what you 
asked, and how to do what you want.

You can use bquote() to get the value of coef[1] into the expression, so
  deriv(bquote(.(coef[1])+.(coef[2])*x+.(coef[3])*x^2+.(coef[4])*x^3),x,
function.arg=x)
will give the derivative.  Bill Venables would tell you to use Horner's Rule 
and write the polynomial as
   coef[1]+x*(coef[2]+x*(coef[3]+x*coef[4]))
to get better speed and numerical stability, but the same trick still works.

However, you really don't need deriv() to differentiate a polynomial, and so 
you can use ordinary lexical scope rather than substitution, for a much tidier 
answer

derivative - function(coef){
function(x) coef[2]+x*(2*coef[3]+x*3*coef[4])
}

-thomas


Thomas Lumley   Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle

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Re: [R] Fastest Way to Divide Elements of Row With Its RowSum

2009-09-17 Thread Thomas Lumley


On Thu, 17 Sep 2009, William Revelle wrote:


At 2:40 PM +0900 9/17/09, Gundala Viswanath wrote:

I have a data frame (dat). What I want to do is for each row,
divide each row  with the sum of its row.

The number of row can be large  1million.
Is there a faster way than doing it this way?

datnorm;
for (rw in 1:length(dat)) {
tmp - dat[rw,]/sum(dat[rw,])
datnorm - rbind(datnorm, tmp);
}


- G.V.



datnorm - dat/rowSums(dat)

this will be faster if dat is a matrix rather than a data.frame.



Even if it's a data frame and he needs a data frame answer it might be faster 
to do
  mat-as.matrix(dat)
  matnorm-mat/rowSums(mat)
  datnorm-as.data.frame(dat)

The other advantage, apart from speed, of doing it with dat/rowSums(dat) rather 
than the loop is he gets the right answer. The loop goes from 1 to the number 
of columns if dat is a data frame and 1 to the number of entries if dat is a 
matrix, not from 1 to the number of rows.

 -thomas

Thomas Lumley   Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle

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[R] How to generate a matrix where each row (or column) is the same vector?

Hi,

I can use the following code to generate a matrix, each column of
which is 'x'. But I have to specify '5' twice in the second command. I
am wondering if there is a better way to do it.

 x=1:10
 matrix(rep(x,5),nc=5)
 t(matrix(rep(x,5),nc=5))

Regards,
Peng

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Re: [R] How to generate a matrix where each row (or column) is the same vector?


On 09/17/2009 04:02 PM, Peng Yu wrote:


Hi,

I can use the following code to generate a matrix, each column of
which is 'x'. But I have to specify '5' twice in the second command. I
am wondering if there is a better way to do it.


x=1:10
matrix(rep(x,5),nc=5)
t(matrix(rep(x,5),nc=5))


Regards,
Peng


This works for me:

 do.call( cbind, rep( list( x ), 5 ) )
 do.call( rbind, rep( list( x ), 5 ) )

Romain

--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/yw8E : New R package : sos
|- http://tr.im/y8y0 : search the graph gallery from R
`- http://tr.im/y8wY : new R package : ant

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Re: [R] How to generate a matrix where each row (or column) is the same vector?


On 09/17/2009 04:02 PM, Peng Yu wrote:


Hi,

I can use the following code to generate a matrix, each column of
which is 'x'. But I have to specify '5' twice in the second command. I
am wondering if there is a better way to do it.


x=1:10
matrix(rep(x,5),nc=5)
t(matrix(rep(x,5),nc=5))


Regards,
Peng


Or this ;

 matrix(rep(x,5),nr=length(x))

Romain

--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://tr.im/yw8E : New R package : sos
|- http://tr.im/y8y0 : search the graph gallery from R
`- http://tr.im/y8wY : new R package : ant

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Re: [R] How to generate a matrix where each row (or column) is the same vector?

2009-09-17 Thread Henrique Dallazuanna

Or:

replicate(5, 1:10)

On Thu, Sep 17, 2009 at 11:02 AM, Peng Yu pengyu...@gmail.com wrote:
 Hi,

 I can use the following code to generate a matrix, each column of
 which is 'x'. But I have to specify '5' twice in the second command. I
 am wondering if there is a better way to do it.

 x=1:10
 matrix(rep(x,5),nc=5)
 t(matrix(rep(x,5),nc=5))

 Regards,
 Peng

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




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

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[R] Why S4 method is not visible from another package?

2009-09-17 Thread Gábor Csárdi

Dear All,

maybe this is something obvious, I seem to be incapable of
understanding how S4 works.

So, in package 'A' I defined a summary method for my class:

setMethod(summary, signature(object=ListHyperGResult),
  function(object, pvalue=pvalueCutoff(object), categorySize=NULL) {
 whatever
  })

ListHyperGResult has a subclass, GOListHyperGResult:

setClass(GOListHyperGResult,
 representation=representation(conditional=logical),
 contains=ListHyperGResult,
 prototype=prototype(testname=GO))

The summary method is exported in the NAMESPACE:

exportMethods(summary)

Package 'B' depends on package 'A', this is stated in the
'DESCRIPTION' file. If I call 'summary' on a 'GOListHyperGResult' in
package B, then the default summary method is called instead of the
correct one, despite that I have

Browse[1] showMethods(summary)
Function: summary (package base)
object=AnnDbBimap
object=ANY
object=Bimap
object=DBIObject
object=HyperGResultBase
object=KEGGHyperGResult
object=LinearMResultBase
object=ListHyperGResult
object=PFAMHyperGResult
object=SQLiteConnection
object=SQLiteDriver
object=SQLiteResult

Browse[1] class(gos[[1]])
[1] GOListHyperGResult

But I still get:

Browse[1] is(gos[[1]], ListHyperGResult)
[1] TRUE
Browse[1] summary(gos[[1]])
Length  Class   Mode
 1 GOListHyperGResult S4

What am I doing wrong?

 sessionInfo()
R version 2.9.0 (2009-04-17)
x86_64-redhat-linux-gnu

locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C

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

other attached packages:
 [1] hgu95av2.db_2.2.12  ALL_1.4.4   ExpressionView_0.1
 [4] caTools_1.9 bitops_1.0-4.1  KEGG.db_2.2.5
 [7] GO.db_2.2.5 RSQLite_0.7-1   DBI_0.2-4
[10] eisa_0.1genefilter_1.24.2   Category_2.10.0
[13] AnnotationDbi_1.6.0 Biobase_2.4.1   isa2_0.1

loaded via a namespace (and not attached):
[1] annotate_1.22.0 graph_1.22.2GSEABase_1.6.0  RBGL_1.20.0
[5] splines_2.9.0   survival_2.35-4 tools_2.9.0 XML_2.6-0
[9] xtable_1.5-5


Thanks,
Gabor

-- 
Gabor Csardi gabor.csa...@unil.ch UNIL DGM

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Re: [R] heatmap.2() problems with re-ordering of rows and columns

Try placing the column names into labCol and the rownames into labRow
e.g.heatmap.2(mat,dendrogram=c(none),
Rowv=F, Colv=F, labRow = seq(-7.5,7.5,by=5), labCol=seq(-3,3,by=2))

Schalk Heunis

On Thu, Sep 17, 2009 at 3:53 PM, bioinformatics_guy wwwhite...@gmail.comwrote:


 Schalk,

 Thats a great function! The only question is, is it as flexible as
 heatmap.2?

 I figured out how to get it from rearranging the rows and columns but I
 can't figure out how to label the rows and columns? What I like about the
 heatmap.2 is that it gives a grid and histogram of the heatmap which is
 nice.

 I'm trying to

 library(gplots)
 data=read.table(filein.txt,header=FALSE)
 mat=as.matrix(data)
 heatmap.2(mat,dendrogram=c(none),trace=c(none),Rowv=F,Colv=F)

 which works but labels my columns V1 thru V21 and rows 1-21.  Id like them
 to be different an under the man pages for heatmap.2 it states:

 # Row/Column Labeling
  margins = c(5, 5),
  ColSideColors,
  RowSideColors,
  cexRow = 0.2 + 1/log10(nr),
  cexCol = 0.2 + 1/log10(nc),
  labRow = NULL,
  labCol = NULL,

 So I'm adding cexRow=30-3(nr) (as I want it to decrement by 3 for each row
 but R spits back an error say nr is not recognized.  I was looking at other
 help pages but couldn't find out how to lable the axis the way I wanted to.

 Schalk Heunis wrote:
 
  Hi bioinformatics_guy
  I think you are looking for the image function:
  image(mat)
 
  The heatmap.2 function does hierarchical clustering on rows and columns
  and
  then orders the rows and columns according to the results of the
  clustering.
   Image simply plots the matrix.
 
  HTH
  Schalk Heunis
 
 
  On Thu, Sep 17, 2009 at 2:23 PM, bioinformatics_guy
  wwwhite...@gmail.comwrote:
 
 
  I have a file of the following form
 
  -11 -10 -9 -8
  -10 -9   -8   NA
  -9   -7NA NA
  -8NA  NA NA
 
  So basically a NxN matrix of log scores.  I want to get a heatmap of
  these
  log scores but I'm having a problem.
 
  I'm using the following code
 
  library(gplots)
  data=read.table(filein.txt,header=FALSE)
  mat=as.matrix(data)
  heatmap.2(mat,dendrogram=c(none))
 
  But on the picture, it rearranges all my row,columns.  I want it the y
  axis
  to be labeled from [10,-10] and the x axis to be the same [-10,10] so
  that
  the bottom left cell is -10,-10 and the top right cell is 10,10 -- which
  is
  the way the matrix is laid out.  Why is it rearranging my cells?
 
  --
  View this message in context:
 
 http://www.nabble.com/heatmap.2%28%29-problems-with-re-ordering-of-rows-and-columns-tp25490249p25490249.html
  Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] comparing random forests and classification trees

2009-09-17 Thread jamesmcc

Greetings tree and forest coders-

I'm interested in comparing randomforests and regression tree/ bagging tree
models. I'd like to propose a basis for doing this, get feedback, and
document this here. I kept it in this thread since that makes sense.

In this case I think it's appropriate to compare the R^2 values as one basic
measure. I'm actually going to compare mean error (ME), mean absolute error
(MAE), root mean squared error (RMSE) as well. This means that I need
estimates from each approach so that I can form residuals. **As I see it,
the important details are in how to set up the models so that I have
comparable estimates, particularly in how the trees/forests are trained and
evaluated.**

For regression/bagging trees, the typical approach for my application is 100
runs of 10-fold CV. In each run all the values are estimated in an
out-of-the-bag sense; each fold is estimated while it is withheld from
fitting, thus fit is not inflated. The estimates are then averaged over the
100 runs at each point to get an average simulation and this is used to
calculate residuals and the measures mentioned above. Somewhat more
specifically, the steps are: I fit a model, I prune it via inspection, I
loop 100 times on xpred.rpart(model,xval=10,cp=cp at bottom of cptable from
pruned fit) to generate the 100 runs (bagging is thus performed while
holding the cp criteria fixed?), I average these pointwise, I calculate the
desired stats/quantities for comparison to other models.

For randomForests, I would want to fit the model in a similar way, ie 100
runs of 10-fold CV. I think the 10-fold part is clear, the 100 runs, maybe
less so. To get 10-fold OOB estimates, I set replace=FALSE,
sampsize=.9*nrow(x). Then I get a randomForest with $predicted being the
average OOB estimates over all trees for which each point was OOB. I would
assume that each tree is constructed with a different 10-fold partitioning
of the data set. Thus the number of runs is really more like the number of
trees constructed. If i wanted to be really thorough, I could fit 100 random
forests and get the $predicted for each and then average these pointwise.
But that seems like over kill; isnt that the lesson of plot.randomForest
that as the # of trees goes up the error converges to some limit. (from what
i've seen).

Thus, my primary concern is in the amount of data used for training and
cross validating the model in an out-of-bag sense; can i meaningfully
compare 10-fold oob estimates sing xpred.rpart to a random forest fit using
90% of the data as sampsize?

Of secondary concern is the number of bagging trees versus then number of
trees in the random forest. As long as the average estimate error is nearing
some limit with the number of bagging trees I'm using, I think this is all
that matters. So this is more of methodological difference to be retained,
similar to differences in pruning under bagging and random forests, though I
should probably specify the node sizes to be similar for each.

Am I overlooking anything of grave consequence?

Any and all thoughts are welcome. If you are aware of any comparisons of
rpart and randomForests in the literature for any field (for regression) of
which I am ignorant, I would appreciate the tip. I have read over Newer
Classification and Regression Tree Techniques: Bagging and Random Forests
for Ecological Prediction by Prasad, Iverson, and Liaw. I may have missed
it, but I did not see discussion of maintaining consistency in the way the
models were trained, though it is a very nice paper overall and contained
many interesting approaches and points.

Thanks in advance,

James

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[R] What does model.matrix() return?

Hi,

I don't understand what the meaning of the following lines returned by
model.matrix(). Can somebody help me understand it? What can they be
used for?

attr(,assign)
[1] 0 1 2 2
attr(,contrasts)
attr(,contrasts)$A
[1] contr.treatment

attr(,contrasts)$B
[1] contr.treatment

Regards,
Peng

 a=2
 b=3
 n=4
 A = rep(sapply(1:a,function(x){rep(x,n)}),b)
 B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), 
 function(x){rep(x,a)}))
 Y = A + B + rnorm(a*b*n)
 fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
 afit=aov(Y ~ A + B,fr)
 model.matrix(afit)
   (Intercept) A2 B2 B3
11  0  0  0
21  0  0  0
31  0  0  0
41  0  0  0
51  1  0  0
61  1  0  0
71  1  0  0
81  1  0  0
91  0  1  0
10   1  0  1  0
11   1  0  1  0
12   1  0  1  0
13   1  1  1  0
14   1  1  1  0
15   1  1  1  0
16   1  1  1  0
17   1  0  0  1
18   1  0  0  1
19   1  0  0  1
20   1  0  0  1
21   1  1  0  1
22   1  1  0  1
23   1  1  0  1
24   1  1  0  1
attr(,assign)
[1] 0 1 2 2
attr(,contrasts)
attr(,contrasts)$A
[1] contr.treatment

attr(,contrasts)$B
[1] contr.treatment

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Re: [R] comparing random forests and classification trees

2009-09-17 Thread Frank E Harrell Jr

The validate.rpart function in the rms package will handle the rpart 
part of this.  It makes sure that the tree is re-built from scratch for 
each re-sample.  It estimates MSE and Somers' Dxy (twice (ROC area -.5)).


Frank

jamesmcc wrote:

Greetings tree and forest coders-

I'm interested in comparing randomforests and regression tree/ bagging tree
models. I'd like to propose a basis for doing this, get feedback, and
document this here. I kept it in this thread since that makes sense.

In this case I think it's appropriate to compare the R^2 values as one basic
measure. I'm actually going to compare mean error (ME), mean absolute error
(MAE), root mean squared error (RMSE) as well. This means that I need
estimates from each approach so that I can form residuals. **As I see it,
the important details are in how to set up the models so that I have
comparable estimates, particularly in how the trees/forests are trained and
evaluated.**

For regression/bagging trees, the typical approach for my application is 100
runs of 10-fold CV. In each run all the values are estimated in an
out-of-the-bag sense; each fold is estimated while it is withheld from
fitting, thus fit is not inflated. The estimates are then averaged over the
100 runs at each point to get an average simulation and this is used to
calculate residuals and the measures mentioned above. Somewhat more
specifically, the steps are: I fit a model, I prune it via inspection, I
loop 100 times on xpred.rpart(model,xval=10,cp=cp at bottom of cptable from
pruned fit) to generate the 100 runs (bagging is thus performed while
holding the cp criteria fixed?), I average these pointwise, I calculate the
desired stats/quantities for comparison to other models.

For randomForests, I would want to fit the model in a similar way, ie 100
runs of 10-fold CV. I think the 10-fold part is clear, the 100 runs, maybe
less so. To get 10-fold OOB estimates, I set replace=FALSE,
sampsize=.9*nrow(x). Then I get a randomForest with $predicted being the
average OOB estimates over all trees for which each point was OOB. I would
assume that each tree is constructed with a different 10-fold partitioning
of the data set. Thus the number of runs is really more like the number of
trees constructed. If i wanted to be really thorough, I could fit 100 random
forests and get the $predicted for each and then average these pointwise.
But that seems like over kill; isnt that the lesson of plot.randomForest
that as the # of trees goes up the error converges to some limit. (from what
i've seen). 


Thus, my primary concern is in the amount of data used for training and
cross validating the model in an out-of-bag sense; can i meaningfully
compare 10-fold oob estimates sing xpred.rpart to a random forest fit using
90% of the data as sampsize? 


Of secondary concern is the number of bagging trees versus then number of
trees in the random forest. As long as the average estimate error is nearing
some limit with the number of bagging trees I'm using, I think this is all
that matters. So this is more of methodological difference to be retained,
similar to differences in pruning under bagging and random forests, though I
should probably specify the node sizes to be similar for each.

Am I overlooking anything of grave consequence?

Any and all thoughts are welcome. If you are aware of any comparisons of
rpart and randomForests in the literature for any field (for regression) of
which I am ignorant, I would appreciate the tip. I have read over Newer
Classification and Regression Tree Techniques: Bagging and Random Forests
for Ecological Prediction by Prasad, Iverson, and Liaw. I may have missed
it, but I did not see discussion of maintaining consistency in the way the
models were trained, though it is a very nice paper overall and contained
many interesting approaches and points. 

Thanks in advance, 


James




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

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[R] How to do a PCA with ordered variables?

2009-09-17 Thread P.Branco


Hi,

I want to do a pca using a set of variables which are ordered. I have used
the dudi.mix method from the ade4 package, but when I do the $index it shows
me that R has considered my variables as quantitative.

What should I do?
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[R] SVM

2009-09-17 Thread Samuel Okoye

Hello,

I have 12 sample each sample has got 1000 observation, i.e I have a matrix X 
with 1000 rows and 12 columns!
 
m - svm(t(X))
p - predict (m)

Can anyone tell me how to use svmtrain() in R!

Many Yhanks,
Samuel



  
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[R] hint required code in Tinn-R

2009-09-17 Thread hentati.sundb


Dear Tinn-R users,
I've a basic question: I don't understand how to configure Tinn-R to hint
about the required elements in a certain code, e.g. when typing mean(   
a box would pop-up just above the code informing that (x, trim = 0, na.rm =
FALSE, ...) is the required info which must be given... 

Anyone who could help me with finding this function?

Best Regards, 
Jonas

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[R] How to separate a function by 2 probabilities

2009-09-17 Thread Marcio Resende


Good Mourning,
I have a function to generate a matrix as I show part of it;

g[j,i]-if (gen[j,i]==0) al1[i,1]+al1[i,1] else ...

However i would like that this function occurred with a probability P and
that another function (another formula to generate g matrix) with
probability P-1

That´s it, if P is .7, i would like that in 70% of the times (for random i
and j) the matrix g was generated according to the formula above and in 30%
of the times with a different formula which i did not write

could anyone help me?
Thank you very much
Márcio

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[R] Problems with the commands FUNCTION and DERIV to build a polynomial

2009-09-17 Thread Noela Sánchez

Hi all,

I need to automate a process in order to prepare a  a big loop in the future
but I have a problem with the *command function*

First I fit a model with lm


model1-lm(data2[,2]~data2[,1]+I(data2[,1]^2)+I(data2[,1]^3)+I(data2[,1]^4))

I extract the coefficients to build the polynomial.

coef-as.matrix(model1$coefficients)

In the next step I need to define the polynomial to derive it. If I write
the coefficients manually (writing the numbers by hand) the deriv command
works fine!

bb-deriv(~2847.22015 -463.06063*x+ 25.43829*x^2 -0.17896*x^3, namevec=x,
function.arg=40)

I would like to automate this step by being able to extract the coefficients
from the linear model and adding them into the polynomial (and not write
them by hand)!

 But if I build the polynomial with the function(x) command calling the *
coef* values, the numeric values are not interpreted, the command function
does not read properly the coefficients from the linear model.

  fun-function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3

 fun

function(x) coef[1]+coef[2]*x+coef[3]*x^2+coef[4]*x^3

How can i avoid to write the values of the coefficients by hand??

I need to do this many many times, this is the reason i need to be able to
automate the process and then build a loop to repeat it many times with
different outputs of a linear model!

Somebody can help me?


-- 
Noela
Grupo de Recursos Marinos y Pesquerías
Universidad de A Coruña

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[R] Simple as.Date question dealing with a timezone offset

2009-09-17 Thread esawdust

I've been trying to understand the as.Date functionality and I have a date
and time stamp field that looks like this:

Tue Sep 15 09:22:09 -0600 2009

and I need to turn it into an R Date object for analysis.

Simple date conversions I have down, no problem:

adate = c(7/30/1959)
as.Date(adate,%m/%d/%Y)
[1] 1959-07-30

But when it comes to the type of date/time string format I have above, I
can't figure out a format string that will work.

The timezone offset is one part that causes problems. Building up to a
working format string for the full time stamp string, I can make it as far
as:

adate = c(Tue Sep 15 09:22:09 -0600 2009)
as.Date(adate,format=%a %b %d %H:%M:%S)
[1] 2009-09-15

(apparently year defaults to current year when it's not specified in the
format string). Because the Year comes after the timezone offset, I have to
deal with the timezone offset in the format string.

But when I get to the timezone offset value I can't use %z or %Z because
those are output only

as.Date(adate,format=%a %b %d %H:%M:%S %z %Y)
[1] NA

I'm close, but can't incorporate the timezone offset field in the date/time
stamp string.

What am I missing? I suppose one workaround is to split the date/time
string into its component parts, reassemble it into a string as.Date can
deal with, but that seems to defeat one of the purposes of as.Date's format
capability.

Any advice for how to translate a Tue Sep 15 09:22:09 -0600 2009 into an R
Date object?

Landon

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Re: [R] SVM

2009-09-17 Thread Steve Lianoglou


Hi,

On Sep 17, 2009, at 7:39 AM, Samuel Okoye wrote:


Hello,

I have 12 sample each sample has got 1000 observation, i.e I have a  
matrix X with 1000 rows and 12 columns!


m - svm(t(X))
p - predict (m)

Can anyone tell me how to use svmtrain() in R!


I guess you're using the svm in the e1071 package?
What's svmtrain?

The call to svm trains the svm.
The call to predict uses it on new data, but you need to give it new  
data to predict on. You have:


p - predict(m)

What exactly do you want your model to do? Predict on what?

Please see the code in the Examples section of ?svm .. it's pretty  
straight forward. Let us know what problems you're having  
understanding those examples and we can try to offer some insight.


-steve

--
Steve Lianoglou
Graduate Student: Computational Systems Biology
  |  Memorial Sloan-Kettering Cancer Center
  |  Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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[R] stableFit

2009-09-17 Thread tzygmund mcfarlane

A quick question about stableFit() in the fBasics package. Is it
possible to constrain the gamma and delta parameters and only estimate
the alpha and beta parameters? I tried:

##
set.seed(1953)
r = rstable(n = 1000, alpha = 1.9, beta = 0.3)
stableFit(r, gamma=1, delta=0, type=c(q, mle), doplot=TRUE, trace=TRUE)
##

but that seems to estimate the gamma and delta as well.

Thanks

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[R] r-inferno.pdf with detailed table of contents and bookmarks

Hi,

I don't find a r-inferno.pdf that has detailed table of contents and
bookmarks. If it is possible, can somebody help generated one and post
it on line?

Regards,
Peng

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Re: [R] How to separate a function by 2 probabilities

Marcio
Define two functions, e.g.
f1-function(i,j) i+j
f2-function(i,j) i-j
then call them based on the probability e.g. 0.7
f - if(runif(1)0.7) f1 else f2
f(1,1)

or more compact
(if(runif(1)0.7) f1 else f2)(1,1)

HTH
Schalk Heunis


On Thu, Sep 17, 2009 at 5:10 PM, Marcio Resende mresende...@yahoo.com.brwrote:


 Good Mourning,
 I have a function to generate a matrix as I show part of it;

 g[j,i]-if (gen[j,i]==0) al1[i,1]+al1[i,1] else ...

 However i would like that this function occurred with a probability P and
 that another function (another formula to generate g matrix) with
 probability P-1

 That´s it, if P is .7, i would like that in 70% of the times (for random i
 and j) the matrix g was generated according to the formula above and in 30%
 of the times with a different formula which i did not write

 could anyone help me?
 Thank you very much
 Márcio

 --
 View this message in context:
 http://www.nabble.com/How-to-separate-a-function-by-2-probabilities-tp25491943p25491943.html
 Sent from the R help mailing list archive at Nabble.com.

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[R] lpSolve constraints don't seem to have an effect

2009-09-17 Thread Adam Lenart


Dear R users,

I would like to optimize a linear approximation of a quadratic function
using lpSolve. My code runs without any error or warning message but the
constraints that I set don't seem to work properly.

Nevertheless, I am certain that my code is somewhere wrong.

I would like to solve the following problem:

max 2x-x^2+y
subject to
2x^2 + 3y^2 = 6
2= x,y = 0

I would split the [0,2] interval into 8 equal parts by defining k=9 points
(0 included) -lambdas in the following code- and observe the value of the
objective function and the constraint on these points and approximate it by
the following program:

(Sorry for the Latex type equations)

max Sum_{j=1}^{2}Sum_{k=0}^{8}f_{kj}lambda_{kj}
subject to:
Sum_{j=1}^{2}Sum_{k=0}^{8}g_{kj}lambda_{kj}+x2 = 6
Sum_{k=0}^{8}lambda_{k1} = 1
Sum_{k=0}^{8}lambda_{k2} = 1
lambda_{k1} = 0
lambda_{k2} = 0
x2 = 0



library(lpSolve)

#  Objective function and constraint
f1 - function(x) 2*x-x^2
f2 - function(y) y
g1 - function(x) 2*x^2
g2 - function(y) 3*y^2

# Setting objective function and constraint values
G1 - g1(seq(0,2,by=0.25))
G2 - g2(seq(0,2,by=0.25))
F1 - f1(seq(0,2,by=0.25))
F2 - f2(seq(0,2,by=0.25))
con - c(G1,G2,1) # slack variable included
obj - c(F1,F2)

# Defining the constraints
lambdasx2 - diag(1,ncol=19,nrow=19)
lambda1x2 - c(rep(1,times=9),rep(0,times=9),0)
lambda2x2 - c(rep(0,times=9),rep(1,times=9),0)
lambdamatrix - rbind(lambda1x2,lambda2x2,lambdasx2)
conlamb - rbind(con,lambdamatrix)

# Defining the right-hand side
rightside - c(6,1,1,rep(0,times=19))

# Defining the directions
direc - c(=,=,=,rep(=,times=19))

# LP
lp(max,obj,conlamb,direc,rightside)$solution

# But I receive [1] 0.00 1.00 5.75 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
#  [12] 0.00 0.00 0.00 0.00 0.00 0.00 0.00
# However, in the constraints the sum of the first 8 values should be 1 and
the second 8 values should be # 1

Thank you very much in advance,

Adam



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Re: [R] How to separate a function by 2 probabilities

2009-09-17 Thread Bert Gunter


Assuming storage is not a problem, first generate two matrices, one by each
method, call these A and B. Then if dim(A) = dim(B) = c(m,n) and k = m*n

z - rbinom(k,1, .7)

result - A*z + B*(1-z)


Bert Gunter
Genentech Nonclinical Biostatistics

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Marcio Resende
Sent: Thursday, September 17, 2009 8:11 AM
To: r-help@r-project.org
Subject: [R] How to separate a function by 2 probabilities


Good Mourning,
I have a function to generate a matrix as I show part of it;

g[j,i]-if (gen[j,i]==0) al1[i,1]+al1[i,1] else ...

However i would like that this function occurred with a probability P and
that another function (another formula to generate g matrix) with
probability P-1

That´s it, if P is .7, i would like that in 70% of the times (for random i
and j) the matrix g was generated according to the formula above and in 30%
of the times with a different formula which i did not write

could anyone help me?
Thank you very much
Márcio

-- 
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http://www.nabble.com/How-to-separate-a-function-by-2-probabilities-tp254919
43p25491943.html
Sent from the R help mailing list archive at Nabble.com.

__
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[R] QQ plotting of various distributions...

2009-09-17 Thread Petar Milin


Hello!
I am trying with this question again:
I would like to test few distributional assumptions for some behavioral 
response data. There are few theories about true distribution of those 
data, like: normal, lognormal, gamma, ex-Gaussian 
(exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would 
be via qq-plot, to show to students differences. First two are trivial:

qqnorm(dat$X)
qqnorm(log(dat$X))
Then, things are getting more hairy. I am not sure how to make plots 
for the rest. I tried gamma with:

qqmath(~ X, data=dat, distribution=function(X)
   qgamma(X, shape, scale))
Which should be the same as:
plot(qgamma(ppoints(dat$X), shape, scale), sort(dat$X))
Shape and scale parameters I got via mhsmm package that has gammafit() 
for shape and scale parameters estimation.
Am I on right track? Does anyone know how to plot the rest: ex-Gaussian 
(exponential-Gaussian), Wald (inverse Gaussian)?


Thanks,
PM

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Re: [R] How to do a PCA with ordered variables?

2009-09-17 Thread Mark Difford


P. Branco wrote:

 I have used the dudi.mix method from the ade4 package, but when I do the
 $index it shows 
 me that R has considered my variables as quantitative. 

 What should I do?

You should make sure that they are encoded as ordered factors, which has
nothing to do with ade4's dudi.mix().

##
?ordered

Mark.



P.Branco wrote:
 
 Hi,
 
 I want to do a pca using a set of variables which are ordered. I have used
 the dudi.mix method from the ade4 package, but when I do the $index it
 shows me that R has considered my variables as quantitative.
 
 What should I do?
 

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[R] Odp: boxplot

2009-09-17 Thread Petr PIKAL

Hi

I do not know rma but from help page boxplot requires as input a formula, 
list (only some list of numerics), data frame or numeric vector. I am not 
sure if your object is one of these. If not you need to convert it to 
object which is acceptable for boxplot. 

Regards
Petr

r-help-boun...@r-project.org napsal dne 17.09.2009 08:29:43:

 Hi,
 
 I m not able to plot normalized data(normalization by rma) using 
boxplot. I
 don't know why?
 basically, object(formed of normalized data) belong to ExpressionSet 
class.
 
 It is showing error
 
 Error in x[!xna] : object of type 'S4' is not subsettable
 In addition: Warning messages:
 1: In is.na(x) : is.na() applied to non-(list or vector) of type 'S4'
 2: In is.na(x) : is.na() applied to non-(list or vector) of type 'S4'
 
 
 Now, how to plot? Should  I have use another function?
 
 By
 Sukhbir Singh Rattan
 
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[R] Dealing with heterogeneity with varComb weights

2009-09-17 Thread RS27


Hi,
I am trying to add multiple variance structures such as the first example
below:

vf1 - varComb(varIdent(form = ~1|Sex), varPower())

However my code below will not work can anybody please advise me?

VFcomb-varComb(varExp(form=~depcptwithextybf),varFixed(form=~FebNAO))

also if you have two variables with the same weights function would you
write that as:

VFcomb-varComb(varExp(form=~depcptwithextybf),varExp(form=~FebNAO))

thanks
Rebecca


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Re: [R] Why S4 method is not visible from another package?

2009-09-17 Thread Martin Morgan

Gábor Csárdi wrote:
 Dear All,
 
 maybe this is something obvious, I seem to be incapable of
 understanding how S4 works.
 
 So, in package 'A' I defined a summary method for my class:
 
 setMethod(summary, signature(object=ListHyperGResult),
   function(object, pvalue=pvalueCutoff(object), categorySize=NULL) {
  whatever
   })
 
 ListHyperGResult has a subclass, GOListHyperGResult:
 
 setClass(GOListHyperGResult,
  representation=representation(conditional=logical),
  contains=ListHyperGResult,
  prototype=prototype(testname=GO))
 
 The summary method is exported in the NAMESPACE:
 
 exportMethods(summary)
 
 Package 'B' depends on package 'A', this is stated in the
 'DESCRIPTION' file. If I call 'summary' on a 'GOListHyperGResult' in

Hi Gabor

It is not S4 alone, but S4 + name spaces that are giving you problems.

You probably want to Import: A rather than depends, and importFrom(A,
summary).

As it stands, inside the B name space, you find base::summary, whereas
you've defined a method on summary that has been promoted to a generic
in one of the packages that A imports (probably AnnotationDbi).

This is a little bit of a guess; at some level it might seem more
appropriate to Import: AnnotationDbi and importFrom(AnnotationDbi,
summary) (or wherever the generic for summary that you are trying to use
is created).

Martin

 package B, then the default summary method is called instead of the
 correct one, despite that I have
 
 Browse[1] showMethods(summary)
 Function: summary (package base)
 object=AnnDbBimap
 object=ANY
 object=Bimap
 object=DBIObject
 object=HyperGResultBase
 object=KEGGHyperGResult
 object=LinearMResultBase
 object=ListHyperGResult
 object=PFAMHyperGResult
 object=SQLiteConnection
 object=SQLiteDriver
 object=SQLiteResult
 
 Browse[1] class(gos[[1]])
 [1] GOListHyperGResult
 
 But I still get:
 
 Browse[1] is(gos[[1]], ListHyperGResult)
 [1] TRUE
 Browse[1] summary(gos[[1]])
 Length  Class   Mode
  1 GOListHyperGResult S4
 
 What am I doing wrong?
 
 sessionInfo()
 R version 2.9.0 (2009-04-17)
 x86_64-redhat-linux-gnu
 
 locale:
 LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base
 
 other attached packages:
  [1] hgu95av2.db_2.2.12  ALL_1.4.4   ExpressionView_0.1
  [4] caTools_1.9 bitops_1.0-4.1  KEGG.db_2.2.5
  [7] GO.db_2.2.5 RSQLite_0.7-1   DBI_0.2-4
 [10] eisa_0.1genefilter_1.24.2   Category_2.10.0
 [13] AnnotationDbi_1.6.0 Biobase_2.4.1   isa2_0.1
 
 loaded via a namespace (and not attached):
 [1] annotate_1.22.0 graph_1.22.2GSEABase_1.6.0  RBGL_1.20.0
 [5] splines_2.9.0   survival_2.35-4 tools_2.9.0 XML_2.6-0
 [9] xtable_1.5-5
 
 Thanks,
 Gabor


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[R] Odp: What is the best way to get a subset of a data.frame?

2009-09-17 Thread Petr PIKAL

Hi

r-help-boun...@r-project.org napsal dne 17.09.2009 04:14:24:

 Hi,
 
 I want to construct a data.frame 'y' by using x$x and x$y. I think
 that there might be better ways to do it (because, for example, we can
 use a_matrix[3:5,] to extract certain rows, where 'a_matrix' is a
 matrix). Can somebody let know what the best way is?
 
  a=data.frame(x=1:10,y=rep(abc,10),z=rep(xyz,10))
  a
 x   y   z
 1   1 abc xyz
 2   2 abc xyz
 3   3 abc xyz
 4   4 abc xyz
 5   5 abc xyz
 6   6 abc xyz
 7   7 abc xyz
 8   8 abc xyz
 9   9 abc xyz
 10 10 abc xyz
  b=data.frame(x=a$x, y=a$y)
  b

what is wrong on

b[1:2,]

Regards
Petr


 x   y
 1   1 abc
 2   2 abc
 3   3 abc
 4   4 abc
 5   5 abc
 6   6 abc
 7   7 abc
 8   8 abc
 9   9 abc
 10 10 abc
 
 
 
 Regards,
 Peng
 
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[R] Help with date specification

2009-09-17 Thread Subodh Acharya

Hi everyone,I have a data daily data (x) for 10 years starting from
04-01-1995 to 03-31-2005.
I was able to get the yearly sum for the ten years using
aggregate(x, years, sum).
But this gave me the yearly sum for 1995 (Apr- Dec); 1996 (Jan-Dec)
-2005 (Jan-Mar).
But I want to get the aggregates for Apr-1995 to Mar 1996, Apr 1996- mar
1997 and so on.

your help will be higly appreciated.
Thanks in advance

-- 
Acharya, Subodh

[[alternative HTML version deleted]]

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Re: [R] Which one shall I use? '=' or '-'?

2009-09-17 Thread David M Smith

I wrote a post on the Revolutions blog a little while back discussing
the difference between the two (and the history of the - operator).
You can find it on blog.revolution-computing.com at:
http://bit.ly/3YWw3R

# David Smith

On Wed, Sep 16, 2009 at 7:58 PM, Peng Yu pengyu...@gmail.com wrote:
 Hi,

 I was told to use  - instead of = in the mailing list. I am
 wondering what the difference between them?

 Regards,
 Peng

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-- 
David M Smith da...@revolution-computing.com
Director of Community, REvolution Computing www.revolution-computing.com
Tel: +1 (206) 577-4778 x3203 (San Francisco, USA)

Check out our upcoming events schedule at www.revolution-computing.com/events

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Re: [R] Help with date specification

Subodh
Assuming the data is ordered by date then you can define
fin.years = (0:(10*12-1)) %/% 12

then use aggregate:
aggregate(x, list(fin.years),sum)

HTH
Schalk Heunis


On Thu, Sep 17, 2009 at 6:11 PM, Subodh Acharya shoeb...@gmail.com wrote:

 Hi everyone,I have a data daily data (x) for 10 years starting from
 04-01-1995 to 03-31-2005.
 I was able to get the yearly sum for the ten years using
 aggregate(x, years, sum).
 But this gave me the yearly sum for 1995 (Apr- Dec); 1996 (Jan-Dec)
 -2005 (Jan-Mar).
 But I want to get the aggregates for Apr-1995 to Mar 1996, Apr 1996- mar
 1997 and so on.

 your help will be higly appreciated.
 Thanks in advance

 --
 Acharya, Subodh

[[alternative HTML version deleted]]

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Re: [R] r-inferno.pdf with detailed table of contents and bookmarks

2009-09-17 Thread Victor Manuel Garcia Guerrero

Hey, check this link: http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
 
Greetings
 
Víctor
 


De: r-help-boun...@r-project.org en nombre de Peng Yu
Enviado el: jue 17/09/2009 10:43
Para: r-h...@stat.math.ethz.ch
Asunto: [R] r-inferno.pdf with detailed table of contents and bookmarks



Hi,

I don't find a r-inferno.pdf that has detailed table of contents and
bookmarks. If it is possible, can somebody help generated one and post
it on line?

Regards,
Peng

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Re: [R] Dealing with heterogeneity with varComb weights

2009-09-17 Thread Alain Zuur




RS27 wrote:
 
 Hi,
 I am trying to add multiple variance structures such as the first example
 below:
 
 vf1 - varComb(varIdent(form = ~1|Sex), varPower())
 
 However my code below will not work can anybody please advise me?
 
 VFcomb-varComb(varExp(form=~depcptwithextybf),varFixed(form=~FebNAO))
 
 
 
 VarFixed won't work if FebNAO has values equal to 0. In fact, I wouldn't
 use varFixed at all. 
 
 
 A bit more info on the error message would be handy...
 
 Alain
 
 
 also if you have two variables with the same weights function would you
 write that as:
 
 VFcomb-varComb(varExp(form=~depcptwithextybf),varExp(form=~FebNAO))
 
 thanks
 Rebecca
 
 
 


-

Dr. Alain F. Zuur
First author of:

1. Analysing Ecological Data (2007).
Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p.

2. Mixed effects models and extensions in ecology with R. (2009).
Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer.

3. A Beginner's Guide to R (2009).
Zuur, AF, Ieno, EN, Meesters, EHWG. Springer


Statistical consultancy, courses, data analysis and software
Highland Statistics Ltd.
6 Laverock road
UK - AB41 6FN Newburgh
Email: highs...@highstat.com
URL: www.highstat.com



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Re: [R] How to colour the tip labels in a phylogenetic tree

2009-09-17 Thread Ben Bolker




Graham Etherington wrote:
 
 Hi,
 Using Ape, I have constructed an object of class phylo, using the 
 method 'nj' (lets call the object 'tree_ja').
 I also have a given subset of 'tree_ja' in a vector (lets call the 
 vector 'subspecies').
 What I want to do, is construct a nj tree - plot(tree_ja) - but have the 
 species in vector 'subspecies' shown as red at the tips of the tree.
 
 The closest I've come is this:
 Given that 'tree_ja$tip.label' provides the following:
   [1] 1_T1  2_T1  3_T1  4_T1  5_T1  6_T1
   [7] 7_T1  8_T1  9_T1 10_T1 11_T1 12_T1
  
 and that my 'subspecies' vector is:
 subspecies - c(1_T1, 2_T1, 3_T1, 4_T1, 6_T1)
 which can also be written as:
 subspecies - c(tree_ja$tip.label[1:4], tree_ja$tip.label[5])
 
 I can construct a method which gives me the following statement:
 
 plot(tree_ja, tip.col = c('red', 'red', 'red', 'red', 'black', 'red', 
 'black', 'black', 'black', 'black', 'black', 'black'))
 
 But this doesn't work (at least not on my full dataset, which as 118 
 tips - reduced to 12 here for brevity) and I'm SURE there must be a 
 better way of doing it.
 
 Could anyone help me with this?
 Many thanks,
 Graham
 
 -- 
 Dr. Graham Etherington
 Post-doctoral Bioinformatician,
 Department of Computational and Systems Biology
 John Innes Centre
 Norwich Research Park
 Colney
 Norwich
 NR4 7UH
 UK
 
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  In general the r-sig-phylo group is better for this kind of question, and
it would
be better to give us a reproducible example, but here's an example that (I
think)
does what you want:

 library(ape)
 set.seed(1001)
 z = rcoal(10)
 z

Phylogenetic tree with 10 tips and 9 internal nodes.

Tip labels:
t1, t7, t6, t10, t3, t9, ...

Rooted; includes branch lengths.
 names(z)
[1] edgeedge.length tip.label   Nnode  
 ss - z$tip.label[c(1,3,5,7)]
 ?plot.phylo
 plot(z,tip.color=ifelse(z$tip.label %in% ss, red,black))

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Re: [R] Help with date specification

2009-09-17 Thread Gabor Grothendieck

Try either of these which convert to year and qtr and then
shift the qtr by one.  The resulting series is labeled by
the year in which the series dates start.  Use
as.integer(...) + 1 if you prefer to label by ending year.

library(zoo)
# test data
DF - data.frame(date = Sys.Date() + 1:1000, value = 1:1000)

# 1. Using a data frame --
# uses as.yearqtr from zoo but no zoo objects

tapply(DF$value, as.integer(as.yearqtr(DF$date) - 0.25), sum)

# 2. or converting the time series to zoo and then aggregating

z - zoo(DF$value, as.Date(DF$date))
aggregate(z, as.integer(as.yearqtr(time(z))-0.25), sum)

In future please provide some of your data using dput (see ?dput)
to make it easier to answer and code if relevant.

On Thu, Sep 17, 2009 at 12:11 PM, Subodh Acharya shoeb...@gmail.com wrote:
 Hi everyone,I have a data daily data (x) for 10 years starting from
 04-01-1995 to 03-31-2005.
 I was able to get the yearly sum for the ten years using
 aggregate(x, years, sum).
 But this gave me the yearly sum for 1995 (Apr- Dec); 1996 (Jan-Dec)
 -2005 (Jan-Mar).
 But I want to get the aggregates for Apr-1995 to Mar 1996, Apr 1996- mar
 1997 and so on.

 your help will be higly appreciated.
 Thanks in advance

 --
 Acharya, Subodh

        [[alternative HTML version deleted]]

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Re: [R] Help with date specification

2009-09-17 Thread Mark Knecht

On Thu, Sep 17, 2009 at 9:11 AM, Subodh Acharya shoeb...@gmail.com wrote:
 Hi everyone,I have a data daily data (x) for 10 years starting from
 04-01-1995 to 03-31-2005.
 I was able to get the yearly sum for the ten years using
 aggregate(x, years, sum).
 But this gave me the yearly sum for 1995 (Apr- Dec); 1996 (Jan-Dec)
 -2005 (Jan-Mar).
 But I want to get the aggregates for Apr-1995 to Mar 1996, Apr 1996- mar
 1997 and so on.

 your help will be higly appreciated.
 Thanks in advance

 --
 Acharya, Subodh

        [[alternative HTML version deleted]]


subset(x, date=Apr-96  date=Mar))

then do the sum?

- Mark

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Re: [R] Fastest Way to Divide Elements of Row With Its RowSum

2009-09-17 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Thomas Lumley
 Sent: Thursday, September 17, 2009 6:59 AM
 To: William Revelle
 Cc: r-h...@stat.math.ethz.ch
 Subject: Re: [R] Fastest Way to Divide Elements of Row With Its RowSum

 On Thu, 17 Sep 2009, William Revelle wrote:

  At 2:40 PM +0900 9/17/09, Gundala Viswanath wrote:
  I have a data frame (dat). What I want to do is for each row,
  divide each row  with the sum of its row.

  The number of row can be large  1million.
  Is there a faster way than doing it this way?

  datnorm;
  for (rw in 1:length(dat)) {
  tmp - dat[rw,]/sum(dat[rw,])
  datnorm - rbind(datnorm, tmp);
  }

  - G.V.

  datnorm - dat/rowSums(dat)

  this will be faster if dat is a matrix rather than a data.frame.

 Even if it's a data frame and he needs a data frame answer it 
 might be faster to do
mat-as.matrix(dat)
matnorm-mat/rowSums(mat)
datnorm-as.data.frame(dat)

If the data.frame has many more rows than columns and the
number of rows is large (e.g., dimensions 10^6 x 20) you may
find that you run out of space converting it to a matrix.  You
can use much less space by looping over the columns, both
to compute the row sums and to do the division.  E.g., the
following should require only 1 (maybe 2) column's worth of
scratch space:

f2 - function(x){
   stopifnot(is.data.frame(x), ncol(x)=1)
   rowsum - x[[1]]
   if(ncol(x)1) for(i in 2:ncol(x))
  rowsum - rowsum + x[[i]]
   for(i in 1:ncol(x))
  x[[i]] - x[[i]] / rowsum
   x
}

For a 10^6 by 20 all numeric data.frame this runs in 13 seconds
on my machine but things like x/rowSums(x) run out of memory.

When working with data.frames it generally pays to think a column
at a time instead of a row at a time.

Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com 

 The other advantage, apart from speed, of doing it with 
 dat/rowSums(dat) rather than the loop is he gets the right 
 answer. The loop goes from 1 to the number of columns if dat 
 is a data frame and 1 to the number of entries if dat is a 
 matrix, not from 1 to the number of rows.

   -thomas

 Thomas Lumley Assoc. Professor, Biostatistics
 tlum...@u.washington.edu  University of Washington, Seattle

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Re: [R] r-inferno.pdf with detailed table of contents and bookmarks

2009-09-17 Thread Patrick Burns


That's a reasonable request that I have
planned for whenever I revise it.
However, I'm not going to be doing that
for some time yet (unit of time is somewhere
in the months to years range).

If someone is keen to do that, I can make
the LyX file available to them.



Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)

Peng Yu wrote:

Hi,

I don't find a r-inferno.pdf that has detailed table of contents and
bookmarks. If it is possible, can somebody help generated one and post
it on line?

Regards,
Peng

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[R] Grouped Logistic (Or conditional Logistic.)

2009-09-17 Thread Noah Silverman


Hi,

I'm not sure of the correct nomenclature or function for what I'm trying 
to do.


I'm interested in calculated a logistic regression on a binary dependent 
variable (True,False).


There are a few ways to easily do this in R.  Both SVM and GLM work easily.

The part that I want to add is group wise awareness.  So that the 
algorithm computes the coefficients to maximize the liklihood of of a 
True label per group.


An toy explanation is probably best.  I've been looking at horse racing 
models as a fun field to learn about statistics and R.


So, for this example, lets assume the following:
100 horses in our stable
10 horses per race
75 races this season (some horses race more than once.)

The independent variables are things about a horse (average speed, 
number of past wins, etc.)

The dependent variable is (Win, Lose) represented by (1,0)

As mentioned above, an SVM or GLM will quickly work to estimate 
coefficients and probability of a Win.  I'd like to take it further and 
estimate the probability of a win but look at the per race.


I'm NOT interested in the group label as a final part of the model.  I 
don't want a separate set of coefficients for each group.  I just want 
the iterative algorithm to work toward maximizing the liklihood PER 
GROUP as an average.


I looked extensively through rseek.org for things like grouped 
logistic and nested logistic.  I couldn't seem to find anything do 
this.  I'm probably naming it wrong.


I assume that a MANUAL iteration concept would be to :
1) Pick a coefficient
2) Calculate the resulting probability for each horse.
3) Measure the strength of the result for each race (sum them 
together or average them?)

4) Adjust coefficient and repeat

Surely there must be some standard function in a library that will do this.

Can any of the stat gurus here offer some suggestions?

Thanks!

--
Noah

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[R] Hi

2009-09-17 Thread MOKGATLHE, L.L. (DR.)

Can I run a lack-of-fit test using R? How do I do that?

Thank you 


[[alternative HTML version deleted]]

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Re: [R] ggplot, ribbon not showing up properly

2009-09-17 Thread Sock Cheng

Hi Thierry,

I tried the code suggested below, but it didn't work fully.  The ribbon
showed up correctly for first and last days, but the days in between
appeared to be ignored.  I tried other ways of feeding geom_ribbon the
summary stats but my ways didn't work either.  

Thanks for trying to help,  

Sock

On Tue, 15 Sep 2009 11:17 +0200, ONKELINX, Thierry
thierry.onkel...@inbo.be wrote:
 Dear Sock,
 
 I'm wondering if that mean_sdl function is return what you are
 expecting. I would calculate the statistics outside ggplot.
 
 RibbonData - ddply(dat.less, Day, function(x){
   mean(x$Y) + c(ymin = -1, ymax = 1) * sd(x$Y)
 })
 p + stat_summary(data=dat.less, aes(group=1), geom=crossbar,
 fun.data=mean_sdl, mult=1) + geom_ribbon(data = RibbonData, aes(group
 = 1, ymin = ymin, ymax = ymax), fill=alpha(blue, 1/10))
 
 HTH,
 
 Thierry 
 
 
 
 
 ir. Thierry Onkelinx
 Instituut voor natuur- en bosonderzoek / Research Institute for Nature
 and Forest
 Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
 methodology and quality assurance
 Gaverstraat 4
 9500 Geraardsbergen
 Belgium
 tel. + 32 54/436 185
 thierry.onkel...@inbo.be
 www.inbo.be
 
 To call in the statistician after the experiment is done may be no more
 than asking him to perform a post-mortem examination: he may be able to
 say what the experiment died of.
 ~ Sir Ronald Aylmer Fisher
 
 The plural of anecdote is not data.
 ~ Roger Brinner
 
 The combination of some data and an aching desire for an answer does not
 ensure that a reasonable answer can be extracted from a given body of
 data.
 ~ John Tukey
  
 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens Sock Cheng
 Verzonden: maandag 14 september 2009 20:58
 Aan: r-help@r-project.org
 Onderwerp: [R] ggplot, ribbon not showing up properly
 
 Hi,
 
 I'm trying to plot a longitudinal data set, using ggplot and adding some
 summary info (eg. mean, 1 sd bounds) using geom=ribbon.  The summary
 info is based on a subset of the original data (eg. less an outlier). 
 But I'm having trouble getting the ribbons to show up correctly.  It's
 probably something obvious that I'm missing as a novice at ggplot2, and
 any help is much appreciated!
  
 Here's a simple example.  I tried several things.  
 - if I use geom=crossbar, everything is ok
 - if Day is set as rep(c(1,2,3,8,9), each=8), then everything is ok,
 which makes me wonder if the problem has to do with the ordering of Day?
  Day is supposed to be numeric.  
 
 Thanks!
 Sock
 
 ### Example data.  Ran using R version 2.9.2, ggplot2 version 0.8.3 ###
 
 set.seed(13)
 
 Day - rep(c(1, 2, 3, 8, 20), each=8)  
 # The plot is ok if Day - rep(c(1,2,3,8,9), each=8) 
 
 ID - rep(LETTERS[1:8], 5)
 Y - rnorm(length(Day), 100, 5)
 dat - data.frame(Day=Day, ID=ID, Y=Y)
 
  # outlier
 dat$Y[dat$ID==A  dat$Day==8] - 150 
 dat.less - dat[!(dat$ID==A  dat$Day==8),]   
 
 # Longitudinal data plot.  Obs for each subject is connected by a line
 over time
 
 p - ggplot(dat, aes(x=Day, y=Y, group=ID)) +
 scale_x_continuous(breaks=sort(unique(dat$Day))) +
 geom_line(colour=alpha(blue, 5/10))
 
 # Adding mean, 1 sd bounds using crossbar geom is ok.  But the same info
 using ribbon geom doesn't work. 
  
 p + stat_summary(data=dat.less, aes(group=1), geom=crossbar,
 fun.data=mean_sdl, mult=1) + stat_summary(data=dat.less, aes(group=1),
 geom=ribbon, fun.data=mean_sdl, mult=1, fill=alpha(blue, 1/10))
 
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[R] define a new family (and a new link function) for gam in gam package

2009-09-17 Thread Topi, Corrado


Dear R list,

is it possible to define a new family (and a new link function) for gam 
in gam package? How?


I read the help for gam, family, gam.model, make.link but I did not find 
a solution.


Regards
--
Corrado Topi

Area 18,Department of Biology
University of York, York, YO10 5YW, UK
Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk

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Re: [R] inline error message

2009-09-17 Thread Duncan Murdoch


On 9/17/2009 9:16 AM, Uwe Ligges wrote:


_ wrote:

Hi all,
I installed the library inline to my default R-enviroment 
(c:\Programme\R.. )
downloaded and installed RTools from 
http://www.murdoch-sutherland.com/Rtools/

to c:\Rtools.
Path variable is set right (with respect to order) but I get still the 
error message

from my R
Error in compileCode(f, code, language, verbose) :
 Compilation ERROR, function(s)/method(s) not created!

Where is my failure ?


If you tell us what library (I guess package?) you mean and how you 
traied to install it and what command gave the error message (including 
a complete logfile perhaps), we may be able to help.


I think the package is inline, but the code that led to the error (or 
a simplified example) would help a lot in diagnosing the error.  I know 
I wouldn't try without that.


Duncan Murdoch

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Re: [R] How to do a PCA with ordered variables?

2009-09-17 Thread Stas Kolenikov

If you are really serious about your variables being ordinal, you
should analyze them using polychoric correlations. See polycor package
by John Fox.

On Thu, Sep 17, 2009 at 10:52 AM, Mark Difford mark_diff...@yahoo.co.uk wrote:
 I have used the dudi.mix method from the ade4 package, but when I do the
 $index it shows
 me that R has considered my variables as quantitative.

 What should I do?

 You should make sure that they are encoded as ordered factors, which has
 nothing to do with ade4's dudi.mix().

 ##
 ?ordered

 Mark.



 P.Branco wrote:

 Hi,

 I want to do a pca using a set of variables which are ordered. I have used
 the dudi.mix method from the ade4 package, but when I do the $index it
 shows me that R has considered my variables as quantitative.

 What should I do?


 --
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-- 
Stas Kolenikov, also found at http://stas.kolenikov.name
Small print: I use this email account for mailing lists only.

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Re: [R] R functions with array arguments

2009-09-17 Thread Maurizio Paolillo


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Re: [R] Grouped Logistic (Or conditional Logistic.)

2009-09-17 Thread Ted Harding

On 17-Sep-09 17:28:16, Noah Silverman wrote:
 Hi,
 I'm not sure of the correct nomenclature or function for what
 I'm trying to do.
 
 I'm interested in calculated a logistic regression on a binary
 dependent variable (True,False).
 
 There are a few ways to easily do this in R. Both SVM and GLM
 work easily.
 
 The part that I want to add is group wise awareness.  So that
 the algorithm computes the coefficients to maximize the liklihood
 of of a True label per group.
 
 An toy explanation is probably best.  I've been looking at horse
 racing models as a fun field to learn about statistics and R.
 
 So, for this example, lets assume the following:
 100 horses in our stable
 10 horses per race
 75 races this season (some horses race more than once.)
 
 The independent variables are things about a horse (average speed, 
 number of past wins, etc.)
 The dependent variable is (Win, Lose) represented by (1,0)
 
 As mentioned above, an SVM or GLM will quickly work to estimate 
 coefficients and probability of a Win. I'd like to take it further
 and estimate the probability of a win but look at the per race.
 
 I'm NOT interested in the group label as a final part of the model.
 I don't want a separate set of coefficients for each group. I just
 want the iterative algorithm to work toward maximizing the liklihood
 PER GROUP as an average.
 
 I looked extensively through rseek.org for things like grouped 
 logistic and nested logistic.  I couldn't seem to find anything
 do this.  I'm probably naming it wrong.
 
 I assume that a MANUAL iteration concept would be to :
  1) Pick a coefficient
  2) Calculate the resulting probability for each horse.
  3) Measure the strength of the result for each race (sum them 
 together or average them?)
  4) Adjust coefficient and repeat
 
 Surely there must be some standard function in a library that will
 do this.
 
 Can any of the stat gurus here offer some suggestions?
 
 Thanks!
 --
 Noah

In the context of your fun example, you have a fundamental problem
in that (if I've understood your statement of it correctly) you will
have more than one of your horses in the same race (apparently 10).

Therefore, one of them winning excludes any of the others winning in
that same race, so their results are not independent of each
other.

Also, at least in real life, the probability that a given horse will
win in a particular race depends not only on the covariates per horse
(such as your average speed, number of past wins, etc.), and indeed
on the condition of the race-course at the time, but also (and usually
strongly) on the characteristics of the other horses in the same race.

So a simple logistic model of the kind you seem to be proposing would
certainly not be realistic!

I would be happier thinking about your problem in the context of a
different kind of example ...

Ted.


E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 17-Sep-09   Time: 19:06:27
-- XFMail --

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[R] dyn.load search path?

2009-09-17 Thread Steve Jaffe


Sorry if this is somewhere in the fine manuals but I've been unable to locate
it.

Does dyn.load use a search path or does it just look in the current
directory for non-fully-qualified filenames? If there is a search path, what
is it?

Thanks for your help
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Re: [R] Simple as.Date question dealing with a timezone offset



On Sep 17, 2009, at 11:25 AM, esawdust wrote:



I've been trying to understand the as.Date functionality and I have  
a date

and time stamp field that looks like this:

Tue Sep 15 09:22:09 -0600 2009

and I need to turn it into an R Date object for analysis.

Simple date conversions I have down, no problem:


adate = c(7/30/1959)
as.Date(adate,%m/%d/%Y)

[1] 1959-07-30




But when it comes to the type of date/time string format I have  
above, I

can't figure out a format string that will work.

The timezone offset is one part that causes problems.  Building up  
to a
working format string for the full time stamp string, I can make it  
as far

as:


adate = c(Tue Sep 15 09:22:09 -0600 2009)
as.Date(adate,format=%a %b %d %H:%M:%S)

[1] 2009-09-15

(apparently year defaults to current year when it's not specified in  
the
format string).  Because the Year comes after the timezone offset, I  
have to

deal with the timezone offset in the format string.

But when I get to the timezone offset value I can't use %z or %Z  
because

those are output only


as.Date(adate,format=%a %b %d %H:%M:%S %z %Y)

[1] NA


You are confusing R Date objects with the the date-time classes. I  
don't think Date classes objects even allow TZ offets.


?DateTimeClasses

 as.POSIXct(as.Date(adate,%m/%d/%Y), origin=1960-01-01, tz=GMT)
[1] 1959-07-29 20:00:00 EDT'

Notice that my TZ (GMT -4) was used. so it was still the prior day in  
New England.




I'm close, but can't incorporate the timezone offset field in the  
date/time

stamp string.

What am I missing?   I suppose one workaround is to split the date/ 
time
string into its component parts, reassemble it into a string as.Date  
can
deal with, but that seems to defeat one of the purposes of as.Date's  
format

capability.

Any advice for how to translate a Tue Sep 15 09:22:09 -0600 2009  
into an R

Date object?

Landon


--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Grouped Logistic (Or conditional Logistic.)

2009-09-17 Thread Noah Silverman


Ted,

Thanks for the reply.

For the example, I'm not looking to predict THE winner, but to find 
the best probabilities of winning.


It would seem that the process of iterating through possible 
coefficients would be the same as a standard GLM, the evalation part 
as you work through them would have to be adjusted to look per group.


I would call this something like grouped maximum liklihood  if I got 
to make up the name.



-N


On 9/17/09 11:06 AM, (Ted Harding) wrote:

On 17-Sep-09 17:28:16, Noah Silverman wrote:
   

Hi,
I'm not sure of the correct nomenclature or function for what
I'm trying to do.

I'm interested in calculated a logistic regression on a binary
dependent variable (True,False).

There are a few ways to easily do this in R. Both SVM and GLM
work easily.

The part that I want to add is group wise awareness.  So that
the algorithm computes the coefficients to maximize the liklihood
of of a True label per group.

An toy explanation is probably best.  I've been looking at horse
racing models as a fun field to learn about statistics and R.

So, for this example, lets assume the following:
100 horses in our stable
10 horses per race
75 races this season (some horses race more than once.)

The independent variables are things about a horse (average speed,
number of past wins, etc.)
The dependent variable is (Win, Lose) represented by (1,0)

As mentioned above, an SVM or GLM will quickly work to estimate
coefficients and probability of a Win. I'd like to take it further
and estimate the probability of a win but look at the per race.

I'm NOT interested in the group label as a final part of the model.
I don't want a separate set of coefficients for each group. I just
want the iterative algorithm to work toward maximizing the liklihood
PER GROUP as an average.

I looked extensively through rseek.org for things like grouped
logistic and nested logistic.  I couldn't seem to find anything
do this.  I'm probably naming it wrong.

I assume that a MANUAL iteration concept would be to :
  1) Pick a coefficient
  2) Calculate the resulting probability for each horse.
  3) Measure the strength of the result for each race (sum them
together or average them?)
  4) Adjust coefficient and repeat

Surely there must be some standard function in a library that will
do this.

Can any of the stat gurus here offer some suggestions?

Thanks!
--
Noah
 

In the context of your fun example, you have a fundamental problem
in that (if I've understood your statement of it correctly) you will
have more than one of your horses in the same race (apparently 10).

Therefore, one of them winning excludes any of the others winning in
that same race, so their results are not independent of each
other.

Also, at least in real life, the probability that a given horse will
win in a particular race depends not only on the covariates per horse
(such as your average speed, number of past wins, etc.), and indeed
on the condition of the race-course at the time, but also (and usually
strongly) on the characteristics of the other horses in the same race.

So a simple logistic model of the kind you seem to be proposing would
certainly not be realistic!

I would be happier thinking about your problem in the context of a
different kind of example ...

Ted.


E-Mail: (Ted Harding)ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 17-Sep-09   Time: 19:06:27
-- XFMail --



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Re: [R] Grouped Logistic (Or conditional Logistic.)



On Sep 17, 2009, at 2:06 PM, (Ted Harding) wrote:


On 17-Sep-09 17:28:16, Noah Silverman wrote:

Hi,
I'm not sure of the correct nomenclature or function for what
I'm trying to do.

I'm interested in calculated a logistic regression on a binary
dependent variable (True,False).

There are a few ways to easily do this in R. Both SVM and GLM
work easily.

The part that I want to add is group wise awareness.  So that
the algorithm computes the coefficients to maximize the liklihood
of of a True label per group.

An toy explanation is probably best.  I've been looking at horse
racing models as a fun field to learn about statistics and R.

So, for this example, lets assume the following:
100 horses in our stable
10 horses per race
75 races this season (some horses race more than once.)

The independent variables are things about a horse (average speed,
number of past wins, etc.)
The dependent variable is (Win, Lose) represented by (1,0)

As mentioned above, an SVM or GLM will quickly work to estimate
coefficients and probability of a Win. I'd like to take it further
and estimate the probability of a win but look at the per race.

I'm NOT interested in the group label as a final part of the model.
I don't want a separate set of coefficients for each group. I just
want the iterative algorithm to work toward maximizing the liklihood
PER GROUP as an average.

I looked extensively through rseek.org for things like grouped
logistic and nested logistic.  I couldn't seem to find anything
do this.  I'm probably naming it wrong.

I assume that a MANUAL iteration concept would be to :
1) Pick a coefficient
2) Calculate the resulting probability for each horse.
3) Measure the strength of the result for each race (sum them
together or average them?)
4) Adjust coefficient and repeat

Surely there must be some standard function in a library that will
do this.

Can any of the stat gurus here offer some suggestions?

Thanks!
--
Noah


In the context of your fun example, you have a fundamental problem
in that (if I've understood your statement of it correctly) you will
have more than one of your horses in the same race (apparently 10).

Therefore, one of them winning excludes any of the others winning in
that same race, so their results are not independent of each
other.

Also, at least in real life, the probability that a given horse will
win in a particular race depends not only on the covariates per  
horse

(such as your average speed, number of past wins, etc.), and indeed
on the condition of the race-course at the time, but also (and usually
strongly) on the characteristics of the other horses in the same race.

So a simple logistic model of the kind you seem to be proposing would
certainly not be realistic!

I would be happier thinking about your problem in the context of a
different kind of example ...


Ted;

Would your set of concerns be addressed if the OP switched to a  
proportional
odds logistic regression framework? Harrell discusses such in his RMS  
text.


--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] dyn.load search path?

2009-09-17 Thread cls59




Steve Jaffe wrote:
 
 Sorry if this is somewhere in the fine manuals but I've been unable to
 locate it.
 
 Does dyn.load use a search path or does it just look in the current
 directory for non-fully-qualified filenames? If there is a search path,
 what is it?
 
 Thanks for your help
 

I believe dyn.load() will attempt to load a library from whatever path you
point it at, like so:

dyn.load('/path/to/dynamic/library')

If your dynamic library references code in other dynamic libraries, I would
imagine environment variables such as LD_LIBRARY_PATH would come into play--
but I'm not an expert on such voodoo.

If you would like a shortcut to loading libraries in certain places that R
knows about, such as the R package library, check out providing arguments to
dyn.load() using system.file() or loading package libraries using
library.dynam().

Hope that helps!

-Charlie

-
Charlie Sharpsteen
Undergraduate
Environmental Resources Engineering
Humboldt State University
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[R] How write the same number of elements in the first line as the rest of the file with write.table()?

Hi,

The first line has less elements than the rest of
'rownames_colnames.write.table.xls'. I am wondering if there is a way
to print an additional '\t' at the beginning of the first line.

$ Rscript write.table.R
 x=matrix(1:20,nc=2)
 rownames(x)=letters[1:10]
 colnames(x)=letters[1:2]
 write.table(x,rownames_colnames.write.table.xls,sep='\t')

$ cat rownames_colnames.write.table.xls
a b
a 1   11
b 2   12
c 3   13
d 4   14
e 5   15
f 6   16
g 7   17
h 8   18
i 9   19
j 10  20

Regards,
Peng

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Re: [R] How to separate a function by 2 probabilities

2009-09-17 Thread Ben Bolker




Marcio Resende wrote:
 
 Good Mourning,
 I have a function to generate a matrix as I show part of it;
 
 g[j,i]-if (gen[j,i]==0) al1[i,1]+al1[i,1] else ...
 
 However i would like that this function occurred with a probability P and
 that another function (another formula to generate g matrix) with
 probability P-1
 
 That´s it, if P is .7, i would like that in 70% of the times (for random i
 and j) the matrix g was generated according to the formula above and in
 30% of the times with a different formula which i did not write
 
 

something along the lines of

if (runif(1)0.7) {first formula} else {second formula}

you can probably make this much more efficient by appropriate vectorization,
but you didn't show enough of what you're doing to make specific
recommendations

  cheers

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http://www.nabble.com/How-to-separate-a-function-by-2-probabilities-tp25491943p25492242.html
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Re: [R] How write the same number of elements in the first line as the rest of the file with write.table()?



On Sep 17, 2009, at 3:23 PM, Peng Yu wrote:


Hi,

The first line has less elements than the rest of
'rownames_colnames.write.table.xls'. I am wondering if there is a way
to print an additional '\t' at the beginning of the first line.

$ Rscript write.table.R

x=matrix(1:20,nc=2)
rownames(x)=letters[1:10]
colnames(x)=letters[1:2]
write.table(x,rownames_colnames.write.table.xls,sep='\t')


If you replace your script file with this it gives the needed padding:

#write.table.R
x=matrix(1:20,nc=2)
rownames(x)=letters[1:10]
colnames(x)=letters[1:2]
sink(rownames_colnames.write.table.xls)
cat(\t)
write.table(x,sep='\t')
sink()





$ cat rownames_colnames.write.table.xls
a   b
a   1   11
b   2   12
c   3   13
d   4   14
e   5   15
f   6   16
g   7   17
h   8   18
i   9   19
j   10  20



--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] generating unordered combinations

2009-09-17 Thread Dan Halligan

Hi,

I am trying to generate all unordered combinations of a set of
numbers / characters, and I can only find a (very) clumsy way of doing
this using expand.grid.  For example, all unordered combinations of
the numbers 0, 1, 2 are:
0, 0, 0
0, 0, 1
0, 0, 2
0, 1, 1
0, 1, 2
0, 2, 2
1, 1, 1
1, 1, 2
1, 2, 2
2, 2, 2

(I have not included, for example, 1, 0, 0, since it is equivalent to
0, 0, 1).

I have found a way to generate this data.frame using expand.grid as
follows:

g - expand.grid(c(0,1,2), c(0,1,2), c(0,1,2))
for(i in 1:nrow(g)) {
g[i,] - sort(as.character(g[i,]))
}
o - order(g$Var1, g$Var2, g$Var3)
unique(g[o,]).

This is obviously quite clumsy and hard to generalise to a greater
number of characters, so I'm keen to find any other solutions.  Can
anyone suggest a better (more general, quicker) method?

Cheers

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Re: [R] What does model.matrix() return?