date:20120220

Hi all, I am having difficulties to understand how R sort strings:

If I do
R) sort(c(X.,X0B))
[1] X.  X0B

So for me, as far as lexicographic order is concerned I can add whatever to
the end, the order will remain the same, but :
R) sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z

Can somebody give me a trick for the order to become lexicographic ?  



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[R] apply with as function ifelse with 2 logical conditions

2012-02-20 Thread Nerak

Hi all,
I have a question concerning using several conditions in an ifelse function
used as the function in apply.
I want to create a new value with the  function ifelse ‘ object which can be
coerced to logical mode  “test[n,] 1  test[n-1,]==0” 
With n I mean the row. I don’t know how I could do this without a loop. I
want to avoid the usage of loops and was thinking about apply. This was what
I was thinking about:

test-data.frame(C=c(0,0,0,0,5,2,0,0,0,15,12,10,6,0,0,0),B=c(0,0,0,0,9,6,2,0,0,24,20,16,2,0,0,0),F=c(0,0,0,0,6,5,1,0,0,18,16,12,10,5,1,0))

test.b-test[-(nrow(test)),]
test.2b-rbind(0,test.b)
result-as.data.frame(apply(test,M=2,function(x)ifelse((test1test.2b==0),1,0)))

But I get 3 times the amount of rows than that I want… 
what I should achieve:
test.result-data.frame(C=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),B=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),F=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0))

Has someone a suggestions about what I’m doing wrong?

Many thanks,
Nerak


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Re: [R] Run function several times changing only one argument - without a loop

2012-02-20 Thread Eik Vettorazzi

Hi Marion,
you can either use any of the *apply-functions or vectorize your
function (which internally uses mapply):

par(las=1)
par(mar=c(5,13,4,2))
barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)

#using sapply
invisible(sapply((1:9)*10,function(x)axis(2,pos=x,tick=T, tcl=F,
labels=F,col=white)))

#using Vectorize
barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
vaxis-Vectorize(axis,pos)
invisible(vaxis(2,pos=(1:9)*10, tick=T, tcl=F, labels=F,col=white))

Cheers!

Am 20.02.2012 11:04, schrieb Marion Wenty:
 Dear people,
 
 I created a plot which looks like this:
 
 Ee1-matrix(c(88,86,74,62,41),ncol=5)
 colnames(Ee1)-c(Lehrer,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister)
 par(las=1)
 par(mar=c(5,13,4,2))
 barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
 axis(2,pos=10, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=20, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=30, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=40, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=50, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=60, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=70, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=80, tick=T, tcl=F, labels=F,col=white)
 axis(2,pos=90, tick=T, tcl=F, labels=F,col=white)
 
 Now I would like to shorten the whole thing - namely use only one step to
 create the 9 axes without having to use a loop.
 
 In general, I would be interested if there is a way to use a function
 several times changing only one argument, without having to use a loop.
 
 Does anyone know how to do that.
 
 Marion
 
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Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
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Re: [R] Sorting strings

On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
 Hi all, I am having difficulties to understand how R sort strings:
 
 If I do
 R) sort(c(X.,X0B))
 [1] X.  X0B
 
 So for me, as far as lexicographic order is concerned I can add whatever to
 the end, the order will remain the same, but :

Hi.

This neednot be true for strings of different length.
For example

  ab
  abc

become by concatenation with z

  abcz
  abz

Petr Savicky.

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[R] chisq.test vs manual calculation - why are different results produced?

2012-02-20 Thread Louise Mair

Hello,

I am trying to fit gamma, negative exponential and inverse power functions
to a dataset, and then test whether the fit of each curve is good. To do
this I have been advised to calculate predicted values for bins of data (I
have grouped a continuous range of distances into 1km bins), and then apply
a chi-squared test. Example:

 data - data.frame(distance=c(1,2,3,4,5,6,7), observed=c(43,13,10,6,2,1),
predicted=c(28, 18, 10, 5 ,3, 1, 1))

 chisq.test(data$observed, data$predicted)

Which gives:

Pearson's Chi-squared test

data:  data$observed and data$predicted
X-squared = 35, df = 25, p-value = 0.0882

Warning message:
In chisq.test(data$observed, data$predicted) :
  Chi-squared approximation may be incorrect

I understand this is due to having observed/predicted values of less than
five, however I am interested to know firstly why R uses such a large
number of degrees of freedom (when by my understanding there should only be
4 df), and secondly whether using the following manual calculation is
therefore inappropriate -

 X2 - sum(((data$observed - data$predicted)^2)/data$predicted)
 1-pchisq(X2,4)
[1] 0.04114223

If chi-squared is unsuitable, what other test can I use to determine
whether my observed and predicted data come from the same distribution? The
frequently recommended fisher's test doesn't seem to be any more
appropriate as it requires values of greater than 5 for contingency tables
larger than 2 x 2.

Thanks for your help.

Louise

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Re: [R] installing the package Rcplex

2012-02-20 Thread Uwe Ligges




On 20.02.2012 01:54, David Winsemius wrote:


On Feb 19, 2012, at 7:45 PM, zheng wei wrote:


I did not know this before. I installed it as you suggested. what to
do next?


Read the Installation Manual?



And don't forget this is a source package for which no CRAN Windows 
binary exists, hence it may be not that straightforward to get it done 
and you wil have to read the INSTALL file from the source package carefully.


Uwe Ligges

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Re: [R] GTK+

2012-02-20 Thread Uwe Ligges

On 19.02.2012 18:58, Idielle Walters wrote:

I am struggling to install GTK+ for Windows 7. RGtk2 needs this
package to load. Does anybody know of a installation file that works?

See the ReadMe for Windows binary packages. For the current R-reelase
this is

http://cran.r-project.org/bin/windows/contrib/2.14/ReadMe

and find that

- Package RGtk2 requires an an installation of Gtk+ aka Gtk2 = 2.20.
For 32-bit R, version 2.20 or later from
http://www.gtk.org/download/win32.php, e.g.

http://ftp.gnome.org/pub/gnome/binaries/win32/gtk+/2.22/gtk+-bundle_2.22.0-20101016_win32.zip
For 64-bit R, version 2.20 or later from
http://www.gtk.org/download/win64.php, e.g.

http://ftp.gnome.org/pub/gnome/binaries/win64/gtk+/2.22/gtk+-bundle_2.22.0-20101016_win64.zip
In each case, unpack the zip file in a suitable empty directory and
put the 'bin' directory in your path. NB: the 32-bit and 64-bit
distributions contain DLLs of the same names, and so you must ensure
that you have the 32-bit version in your path when running 32-bit R
and the 64-bit version when running 64-bit R - and the error messages
you get with the wrong version are confusing.

Uwe Ligges

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Re: [R] Sorting strings

2012-02-20 Thread Keith Jewell


Petr Savicky savi...@cs.cas.cz wrote in message 
news:20120220105153.gc21...@cs.cas.cz...
 On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:
 Hi all, I am having difficulties to understand how R sort strings:

 If I do
 R) sort(c(X.,X0B))
 [1] X.  X0B

 So for me, as far as lexicographic order is concerned I can add whatever 
 to
 the end, the order will remain the same, but :

 Hi.

 This neednot be true for strings of different length.
 For example

  ab
  abc

 become by concatenation with z

  abcz
  abz

 Petr Savicky.


That's not the explanation in this case.

The OP isn't telling us everything.
I get [R version 2.14.1 Platform: i386-pc-mingw32/i386 (32-bit)]:
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X.Z   X0B.Z

KJ

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[R] Problems with installation of lsa package

2012-02-20 Thread A J


Hi everybody!
First of all, I would like to point that I am newebie in using R. The issue is 
that I need to install lsa package in R. In theory, I have downloaded and 
installed all necessary packages to run lsa library, but when I am going to 
load it I get this message:
 library(lsa)Loading required package: SnowballError : .onLoad failed in 
 loadNamespace() for 'Snowball', details:  call: NULL  error: .onLoad failed 
 in loadNamespace() for 'rJava', details:  call: stop(No CurrentVersion entry 
 in ', key, '! Try re-installing Java and make sure R and Java have matching 
 architectures.)  error: objeto 'key' no encontradoError: package Snowball 
 could not be loaded
I have already installed Snowball and rJava, but there is no way to load lsa 
library. Any idea?
Thank you in advance.
Best,
AJ
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Re: [R] GTK

2012-02-20 Thread Yvonnick Noel




I am struggling to install GTK+ for Windows 7.  RGtk2 needs this
package to load.  Does anybody know of a installation file that works?




GTK+ is automatically installed when you install the RGtk2 package 
(you'll be asked about it during installation). As of R-2.14.1, it is 
installed under the R tree, so if you had write access when installing R 
itself, you should have no problem.


HTH,

Yvonnick Noel
University of Brittany, Rennes 2
France

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Re: [R] apply with as function ifelse with 2 logical conditions

2012-02-20 Thread Berend Hasselman


On 20-02-2012, at 10:40, Nerak wrote:

 Hi all,
 I have a question concerning using several conditions in an ifelse function
 used as the function in apply.
 I want to create a new value with the  function ifelse ‘ object which can be
 coerced to logical mode  “test[n,] 1  test[n-1,]==0” 
 With n I mean the row. I don’t know how I could do this without a loop. I
 want to avoid the usage of loops and was thinking about apply. This was what
 I was thinking about:
 
 test-data.frame(C=c(0,0,0,0,5,2,0,0,0,15,12,10,6,0,0,0),B=c(0,0,0,0,9,6,2,0,0,24,20,16,2,0,0,0),F=c(0,0,0,0,6,5,1,0,0,18,16,12,10,5,1,0))
 
 test.b-test[-(nrow(test)),]
 test.2b-rbind(0,test.b)
 result-as.data.frame(apply(test,M=2,function(x)ifelse((test1test.2b==0),1,0)))
 

You are not using the argument x  in the ifelse. 
For every column (you have 3) you are evaluating the ifelse condition.

 But I get 3 times the amount of rows than that I want… 
 what I should achieve:
 test.result-data.frame(C=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),B=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),F=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0))
 
 Has someone a suggestions about what I’m doing wrong?

Simplify. You don't need apply. This'll do it

result - ifelse((test1)  (test.2b==0),1,0)

Berend

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Re: [R] Sorting strings

2012-02-20 Thread Enrico Schumann



See ?Comparison, which holds some warnings about what to expect when 
sorting strings.



Am 20.02.2012 11:51, schrieb Petr Savicky:

On Mon, Feb 20, 2012 at 02:18:42AM -0800, statquant2 wrote:

Hi all, I am having difficulties to understand how R sort strings:

If I do
R) sort(c(X.,X0B))
[1] X.  X0B

So for me, as far as lexicographic order is concerned I can add whatever to
the end, the order will remain the same, but :


Hi.

This neednot be true for strings of different length.
For example

   ab
   abc

become by concatenation with z

   abcz
   abz

Petr Savicky.

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--
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Lucerne, Switzerland
http://nmof.net/

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Re: [R] Sorting strings

Ok so it changed from 2.12.2 to 2.14.1 ??
Can somebody tell me how to modify my sort or whatever to get the save
resilt that I would get in 2.14.1 ?

Cheers

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[R] How to determine a subset of a binary strings?

2012-02-20 Thread jing tang

Hi,

I need some neat ways of determing a subset of binary strings. For example,
x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
subset of y, but x is not a subset of z.

I tried to search R functions and packages but no hits. Any ideas?

 

Best,

Jing

 

--

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Senior Researcher

Finnish Institute of Molecular Medicine (FIMM)

FI-00014 University of Helsinki

Finland

 

 


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[R] How to inset a cell into a matrix?

2012-02-20 Thread xushaw

hi
i'm facing a problem
How to inset a cell into a matrix?
for example i have a matrix:
 
332   244   332244 
665   332   665332
785   785   785665

i want to covert the matrix like this:

02440244
332   332332332
6650665665
785785   785 0   

i find the minimum value from the 1st row, if any value greater than that, i
will inset a 0 in that column and move the whole column down.
but how to inset a cell?



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[R] Time taken to process a file after a socket connection was made

2012-02-20 Thread Aniruddha Mukherjee

Hello R people,

I have created a '.csv' file of 100 rows by 20 columns whose each cell 
contains a random numbers between 0  1, thru a Java program. Once that is 
created a signal (just a letter) is send to the port of a socket 
connection at localhost, which was earlier started by an R session. 
Now the R reads the '.csv' file into a data frame and calculates the 
average of 2000 numbers. This mean value was then written to the socket 
connection and Java received that successfully. The duration of this 
processing at R session took 1 min 03 seconds.
My question : Is the time duration of 63 seconds OK for only this set of 
activities or it should be less? If so how it can be reduced? Point to be 
noted that both the Java and R were running on the same PC and the 62 
seconds were consumed for the readLines command.

Thanks and regards.
Aniruddha.
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Re: [R] Run function several times changing only one argument - without a loop

2012-02-20 Thread Marion Wenty

Hi Enrico,
Yes, you were right, I just wanted to draw white vertical lines and with
your command it's much simpler. Thanks a lot!

Hi Eik,
I learned a lot from your tips about sapply and Vectorize!
I had only used apply, lapply and mapply but not yet sapply.
I find the Vectorize function very fascinating: being able to change a
function - this is something I hadn't come accross yet.
By checking out the command invisible I also learned a new command.
Thank you very much!

Marion


Hi Marion,

is all you want the white vertical lines? Then try

abline(v = seq(10, 90, by = 10), col = white)

instead of your axis commands.


Regards,
Enrico

2012/2/20 Eik Vettorazzi e.vettora...@uke.de

 Hi Marion,
 you can either use any of the *apply-functions or vectorize your
 function (which internally uses mapply):

 par(las=1)
 par(mar=c(5,13,4,2))
 barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)

 #using sapply
 invisible(sapply((1:9)*10,function(x)axis(2,pos=x,tick=T, tcl=F,
 labels=F,col=white)))

 #using Vectorize
 barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
 vaxis-Vectorize(axis,pos)
 invisible(vaxis(2,pos=(1:9)*10, tick=T, tcl=F, labels=F,col=white))

 Cheers!

 Am 20.02.2012 11:04, schrieb Marion Wenty:
  Dear people,
 
  I created a plot which looks like this:
 
  Ee1-matrix(c(88,86,74,62,41),ncol=5)
 
 colnames(Ee1)-c(Lehrer,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister)
  par(las=1)
  par(mar=c(5,13,4,2))
  barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
  axis(2,pos=10, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=20, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=30, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=40, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=50, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=60, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=70, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=80, tick=T, tcl=F, labels=F,col=white)
  axis(2,pos=90, tick=T, tcl=F, labels=F,col=white)
 
  Now I would like to shorten the whole thing - namely use only one step to
  create the 9 axes without having to use a loop.
 
  In general, I would be interested if there is a way to use a function
  several times changing only one argument, without having to use a loop.
 
  Does anyone know how to do that.
 
  Marion
 
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  R-help@r-project.org mailing list
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  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.


 --
 Eik Vettorazzi
 Institut für Medizinische Biometrie und Epidemiologie
 Universitätsklinikum Hamburg-Eppendorf

 Martinistr. 52
 20246 Hamburg

 T ++49/40/7410-58243
 F ++49/40/7410-57790

 --
 Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
 Genossenschaftsregister sowie das Unternehmensregister (EHUG):

 Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen
 Rechts; Gerichtsstand: Hamburg

 Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden),
 Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus



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Re: [R] proto: make a parameter persist

2012-02-20 Thread Ben quant

I like it better. Thanks!

Ben

On Fri, Feb 17, 2012 at 11:38 PM, Gabor Grothendieck 
ggrothendi...@gmail.com wrote:

 On Sat, Feb 18, 2012 at 12:44 AM, Ben quant ccqu...@gmail.com wrote:
  The code below works as expected but:
  Using the proto package, is this the best way to 1) make a parameter
  persist if the parameter is passed
  in with a value, 2) allow for calling the bias() function without a
  parameter assignment, 3) have
  the x2 value initialize as 5? Thanks for your feedback. Giving the
  proto package a test beat and
  establishing some templates for myself.
 
  oo - proto(expr = {x = c(10, 20, 15, 19, 17) x2 =
 5 # so x2 initializes as 5, but can be overwritten with param assignment
   bias - function(.,x2=.$x2) { # x2=.$x2 so no default
 param is needed .$x2 = x2 # so x2 persists in the
 env .$x - .$x + x2 } }) o =
 oo$proto() o$x # [1] 10 20 15 19 17 o$x2 #[1] 5 o$bias(x2 = 100) o$x2 #
 [1] 100 o$x # [1] 110 120 115 119 117
 

 This is not very different from what you have already but here it is
 for comparison.  Note that the with(...) line has the same meaning as
 .$x - .$x + .$x2 :

 oo - proto(
   x = c(10, 20, 15, 19, 17),
   x2 = 5,
   bias = function(., x2) {
  if (!missing(x2)) .$x2 - x2
  with(., x - x + x2)
   }
 )

 --
 Statistics  Software Consulting
 GKX Group, GKX Associates Inc.
 tel: 1-877-GKX-GROUP
 email: ggrothendieck at gmail.com


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[R] Triangular Test

2012-02-20 Thread kende jan

Hello,

I would like to perform triangular test for clinical trial with R.
can you help me please ?

Jan

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Re: [R] How to determine a subset of a binary strings?

2012-02-20 Thread Berend Hasselman


On 20-02-2012, at 14:15, jing tang wrote:

 Hi,
 
 I need some neat ways of determing a subset of binary strings. For example,
 x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
 subset of y, but x is not a subset of z.
 
 I tried to search R functions and packages but no hits. Any ideas?

x - c(0,0,1)
y - c(0,1,1)
z - c(0,1,0)

any(x  y)
any(z  y)
any(x  z)

Berend

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Re: [R] coxme: model simplification using LR-test?

2012-02-20 Thread Terry Therneau

Summary of the query: update does not work on a coxme object

 I ran into this bug myself 2 days ago -- I rarely use update() so
hadn't encountered it before.  The problem is that coxme breaks the
formula into fixed and random portions, and this confuses the default
method for formula.

Solution: add the following function:
  formula.coxme - function(x, ...) x$call$formula

 This method will be included in my next update of coxme.

Terry Therneau

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Re: [R] Finicky factor comparison operators



On Feb 20, 2012, at 1:45 AM, johnmark wrote:


MIchael -

Thanks for your insight.  I think I see where you're going with this.

To make '==' comparisons for subsetting against an ordered factor,  
I've had
to create a lookup table for all possible values I'd ever want to  
compare
against (all dates covered by the quarters in question, in this  
case) that
maps into the ordered factors values.  This is wrapped by a function  
that

returns an ordered factor, which allows me to write:

/(opps$close_quarter == which.quarter.end(2010-10-20)/

Otherwise if I try to create an ordered factor from the constant  
just for
the purposes of comparison, the error tells me that ordered factors  
from

different sources cannot be compared:

/(opps$close_quarter == factor(2007-10-20, ordered=T)
Error in Ops.factor(factor(2007-10-30, ordered = T),  
quarter.factors[1,

2]) :
 level sets of factors are different/


Actually it is telling you that you cannot compare ordered factors  
which have different levels. That makes perfect sense for the same  
reasons that you are not allowed to compare Dates to ordered factors.  
If the factors from different sources had the same levels you should  
have succeeded.


 z - factor(LETTERS[3:1], ordered = TRUE)
 z3 - factor(LETTERS[1:3] , ordered=TRUE)
 z[2] == z3[2]
[1] TRUE




That makes sense, since internally factors are integers -- enums  
in other

terms.

But what I want to avoid -- and what I don't see as necessary is  
explicitly
coercing the terms to a common representation that mimics their  
print form:


/as.character(2007-10-20)== as.character(factor(2007-10-20,  
ordered=T))

/
I don't think there should be confusion since the conversion to  
print form
is obvious -- but it does conflict with the conversion rules for  
creating

vectors by c():

/c(2011-10-20, factor(2007-10-20, ordered=T))
[1] 2011-10-20 1 /

where the factor is converted to its internal enum representation,  
then to

a character.


That just an example of the need to use as.character when converting  
data out of factor class.




Having given this some more thought to what motivated the original  
question,

one could use which() to invert the factor's levels vector:

/which(2008-04-30 == levels(quarter.factors[,2]))
[1] 3 /

Its still not clear to me what exactly are the implicit conversion  
rules for

factors.


In your last case you are comparing a character to a character value  
and getting the expected result. (Since levels(quarter.factors) is NOT  
a factor.)  You should also succeed when testing equality between  
ordered factor and character types. You have still not provided an  
example for testing so this may suffice.


 z - factor(LETTERS[3:1], ordered = TRUE)
 z == A
[1] FALSE FALSE  TRUE

You should be able to assemble a list of valid candidate (character)  
values with levels(fac). Or if you want them in factor representation  
then use unique(fac).



--
David Winsemius, MD
West Hartford, CT

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Re: [R] Help on lattice barchart ploting



On Feb 20, 2012, at 5:34 AM, MLSC wrote:


Hi friends,

I have following data and would like to plot this with barchart()  
availble

with lattice package.

  RsID  Freqs Genotype
AAA 63.636  1/1
AAA 32.727  1/2
AAA  3.636  2/2
BBB 85.965  2/2
BBB 14.035  2/1
CCC 63.158  1/1
CCC 21.053  1/2
CCC 15.789  2/2
DDD 26.786  2/2
DDD 46.429  2/1
DDD 26.786  1/1
EEE 32.759  2/2
EEE 43.103  2/1
EEE 24.138  1/1
EEE 37.931  1/1
EEE 51.724  1/2
EEE 10.345  2/2
FFF 23.214  2/2
FFF 53.571  2/1
FFF 23.214  1/1
GGG 46.552  1/1
GGG 44.828  1/2
GGG  8.621  2/2
HHH 65.517  2/2
HHH 32.759  2/1
HHH  1.724  1/1

Following is the code which I have written to get the plot.

barchart(Genotype~Freqs | RsID, data=gDataFr,layout=c(4,6),
   main=Genotype Frequency,
   ylab=Genotype, xlab=Frequency,
   scales=list(x=list(alternating=c(1, 1, 1))),
   panel=function(x,y,...){
   panel.fill(col=white)
   panel.grid(-1,0,lty=3,col=black)
   panel.barchart(x,y,col=c(blue,green,red),...)
   }
)

but in plot, x-axis scale ranges from 0 to 30 though I have Freq  
column
values in range of 0 to 100. I have tried many ways to get x-axis  
scale to

0 to 100, including xlim=c(1:100), but still bars in plot are not
propotional to the range 0 to 100. hence can somebody let me know  
how get

bars in the plot propotional to x-axis range 0 to 100?



Check the class of Genotype. I do not see what you are describing when  
I create a dataset from your example. I suspect you had implicit  
coercion of your Genotype column to factor class. Or that you  
recognized that and forgot to tell us that you made the newbie error  
of using as.numeric() without first converting using as.character()  
from factor.



--
David Winsemius, MD
West Hartford, CT

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Re: [R] How to determine a subset of a binary strings?

2012-02-20 Thread Enrico Schumann



Hi Jing,

I am not sure I got your definition of a subset right, but maybe this 
helps.


Regards,
Enrico

x - c(0,0,1)
y - c(0,1,1)
z - c(0,1,0)

## is x a 'subset' of y?
isSubset - function(x, y) {
x - as.logical(x)
y - as.logical(y)
all(y[x] == TRUE)
}

isSubset(x, y)
isSubset(z, y)
isSubset(x, z)

## create all subsets
exampleVec - c(0,1,1,0,1)
fun - function(t)
if (t) 0:1 else 0
expand.grid(lapply(as.list(exampleVec), fun))



Am 20.02.2012 14:15, schrieb jing tang:

Hi,

I need some neat ways of determing a subset of binary strings. For example,
x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
subset of y, but x is not a subset of z.

I tried to search R functions and packages but no hits. Any ideas?



Best,

Jing



--

Jing Tang, PhD

Senior Researcher

Finnish Institute of Molecular Medicine (FIMM)

FI-00014 University of Helsinki

Finland






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--
Enrico Schumann
Lucerne, Switzerland
http://nmof.net/

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Re: [R] Run function several times changing only one argument - without a loop



On Feb 20, 2012, at 5:04 AM, Marion Wenty wrote:


Dear people,

I created a plot which looks like this:

Ee1-matrix(c(88,86,74,62,41),ncol=5)
colnames(Ee1)- 
c 
(Lehrer 
,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister)

par(las=1)
par(mar=c(5,13,4,2))
barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
axis(2,pos=10, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=20, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=30, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=40, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=50, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=60, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=70, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=80, tick=T, tcl=F, labels=F,col=white)
axis(2,pos=90, tick=T, tcl=F, labels=F,col=white)

Now I would like to shorten the whole thing - namely use only one  
step to

create the 9 axes without having to use a loop.


See if grid() is helpful:

barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
grid(nx=9, ny=NA, col=white)

Somehow using axis to draw elements inside a graph seems wrong,  
especially when you are using it for what is basically  side-effect as  
you have been attempting. Could also have used abline which accepts a  
vector for its v argument.


 barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F)
 abline(v=seq(10,90,by=10), col=white)




In general, I would be interested if there is a way to use a function
several times changing only one argument, without having to use a  
loop.


A large number of functions will offer that facility using seq() as  
the functional input argument.


Does anyone know how to do that.

Marion

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You might consider working in the settings for your mail client. I  
know that gmail offers that option.


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West Hartford, CT

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Re: [R] chisq.test vs manual calculation - why are different results produced?



On Feb 20, 2012, at 5:57 AM, Louise Mair wrote:


Hello,

I am trying to fit gamma, negative exponential and inverse power  
functions
to a dataset, and then test whether the fit of each curve is good.  
To do
this I have been advised to calculate predicted values for bins of  
data (I
have grouped a continuous range of distances into 1km bins), and  
then apply

a chi-squared test. Example:

data - data.frame(distance=c(1,2,3,4,5,6,7),  
observed=c(43,13,10,6,2,1),

predicted=c(28, 18, 10, 5 ,3, 1, 1))


There's an error with that code.




chisq.test(data$observed, data$predicted)


Which gives:

   Pearson's Chi-squared test

data:  data$observed and data$predicted
X-squared = 35, df = 25, p-value = 0.0882

Warning message:
In chisq.test(data$observed, data$predicted) :
 Chi-squared approximation may be incorrect

I understand this is due to having observed/predicted values of less  
than

five, however I am interested to know firstly why R uses such a large
number of degrees of freedom (when by my understanding there should  
only be

4 df), and secondly whether using the following manual calculation is
therefore inappropriate -


Read the help page Details section  end of second paragraph.

You probably wanted:

chisq.test(cbind(data$observed, data$predicted))




X2 - sum(((data$observed - data$predicted)^2)/data$predicted)
1-pchisq(X2,4)

[1] 0.04114223

If chi-squared is unsuitable, what other test can I use to determine
whether my observed and predicted data come from the same  
distribution? The

frequently recommended fisher's test doesn't seem to be any more
appropriate as it requires values of greater than 5 for contingency  
tables

larger than 2 x 2.

Thanks for your help.

Louise

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Plain text is requested as the mail format.


David Winsemius, MD
West Hartford, CT

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Re: [R] Sorting strings

I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?

--
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Re: [R] Reporting Kaplan-Meier / Cox-Proportional Hazard Standard Error, km.coxph.plot, survfit.object

2012-02-20 Thread Frank Harrell

The 10% rule does not provide a unique answer. Should it apply to the
cumulative probability, its logarithm, or log-log (log hazard scale)? Many
studies are too small to achieve 10% at any time point. I think it is more
traditional (but not without bias) to stop where fewer than 10 subjects are
still being followed. There's room for many other choices though.
Sometimes I think that the curve should go to the max but be accompanied by
confidence bands.
Frank

Paul Johnston wrote

What is the best way to report the standard error when publishing
Kaplan-Meier plots? In my field (Vascular Surgery), practitioners
loosely refer to the 10% error cutoff as the point at which to stop
drawing the KM curve. I am interpreting this as the *standard error
of the cumulative hazard*, although I'm having a difficult time
finding some guidelines about this (perhaps I am not searching the
correct terms or references). My KM figures contain typically two
curves that I am comparing using the logrank test. Inspecting the
?survfit.object yields the std.err field that gives the standard error
for each timepoint on the curve.

Is it recommended that I just name the timepoint at which the standard
error exceeds 0.1 in the figure legend? For example, The standard
error exceeds 10% at time points beyond 394 days. I have seen this
strategy in other publications.

What is your approach?

Thanks for your help,
PCJ

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-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] Help on lattice barchart ploting

2012-02-20 Thread Richard M. Heiberger

I do not see the restricted range that you report.
There is probably some masking as David pointed out.
Try again in a fresh R session with --vanilla.

You probably want three additional arguments.
update(.Last.value, xlim=c(0, 100), between=list(x=1, y=1), origin = 0)
to the barchart function call.

The origin=0 makes the bars start at 0, not at the left margin of the panel.
The between visually separates the panels.
If you want the values 0 and 100 to appear on the x-axis, then you will
need to use
xlim=x(-5, 105)
On Mon, Feb 20, 2012 at 5:34 AM, MLSC mlscm...@gmail.com wrote:

 Hi friends,

 I have following data and would like to plot this with barchart() availble
 with lattice package.

   RsID  Freqs Genotype
 AAA 63.636  1/1
 AAA 32.727  1/2
 AAA  3.636  2/2
 BBB 85.965  2/2
 BBB 14.035  2/1
 CCC 63.158  1/1
 CCC 21.053  1/2
 CCC 15.789  2/2
 DDD 26.786  2/2
 DDD 46.429  2/1
 DDD 26.786  1/1
 EEE 32.759  2/2
 EEE 43.103  2/1
 EEE 24.138  1/1
 EEE 37.931  1/1
 EEE 51.724  1/2
 EEE 10.345  2/2
 FFF 23.214  2/2
 FFF 53.571  2/1
 FFF 23.214  1/1
 GGG 46.552  1/1
 GGG 44.828  1/2
 GGG  8.621  2/2
 HHH 65.517  2/2
 HHH 32.759  2/1
 HHH  1.724  1/1

 Following is the code which I have written to get the plot.

 barchart(Genotype~Freqs | RsID, data=gDataFr,layout=c(4,6),
main=Genotype Frequency,
ylab=Genotype, xlab=Frequency,
scales=list(x=list(alternating=c(1, 1, 1))),
panel=function(x,y,...){
panel.fill(col=white)
panel.grid(-1,0,lty=3,col=black)
panel.barchart(x,y,col=c(blue,green,red),...)
}
 )

 but in plot, x-axis scale ranges from 0 to 30 though I have Freq column
 values in range of 0 to 100. I have tried many ways to get x-axis scale to
 0 to 100, including xlim=c(1:100), but still bars in plot are not
 propotional to the range 0 to 100. hence can somebody let me know how get
 bars in the plot propotional to x-axis range 0 to 100?

 Regards,
 mlsc

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Re: [R] Sorting strings

I don't *think* it's version specific, but rather it depends on your
(still unstated) locale, as the documentation goes to great lengths to
point out. Change that and you might see different behaviors.

Michael

On Mon, Feb 20, 2012 at 8:55 AM, statquant2 statqu...@gmail.com wrote:
 I did, but this does not give the answer to my question...
 Anybody knows how to tweack the behaviour of sort or how to do ?

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4404091.html
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[R] Confused: Inconsistent result?

2012-02-20 Thread Ajay Askoolum

This is copy  paste from my session:

 xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
 dim(xyz)-c(length(xyz)/2,2)
 
 allobj-function(){
+ xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+ }
 xyz
  [,1]  [,2]   
 [1,] a   character
 [2,] aa  character
 [3,] abc character
 [4,] AirPassengers   character
 [5,] allobj  character
 [6,] allObjects  character
 [7,] allObjects2 character
 [8,] arrayFromAPL    character
 [9,] classes character
[10,] myCharVector    character
[11,] myDateVector    character
[12,] myNumericVector character
[13,] newArrayFromAPL character
[14,] obj character
[15,] objClass    character
[16,] x   character
[17,] xyz character
[18,] y   character
 allobj()
 [,1] [,2]
 

As far as I can see, the function allobj has the same expressions as those 
executed from the command line. Why are the results different?
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[R] stats on transitions from one state to another

2012-02-20 Thread Murali.Menon

Folks,

I'm trying to get stats from a matrix for each transition from one state to 
another.

I have a matrix x as below.

structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0, 
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19, 
-0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38, 
0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66, 
0.9, -0.94, -1.29, -0.23, -0.92, -0.21, 1.02, -0.77, -0.68, -0.33, 
0.04), .Dim = c(20L, 3L), .Dimnames = list(NULL, c(State, V1, 
V2)))

Is it possible to get, say, mean values of each variable in state 1 when the 
previous state was 0, in state 2 when the previous state was 0, and so on with 
all available transitions between states 0, 1, 2?

In the above case, mean of V1 in state 2 when previous state was 0 would be 

mean(c(-0.57, -0.59, 0.16, 0.06, 0.84)) = -0.02

while the mean of V1 in state 0 when previous state was 2 would be:

mean(c(1.62, 0.18, -0.81)) = 0.33

If I try something like

by(x[, 2:3], x[, 1], FUN = colMeans)

I get the means for each state. I'm not sure how to get the split by transition?

Thanks,

Murali

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Re: [R] Confused: Inconsistent result?



On Feb 20, 2012, at 10:07 AM, Ajay Askoolum wrote:


This is copy  paste from my session:


xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
dim(xyz)-c(length(xyz)/2,2)

allobj-function(){

+ xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+ }

xyz

  [,1]  [,2]
 [1,] a   character
 [2,] aa  character
 [3,] abc character
 [4,] AirPassengers   character
 [5,] allobj  character
 [6,] allObjects  character
 [7,] allObjects2 character
 [8,] arrayFromAPLcharacter
 [9,] classes character
[10,] myCharVectorcharacter
[11,] myDateVectorcharacter
[12,] myNumericVector character
[13,] newArrayFromAPL character
[14,] obj character
[15,] objClasscharacter
[16,] x   character
[17,] xyz character
[18,] y   character

allobj()

 [,1] [,2]




As far as I can see, the function allobj has the same expressions as  
those executed from the command line. Why are the results different?


The ls function looks only in the local environment if not supplied  
with specific directions about where to look.



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David Winsemius, MD
West Hartford, CT

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Re: [R] Confused: Inconsistent result?

Sorry, just checked it and you need to add .GlobalEnv to both ls() calls.


Michael

On Mon, Feb 20, 2012 at 10:17 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 Short answer, environments -- ls() looks (by default) in its current
 environment, which is not the same as the global environment when
 being called inside a function.

 This would (I think) give the same answer but I haven't checked it. :

 allobj-function(){
 + xyz-as.vector(c(ls(.GlobalEnv),as.matrix(lapply(ls(),class;
 + dim(xyz)-c(length(xyz)/2,2);
 + return(xyz)
 + }

 On Mon, Feb 20, 2012 at 10:07 AM, Ajay Askoolum aa2e...@yahoo.co.uk wrote:
 This is copy  paste from my session:

 xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
 dim(xyz)-c(length(xyz)/2,2)

 allobj-function(){
 + xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
 + dim(xyz)-c(length(xyz)/2,2);
 + return(xyz)
 + }
 xyz
   [,1]  [,2]
  [1,] a   character
  [2,] aa  character
  [3,] abc character
  [4,] AirPassengers   character
  [5,] allobj  character
  [6,] allObjects  character
  [7,] allObjects2 character
  [8,] arrayFromAPL    character
  [9,] classes character
 [10,] myCharVector    character
 [11,] myDateVector    character
 [12,] myNumericVector character
 [13,] newArrayFromAPL character
 [14,] obj character
 [15,] objClass    character
 [16,] x   character
 [17,] xyz character
 [18,] y   character
 allobj()
  [,1] [,2]


 As far as I can see, the function allobj has the same expressions as those 
 executed from the command line. Why are the results different?
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Re: [R] Confused: Inconsistent result?

Short answer, environments -- ls() looks (by default) in its current
environment, which is not the same as the global environment when
being called inside a function.

This would (I think) give the same answer but I haven't checked it. :

 allobj-function(){
+ xyz-as.vector(c(ls(.GlobalEnv),as.matrix(lapply(ls(),class;
+ dim(xyz)-c(length(xyz)/2,2);
+ return(xyz)
+ }

On Mon, Feb 20, 2012 at 10:07 AM, Ajay Askoolum aa2e...@yahoo.co.uk wrote:
 This is copy  paste from my session:

 xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
 dim(xyz)-c(length(xyz)/2,2)

 allobj-function(){
 + xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
 + dim(xyz)-c(length(xyz)/2,2);
 + return(xyz)
 + }
 xyz
   [,1]  [,2]
  [1,] a   character
  [2,] aa  character
  [3,] abc character
  [4,] AirPassengers   character
  [5,] allobj  character
  [6,] allObjects  character
  [7,] allObjects2 character
  [8,] arrayFromAPL    character
  [9,] classes character
 [10,] myCharVector    character
 [11,] myDateVector    character
 [12,] myNumericVector character
 [13,] newArrayFromAPL character
 [14,] obj character
 [15,] objClass    character
 [16,] x   character
 [17,] xyz character
 [18,] y   character
 allobj()
  [,1] [,2]


 As far as I can see, the function allobj has the same expressions as those 
 executed from the command line. Why are the results different?
        [[alternative HTML version deleted]]


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[R] Computing plot size in Sweave

2012-02-20 Thread BXC (Bendix Carstensen)

Sometimes you want to compute the physical size of a plot based on data.
In R itself this is no problem.

But is there a way to compute the values of height and width in S-weave, say:

graph,fig=TRUE,height=xx,width=yy=

where xx and yy are computed and not physically written in the document?

Bendix
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Bendix Carstensen 
Senior Statistician
Epidemiology

Steno Diabetes Center A/S
Niels Steensens Vej 2-4
DK-2820 Gentofte
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+45 30 75 87 38 (mobile)
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This e-mail (including any attachments) is intended for ...{{dropped:8}}

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Re: [R] Confused: Inconsistent result?

2012-02-20 Thread Petr PIKAL

Hi
 
 This is copy  paste from my session:
 
  xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class
  dim(xyz)-c(length(xyz)/2,2)
  
  allobj-function(){
 + xyz-as.vector(c(ls(),as.matrix(lapply(ls(),class;
 + dim(xyz)-c(length(xyz)/2,2);
 + return(xyz)
 + }
  xyz
   [,1]  [,2]   
  [1,] a   character
  [2,] aa  character
  [3,] abc character
  [4,] AirPassengers   character
  [5,] allobj  character
  [6,] allObjects  character
  [7,] allObjects2 character
  [8,] arrayFromAPLcharacter
  [9,] classes character
 [10,] myCharVectorcharacter
 [11,] myDateVectorcharacter
 [12,] myNumericVector character
 [13,] newArrayFromAPL character
 [14,] obj character
 [15,] objClasscharacter
 [16,] x   character
 [17,] xyz character
 [18,] y   character
  allobj()
  [,1] [,2]
  
 
 As far as I can see, the function allobj has the same expressions as 
those
 executed from the command line. Why are the results different?

Probably due to environment handling. 

Do you really want to check if ls behaves as is intended and that it 
produces character vector? Or your intention is a little bit more 
ambitious and you want to know what objects do you have?

If the later, I recommend to use this function:

function (pos = 1, pattern, order.by) 
{
napply - function(names, fn) sapply(names, function(x) fn(get(x, 
pos = pos)))
names - ls(pos = pos, pattern = pattern)
obj.class - napply(names, function(x) as.character(class(x))[1])
obj.mode - napply(names, mode)
obj.type - ifelse(is.na(obj.class), obj.mode, obj.class)
obj.size - napply(names, object.size)
obj.dim - t(napply(names, function(x) as.numeric(dim(x))[1:2]))
vec - is.na(obj.dim)[, 1]  (obj.type != function)
obj.dim[vec, 1] - napply(names, length)[vec]
out - data.frame(obj.type, obj.size, obj.dim)
names(out) - c(Type, Size, Rows, Columns)
if (!missing(order.by)) 
out - out[order(out[[order.by]]), ]
out
}







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Re: [R] stats on transitions from one state to another



On Feb 20, 2012, at 10:11 AM, murali.me...@avivainvestors.com wrote:


Folks,

I'm trying to get stats from a matrix for each transition from one  
state to another.


I have a matrix x as below.

structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19,
-0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38,
0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66,
0.9, -0.94, -1.29, -0.23, -0.92, -0.21, 1.02, -0.77, -0.68, -0.33,
0.04), .Dim = c(20L, 3L), .Dimnames = list(NULL, c(State, V1,
V2)))

Is it possible to get, say, mean values of each variable in state 1  
when the previous state was 0, in state 2 when the previous state  
was 0, and so on with all available transitions between states 0, 1,  
2?


In the above case, mean of V1 in state 2 when previous state was 0  
would be


mean(c(-0.57, -0.59, 0.16, 0.06, 0.84)) = -0.02

while the mean of V1 in state 0 when previous state was 2 would be:

mean(c(1.62, 0.18, -0.81)) = 0.33

If I try something like

by(x[, 2:3], x[, 1], FUN = colMeans)

I get the means for each state. I'm not sure how to get the split by  
transition?


Add an extra column of previous states: and tabulate:

 sss -cbind(sss,  c(NA, sss[,State][-nrow(sss)]) )

 table(sss[,State], sss[,4])

0 1 2
  0 3 1 1
  1 1 5 1
  2 2 1 4

The requested means for V1 by transition types:
 tapply(sss[,V1], INDEX=interaction(sss[,State], sss[,4]), mean)
   0.01.02.00.11.12.10.21.22.2
-0.670 -0.190 -0.255  0.390 -0.640  2.790 -1.620  1.630  0.205

The counts on which those means are based:
 tapply(sss[,V1], INDEX=interaction(sss[,State], sss[,4]), length)
0.0 1.0 2.0 0.1 1.1 2.1 0.2 1.2 2.2
  3   1   2   1   5   1   1   1   4

--

David Winsemius, MD
West Hartford, CT

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Re: [R] How to resample matrices to test for the robustness of their correlation

2012-02-20 Thread camilleislande

Thank you Chuck,

Here is the head of my data set (tjornres):
   Fish.1 Fish.2  MORPHO  DIET
1 1  20.03768   0.1559250
2 1  30.05609   0.7897060
3 1  40.03934   0.4638010
4 1  50.03363   0.1200480
5 1  60.05629   0.4390760
6 1  80.08366   0.1866750
7 1  90.04892   0.0988235
8 1 10   0.04427   0.2637140

MORPHO and DIET refer to the morphological and diet distances between fish 1
and fish 2. My original data set has over 2400 pairs of fish. My goal  is to
resample this dataste by selecting only 435.
I would like to do this 999 times and get a distribution of the correlation
coefficients MORPHO~DIET. 

I went on and wrote this code:

head(tjornres)

essayres = tjornres  # copy of the data 
R = 999 # the number of replicates  
   
cor.values = numeric(R) # store the data 
for (i in 1:R) {  # loop 
+ group1 = sample(essayres, size=435, replace=F)
+ group2 = sample(essayres, size=435, replace=F)
+ cor.values[i] = cor.test(group1,group2)$cor
+ }

I have a syntax error in this code. 

Also if I run one resampling, sample(essayres, size=435, replace=F), I get
this error message: Error in `[.data.frame`(x, .Internal(sample(length(x),
size, replace,  : cannot take a sample larger than the population when
'replace = FALSE'.

Does anyone know why this code is not working? Are there any other ways to
resample (without replacement) ?
Thank you for your help,







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Re: [R] How to determine a subset of a binary strings?

On Mon, Feb 20, 2012 at 03:15:53PM +0200, jing tang wrote:
 Hi,
 
 I need some neat ways of determing a subset of binary strings. For example,
 x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
 subset of y, but x is not a subset of z.
 
 I tried to search R functions and packages but no hits. Any ideas?

Hi.

Try this

  all(x = y) # [1] TRUE
  all(z = y) # [1] TRUE
  all(x = z) # [1] FALSE

Hope this helps.

Petr Savicky.

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Re: [R] Sorting strings

On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
 I did, but this does not give the answer to my question...
 Anybody knows how to tweack the behaviour of sort or how to do ?

Hi.

Try this

  Sys.setlocale(LC_COLLATE, C) 


This comes from ?locale and reads there

 Sys.setlocale(LC_COLLATE, C)   # turn off locale-specific sorting,
#  usually

See also ?sort

 The sort order for character vectors will depend on the collating
 sequence of the locale in use: see ‘Comparison’.

?Comparison

 Comparison of strings in character vectors is lexicographic within
 the strings using the collating sequence of the locale in use: see
 ‘locales’.  The collating sequence of locales such as ‘en_US’ is
 normally different from ‘C’ (which should use ASCII) and can be
 surprising.  Beware of making _any_ assumptions about the
 collation order: ...

Hope this helps.

Petr Savicky.

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Re: [R] Sorting strings

2012-02-20 Thread Rui Barradas

Hello,


statquant2 wrote
 
 Ok so it changed from 2.12.2 to 2.14.1 ??
 Can somebody tell me how to modify my sort or whatever to get the save
 resilt that I would get in 2.14.1 ?
 
 Cheers
 

I don't know about 2.12.2 but for 2.12.0 I get:

 R.version
   _
platform   i386-pc-mingw32  
arch   i386 
os mingw32  
system i386, mingw32
status  
major  2
minor  12.0 
year   2010 
month  10   
day15   
svn rev53317
language   R
version.string R version 2.12.0 (2010-10-15)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z)) 
[1] X.Z   X0B.Z

And the same for 2.14.1:

 R.version
   _
platform   i386-pc-mingw32
[... deleted...]
version.string R version 2.14.1 (2011-12-22)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z)) 
[1] X.Z   X0B.Z

Could it be OS related?

Rui Barradas.

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[R] Metafor: Moderator variables for each study arm

2012-02-20 Thread Faiz Ahmad Khan

Hello,

I am trying to do a meta-analysis where each study has two arms, similar to the 
BCG data set. However, follow-up duration was different for each study arm, so 
I would like to fit a model that uses the length of follow-up in each arm as a 
moderator. Is this possible?

For example, if this was the BCG data, each study would have 2 follow up 
variables: 1 for the treatment group, and 1 for the control
group. 

thanks for the help,
Faiz

.
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[R] Column wise matrix multiplication

2012-02-20 Thread Graziano Mirata

Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting 
vector with length equal to the number of rows of the original matrix. In short 
I would like to do what prod(.) function in Matlab does, i.e.

A -matrix(c(1:10),5,2)

V = A[,1]*A[,2]

Thank you

Graziano



[[alternative HTML version deleted]]

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Re: [R] Sorting strings

Ok I have :

R) str(R.Version())
List of 13
 $ platform  : chr x86_64-unknown-linux-gnu
 $ arch  : chr x86_64
 $ os: chr linux-gnu
 $ system: chr x86_64, linux-gnu
 $ status: chr 
 $ major : chr 2
 $ minor : chr 12.2
 $ year  : chr 2011
 $ month : chr 02
 $ day   : chr 25
 $ svn rev   : chr 54585
 $ language  : chr R
 $ version.string: chr R version 2.12.2 (2011-02-25)

R) sort(c(X.,X0B))
[1] X.  X0B
R) sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z  

I am using a linux redHat 
$ uname -a
Linux 2.6.18-238.9.1.el5 #1 SMP Fri Mar 18 12:42:39 EDT 2011 x86_64 x86_64
x86_64 GNU/Linux


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[R] FW: Problems with R

2012-02-20 Thread Thorsten Pöllinger



 





Dears,
 
I am a new R user and I am trying to analyze my data sets, R gives me a default 
when I type in the regression formula as following:
 
fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue$revenue.x,family=NO)
Fehler in model.frame.default(formula = tot_remun_revenue$tot_remun.y ~  : 
  ungültiger Typ (NULL) für die Variable 'tot_remun_revenue$tot_remun.y' Is 
there anything wrong in my data set, as the instruction is directly copied from 
lectures I had about R? Attached please find the data set as .csv. Best regards 
and thanks in advance! Thorsten Poellinger__
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Re: [R] Column wise matrix multiplication

2012-02-20 Thread Dimitris Rizopoulos


Try

apply(A, 1, prod)


I hope it helps.

Best,
Dimitris


On 2/20/2012 4:21 PM, Graziano Mirata wrote:

Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting 
vector with length equal to the number of rows of the original matrix. In short 
I would like to do what prod(.) function in Matlab does, i.e.

A-matrix(c(1:10),5,2)

V = A[,1]*A[,2]

Thank you

Graziano



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

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/

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Re: [R] stats on transitions from one state to another

2012-02-20 Thread Berend Hasselman


On 20-02-2012, at 16:11, murali.me...@avivainvestors.com 
murali.me...@avivainvestors.com wrote:

 Folks,
 
 I'm trying to get stats from a matrix for each transition from one state to 
 another.
 
 I have a matrix x as below.
 
 structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0, 
 0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19, 
 -0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38, 
 0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66, 
 0.9, -0.94, -1.29, -0.23, -0.92, -0.21, 1.02, -0.77, -0.68, -0.33, 
 0.04), .Dim = c(20L, 3L), .Dimnames = list(NULL, c(State, V1, 
 V2)))
 
 Is it possible to get, say, mean values of each variable in state 1 when the 
 previous state was 0, in state 2 when the previous state was 0, and so on 
 with all available transitions between states 0, 1, 2?
 
 In the above case, mean of V1 in state 2 when previous state was 0 would be 
 
 mean(c(-0.57, -0.59, 0.16, 0.06, 0.84)) = -0.02
 

What do you mean by previous state.
I do this 

 x - cbind(c(NA, x[,State][-nrow(x)]), x)
 colnames(x)[1] - State.prev
 x
  State.prev StateV1V2
 [1,] NA 0  0.21  0.51
 [2,]  0 2 -0.57 -1.00
 [3,]  2 2 -0.59 -1.29
 [4,]  2 2  0.16  2.15
 [5,]  2 0 -1.62  0.39
 [6,]  0 0  0.18  0.78
 [7,]  0 0 -0.81  0.85
 [8,]  0 1 -0.19  1.18
 [9,]  1 1 -0.76  1.66
[10,]  1 1  0.74  0.90
[11,]  1 1 -1.51 -0.94
[12,]  1 2  2.79 -1.29
[13,]  2 2  0.41 -0.23
[14,]  2 1  1.63 -0.92
[15,]  1 1 -0.86 -0.21
[16,]  1 1 -0.81  1.02
[17,]  1 0  0.39 -0.77
[18,]  0 0 -1.38 -0.68
[19,]  0 2  0.06 -0.33
[20,]  2 2  0.84  0.04

The mean of V1 in state 2 when the previous state is 0 would be in my 
interpretation

mean(c(-0.57, 0.06))

 while the mean of V1 in state 0 when previous state was 2 would be:
 
 mean(c(1.62, 0.18, -0.81)) = 0.33
 

Your second case would be the mean of -1.62

Berend

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Re: [R] Sorting strings

2012-02-20 Thread De-Jian Zhao

It seems OS-dependent. I got different results when trying it on windows 
xp and Redhat linux.



 R.version
   _
platform   x86_64-unknown-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  9.1
year   2009
month  06
day26
svn rev48839
language   R
version.string R version 2.9.1 (2009-06-26)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X.Z   X0B.Z


 R.version
   _
platform   x86_64-unknown-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  9.1
year   2009
month  06
day26
svn rev48839
language   R
version.string R version 2.9.1 (2009-06-26)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z


On 2012-2-20 23:27, statquant2 wrote:

Ok I have :

R) str(R.Version())
List of 13
  $ platform  : chr x86_64-unknown-linux-gnu
  $ arch  : chr x86_64
  $ os: chr linux-gnu
  $ system: chr x86_64, linux-gnu
  $ status: chr 
  $ major : chr 2
  $ minor : chr 12.2
  $ year  : chr 2011
  $ month : chr 02
  $ day   : chr 25
  $ svn rev   : chr 54585
  $ language  : chr R
  $ version.string: chr R version 2.12.2 (2011-02-25)

R) sort(c(X.,X0B))
[1] X.  X0B
R) sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z

I am using a linux redHat
$ uname -a
Linux 2.6.18-238.9.1.el5 #1 SMP Fri Mar 18 12:42:39 EDT 2011 x86_64 x86_64
x86_64 GNU/Linux


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Re: [R] Sorting strings

On Mon, Feb 20, 2012 at 04:56:21PM +0100, Petr Savicky wrote:
 On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
  I did, but this does not give the answer to my question...
  Anybody knows how to tweack the behaviour of sort or how to do ?
 
 Hi.
 
 Try this
 
   Sys.setlocale(LC_COLLATE, C) 
 
 
 This comes from ?locale and reads there

This is not in ?locale, but in ?locales

  Sys.setlocale(LC_COLLATE, C)   # turn off locale-specific sorting,
 #  usually

This in the example section at the end.

Try also to see

  Sys.getlocale()

Relevant can also be LC_CTYPE

  Sys.setlocale(LC_CTYPE, C)

Hope this helps.

Petr Savicky.

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Re: [R] Sorting strings

2012-02-20 Thread De-Jian Zhao


Sorry, just made a mistake. This is the result from windows xp.

 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X.Z   X0B.Z
 R.version
   _
platform   i386-pc-mingw32
arch   i386
os mingw32
system i386, mingw32
status
major  2
minor  13.0
year   2011
month  04
day13
svn rev55427
language   R
version.string R version 2.13.0 (2011-04-13)


On 2012-2-21 0:13, De-Jian Zhao wrote:
It seems OS-dependent. I got different results when trying it on 
windows xp and Redhat linux.



 R.version
   _
platform   x86_64-unknown-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  9.1
year   2009
month  06
day26
svn rev48839
language   R
version.string R version 2.9.1 (2009-06-26)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X.Z   X0B.Z


 R.version
   _
platform   x86_64-unknown-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  9.1
year   2009
month  06
day26
svn rev48839
language   R
version.string R version 2.9.1 (2009-06-26)
 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z


On 2012-2-20 23:27, statquant2 wrote:

Ok I have :

R) str(R.Version())
List of 13
  $ platform  : chr x86_64-unknown-linux-gnu
  $ arch  : chr x86_64
  $ os: chr linux-gnu
  $ system: chr x86_64, linux-gnu
  $ status: chr 
  $ major : chr 2
  $ minor : chr 12.2
  $ year  : chr 2011
  $ month : chr 02
  $ day   : chr 25
  $ svn rev   : chr 54585
  $ language  : chr R
  $ version.string: chr R version 2.12.2 (2011-02-25)

R) sort(c(X.,X0B))
[1] X.  X0B
R) sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z

I am using a linux redHat
$ uname -a
Linux 2.6.18-238.9.1.el5 #1 SMP Fri Mar 18 12:42:39 EDT 2011 x86_64 
x86_64

x86_64 GNU/Linux


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View this message in context: 
http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4404298.html

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Re: [R] FW: Problems with R



On Feb 20, 2012, at 10:26 AM, Thorsten Pöllinger wrote:


Dears,

I am a new R user and I am trying to analyze my data sets, R gives  
me a default when I type in the regression formula as following:


fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue 
$revenue.x,family=NO)


Most regression functions work best with a data argument and then a  
formula that uses just column names.



Fehler in model.frame.default(formula = tot_remun_revenue 
$tot_remun.y ~  :
 ungültiger Typ (NULL) für die Variable 'tot_remun_revenue 
$tot_remun.y' Is there anything wrong in my data set, as the  
instruction is directly copied from lectures I had about R? Attached  
please find the data set as .csv. Best regards and thanks in  
advance! Thorsten Poellinger 	


You mailer probably did not properly indicate that the .csv file was  
an acceptable file type. You may want to send a followup after  
renaming the file to somthing.txt before attaching,... and you  
should include the data input commands you used.


--
David Winsemius, MD
West Hartford, CT

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Re: [R] Non-parametric test for repeated measures and post-hoc single comparisons in R?

2012-02-20 Thread Meyners, Michael

No, the authors are correct: the individuals (i.e. the 17 individuals) you have 
need to be independent (i.e. no correlation between them, let alone any 
individual running through your temporal experiment more than once, as 
indicated in the citation), while the *observations* are of course dependent as 
they are within the same subject (individual -- they have the same subject 
effect). Think of Friedman as a non-parametric 2-way ANOVA with one of the 
factors being subject; observations of the same subject are dependent, but once 
you include the subject effect, the errors are assumed to be independent (which 
implies that subjects need to be independent and should, e.g., not work on the 
assessment together).
The imprecision is in your interpretation of individuals vs. observations. 
HTH, Michael

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of saschav...@gmail.com
 Sent: Montag, 20. Februar 2012 09:59
 To: peter dalgaard
 Cc: r-help@r-project.org
 Subject: Re: [R] Non-parametric test for repeated measures and post-hoc
 single comparisons in R?
 
 Thanks, I got it! (And I think I should have googled what replicated
 means!) However, then Bortz, Lienert, Boehnke are imprecise, if not
 wrong: Der Friedman-Test setzt voraus, dass die N Individuen
 wechselseitig unabhängig sind, dass also nicht etwa ein und dasselbe
 Individuum zweimal oder mehrmals im Untersuchungsplan auftritt (p.
 271). Which I (hope to) translate: The Friedman test requires the N
 individuals to be reciprocally independent, which means that one
 individual cannot occur twice or more times in the research design.
 
 *S*
 
 On 19.02.12 22:04, peter dalgaard wrote:
 
  Repeated measures means that you have multiple measurements on the
 same individual. Usually, the same person measured at different time
 points. So if you have N individuals and T times, then you can place
 your observations in an N*T layout.
 
  In this layout, you can have 1 observation per cell or R  1
 observations. In the former case, the design is referred to as
 unreplicated.  Got it?
 
  -pd
 
 
  On Feb 19, 2012, at 19:25 , saschav...@gmail.com wrote:
 
  Some attribute x from 17 individuals was recorded repeatedly on 6
 time points using a Likert scale with 7 distractors. Which statistical
 test(s) can I apply to check whether the changes along the 6 time
 points were significant?
 
  set.seed( 123 )
  x- matrix( sample( 1:7, 17*6, repl=T ),
nrow = 17, byrow = TRUE,
dimnames = list(1:17, paste( 'T', 1:6, sep='' ))
  )
 
  I found the Friedman test and the Quade test for testing the overall
 hypothesis.
 
  friedman.test( x )
  quade.test( x )
 
  However, the R help files, my text books (Bortz, Lienert and
 Boehnke, 2008; Köhler, Schachtel and Voleske, 2007; both German), and
 the Wikipedia texts differ in what they propose as requirements for the
 tests. R says that data need to be unreplicated. I read 'unreplicated'
 as 'not-repeated', but is that right? If so, the example, in contrast,
 in friedman.test() appears to use indeed repeated measures. Yet,
 Wikipedia says the contrary that is to say the test is good especially
 if data represents repeated measures. The text books say either (in the
 same paragraph, which is very confusing). What is right?
 
  In addition, what would be an appropriate test for post-hoc single
 comparisons for the indication which column differs from others
 significantly?
 
  Bortz, Lienert, Boehnke (2008). Verteilungsfreie Methoden in der
  Biostatistik. Berlin: Springer Köhler, Schachtel, Voleske (2007).
  Biostatistik: Eine Einführung für Biologen und Agrarwissenschaftler.
  Berlin: Springer
 
  --
  Sascha Vieweg, saschav...@gmail.com
 
  __
  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.
 
 
 
 --
 Sascha Vieweg, saschav...@gmail.com
 
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Re: [R] coxme: model simplification using LR-test?

2012-02-20 Thread Simon Tragust


Many thanks for the suggestion.

I tried it already, but as I've never wrote a function, I had no luck. 
If it's not asked too much, maybe you or somebody else could help me 
getting the code for the function right.


Regards
Simon


Am 20/02/2012 14:52, schrieb Terry Therneau:

Summary of the query: update does not work on a coxme object

  I ran into this bug myself 2 days ago -- I rarely use update() so
hadn't encountered it before.  The problem is that coxme breaks the
formula into fixed and random portions, and this confuses the default
method for formula.

Solution: add the following function:
   formula.coxme- function(x, ...) x$call$formula

  This method will be included in my next update of coxme.

Terry Therneau





--
Simon Tragust
Animal Ecology I
University of Bayreuth

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Re: [R] Sorting strings

2012-02-20 Thread De-Jian Zhao


On 2012-2-20 23:15, Rui Barradas wrote:

Could it be OS related?


Yes, it seems. I tried it on my local windows xp and redhat linux 
server, and got different results. Hope it will be fixed in the future 
versions. Maybe we should keep alert to check whether the results are 
consistent when transferring our code from one platform to another.



 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X0B.Z X.Z
 R.version
   _
platform   x86_64-unknown-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  9.1
year   2009
month  06
day26
svn rev48839
language   R
version.string R version 2.9.1 (2009-06-26)



 sort(c(X.,X0B))
[1] X.  X0B
 sort(c(X.Z,X0B.Z))
[1] X.Z   X0B.Z
 R.version
   _
platform   i386-pc-mingw32
arch   i386
os mingw32
system i386, mingw32
status
major  2
minor  13.0
year   2011
month  04
day13
svn rev55427
language   R
version.string R version 2.13.0 (2011-04-13)

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

Re: [R] overlay of two sets of boxplots

2012-02-20 Thread John Kane

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

At a rough guess you may want to have a look at the mfrow in ?par  but without 
some sample data and a bit more information about what you need it is difficult 
to suggest more.

By the way dput (see ?dput) is a handy way to supply a sample data set here.

John Kane
Kingston ON Canada


 -Original Message-
 From: ap...@neuro.mpg.de
 Sent: Mon, 20 Feb 2012 10:27:11 +
 To: r-help@r-project.org
 Subject: [R] overlay of two sets of boxplots
 
 Hello,
 I am new to R and currently have the following problem:
 I have successfully loaded my data in R which consists of two numeric
 columns (LI_F and female) and one character column (Strain). So far I can
 plot two different set of boxplots for each of the numeric columns
 plotted by the groups of the character column and the commands look like
 that:
 
 boxplot(LI_F~Strain, ylab=LI_F, xlab=Strain, data=pain)
 boxplot(female~Strain, ylab=female, xlab=Strain, data=pain)
 
 How can I overlay the two set of boxplots (preferably in different
 colors), so that I can compare them one by one, meaning two boxplots
 corresponding to the same character in Strain are directly above each
 other?
 I have tried a lot of things and would greatly appreciate your help.
 
 Best,
 Mirjam
 
   [[alternative HTML version deleted]]
 
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Re: [R] Sorting strings

2012-02-20 Thread Ted Harding

On 20-Feb-2012 Petr Savicky wrote:
 On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
 I did, but this does not give the answer to my question...
 Anybody knows how to tweack the behaviour of sort or how to do ?
 
 Hi.
 Try this
 
   Sys.setlocale(LC_COLLATE, C) 
 
 This comes from ?locale and reads there
 
   Sys.setlocale(LC_COLLATE, C)   # turn off locale-specific sorting,
  #  usually
 
 See also ?sort
 
  The sort order for character vectors will depend on the
  collating sequence of the locale in use: see 'Comparison'.
 
 ?Comparison
 
  Comparison of strings in character vectors is lexicographic
  within the strings using the collating sequence of the locale
  in use: see 'locales'. The collating sequence of locales such
  as 'en_US' is normally different from 'C' (which should use
  ASCII) and can be surprising. Beware of making _any_ assumptions
  about the collation order: ...
 
 Hope this helps.
 Petr Savicky.

I've been following this thread with interest. I had begun composing
a reply on similar lines to Petr's above, but put it on one side
while waiting to see how the thread would evolve.

In view of the tangle of mixed experiences reported by different
users, I now wonder whether we should have something like lc_collate
as a specific parameter for sort(), e.g. so that one can set, for a
particular sorting operation,

   sort(c(X.,X0B),lc_collate=C)

without affecting the system LC_COLLATE setting (i.e. the change
takes effect only within the execution of that sort() command).

Ted.

-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 20-Feb-2012  Time: 17:16:47
This message was sent by XFMail

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Re: [R] ggplot rank stack bar automatically.

2012-02-20 Thread Ista Zahn

Hi,

That is not what I get. After running your code I get

 levels(df.m$Period)
 [1] 1991-00 1901-10 1981-90 2001-06 1911-20 1881-90 1971-80
 [8] 1921-30 1891-00 1961-70 1871-80 1851-60 1951-60 1861-70
[15] 1841-50 1941-50 1831-40 1931-40 1820-30

this is my version info:

sessionInfo()
R version 2.14.1 (2011-12-22)
Platform: x86_64-unknown-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C  
 [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8   
 [7] LC_PAPER=C LC_NAME=C 
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C   

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

other attached packages:
[1] ggplot2_0.9.0 reshape_0.8.4 plyr_1.7.1   

loaded via a namespace (and not attached):
 [1] colorspace_1.1-1   dichromat_1.2-4digest_0.5.1   grid_2.14.1   
 [5] MASS_7.3-16memoise_0.1munsell_0.3proto_0.3-9.2 
 [9] RColorBrewer_1.0-5 reshape2_1.2.1 scales_0.1.0   stringr_0.6

what is yours?

Best,
Ista

On Sunday, February 19, 2012 11:59:30 PM vd3000 wrote:
  df - structure(c(106487, 495681, 1597442,
 
  2452577, 2065141, 2271925, 4735484, 3555352,
  8056040, 4321887, 2463194, 347566, 621147,
  1325727, 1123492, 800368, 761550, 1359737,
  1073726, 36, 53, 141, 41538, 64759, 124160,
  69942, 74862, 323543, 247236, 112059, 16595,
  37028, 153249, 427642, 1588178, 2738157,
  2795672, 2265696, 11951, 33424, 62469,
  74720, 166607, 404044, 426967, 38972, 361888,
  1143671, 1516716, 160037, 354804, 996944,
  1716374, 1982735, 3615225, 4486806, 3037122,
  17, 54, 55, 210, 312, 358, 857, 350, 7368,
  8443, 6286, 1750, 7367, 14092, 28954, 80779,
  176893, 354939, 446792, 3, 69911, 53144,
  29169, 18005, 11704, 13363, 18028, 46547,
  14574, 8954, 2483, 14693, 25467, 25215,
  41254, 46237, 98263, 185986), .Dim = c(19,
  5), .Dimnames = list(c(1820-30, 1831-40,
  1841-50, 1851-60, 1861-70, 1871-80,
  1881-90, 1891-00, 1901-10, 1911-20,
  1921-30, 1931-40, 1941-50, 1951-60,
  1961-70, 1971-80, 1981-90, 1991-00,
  2001-06), c(Europe, Asia, Americas,
  Africa, Oceania)))
 
  df.m - melt(df)
  df.m - rename(df.m, c(X1 = Period, X2 = Region))
  
  df.m - transform(df.m, Period = reorder(Period, -1*value))
  
  ggplot(df.m, aes(x = Period, y = value/1e+06, fill = Region))  +
 
   geom_bar(stat = identity, position = stack)
 
 =
 
  levels(df.m$Period)
 
  [1] 1820-30 1831-40 1841-50 1851-60 1861-70 1871-80 1881-90
  [8] 1891-00 1901-10 1911-20 1921-30 1931-40 1941-50 1951-60
 [15] 1961-70 1971-80 1981-90 1991-00 2001-06
 =
 I think after reordering, the levels changed, but the fact was nothing has
 got change _!!!
 
 I found I have many commands happen like this. especially drawing
 graph...ordering doesn't work...
 [[elided Yahoo spam]]
 it is crazy, R teases me.
 
 
 
 
 
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 View this message in context:
 http://r.789695.n4.nabble.com/ggplot-rank-stack-bar-automatically-tp4391042
 p4403442.html Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] help updating package rJava (on ubuntu)

2012-02-20 Thread Karl Brand


Simon,

Thanks alot for the further clarification. As i said some where in my 
lengthy explanation - i don't what of the myriad steps were needed, only 
that they were performed and i now have an up-to-date rJava.


Next time (actually coming up soon) i'll certainly be following the 
couple simple steps you suggest. Also good to know that Java 1.7 can be 
used with the latest R (which is all i intend to install).


Thanks again for the follow up,

Karl

On 02/20/2012 03:37 AM, Simon Urbanek wrote:

On Feb 19, 2012, at 4:31 PM, Karl Brand wrote:


Hi Hasan,

Success. For myself and FWIW to other useR's here's how i spent the sunny half 
of my sunday to achieve it :/

Many thanks for your and Simon's input,



FWIW you should not need to set any custom settings if you system is properly 
configured (if you use Java 1.7 you may need R 2.14.0 or higher which works 
around some bugs in 1.7 binaries). On Debian/Ubuntu you just install your 
favorite JDK (Sun/Oracle or OpenJDK) and use 'sudo update-alternatives --config 
java' or 'sudo update-java-alternatives' (depending on the age of your system) 
to select the right one.

The fact you you had to tweak PATH means that something is seriously broken or 
you did setup the alternatives configuration correctly.

Cheers,
Simon




Karl

Since:

$ javac -version

returned nothing i believe you (and Simon) were right, i.e, it (and JDK) were 
missing on my system. Furthermore:

$ sudo R CMD javareconf
Java interpreter : /usr/bin/java
Java version : 1.6.0_23
Java home path   : /usr/lib/jvm/java-6-openjdk/jre
Java compiler: not present
Java headers gen.:
Java archive tool:
Java library path: 
$(JAVA_HOME)/lib/amd64/server:$(JAVA_HOME)/lib/amd64:$(JAVA_HOME)/../lib/amd64:/usr/java/packages/lib/amd64:/usr/lib/jni:/lib:/usr/lib
JNI linker flags : -L$(JAVA_HOME)/lib/amd64/server -L$(JAVA_HOME)/lib/amd64 
-L$(JAVA_HOME)/../lib/amd64 -L/usr/java/packages/lib/amd64 -L/usr/lib/jni 
-L/lib -L/usr/lib -ljvm
JNI cpp flags:

Updating Java configuration in /etc/R
Done.

Some things are clearly absent i.e.,
Java compiler: not present
Java headers gen.:
Java archive tool:

Now i have:

$ javac -version
javac 1.6.0_23

$ sudo R CMD javareconf
Java interpreter : /usr/bin/java
Java version : 1.6.0_23
Java home path   : /usr/lib/jvm/java-6-openjdk/jre
Java compiler: /usr/bin/javac
Java headers gen.: /usr/bin/javah
Java archive tool: /usr/bin/jar
Java library path: 
$(JAVA_HOME)/lib/amd64/server:$(JAVA_HOME)/lib/amd64:$(JAVA_HOME)/../lib/amd64:/usr/java/packages/lib/amd64:/usr/lib/jni:/lib:/usr/lib
JNI linker flags : -L$(JAVA_HOME)/lib/amd64/server -L$(JAVA_HOME)/lib/amd64 
-L$(JAVA_HOME)/../lib/amd64 -L/usr/java/packages/lib/amd64 -L/usr/lib/jni 
-L/lib -L/usr/lib -ljvm
JNI cpp flags: -I$(JAVA_HOME)/../include

Updating Java configuration in /etc/R
Done.

Certainly i don't know exactly what was needed to achieve this. BUT - for 
posterities sake, this is what i did:

## purge and reinstall openjdk-6-jdk which turned outwasn't installed!
## how this can be when i have a /usr/lib/jvm/java-6-openjdk full of
## files i don't understand. Moreover since i had the previosu version
## of rJava running fine!!!
$ sudo apt-get purge openjdk-6-jdk
snip
Package openjdk-6-jdk is not installed, so not removed
0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.

## install openjdk-6-jdk
$ sudo apt-get install openjdk-6-jdk

## Per- http://ubuntuforums.org/showthread.php?t=1491846
## added 2 entries to /etc/bash.bashrc file
$ gksudo gedit /etc/bash.bashrc

## entries were:
export JAVA_HOME=/usr/lib/jvm/java-6-openjdk/jre
export PATH=$PATH:$JAVA_HOME/bin

## note the addition of /jre for JAVA_HOME per-
## http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing

saved bash.bashrc file then reloaded

$ source /etc/bash.bashrc

## confirmed
$ echo $JAVA_HOME
/usr/lib/jvm/java-6-openjdk/jre

## configured java
$ sudo R CMD javareconf

I was then able to update rJava in a new R-session.

Also note that per- 
http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing

$ apt-get install r-cran-rjava

did NOT enable me to update my rJava. Unless i buggered up something else along 
the way :/




On 02/18/2012 10:20 PM, Hasan Diwan wrote:

On 18 February 2012 13:13, Karl Brandk.br...@erasmusmc.nl   wrote:

Thanks for yout fast response. Thing is - i managed to get Version 0.9-1
installed and fully functional. And

$ locate jdk
returns too many entries to post here, so i'm pretty sure its on the
machine.



What you want to look for is javac, not jdk. On my ubuntu system, this
is to be found at /usr/lib/jvm/java-6-openjdk-i386/bin/javac


So i'd like to know how i can ensure it's registered in R. This i have no
idea how to do. I simply don't have enough R and linux experience. Also the
thread i mentioned
http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing
seems to be about pointing R at the right location of certain

Re: [R] coxme: model simplification using LR-test?



On Feb 20, 2012, at 11:41 AM, Simon Tragust wrote:


Many thanks for the suggestion.

I tried it already,


it means .. what exactly?


but as I've never wrote a function, I had no luck.


no luck means ... what?

If it's not asked too much, maybe you or somebody else could help me  
getting the code for the function right.


He gave you a perfectly formed function. Were you thinking you had to  
fill in some blanks? Not so. That ... is a meaningful construct in  
R. It says to take whatever arguments follow and pass them on to the  
next function.  So when you enter this at the console:


 m-coxme(Surv(day,status) ~ condition*infection +
(1|infection/population), data=all)
 formula.coxme- function(x, ...) x$call$formula
 m1-update(m, ~.-condition:infection)

 what happens?

--
david.


Am 20/02/2012 14:52, schrieb Terry Therneau:

Summary of the query: update does not work on a coxme object

 I ran into this bug myself 2 days ago -- I rarely use update() so
hadn't encountered it before.  The problem is that coxme breaks the
formula into fixed and random portions, and this confuses the default
method for formula.

Solution: add the following function:
  formula.coxme- function(x, ...) x$call$formula

 This method will be included in my next update of coxme.

Terry Therneau




David Winsemius, MD
West Hartford, CT

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Re: [R] Computing plot size in Sweave

2012-02-20 Thread Yihui Xie

I guess that is not possible with Sweave, but it is possible in the
knitr package (an alternative to Sweave). You can set

opts_knit$set(eval.opts = c('fig.height', 'fig.width'))

so that these two options will be evaluated as R expressions (e.g.
fig.height=x means it takes value from a variable x).

Note width/height have been renamed to fig.width/fig.height in knitr.
See http://yihui.name/knitr/

Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA



On Mon, Feb 20, 2012 at 9:15 AM, BXC (Bendix Carstensen) b...@steno.dk wrote:
 Sometimes you want to compute the physical size of a plot based on data.
 In R itself this is no problem.

 But is there a way to compute the values of height and width in S-weave, say:

 graph,fig=TRUE,height=xx,width=yy=

 where xx and yy are computed and not physically written in the document?

 Bendix
 __

 Bendix Carstensen
 Senior Statistician
 Epidemiology

 Steno Diabetes Center A/S
 Niels Steensens Vej 2-4
 DK-2820 Gentofte
 Denmark
 +45 44 43 87 38 (direct)
 +45 30 75 87 38 (mobile)
 b...@steno.dk    http://BendixCarstensen.com
 www.steno.dk

 This e-mail (including any attachments) is intended for ...{{dropped:8}}

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Re: [R] Column wise matrix multiplication

2012-02-20 Thread Ted Harding

On 20-Feb-2012 Graziano Mirata wrote:
 Hi all,
 I am trying to multiply each column of a matrix such to have
 a unique resulting vector with length equal to the number of
 rows of the original matrix. In short I would like to do what
 prod(.) function in Matlab does, i.e.
 
 A -matrix(c(1:10),5,2)
 
 V = A[,1]*A[,2]
 
 Thank you
 
 Graziano

The Matlab prod(A,2) function computes the products along the
rows of the matrix A and returns the result as a column vector,
of length equal to the number of rows in A, which seems to be
what you describe.

Your code above does this for your 2-column example, but the
result is a simple R vector which is not an array (and in
particular is not a column vector):

  A[,1]*A[,2]
  # [1]  6 14 24 36 50

  dim(A[,1]*A[,2])
  # NULL

For a matrix A with arbitrary number of columns, if you wanted
the row sums rather than the row products, you could use the
R function rowSums():

  rowSums(A)
  # [1]  7  9 11 13 15

This is still a dimensionless simple R vector:

  dim(rowSums(A))
  # NULL

Unfortunately, there seems to be no equivalent for products
(e.g. rowProds). But you can define one:

  rowProds - function(X){ apply(X,1,FUN=prod) }

  rowProds(A)
  # [1]  6 14 24 36 50

Even then, the result is a simple R vector, without dimensions:

  dim(rowProds(A))
  # NULL

If you need an array (row) vector then you can apply t():

  t(rowProds(A))
  #  [,1] [,2] [,3] [,4] [,5]
  # [1,]6   14   24   36   50

or t(t()) for a column vector:

  t(t(rowProds(A)))
  #  [,1]
  # [1,]6
  # [2,]   14
  # [3,]   24
  # [4,]   36
  # [5,]   50

Ted.

-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 20-Feb-2012  Time: 17:54:13
This message was sent by XFMail

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[R] How to change the objective function of quantile regression?

2012-02-20 Thread Nathalie Sanches






Dear all,

I need to run a quantile regression but considering a different objective 
function to be minimized: instead of finding the parameters that minimize 
rho(t) = u*(t - I(u0)), I need to find the parameters beta that minimize a sum 
of two rho functions rho(t1) = u1*(t1 - I(u10)) and rho(t2) = u2*(t2 - 
I(u20)) where t1 and t2 are different quantile levels and u1 = y1 - b'x1 and 
u2 = y2 - b'x2. 

y1 and x1 are response variable and regressors, respectively, that will be 
weighted by t1 and similar for t2.

The problem is that I do not know how to change the function rq() in R because 
it only accepts two arguments, y and x. In my case, I think I also have two 
arguments, that is  x - rbind(x1,x2) and y - rbind(y1,y2) , but I need to 
split the residuals to be minimized, weighting u1 by t1 and u2 by t2.  I am 
trying my best, but I cannot do it. 

Does anybody have any suggestion? 

I appreciate any kind of suggestion and help.

All the best,

Nathalie

  
[[alternative HTML version deleted]]

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[R] Disabling shell access through system() possible?

2012-02-20 Thread lists

Hi! I'm deploying R behind a web-app on a linux-server and I don't want 
to grant the users shell access through the system() function for 
security reasons.


Is there any safe way to deny a user access to the function?

I tried
a) alter the function in the R-Sources before compiling them. Doesn't 
seem to work because system() is apparently needed during the build.
b) Redefine system() in the global Rprofile. Leads to of course to some 
warning messages at any session start and feels generally kludgy and 
unsafe.


Best Regards
M.W.

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Re: [R] Sorting strings

NICE DDE
It solves my problem !
Awesome stuff

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[R] how to How to compare the concordance index between two nomograms with R

2012-02-20 Thread lijundfgd

I have got two kinds of nomogram for the same data set analyzed by Cox
regression. My question is how to compare the concordance index between
these two nomograms with R?

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

[R] Time Series - Trend Line

2012-02-20 Thread anaraster

How can I had a trend line to my plot?

My data looks like this:

Date=seq(as.Date(1910/1/1), as.Date(1920/1/1), day)
Values=runif(length(Date), min=-5, max=5)

dataset=data.frame(Values,as.Date(Date))

I just want to add a linear trend line to this
plot(dataset,col=rgb(1,0,0,1/8),cex=0.5,pch=19)



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

Re: [R] overlay of two sets of boxplots

2012-02-20 Thread Mirjam Appel


Hello John,
Thanks for your fast answer. I will try to be clearer and more detailed this 
time. At the moment I am importing a dataframe like below as a '.csv file'. I 
want to generate a boxplot for M and F values grouped by X whereby boxplots for 
M and F should be above or very close to each other.  Please see attachment for 
example figure. Till now I have tried a code that looks like the one below the 
frame (I found it in the graph help of R for boxes close to each other):

X   MF
Exp1-90.0273224 -77.66531714
Exp1-73.-96.36363636
Exp1-98.24561404-91.57848325
Exp1-65.2173913 -59.18367347
Exp1-98.7654321 -92.42424242
Exp1-96.42857143-89.25925926
Exp2-84.61538462 -47.36842105
Exp2-43.63636364 -22.37762238
Exp2-47.5  -33.
Exp2-55.49450549-66.6667

boxplot(
formula = F~Strain,
data= pain,
boxwex  = 0.25,
at  = 1:38 - 0.1,
subset  =supp== F,
col = red,
xlab= Strain,
ylab=  F
 )
boxplot(
formula = M~Strain,
data= pain,
boxwex  = 0.25,
at  = 1:38 + 0.1,
subset  =supp== M,
col = orange,
add = TRUE )

...but I did cannot find anywhere what supp means and R cannot read it. 
I hope this is explained ok. Any kind of answer would help!
Best
Mirjam

-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com] 
Sent: Monday, February 20, 2012 6:00 PM
To: Mirjam Appel; r-help@r-project.org
Subject: RE: [R] overlay of two sets of boxplots

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

At a rough guess you may want to have a look at the mfrow in ?par  but without 
some sample data and a bit more information about what you need it is difficult 
to suggest more.

By the way dput (see ?dput) is a handy way to supply a sample data set here.

John Kane
Kingston ON Canada


 -Original Message-
 From: ap...@neuro.mpg.de
 Sent: Mon, 20 Feb 2012 10:27:11 +
 To: r-help@r-project.org
 Subject: [R] overlay of two sets of boxplots
 
 Hello,
 I am new to R and currently have the following problem:
 I have successfully loaded my data in R which consists of two numeric 
 columns (LI_F and female) and one character column (Strain). So far I 
 can plot two different set of boxplots for each of the numeric columns 
 plotted by the groups of the character column and the commands look 
 like
 that:
 
 boxplot(LI_F~Strain, ylab=LI_F, xlab=Strain, data=pain) 
 boxplot(female~Strain, ylab=female, xlab=Strain, data=pain)
 
 How can I overlay the two set of boxplots (preferably in different 
 colors), so that I can compare them one by one, meaning two boxplots 
 corresponding to the same character in Strain are directly above 
 each other?
 I have tried a lot of things and would greatly appreciate your help.
 
 Best,
 Mirjam
 
   [[alternative HTML version deleted]]
 
 __
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overlayed boxplots.pdf
Description: overlayed boxplots.pdf
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Re: [R] overlay of two sets of boxplots



On Feb 20, 2012, at 1:59 PM, Mirjam Appel wrote:



Hello John,
Thanks for your fast answer. I will try to be clearer and more  
detailed this time. At the moment I am importing a dataframe like  
below as a '.csv file'. I want to generate a boxplot for M and F  
values grouped by X whereby boxplots for M and F should be above or  
very close to each other.  Please see attachment for example figure.  
Till now I have tried a code that looks like the one below the frame  
(I found it in the graph help of R for boxes close to each other):


I think you must have missed the spot in whatever example code you are  
following for your homework task where they reshaped this data so that  
it was in long format. Go back into your text and read it more  
carefully and if it's not clear, then pose further questions to your  
instructor.




X   MF
Exp1-90.0273224 -77.66531714
Exp1-73.-96.36363636
Exp1-98.24561404-91.57848325
Exp1-65.2173913 -59.18367347
Exp1-98.7654321 -92.42424242
Exp1-96.42857143-89.25925926
Exp2-84.61538462 -47.36842105
Exp2-43.63636364 -22.37762238
Exp2-47.5  -33.
Exp2-55.49450549-66.6667


boxplot(

   formula = F~Strain,
   data= pain,
   boxwex  = 0.25,
   at  = 1:38 - 0.1,
   subset  =supp== F,


Clearly the strain variable was created in the process of reshaping  
and the supp variable as well.


--
david.


   col = red,
   xlab= Strain,
   ylab=  F
 )
boxplot(
   formula = M~Strain,
   data= pain,
   boxwex  = 0.25,
   at  = 1:38 + 0.1,
   subset  =supp== M,
   col = orange,
   add = TRUE )

...but I did cannot find anywhere what supp means and R cannot read  
it.

I hope this is explained ok. Any kind of answer would help!
Best
Mirjam

-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com]
Sent: Monday, February 20, 2012 6:00 PM
To: Mirjam Appel; r-help@r-project.org
Subject: RE: [R] overlay of two sets of boxplots

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

At a rough guess you may want to have a look at the mfrow in ?par   
but without some sample data and a bit more information about what  
you need it is difficult to suggest more.


By the way dput (see ?dput) is a handy way to supply a sample data  
set here.


John Kane
Kingston ON Canada



-Original Message-
From: ap...@neuro.mpg.de
Sent: Mon, 20 Feb 2012 10:27:11 +
To: r-help@r-project.org
Subject: [R] overlay of two sets of boxplots

Hello,
I am new to R and currently have the following problem:
I have successfully loaded my data in R which consists of two numeric
columns (LI_F and female) and one character column (Strain). So far I
can plot two different set of boxplots for each of the numeric  
columns

plotted by the groups of the character column and the commands look
like
that:

boxplot(LI_F~Strain, ylab=LI_F, xlab=Strain, data=pain)
boxplot(female~Strain, ylab=female, xlab=Strain, data=pain)

How can I overlay the two set of boxplots (preferably in different
colors), so that I can compare them one by one, meaning two boxplots
corresponding to the same character in Strain are directly above
each other?
I have tried a lot of things and would greatly appreciate your help.

Best,
Mirjam

[[alternative HTML version deleted]]



David Winsemius, MD
West Hartford, CT

__
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Re: [R] Disabling shell access through system() possible?

On Mon, Feb 20, 2012 at 06:17:23PM +0100, li...@mwoywod.de wrote:
 Hi! I'm deploying R behind a web-app on a linux-server and I don't want 
 to grant the users shell access through the system() function for 
 security reasons.
 
 Is there any safe way to deny a user access to the function?

Hi.

Disabling system() is not sufficient. The user may call directly

  .Internal(system(, ))

Other dangerous things may be opening connections with write access
or manipulation with files. In particular, pipe() can also run
a command.

It could be better to run R under a user name with restricted
permissions.

Petr Savicky.

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Re: [R] JGR Segmentation fault

2012-02-20 Thread dberringer

The  http://mxcl.github.com/homebrew/ Homebrew package manager  is an exelent
way to manage packages like R on osx.  I highly recommend it to any
developer.  I am hoping there are others in the R community that are
interested in getting this working too.  Does anyone else run a Homebrew
install of R?  If so have you been able to get JGR / Deducer running?

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[R] error with persp()- increasing 'x' and 'y' values expected

2012-02-20 Thread adick

I want to use persp to graph my data and it keeps giving me the error 
increasing 'x' and 'y' values expected, even though my data is in increasing
order with respect to x and y.

Here is the code I'm currently using: 
bob= scan (SBA3dataTaxonB.txt,what=char)
labels = bob[1:3]
bob=bob[-c(1,2,3)]
bob=as.numeric(bob)
bob=array(bob,dim=c(3, length(bob)/3))
bob=t(bob)
data.frame(bob) - bob
bob = bob[order(bob[,1], bob[,3]),]
persp(bob[,1], bob[,3], bob[,2], log=bob[,1])

The first few lines of the txt file SBA3dataTaxonB.txt look like this:
variableTaxonB  central
10  19  0.002
10  25  0.002
10  26  0.002
1   17  0.002
1   23  0.002
1   16  0.002
0.1 48  0.002
0.1 47  0.002
0.1 57  0.002
10  15  0.004
10  22  0.004
10  22  0.004
1   21  0.004
1   22  0.004
1   27  0.004
0.1 73  0.004
0.1 62  0.004

Can anyone help me figure out why I am getting this error? And also a
possible solution would be greatly appreciated.
-Thanks, Amanda

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Re: [R] overlay of two sets of boxplots

2012-02-20 Thread John Kane

Does something like the code below give you want you want?

It requires the ggplot2 package so you will likely have to install it.

John Kane
Kingston ON Canada




# sample data converted using dput
xx - structure(list(X = c(Exp1, Exp1, Exp1, Exp1, Exp1, 
Exp1, Exp2, Exp2, Exp2, Exp2), M = c(-90.0273224, -73., 
-98.24561404, -65.2173913, -98.7654321, -96.42857143, -84.61538462, 
-43.63636364, -47.5, -55.49450549), F = c(-77.66531714, -96.36363636, 
-91.57848325, -59.18367347, -92.42424242, -89.25925926, -47.36842105, 
-22.37762238, -33., -66.6667)), .Names = c(X, M, 
F), class = data.frame, row.names = c(NA, -10L))

# melt data using reshape --automatically loads with ggplot2
library(ggplot2)
mm - melt(xx, id=X)

ggplot(mm, aes(variable, value, fill=variable)) + geom_boxplot() + 
 opts(legend.position=none)  + facet_grid(. ~ X) 
 



 -Original Message-
 From: ap...@neuro.mpg.de
 Sent: Mon, 20 Feb 2012 18:59:49 +
 To: jrkrid...@inbox.com, r-help@r-project.org
 Subject: RE: [R] overlay of two sets of boxplots
 
 
 Hello John,
 Thanks for your fast answer. I will try to be clearer and more detailed
 this time. At the moment I am importing a dataframe like below as a '.csv
 file'. I want to generate a boxplot for M and F values grouped by X
 whereby boxplots for M and F should be above or very close to each other.
 Please see attachment for example figure. Till now I have tried a code
 that looks like the one below the frame (I found it in the graph help of
 R for boxes close to each other):
 
 X MF
 Exp1  -90.0273224 -77.66531714
 Exp1  -73.-96.36363636
 Exp1  -98.24561404-91.57848325
 Exp1  -65.2173913 -59.18367347
 Exp1  -98.7654321 -92.42424242
 Exp1  -96.42857143-89.25925926
 Exp2  -84.61538462 -47.36842105
 Exp2  -43.63636364 -22.37762238
 Exp2  -47.5  -33.
 Exp2  -55.49450549-66.6667
 
 boxplot(
 formula = F~Strain,
 data= pain,
 boxwex  = 0.25,
 at  = 1:38 - 0.1,
 subset  =supp== F,
 col = red,
 xlab= Strain,
 ylab=  F
)
 boxplot(
 formula = M~Strain,
 data= pain,
 boxwex  = 0.25,
 at  = 1:38 + 0.1,
 subset  =supp== M,
 col = orange,
 add = TRUE )
 
 ...but I did cannot find anywhere what supp means and R cannot read it.
 I hope this is explained ok. Any kind of answer would help!
 Best
 Mirjam
 
 -Original Message-
 From: John Kane [mailto:jrkrid...@inbox.com]
 Sent: Monday, February 20, 2012 6:00 PM
 To: Mirjam Appel; r-help@r-project.org
 Subject: RE: [R] overlay of two sets of boxplots
 
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 At a rough guess you may want to have a look at the mfrow in ?par  but
 without some sample data and a bit more information about what you need
 it is difficult to suggest more.
 
 By the way dput (see ?dput) is a handy way to supply a sample data set
 here.
 
 John Kane
 Kingston ON Canada
 
 
 -Original Message-
 From: ap...@neuro.mpg.de
 Sent: Mon, 20 Feb 2012 10:27:11 +
 To: r-help@r-project.org
 Subject: [R] overlay of two sets of boxplots
 
 Hello,
 I am new to R and currently have the following problem:
 I have successfully loaded my data in R which consists of two numeric
 columns (LI_F and female) and one character column (Strain). So far I
 can plot two different set of boxplots for each of the numeric columns
 plotted by the groups of the character column and the commands look
 like
 that:
 
 boxplot(LI_F~Strain, ylab=LI_F, xlab=Strain, data=pain)
 boxplot(female~Strain, ylab=female, xlab=Strain, data=pain)
 
 How can I overlay the two set of boxplots (preferably in different
 colors), so that I can compare them one by one, meaning two boxplots
 corresponding to the same character in Strain are directly above
 each other?
 I have tried a lot of things and would greatly appreciate your help.
 
 Best,
 Mirjam
 
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Re: [R] error with persp()- increasing 'x' and 'y' values expected



On Feb 20, 2012, at 2:15 PM, adick wrote:


I want to use persp to graph my data and it keeps giving me the error
increasing 'x' and 'y' values expected, even though my data is in  
increasing

order with respect to x and y.



Do you have missing entries? (Running your data fragment through your  
code produces the same error.) You may want to use rep with each and  
times arguments to construct a regular grid and then population the  
missing entries in  matrix with NA's.




Here is the code I'm currently using:
bob= scan (SBA3dataTaxonB.txt,what=char)
labels = bob[1:3]
bob=bob[-c(1,2,3)]
bob=as.numeric(bob)
bob=array(bob,dim=c(3, length(bob)/3))
bob=t(bob)
data.frame(bob) - bob
bob = bob[order(bob[,1], bob[,3]),]
persp(bob[,1], bob[,3], bob[,2], log=bob[,1])


Any way  The z argument to `persp` is supposed to be a matrix of  
dimension == length(x)*length(y):


?matrix



The first few lines of the txt file SBA3dataTaxonB.txt look like this:
variableTaxonB  central
10  19  0.002
10  25  0.002
10  26  0.002
1   17  0.002
1   23  0.002
1   16  0.002
0.1 48  0.002
0.1 47  0.002
0.1 57  0.002
10  15  0.004
10  22  0.004
10  22  0.004
1   21  0.004
1   22  0.004
1   27  0.004
0.1 73  0.004
0.1 62  0.004

Can anyone help me figure out why I am getting this error? And also a
possible solution would be greatly appreciated.
-Thanks, Amanda

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David Winsemius, MD
West Hartford, CT

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Re: [R] Computing plot size in Sweave

2012-02-20 Thread Marc Schwartz

On Feb 20, 2012, at 9:15 AM, BXC (Bendix Carstensen) wrote:

 Sometimes you want to compute the physical size of a plot based on data.
 In R itself this is no problem.
 
 But is there a way to compute the values of height and width in S-weave, say:
 
 graph,fig=TRUE,height=xx,width=yy=
 
 where xx and yy are computed and not physically written in the document?
 
 Bendix

Bendix,

By default, Sweave.sty sets:

\setkeys{Gin}{width=0.8\textwidth}

which modifies the default \includegraphics LaTeX command auto-generated during 
Sweave processing. This means that irrespective of the 'height' and 'width' 
arguments in the figure chunk header, which do control the size of the PDF/EPS 
files created, the actual size of the graphic as included in the resultant 
document will ALWAYS be 80% of the current text width and the height will be 
scaled accordingly.

In general, if you wanted to control the actual height and width of the figure 
in the resultant document, you could set:

FigureChunkName,include=false,echo=false,fig=true,height=YourHeight,width=YourWidth=

Plot Code Here

@

\begin{figure}[tbp]
\centering
\includegraphics{RnwFileName-FigureChunkName}
\caption[LOF Caption]{Some Longer Caption}
\end{figure}

'RnwFileName' is the name of your working .Rnw Sweave file and 
'FigureChunkName' is the name of the figure chunk and as a result, the graphic 
file name being created by Sweave, separated by the hyphen ('-').

By setting the 'include' option to false, Sweave does not auto-generate the 
\includegraphics line with the width=0.8\textwidth argument, and you then 
explicitly include it in the LaTeX code following the figure chunk. The plot 
file(s) would then be created with the height and width parameters in the 
figure chunk header and the resultant document will have a figure of the size 
you desire, overriding the default behavior.

In my .Rnw files, I actually set:

  \usepackage[nogin]{Sweave}

in my preamble, which overrides the default 'Gin' behavior. Then the height and 
width parameters in the figure chunks are reflected in the resultant document, 
but of course, I need to explicitly pre-define those.

If you want to calculate the figure's height and width at run-time, I suspect 
that the only way to do that would be to have your R code generate all of the 
LaTeX code output at runtime as well. So something like the following:

CodeChunkName,echo=false=

# Plot Size Calculations Here
Height - ResultOfCalcs
Width - ResultOfCalcs

pdf(MyPlotFileName.pdf, height = Height, width = Width)

Plot Code Here

dev.off()

cat(\\begin{figure}[tbp]\n)
cat(\\centering\n)
cat(\\includegraphics{MyPlotFileName}\n) 
cat(\\caption[LOF Caption]{Some Longer Caption}\n)
cat(\\end{figure}\n)

@


The result of the cat() function calls will be to output the included character 
vectors to the .tex file being created by Sweave at run-time. So you are using 
R in a normal code chunk to generate LaTeX code.

If you need an EPS file either in place of the PDF (because you are using 
postscript stuff like pstricks) or in addition to the PDF, you can replace the 
pdf() call with postscript() or run a second iteration of the plotting code 
using postscript()/dev.off() as well.

HTH,

Marc Schwartz

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Re: [R] Column wise matrix multiplication

2012-02-20 Thread Ted Harding

[See at end]

On 20-Feb-2012 Ted Harding wrote:
 On 20-Feb-2012 Graziano Mirata wrote:
 Hi all,
 I am trying to multiply each column of a matrix such to have
 a unique resulting vector with length equal to the number of
 rows of the original matrix. In short I would like to do what
 prod(.) function in Matlab does, i.e.
 
 A -matrix(c(1:10),5,2)
 
 V = A[,1]*A[,2]
 
 Thank you
 
 Graziano
 
 The Matlab prod(A,2) function computes the products along the
 rows of the matrix A and returns the result as a column vector,
 of length equal to the number of rows in A, which seems to be
 what you describe.
 
 Your code above does this for your 2-column example, but the
 result is a simple R vector which is not an array (and in
 particular is not a column vector):
 
   A[,1]*A[,2]
   # [1]  6 14 24 36 50
 
   dim(A[,1]*A[,2])
   # NULL
 
 For a matrix A with arbitrary number of columns, if you wanted
 the row sums rather than the row products, you could use the
 R function rowSums():
 
   rowSums(A)
   # [1]  7  9 11 13 15
 
 This is still a dimensionless simple R vector:
 
   dim(rowSums(A))
   # NULL
 
 Unfortunately, there seems to be no equivalent for products
 (e.g. rowProds). But you can define one:
 
   rowProds - function(X){ apply(X,1,FUN=prod) }
 
   rowProds(A)
   # [1]  6 14 24 36 50
 
 Even then, the result is a simple R vector, without dimensions:
 
   dim(rowProds(A))
   # NULL
 
 If you need an array (row) vector then you can apply t():
 
   t(rowProds(A))
   #  [,1] [,2] [,3] [,4] [,5]
   # [1,]6   14   24   36   50
 
 or t(t()) for a column vector:
 
   t(t(rowProds(A)))
   #  [,1]
   # [1,]6
   # [2,]   14
   # [3,]   24
   # [4,]   36
   # [5,]   50
 
 Ted.
 
 -

Further to the above: I have managed to track down a
function rowProds in the matrixStats package:

http://finzi.psych.upenn.edu/R/library/matrixStats/html/rowProds.html

http://www.stats.bris.ac.uk/R/web/packages/matrixStats/matrixStats.pdf

Note that:

  Details
   Internally the product is calculated via the logarithmic
   transform, treating zeros and negative values specially.

In view of this, which strikes me as potentially getting
close to thin ice, plus the overhead of loading a whole
package just for one function, it may be more straightforward
(and perhaps safer) to define one's own function (as above).
Also (see the PDF reference manual) it is apparently work
in progress and also has dependencies on other packages:
see the description at

http://www.stats.bris.ac.uk/R/web/packages/matrixStats/index.html

Ted.

-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 20-Feb-2012  Time: 21:33:25
This message was sent by XFMail

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Re: [R] Time Series - Trend Line