Re: [R] R Error, very odd....

2009-05-11 Thread Dieter Menne



Katie2009 wrote:
 
 I'm trying to analyse some excel data in R.  The problem is that when i
 input the data with the first column as absolute values, everything works
 fine, can analyse as normal.  When I leave the first column unchanged to
 import negative numbers as well I get:
 
 Error in storage.mode(y) - double : 
   invalid to change the storage mode of a factor
 In addition: Warning message:
 In model.response(mf, numeric) :
   using type=numeric with a factor response will be ignored
 
 

You did not tell us anything how you got the data from Excel, so I have to
guess. Try to re-arrange your Excel row so that the first (3 ? check the
docs; which docs? your unknown function's) lines contain non-missing data.

Dieter
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[R] R Error, very odd....

2009-05-11 Thread Katie2009

I'm trying to analyse some excel data in R.  The problem is that when i input
the data with the first column as absolute values, everything works fine,
can analyse as normal.  When I leave the first column unchanged to import
negative numbers as well I get:


Error in storage.mode(y) - double : 
  invalid to change the storage mode of a factor
In addition: Warning message:
In model.response(mf, numeric) :
  using type=numeric with a factor response will be ignored

What can I do to get around this? I would like to analyse the data in R that
does include the negative numbers.

Thanks!!

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[R] working with groups of labels?

2009-05-11 Thread Phillip Porter

Good Morning,
I have a graph with groups of variables.  I have include the group 
names as variables so that I can have them positioned correctly. 
Unfortunately this means that the group names have to follow all of 
the same rules as the variables within the groups.  I would rather 
have those group names left justified and bolded.


I know that I can do this in an mtext() function using font=2 and 
adj=0, but I can never get the labels lined up exactly right.


Is there a simpler way of getting one group of labels right justified 
and another group of labels left justified and still have them all 
automatically positioned correctly?


Thanks,
Phillip

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Re: [R] R Error, very odd....

2009-05-11 Thread Dieter Menne



Katie2009 wrote:
 
 hi dieter,
 
 the method i'm using is in excel, copying the data, then in r
 
 w-read.delim(clipboard)
 w-as.data.frame(w)
  
 i've been  doing a bit more fiddling, and have identified the 'class' of
 the column that i'm having trouble with, is classified as 'factor' whilst
 the rest are numeric.
 
 

read.delim internally uses the same function as read.table, so you might
consult that documentation:

The number of data columns is determined by looking at the first five lines
of input (or the whole file if it has less than five lines), or from the
length of col.names if it is specified and is longer. This could conceivably
be wrong if fill or blank.lines.skip are true, so specify col.names if
necessary.

So if the first lines are unusual, many things can happen. However, reading
from the clipboard
is not a very stable way to get Excel data; I would suggest to use
library(RODBC) instead,
or one of the other Excel readers. With RODBC, use named ranges, not
worksheet names
for a stable import.

Dieter





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Re: [R] R package for estimating markov transition matrix from observations + confidence?

2009-05-11 Thread Michael Dewey

At 19:18 09/05/2009, U.H wrote:

Dear R gurus,

I have data for which I want to estimate the markov transition matrix
that generated the sequence, and preferably obtain some measure of
confidence for that estimation.


 RSiteSearch(markov, restrict = function)
shows that there are numerous packages on CRAN with Markov in their 
name or documentation. Does any of them help?




e.g., for a series such as
 1 3 4 1 2 3 1 2 1 3 4 3 2 4 2 1 4 1 2 4 1 2 4 1 2 1 2 1 3 1

I would want to get an estimate of the matrix that generated it

[[originally:
[,1] [,2] [,3] [,4]
[1,] 0.00 0.33 0.33 0.33
[2,] 0.33 0.00 0.33 0.33
[3,] 0.33 0.33 0.00 0.33
[4,] 0.33 0.33 0.33 0.00
]]

and the confidence in that estimation.

I know that generating the cross--tab matrix is trivial, but if there
is a package that does that  and provides a likelihood as well, I'd
appreciate knowing about it.

Best,
Uri


Michael Dewey
http://www.aghmed.fsnet.co.uk

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Re: [R] R Error, very odd....

2009-05-11 Thread Katie2009

hi dieter,

the method i'm using is in excel, copying the data, then in r

 w-read.delim(clipboard)
 w-as.data.frame(w)
 
i've been  doing a bit more fiddling, and have identified the 'class' of the
column that i'm having trouble with, is classified as 'factor' whilst the
rest are numeric.

if i just change that column to as.numeric of the column, this has appeared
to have solved the problem.  Has this changed that data at all though?

Thanks.


Dieter Menne wrote:
 
 
 
 Katie2009 wrote:
 
 I'm trying to analyse some excel data in R.  The problem is that when i
 input the data with the first column as absolute values, everything works
 fine, can analyse as normal.  When I leave the first column unchanged to
 import negative numbers as well I get:
 
 Error in storage.mode(y) - double : 
   invalid to change the storage mode of a factor
 In addition: Warning message:
 In model.response(mf, numeric) :
   using type=numeric with a factor response will be ignored
 
 
 
 You did not tell us anything how you got the data from Excel, so I have to
 guess. Try to re-arrange your Excel row so that the first (3 ? check the
 docs; which docs? your unknown function's) lines contain non-missing data.
 
 Dieter
 

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Re: [R] working with groups of labels?

2009-05-11 Thread Richard . Cotton
 I have a graph with groups of variables.  I have include the group 
 names as variables so that I can have them positioned correctly. 
 Unfortunately this means that the group names have to follow all of 
 the same rules as the variables within the groups.  I would rather 
 have those group names left justified and bolded.
 
 I know that I can do this in an mtext() function using font=2 and 
 adj=0, but I can never get the labels lined up exactly right.
 
 Is there a simpler way of getting one group of labels right justified 
 and another group of labels left justified and still have them all 
 automatically positioned correctly?

mtext accepts vector inputs, so you can specify different justifications 
for different labels, e.g.

plot(1:10)
mtext(c(foo, bar, baz), adj=c(0, 0.5, 1), side=1)

Regards,
Richie.

Mathematical Sciences Unit
HSL




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This message contains privileged and confidential inform...{{dropped:20}}

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[R] Unable to install R package

2009-05-11 Thread anupam sinha
Hi all,
 I am unable to install R package on RHEL version 5.1. While running
* ./configure* command following is printed on to the screen and no *
makefile* is created. Can anyone help me out ??? Thanks in advance.



./configure
checking build system type... x86_64-unknown-linux-gnu
checking host system type... x86_64-unknown-linux-gnu
loading site script './config.site'
loading build specific script './config.site'
checking for pwd... /bin/pwd
checking whether builddir is srcdir... yes
checking for working aclocal... missing
checking for working autoconf... missing
checking for working automake... missing
checking for working autoheader... missing
checking for gawk... gawk
checking for grep that handles long lines and -e... /bin/grep
checking for egrep... /bin/grep -E
checking whether ln -s works... yes
checking for bison... no
checking for byacc... no
checking for ar... ar
checking for a BSD-compatible install... /usr/bin/install -c
checking for sed... /bin/sed
checking for less... /usr/bin/less
checking for perl... /usr/bin/perl
checking whether perl version is at least 5.8.0... yes
checking for dvips... /usr/bin/dvips
checking for tex... /usr/bin/tex
checking for latex... /usr/bin/latex
checking for makeindex... /usr/bin/makeindex
checking for pdftex... /usr/bin/pdftex
checking for pdflatex... /usr/bin/pdflatex
checking for makeinfo... /usr/bin/makeinfo
checking whether makeinfo version is at least 4.7... yes
checking for texi2dvi... no
checking for unzip... /usr/bin/unzip
checking for zip... /usr/bin/zip
checking for gzip... /usr/bin/gzip
checking for firefox... /usr/bin/firefox
using default browser ... /usr/bin/firefox
checking for acroread... no
checking for acroread4... no
checking for evince... /usr/bin/evince
checking for pkg-config... /usr/bin/pkg-config
checking for gcc... gcc
checking for C compiler default output file name... a.out
checking whether the C compiler works... yes
checking whether we are cross compiling... no
checking for suffix of executables...
checking for suffix of object files... o
checking whether we are using the GNU C compiler... yes
checking whether gcc accepts -g... yes
checking for gcc option to accept ISO C89... none needed
checking how to run the C preprocessor... gcc -E
checking whether gcc needs -traditional... no
checking how to run the C preprocessor... gcc -E
checking for gfortran... no
checking for f95... no
checking for fort... no
checking for xlf95... no
checking for ifort... no
checking for ifc... no
checking for efc... no
checking for pgf95... no
checking for lf95... no
checking for gfortran... no
checking for ftn... no
checking for g95... no
checking for f90... no
checking for xlf90... no
checking for pgf90... no
checking for pghpf... no
checking for epcf90... no
checking for g77... g77
checking whether we are using the GNU Fortran 77 compiler... yes
checking whether g77 accepts -g... yes
checking for g++... no
checking for c++... no
checking for gpp... no
checking for aCC... no
checking for CC... no
checking for cxx... no
checking for cc++... no
checking for cl.exe... no
checking for FCC... no
checking for KCC... no
checking for RCC... no
checking for xlC_r... no
checking for xlC... no
checking whether we are using the GNU C++ compiler... no
checking whether g++ accepts -g... no
checking how to run the C++ preprocessor... /lib/cpp
configure: error: C++ preprocessor /lib/cpp fails sanity check
See `config.log' for more details.



Regards,

Anupam

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Re: [R] ade4 and inertia of axis 2

2009-05-11 Thread Stéphane Dray

Dear John,

In RLQ (similarly to other methods in ade4), inertia are given by 
eigenvalues. Thus, the percentage of inertia are obtained simply by:


rlq1$eig/sum(rlq1$eig)*100

For more details on outputs or RLQ, you can have a look at 
http://listes.univ-lyon1.fr/wws/arc/adelist/2009-03/msg1.html


Lastly, if you have questions about ade4, please subscribe and send an 
email to adelist (http://listes.univ-lyon1.fr/wws/info/adelist)


Cheers.


John Poulsen wrote:


Hello,

Could someone please tell me how to find the estimate of inertia for 
the second axis of an RLQ analysis using ade4? Using the example from 
the ade4 package, I *suspect* that the inertia for the 2nd axis of the 
R table would be 4.139332 (see below results summary from rlq1).  
However, the row is labeled 12 not 2 which suggests this is not 
correct.  In addition, this would mean that axis 2 of the RLQ would 
explain more than 100% of the inertia of the separate ordination for 
dudimil ((100*4.139)/dudimil$eig[2]=176%) which I believe is not 
supposed to occur.


Could anyone set me straight by either letting me know how to get the 
inertia for axis 2 or that my thinking is perhaps wrong?


Thanks,
John

data(aviurba)
 coa1 - dudi.coa(aviurba$fau, scannf = FALSE, nf = 2)
 dudimil - dudi.hillsmith(aviurba$mil, scannf = FALSE, nf = 2, row.w 
= coa1$lw)
 duditrait - dudi.hillsmith(aviurba$traits, scannf = FALSE, nf = 2, 
row.w = coa1$cw)

 rlq1 - rlq(dudimil, coa1, duditrait, scannf = FALSE, nf = 2)
 plot(rlq1)
 summary(rlq1)
 randtest.rlq(rlq1)


   summary(rlq1)


Eigenvalues decomposition:
  eig covar  sdR  sdQ  corr
1 0.4782826 0.6915798 1.558312 1.158357 0.3831293
2 0.1418508 0.3766308 1.308050 1.219367 0.2361331

Inertia  coinertia R:
  inertia  max ratio
1  2.428337 2.996911 0.8102800
12 4.139332 5.345110 0.7744148

Inertia  coinertia Q:
  inertia  max ratio
1  1.341791 2.603139 0.5154512
12 2.828648 4.202981 0.6730098

Correlation L:
 corr   max ratio
1 0.3831293 0.6435487 0.5953384
2 0.2361331 0.5220054 0.4523576

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--
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Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I
43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France
Tel: 33 4 72 43 27 57   Fax: 33 4 72 43 13 88
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[R] 3d Scatterplot using rgl

2009-05-11 Thread Zoe Hoare
Hi,
I am trying to plot a 3d scatter plot using the rgl package but am having 
trouble.
I have a matrix containing x, y and z co-ordinates and a fourth element related 
to a count variable. 
I would like to plot the points using plot3d with the sizes of the points 
related to the fourth element in the matrix. Is this possible with plot3d or do 
I need to use something else?

I include a small example of what I am trying to do

 swadt
 A   B  Crun1 run2 run3 run4 run5 run6
  93  96 51  0.1   NA  0.1   NA   NA   NA
  93  97 50  0.3   NA   NA  0.1   NA   NA
  93  98 49  0.1   NA   NA   NA   NA   NA
  94  95 51   NA   NA   NA   NA   NA  0.1
  94 100 46  0.1   NA   NA   NA   NA   NA
  94  96 50  2.1  0.1  1.0  0.7  0.8  0.4
  94  97 49  3.2  0.9  1.9  1.9  1.6  1.6
  94  98 48  0.9   NA  0.3  0.4  0.5  0.1
  95  94 51   NA   NA   NA  0.1  0.1   NA
  95  95 50  3.4  0.7  2.8  2.4  2.9  2.3
  95  96 49 12.5 13.4 13.6 15.0 15.3 17.1
  95  97 48 10.2 10.8  8.9 10.8  8.7  9.3
  95  98 47  1.5   NA  0.9  0.7  0.8  0.1
  96  93 51  0.1   NA   NA   NA   NA   NA

If I use
plot3d(swadt[,1],swadt[,2],swadt[,3],size=swadt[,4])
I get the following error
Error in rgl.numeric(size) : size must be a single numeric value

Is there a way of plotting this?
Thanks
Zoe Hoare


  
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[R] R-2.9.0 on CentOS 5 installation?

2009-05-11 Thread Patrick Connolly
Has anyone else had experience with CentOS and the latest version of R?

$uname -r
 2.6.18-92.1.22.el5

On occasion, I need to have the LD_LIBRARY_PATH environment variable
which I set in ~/.bash_profile.  Until R-2.9.0 that has worked fine,
but now when I check with Sys.getenv(LD_LIBRARY_PATH), I see it's
not set.  It appears to me that it's not something that can be set
after R has started.

The problem doesn't exist with Fedora 8, nor with Mepis 6.5 in my
experience, and I've been informed that Fedora 9 doesn't have the
problem either.

Has anyone else seen it?  Particularly with CentOS.

TIA

-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~}   Great minds discuss ideas
 _( Y )_ Average minds discuss events 
(:_~*~_:)  Small minds discuss people  
 (_)-(_)  . Eleanor Roosevelt
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] Function recommendation for this study...

2009-05-11 Thread Paul Heinrich Dietrich

Thank you for the suggestions, I'm sure irr is what I'm looking for.  Thanks
again.


spencerg wrote:
 
   To look up observer agreement, you might consider the 
 RSiteSearch package.  The RSiteSearch.function looks only for 
 matches in help pages of contributed packages. 
 
 
 library(RSiteSearch)
 oa - RSiteSearch.function(observer agreement)
 attr(oa, hits) # 4 functions matching this term
 HTML(oa) # to open the results as a table in a web browser. 
 
 
   An alternative search term for this is interrater reliability. 
 
 
 ir.r - RSiteSearch.function(inter-rater reliability)
 attr(ir.r, hits) # 1
 irr - RSiteSearch.function(interrater reliability)
 attr(irr, hits) # 19
 ir..r - RSiteSearch.function(inter rater reliability)
 attr(ir..r, hits) # 1
 
 
   In particular, this identified a package irr for interrater 
 reliability. 
 
 
   The development version of the RSiteSearch package on R-Forge 
 includes a unionRSiteSearch function that allows one to combine all 
 these searches.  You can get this version using 
 'install.packages(RSiteSearch, 
 repos=http://R-Forge.R-project.org;)';  if this gives you version 
 1.0-0, please wait 24 hours and try again, because this version contains 
 a bug that I just fixed.  The new version 1.0-1 should be available in 
 24 hours.  With it, the following just worked for me: 
 
 
 IRR - unionRSiteSearch(oa, unionRSiteSearch(ir.r,
  unionRSiteSearch(irr, ir..r) ) )
 attr(IRR, hits)
 HTML(IRR)
 
 
   Hope this helps. 
   Spencer Graves
 
 Murray Cooper wrote:
 Paul,

 I suggest looking up observer agreement. The description of your 
 study sounds like a classical
 categorical observer agreement problem. I can't give
 a reference off the top of my head, but if you get
 stuck, e-mail me and I'll try and find a ref to get you started.

 Murray M Cooper, Ph.D.
 Richland Statistics
 9800 N 24th St
 Richland, MI, USA 49083
 Mail: richs...@earthlink.net

 - Original Message - From: Paul Heinrich Dietrich 
 paul.heinrich.dietr...@gmail.com
 To: r-help@r-project.org
 Sent: Sunday, May 10, 2009 8:25 AM
 Subject: [R] Function recommendation for this study...



 Hi,
 I'm not used to thinking along these lines, and wanted to ask your 
 advice:

 Suppose you have a sample of around 100, consisting of patients 
 according to
 doctors, in which patients and doctors are given a questionnaire with
 categorical responses.  Each patient somehow has roughly 3 doctors, or 3
 rows of data.  The goal is to assess by category of each question or 
 DV the
 agreement between the patient and 3 doctors.  For example, a question 
 may be
 asked about how well the treatment is understood by the patient, and the
 patient answers with their perception, while the 3 doctors each 
 answer with
 their perception.

 The person currently working on this has used a Wilcoxon Sign Rank 
 test, and
 asked what I thought.  Personally, I shy away from nonparametrics and 
 prefer
 parametric Bayesian methods, but of course am up for whatever is most
 appropriate.  I was concerned about using multiple Wilcoxon tests, 
 one for
 each question, and wondering if there is a parametric method in R for
 something like this, and a method which is multivariate?  Thanks for any
 suggestions.
 -- 
 View this message in context: 
 http://www.nabble.com/Function-recommendation-for-this-study...-tp23469646p23469646.html
  

 Sent from the R help mailing list archive at Nabble.com.

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[R] Using ACE in the ape package

2009-05-11 Thread Andrew RODRIGUES
Having a problem running the ace command in ape.  

After reading my table into R and then putting the names of the rows in the
table in same order as the tree I can't get ace to run and get the message

Erreur dans as.matrix(x) : 
  dims [produit 25] ne correspond pas à la longueur de l'objet [0]

This is the command I am using and each of the vectors used in the
snalysis.  

ERreconstruction - ace(car, tr, type=discrete, model=SYM)

car [1] 4 1 7 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 1 2 1

 tr

Phylogenetic tree with 32 tips and 31 internal nodes.

Tip labels:
Batis_maritima, Pentadiplandra_brazzeana, Tovaria_pendula, Reseda_lutea,
Tropaeolum_majus, Carica_sp, ...

Rooted; no branch lengths.


Thanks for any help.

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Re: [R] 3d Scatterplot using rgl

2009-05-11 Thread Olivier ETERRADOSSI

Hi Zoé,
try
plot3d(swadt[,1],swadt[,2],swadt[,3],radius=swadt[,4],type=s)

Hope this helps. regards. Olivier


Zoe Hoare wrote:
 
 Hi,
 I am trying to plot a 3d scatter plot using the rgl package but am having
 trouble.
 I have a matrix containing x, y and z co-ordinates and a fourth element
 related to a count variable. 
 I would like to plot the points using plot3d with the sizes of the points
 related to the fourth element in the matrix. Is this possible with plot3d
 or do I need to use something else?
 
 I include a small example of what I am trying to do
 
  swadt
  A   B  Crun1 run2 run3 run4 run5 run6
   93  96 51  0.1   NA  0.1   NA   NA   NA
   93  97 50  0.3   NA   NA  0.1   NA   NA
   93  98 49  0.1   NA   NA   NA   NA   NA
   94  95 51   NA   NA   NA   NA   NA  0.1
   94 100 46  0.1   NA   NA   NA   NA   NA
   94  96 50  2.1  0.1  1.0  0.7  0.8  0.4
   94  97 49  3.2  0.9  1.9  1.9  1.6  1.6
   94  98 48  0.9   NA  0.3  0.4  0.5  0.1
   95  94 51   NA   NA   NA  0.1  0.1   NA
   95  95 50  3.4  0.7  2.8  2.4  2.9  2.3
   95  96 49 12.5 13.4 13.6 15.0 15.3 17.1
   95  97 48 10.2 10.8  8.9 10.8  8.7  9.3
   95  98 47  1.5   NA  0.9  0.7  0.8  0.1
   96  93 51  0.1   NA   NA   NA   NA   NA
 
 If I use
 plot3d(swadt[,1],swadt[,2],swadt[,3],size=swadt[,4])
 I get the following error
 Error in rgl.numeric(size) : size must be a single numeric value
 
 Is there a way of plotting this?
 Thanks
 Zoe Hoare
 
 
   
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[R] readBin: read from defined offset TO defined offset?

2009-05-11 Thread Johannes Graumann
Hello,

With the help of seek I can start readBin from any byte offset within my 
file that I deem appropriate.
What I would like to do is to be able to define the endpoint of that read as 
well. Is there any solution to that already out there?

Thanks for any hints, Joh

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Re: [R] working with groups of labels?

2009-05-11 Thread Jim Lemon

Phillip Porter wrote:

Good Morning,
I have a graph with groups of variables.  I have include the group 
names as variables so that I can have them positioned correctly. 
Unfortunately this means that the group names have to follow all of 
the same rules as the variables within the groups.  I would rather 
have those group names left justified and bolded.


I know that I can do this in an mtext() function using font=2 and 
adj=0, but I can never get the labels lined up exactly right.


Is there a simpler way of getting one group of labels right justified 
and another group of labels left justified and still have them all 
automatically positioned correctly?



Hi Phillip,
I'm not exactly sure how you are positioning the labels. Is it possible 
to give us an example with some data that will produce a plot to show 
what you want? It shouldn't be too hard to do.


Jim

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Re: [R] Partial Derivatives in R

2009-05-11 Thread Paul Heinrich Dietrich

Hi Spencer,
Thanks for suggesting the genD function.  In attempting it, I have
rearranged my function from phat1 ~ ... to ... - 1, it apparently doesn't
like the first one :)  But when I run it, it tells me the partials are all
zero.  I'm trying out a simple MNL equation before I expand it to what I'm
looking for.  Here is what I tried (and I get different answers from a
textbook solution, deriv(), and genD()):

 ### Variables for an observation
 x01 - rnorm(1,0,1)
 x02 - rnorm(1,0,1)
 ### Parameters for an observation
 b00.1 - rnorm(1,0,1)
 b00.2 - rnorm(1,0,1)
 b00.3 - 0
 b01.1 - rnorm(1,0,1)
 b01.2 - rnorm(1,0,1)
 b01.3 - 0
 b02.1 - rnorm(1,0,1)
 b02.2 - rnorm(1,0,1)
 b02.3 - 0
 ### Predicted Probabilities for an observation
 phat1 - 0.6
 phat2 - 0.3
 phat3 - 0.1
 ### Correct way to calculate a partial derivative
 partial.b01.1 - phat1 * (b01.1 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b01.2 - phat2 * (b01.2 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b01.3 - phat3 * (b01.3 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b01.1; partial.b01.2; partial.b01.3
[1] 0.04288663
[1] -0.1804876
[1] 0.1376010
 
 partial.b02.1 - phat1 * (b02.1 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b02.2 - phat2 * (b02.2 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b02.3 - phat3 * (b02.3 - (b01.1*phat1+b01.2*phat2+b01.3*phat3))
 partial.b02.1; partial.b02.2; partial.b02.3
[1] 0.8633057
[1] 0.3171978
[1] 0.1376010
 ### Derivatives for MNL
 dp1.dx - deriv(phat1 ~ exp(b00.1+b01.1*x01+b02.1*x02) /
+ (exp(b00.1+b01.1*x01+b02.1*x02)+exp(b00.2+b01.2*x01+b02.2*x02)+
+ exp(b00.3+b01.3*x01+b02.3*x02)), c(x01,x02))
 dp2.dx - deriv(phat2 ~ exp(b00.2+b01.2*x01+b02.2*x02) /
+ (exp(b00.1+b01.1*x01+b02.1*x02)+exp(b00.2+b01.2*x01+b02.2*x02)+
+ exp(b00.3+b01.3*x01+b02.3*x02)), c(x01,x02))
 dp3.dx - deriv(phat3 ~ exp(b00.3+b01.3*x01+b02.3*x02) /
+ (exp(b00.1+b01.1*x01+b02.1*x02)+exp(b00.2+b01.2*x01+b02.2*x02)+
+ exp(b00.3+b01.3*x01+b02.3*x02)), c(x01,x02))
 attr(eval(dp1.dx), gradient)
 x01  x02
[1,] -0.01891354 0.058918
 attr(eval(dp2.dx), gradient)
x01 x02
[1,] -0.1509395 -0.06258685
 attr(eval(dp3.dx), gradient)
  x01 x02
[1,] 0.169853 0.003668849
 library(numDeriv)
 dp1.dx - function(x) {exp(b00.1+b01.1*x01+b02.1*x02) /
+ (exp(b00.1+b01.1*x01+b02.1*x02)+exp(b00.2+b01.2*x01+b02.2*x02)+
+ exp(b00.3+b01.3*x01+b02.3*x02)) - phat1}
 test - genD(dp1.dx, c(phat1,b00.1,b01.1,b02.1,x01,x02)); test
$D
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,]000000000 0 0 0 0
0
 [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,] 0 0 0 0 0 0 0 0 0 0 0 0
 [,27]
[1,] 0

$p
[1] 6

$f0
[1] 0.05185856

$func
function(x) {exp(b00.1+b01.1*x01+b02.1*x02) /
(exp(b00.1+b01.1*x01+b02.1*x02)+exp(b00.2+b01.2*x01+b02.2*x02)+
exp(b00.3+b01.3*x01+b02.3*x02)) - phat1}

$x
[1]  0.6  1.401890082 -1.304531849  0.062833294 -0.143141379
[6] -0.005995477

$d
[1] 1e-04

$method
[1] Richardson

$method.args
$method.args$eps
[1] 1e-04

$method.args$d
[1] 1e-04

$method.args$zero.tol
[1] 1.781029e-05

$method.args$r
[1] 4

$method.args$v
[1] 2


attr(,class)
[1] Darray
 





spencerg wrote:
 
   Have you considered genD{numDeriv}? 
 
   If this does not answer your question, I suggest you try the 
 RSiteSearch package.  The following will open a list of options in a 
 web browser, sorted by package most often found with your search term: 
 
 
 library(RSiteSearch)
 pd - RSiteSearch.function('partial derivative')
 pds - RSiteSearch.function('partial derivatives')
 attr(pd, 'hits') # 58
 attr(pds, 'hits')# 52
 summary(pd)
 HTML(pd)
 HTML(pds)
 

   The development version available via 
 'install.packages(RSiteSearch, repos=http://R-Forge.R-project.org;)' 
 also supports the following: 
  
 
 pd. - unionRSiteSearch(pd, pds)
 attr(pd., 'hits')# 94
 HTML(pd.)
 
 
   Hope this helps. 
   Spencer Graves
 
 Paul Heinrich Dietrich wrote:
 Quick question:

 Which function do you use to calculate partial derivatives from a model
 equation?

 I've looked at deriv(), but think it gives derivatives, not partial
 derivatives.  Of course my equation isn't this simple, but as an example,
 I'm looking for something that let's you control whether it's a partial
 or
 not, such as:

 somefunction(y~a+bx, with respect to x, partial=TRUE)

 Is there anything like this in R?

 
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[R] doubt

2009-05-11 Thread Rangasamy, Shanthi
Hi Sir,

I want to install the package cluster in R language.
 I could be installing, but I cant load that package..

I have faced the following warning in installing time..

Warning: cannot remove prior installation of package 'cluster'

So, I tried to update the package, but again I faced the following issue



update.packages()
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://cran.uk.r-project.org/bin/windows/contrib/2.9
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.9
Warning message:
In open.connection(con, r) :
  no fue posible conectarse a 'cran.r-project.org' en el pueto 80.



Sample:

 utils:::menuInstallLocal()
package 'cluster' successfully unpacked and MD5 sums checked
Warning: cannot remove prior installation of package 'cluster'
updating HTML package descriptions
 load(cluster)
Error en load(cluster) : objeto 'cluster' no encontrado
 library(cluster)
Error en library(cluster) : there is no package called 'cluster'
 library(clValid)
Loading required package: cluster
Error: package 'cluster' could not be loaded
Además: Warning message:
In library(pkg, character.only = TRUE, logical.return = TRUE, lib.loc = 
lib.loc) :
  there is no package called 'cluster'
 update.packages()
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://cran.uk.r-project.org/bin/windows/contrib/2.9
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.9
Warning message:
In open.connection(con, r) :
  no fue posible conectarse a 'cran.r-project.org' en el pueto 80.


Can you tell me for any suggestions...



Thank you,


Shanthi Rangasamy
Ph. D. Student
Electronics and Communications Department
CEIT - IK4 Research Alliance
Pº. Manuel Lardizábal, 15
20.018 Donostia-San Sebastián
SPAIN
Tel.  +34 943 212 800  (Ext. 2804)
Fax: +34 943 213 076
http://www.ceit.es/


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[R] aggregate over x cases

2009-05-11 Thread Jens Bölte

Hello,

I have been struggling for quite some time to find a solution for the following 
problem. I have a data frame which is organized by block and trial. Each trial 
is represented across several rows in this data frame. I'd like to extract the 
first x rows per trial and block.

For example
block   trial   x   y
1   1   1   605 150
2   1   1   603 148
3   1   1   604 140
4   1   1   600 140
5   1   1   590 135
6   1   1   580 135
7   1   2   607 148
8   1   2   605 152
10  1   2   600 158
.

Selecting only the the first two rows per trial should result in 

block trial x y 
1	1	605	150

1   1   603 148
1   2   607 148
1   2   605 152

The data I am dealing with a x-y coordinates (samples) from an eye-tracking 
experiment. I receive the data in this format and need to eliminate unwanted 
samples.

Thanks Jens Bölte
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[R] Amsterdam course - Pharmacokinetic and pharmacodynamic modeling and simulation

2009-05-11 Thread Dick Verkerk

LS,

We would like to inform you about the new course Pharmacokinetic
and pharmacodynamic modeling and simulation by Dr. Jan Freijer
in Amsterdam, November 28th.

The course is meant for users of R or S-PLUS in the
bio-pharmaceutical sciences who would like to apply it to
clinical trial simulations.

You can find the full information on the course via:
http://www.can.nl/events/details.php?id=57

Price for the course is EURO 395,- VAT excluded and you can
subscribe via our web site, email, fax (+31 (0)20 5608448) or
post (CAN - Nieuwpoortkade 25 - 1055 RX Amsterdam).

If you have any questions, please feel free to contact me.

Kind Regards,

Dick Verkerk


Check out our complete event list at:
http://www.can.nl/events/

_

CANdiensten, Nieuwpoortkade 23-25, NL-1055 RX Amsterdam
voice: +31 20 5608400 fax: +31 20 5608448 i...@can.nl
_

Your Partner in Finance, Mathematics and Statistics!

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Re: [R] aggregate over x cases

2009-05-11 Thread baptiste auguie

try this,


library(plyr)
ddply(d, .(block, trial), function(.d) .d[1:2, ])


  block trial   x   y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152


HTH,

baptiste


On 11 May 2009, at 13:49, Jens Bölte wrote:


Hello,

I have been struggling for quite some time to find a solution for  
the following problem. I have a data frame which is organized by  
block and trial. Each trial is represented across several rows in  
this data frame. I'd like to extract the first x rows per trial and  
block.


For example
block   trial   x   y
1   1   1   605 150
2   1   1   603 148
3   1   1   604 140
4   1   1   600 140
5   1   1   590 135
6   1   1   580 135
7   1   2   607 148
8   1   2   605 152
10  1   2   600 158
.

Selecting only the the first two rows per trial should result in

block trial x y
1   1   605 150
1   1   603 148
1   2   607 148
1   2   605 152

The data I am dealing with a x-y coordinates (samples) from an eye- 
tracking experiment. I receive the data in this format and need to  
eliminate unwanted samples.


Thanks Jens Bölte
ATT1.txt


_

Baptiste Auguié

School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag

__
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Re: [R] aggregate over x cases

2009-05-11 Thread Gabor Grothendieck
Try this:

do.call(rbind, by(DF, DF[1:2], head, 2))

On Mon, May 11, 2009 at 7:49 AM, Jens Bölte boe...@psy.uni-muenster.de wrote:
 Hello,

 I have been struggling for quite some time to find a solution for the
 following problem. I have a data frame which is organized by block and
 trial. Each trial is represented across several rows in this data frame. I'd
 like to extract the first x rows per trial and block.

 For example
        block   trial   x       y
 1       1       1       605     150
 2       1       1       603     148
 3       1       1       604     140
 4       1       1       600     140
 5       1       1       590     135
 6       1       1       580     135
 7       1       2       607     148
 8       1       2       605     152
 10      1       2       600     158
 .

 Selecting only the the first two rows per trial should result in
 block trial x y 1       1       605     150
 1       1       603     148
 1       2       607     148
 1       2       605     152

 The data I am dealing with a x-y coordinates (samples) from an eye-tracking
 experiment. I receive the data in this format and need to eliminate unwanted
 samples.

 Thanks Jens Bölte

 __
 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.



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Re: [R] aggregate over x cases

2009-05-11 Thread jim holtman
Here is a way of doing it:

 x
   block trial   x   y
1  1 1 605 150
2  1 1 603 148
3  1 1 604 140
4  1 1 600 140
5  1 1 590 135
6  1 1 580 135
7  1 2 607 148
8  1 2 605 152
10 1 2 600 158
 do.call(rbind, lapply(split(x, list(x$block, x$trial), drop=TRUE), head,
2))
  block trial   x   y
1.1.1 1 1 605 150
1.1.2 1 1 603 148
1.2.7 1 2 607 148
1.2.8 1 2 605 152


On Mon, May 11, 2009 at 7:49 AM, Jens Bölte boe...@psy.uni-muenster.dewrote:

 Hello,

 I have been struggling for quite some time to find a solution for the
 following problem. I have a data frame which is organized by block and
 trial. Each trial is represented across several rows in this data frame. I'd
 like to extract the first x rows per trial and block.

 For example
block   trial   x   y
 1   1   1   605 150
 2   1   1   603 148
 3   1   1   604 140
 4   1   1   600 140
 5   1   1   590 135
 6   1   1   580 135
 7   1   2   607 148
 8   1   2   605 152
 10  1   2   600 158
 .

 Selecting only the the first two rows per trial should result in
 block trial x y 1   1   605 150
 1   1   603 148
 1   2   607 148
 1   2   605 152

 The data I am dealing with a x-y coordinates (samples) from an eye-tracking
 experiment. I receive the data in this format and need to eliminate unwanted
 samples.

 Thanks Jens Bölte

 __
 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.htmlhttp://www.r-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




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

What is the problem that you are trying to solve?

[[alternative HTML version deleted]]

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Re: [R] readBin: read from defined offset TO defined offset?

2009-05-11 Thread jim holtman
Can you be more specific on how you want to define the endpoint of that
read.  What is the criteria you want to use?  Can you read in a block and
then search of the pattern?

On Mon, May 11, 2009 at 7:05 AM, Johannes Graumann johannes_graum...@web.de
 wrote:

 Hello,

 With the help of seek I can start readBin from any byte offset within
 my
 file that I deem appropriate.
 What I would like to do is to be able to define the endpoint of that read
 as
 well. Is there any solution to that already out there?

 Thanks for any hints, Joh

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




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

What is the problem that you are trying to solve?

[[alternative HTML version deleted]]

__
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Re: [R] aggregate over x cases

2009-05-11 Thread baptiste auguie

good point, i forgot about head (!),

library(plyr)
ddply(d, .(block, trial), head, 2)

  block trial   x   y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152


On 11 May 2009, at 14:04, Gabor Grothendieck wrote:


Try this:

do.call(rbind, by(DF, DF[1:2], head, 2))

On Mon, May 11, 2009 at 7:49 AM, Jens Bölte boe...@psy.uni-muenster.de 
 wrote:

Hello,

I have been struggling for quite some time to find a solution for the
following problem. I have a data frame which is organized by block  
and
trial. Each trial is represented across several rows in this data  
frame. I'd

like to extract the first x rows per trial and block.

For example
   block   trial   x   y
1   1   1   605 150
2   1   1   603 148
3   1   1   604 140
4   1   1   600 140
5   1   1   590 135
6   1   1   580 135
7   1   2   607 148
8   1   2   605 152
10  1   2   600 158
.

Selecting only the the first two rows per trial should result in
block trial x y 1   1   605 150
1   1   603 148
1   2   607 148
1   2   605 152

The data I am dealing with a x-y coordinates (samples) from an eye- 
tracking
experiment. I receive the data in this format and need to eliminate  
unwanted

samples.

Thanks Jens Bölte

__
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.




__
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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.


_

Baptiste Auguié

School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag

__
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Re: [R] Select the rows in a dataframe that matches a criteria in another dataframe

2009-05-11 Thread David Winsemius


On May 10, 2009, at 7:54 PM, David Winsemius wrote:



?subset
?%in%

(I have gotten tired of converting dataframes that are presented in  
a non-executable form, such as is supported by the dput function.   
So, ... you should read those help pages and take the obvious path  
to success.)


Something along the lines of:

subset(df1, Firm %in% df2[df2$Audited==yes, Firm] )  #untested


Looking at my untested code I can see one error already. (Perhaps)  
should be:


subset(df1, Firm %in% df2[df2$Audited==yes, Firm] )



On May 10, 2009, at 5:52 PM, Cecilia Carmo wrote:

Hi everyone! Thank you for the help you have been given to me, and  
here I'm with another problem with my dataframes:

I have two dataframes (with much more observations), like these:
Dataframe1
Firm Year  cash
5004002002007 100
5004002002006 200
5004002002005 400
5004003002007 300
5004003002006 240
5004003002005 120
5004004002007 340
5004004002006 890
5004004002005 250

Dataframe 2
FirmAudited consolidate
500400200 yes   no
500400300 yes  yes
500400400 nono

I want to make another dataframe equal to the dataframe1, but just  
with the firms «audited», or with the firms «audited» and  
«consolidate». For example, with the audited and consolidated, the  
output would be just firm 500400300, like this:

Firm Year  cash
5004003002007 300
5004003002006 240
5004003002005 120

I’ve tried intersect () but it gives me just the number of the  
firm, and it is not what I want. What I want is a dataframe with  
all the information, but just the firms that match my criteria.




David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] doubt

2009-05-11 Thread David Winsemius
You might try again, assuming you are a Windows user, which was  
suggested by your choice of repositories but not actually stated by  
you. The repository may have been temporarily unavailable. It seems to  
be open for service now with a version of cluster that is 1.11.13.


--
David Winsemius
On May 11, 2009, at 7:22 AM, Rangasamy, Shanthi wrote:


Hi Sir,

I want to install the package cluster in R language.
I could be installing, but I cant load that package..

I have faced the following warning in installing time..

Warning: cannot remove prior installation of package 'cluster'

So, I tried to update the package, but again I faced the following  
issue




update.packages()
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://cran.uk.r-project.org/bin/windows/contrib/2.9
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.9
Warning message:
In open.connection(con, r) :
 no fue posible conectarse a 'cran.r-project.org' en el pueto 80.



Sample:


utils:::menuInstallLocal()

package 'cluster' successfully unpacked and MD5 sums checked
Warning: cannot remove prior installation of package 'cluster'
updating HTML package descriptions

load(cluster)

Error en load(cluster) : objeto 'cluster' no encontrado

library(cluster)

Error en library(cluster) : there is no package called 'cluster'

library(clValid)

Loading required package: cluster
Error: package 'cluster' could not be loaded
Además: Warning message:
In library(pkg, character.only = TRUE, logical.return = TRUE,  
lib.loc = lib.loc) :

 there is no package called 'cluster'

update.packages()

--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository 
http://cran.uk.r-project.org/bin/windows/contrib/2.9
Warning: unable to access index for repository 
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.9
Warning message:
In open.connection(con, r) :
 no fue posible conectarse a 'cran.r-project.org' en el pueto 80.


Can you tell me for any suggestions...



Thank you,



David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Comparing COXPH models, one with age as a continuous variable

2009-05-11 Thread Terry Therneau

  The anova process is only statistically valid for nested models, i.e., where
one includes the other as a proper subset.  A model with continuous age and one
with discrete age are not nested; your p-value will be meaningless.

 Terry Therneau

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Re: [R] readBin: read from defined offset TO defined offset?

2009-05-11 Thread Duncan Murdoch

On 11/05/2009 7:05 AM, Johannes Graumann wrote:

Hello,

With the help of seek I can start readBin from any byte offset within my 
file that I deem appropriate.
What I would like to do is to be able to define the endpoint of that read as 
well. Is there any solution to that already out there?


Sure:  just use the n parameter to readBin.  This will work for any 
type except character, but it doesn't make sense to read those in a 
given range: what do you do if the last one is not zero terminated?  If 
you want the end of block to count as a terminator, then read the block 
as raw, add a null to the end, and then re-read the raw vector.


Duncan Murdoch

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[R] Import Visual FoxPro (.dbf)

2009-05-11 Thread Andy Choens
I have a large series of Visual FoxPro (.dbf) files that I want to
convert to a plain text (.csv) file. I wrote a quick script to do all of
this for me. With a little tweaking it seemed to work like a charm. But,
when I looked at my newly created text files carefully, I discovered a
problem. Every single file produced by my script had a few extra columns
in it. 

To better understand this, I imported a single file. Here are my
results:


 library(foreign)
 temp.df - read.dbf(Activity.dbf)
ilp.dbc changed to: X.ilp.dbc
Field name:  changed to: X
Field name:  changed to: X.1
Field name:  changed to: X.2
Field name:  changed to: X.3
Field name:  changed to: X.4
Field name:  changed to: X.5
Field name:  changed to: X.6


Based on the table definitions I received from the state, none of these
tables have a column called ilp.dbc. When I open these same .dbf files
in OpenOffice.org, individually, I do not see these columns. When I look
at the temp.df in R, these mystery columns are all empty. Otherwise, the
data appears to be fine.

I looked at ?read.dbf and I did not see any options to play with that
would seem to help.

At the moment, I have a work-around in my script that prevents these
extra columns from being written to the csv files, but I am curious to
learn what I am doing wrong. The problem may be entirely of my own
creation. 

Most of the info in these files is protected by HIPAA, which limits my
ability to provide example files. If anyone really wants one, I can look
to see if there are any files that I can release. Or I can try deleting
all of the info and see if the problem persists.


Version Information:
r-project: 2.8.1 
foreign:   0.8.30-1 
Ubuntu:9.04 (32-bit)

Thanks for any thoughts.
--andy


-- 
This is the price and the promise of citizenship.
- Barack Obama

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[R] Problems to run SVM regression with e1071

2009-05-11 Thread Marlene Marchena
Hi R users,

 

I'm trying to run a SVM - regression using e1071 package but the function svm() 
all the time apply a classification method rather than a regression.

 

svm.m1 - svm(st ~ ., data = train, cost = 1000, gamma = 1e-03)  

 

Parameters:

   SVM-Type:  C-classification 

 SVM-Kernel:  radial 

   cost:  1000 

  gamma:  0.001 

 

Number of Support Vectors:  209

 

 When I specify the method =eps-regression is the same  

 

svm.m1 - svm(st ~ ., data = train, method=eps-regression, cost = 1000, gamma 
= 1e-03)  

 

Call:

svm(formula = st ~ ., data = train, method = eps-regression, cost = 1000, 

gamma = 0.001)

 

Parameters:

   SVM-Type:  C-classification 

 SVM-Kernel:  radial 

   cost:  1000 

  gamma:  0.001 

 

Number of Support Vectors:  209

 

I know that it is wrong because when I do prediction appear levels.

 

I'm working with normalized data [0,1] (249 points) . I don't have idea what it 
is wrong. 

 

Somebody can help me? 

 

h_aspire

 

dados=read.table(svmdata.txt,header=TRUE)

index=1:nrow(d)

test=d[210:249,]  

train=d[1:209,]

require(e1071)

tuneobj = tune.svm(st ~ ., data = train, gamma = 10^(-6:-3), cost = 
10^(1:3summary(tuneobj)

svm.m1 - svm(st ~ ., data = train, cost = 1000, gamma = 1e-03)  

svm.pred - predict(svm.m1, test)

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Re: [R] Plot bivariate joint pdf

2009-05-11 Thread Ben Bolker



beetle2 wrote:
 
 For a homework question.
 I was wondering if rcmdr has a function to plot a graph of a bivariate
 function of X and Y.
 I have a function with joint pdf
 
 fX,Y(x,y) = x+y for 0x1  , 0y1
 
 I've tried
 x - seq(0,1,.001) 
 y - seq(0,1,.001) 
 r = x+y
 plot(r)
 
 but it seems to just add them together say .2+.2 .3+.3  not other
 possibilities like
 
 .9 + .1
 
 

We don't do homework questions here, but a few hints:

(1) the question doesn't seem to be fully specified.  Are x and y supposed
to be uniform?

(2) you could get a reasonable representation of the solution (although
not necessarily one your prof would be happy with) by adding
runif(1,0,1) and runif(1,0,1)

(3) to do this analytically, think about convolutions and/or moment
generating
functions.

  Ben Bolker

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Re: [R] Citing R/Packages Question

2009-05-11 Thread Ravi Varadhan
It would be nice if each package went through a peer-review and had a
related publication (either in R-news or J Stat Soft).  This publication can
then be used as the official citation for the package.  However, this still
would not address updates and versions of the package. 

Ravi.



---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvarad...@jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of roger koenker
Sent: Saturday, May 09, 2009 8:36 AM
To: Derek Ogle
Cc: r-help@r-project.org
Subject: Re: [R] Citing R/Packages Question

I've had an email exchange with the authors of a recent paper in Nature who
also made a good faith effort to cite both R and the quantreg package, and
were told that the Nature house style didn't allow such citations so they
were dropped from the published paper and the supplementary material
appearing on the Nature website.

Since the CRAN website makes a special effort to make prior versions of
packages available, it would seem to me to be much more useful to cite
version numbers than access dates.  There  are serious questions about the
ephemerality of url citations, not all of which are adequately resolved by
the Wayback machine, and access dating, but it would be nice to have some
better standards for such contingent citations rather than leave authors at
the mercy of copy editors.  I would also be interested in suggestions by
other contributors.


url:www.econ.uiuc.edu/~rogerRoger Koenker
email   rkoen...@uiuc.edu   Department of Economics
vox:217-333-4558University of Illinois
fax:217-244-6678Champaign, IL 61820


On May 8, 2009, at 5:27 PM, Derek Ogle wrote:

 I used R and the quantreg package in a manuscript that is currently in 
 the proofs stage.  I cited both R and quantreg as suggested by
 citation() and noted the version of R and quantreg that I used in the 
 main text as



  All tests were computed with the R v2.9.0 statistical programming 
 language (R Development Core 2008).  Quantile regressions were 
 conducted with the quantreg v4.27 package (Koenker 2008) for R.



 The editor has asked me to also provide the date when the webpage was
 accessed for both R and quantreg.



 This does not seem like an appropriate request to me as both R and the
 quantreg package are versioned.  This request seems to me to be the  
 same
 as asking someone when they purchased commercial package X version Y
 (which I don't think would be asked).



 Am I thinking about this correctly or has the editor made a valid
 request?



 I would be interested in any comments or opinions.



 Dr. Derek H. Ogle

 Associate Professor of Mathematical Sciences and Natural Resources

 Northland College

 1411 Ellis Avenue

 Box 112

 Ashland, WI

 715.682.1300

 www.ncfaculty.net/dogle/




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[R] Ancestral Character Estimation

2009-05-11 Thread Andrew RODRIGUES
Hello, I was just wondering whether with the ace command in ape it was
possible to incorporate polymorphic characters in ancestral character
estimation.  I am currently working with distibutions and often
distributions are not restricted to a single geographic area.  Thanks for
any help.

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[R] dimnames do not match array extent

2009-05-11 Thread Penner, Johannes
Dear adegenet users,

I am trying to calculate geographical barriers using monmonier.
However, I consistently get the error dimnames do not match array
extent.

- import of the distance matrix: read.table
- conversion 1: as.dist(data, diag=TRUE, upper =TRUE)
- conversion 2: as.matrix(data)

- is.numeric(data) are answered for the geographic and the species
distances are answered TRUE
- is.matrix(data) is also answered TRUE for both

The network is calculated but when setting the threshold I always get
the above error.

However, I checked the names and numbers of rows and columns now several
times in Excel and I can find no differences... What would be the R way?
Or is there a simple way to adopt the names from a master file?

Trying to get around the problem I calculated everything without any
row- or column names. I than get the error dimnames do not match array
extent when using optimize.monmonier

Any solution or suggestions are highly appreciated.

Thanks in advance and best regards
Johannes

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[R] Searching within a ch. string

2009-05-11 Thread RON70

Hi all, is there any function to find some words in a character-string? For
example suppose the string is : gdfsa-sdhchc-88, now I want to find
whether this string contains sdhch. Is there any R function to do that?

Regards,
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Re: [R] Using ACE in the ape package

2009-05-11 Thread Ben Bolker



Andrew RODRIGUES wrote:
 
 Having a problem running the ace command in ape.  
 
 After reading my table into R and then putting the names of the rows in
 the
 table in same order as the tree I can't get ace to run and get the message
 
 Erreur dans as.matrix(x) : 
   dims [produit 25] ne correspond pas à la longueur de l'objet [0]
 
 This is the command I am using and each of the vectors used in the
 snalysis.  
 
 ERreconstruction - ace(car, tr, type=discrete, model=SYM)
 
car [1] 4 1 7 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 1 2 1
 
 tr
 
 Phylogenetic tree with 32 tips and 31 internal nodes.
 
 Tip labels:
   Batis_maritima, Pentadiplandra_brazzeana, Tovaria_pendula, Reseda_lutea,
 Tropaeolum_majus, Carica_sp, ...
 
 Rooted; no branch lengths.
 
 

(1) you haven't actually given us a reproducible example here; try giving us
the
results of write.tree() on your tree.

(2) you will have better luck with this question, and your related more
recent one,
on the r-sig-phylo (phylogenetics special interest group) mailing list
rather than
this  list, where only a few people know about comparative methods, the ape
package, etc.

  good luck
Ben Bolker

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Re: [R] R Error, very odd....

2009-05-11 Thread Don MacQueen


At 12:37 AM -0700 5/11/09, Katie2009 wrote:

hi dieter,

the method i'm using is in excel, copying the data, then in r


 w-read.delim(clipboard)
 w-as.data.frame(w)


i've been  doing a bit more fiddling, and have identified the 'class' of the
column that i'm having trouble with, is classified as 'factor' whilst the
rest are numeric.

if i just change that column to as.numeric of the column, this has appeared
to have solved the problem.  Has this changed that data at all though?


Well, you can look at the data to find out. Either just print it, or 
perhaps use simple exploration techniques, such as

  unique( column)
  table(column)
  summary(column)
compare before and after converting to numeric

In general, changing a factor to numeric can change the data relative 
to what you might think the data actually is.  Here is an example:



 tmp1 - c(-1,0,1)
 tmp2 - factor(tmp1)
 tmp3 - as.numeric(tmp2)

 tmp1

[1] -1  0  1

 tmp2

[1] -1 0  1
Levels: -1 0 1

 tmp3

[1] 1 2 3

tmp3 is  clearly not the same as tmp1.


 tmp4 - as.numeric(format(tmp2))
 tmp4

[1] -1  0  1


Evidently, R is creating a factor from a column that you think is 
numeric. Your job is to find out why. I would guess that you have, 
somewhere in the column, an entry that isn't a number. Perhaps a 
typographical error in the spreadsheet column. The effect of 
inputting as absolute values vs with negative numbers should give a 
clue to that.


Try
   tmp -  as.numeric(format( column ))
   any(is.na(tmp))
and if that is TRUE, then
   column[is.na(tmp)]

-Don



Thanks.


Dieter Menne wrote:




 Katie2009 wrote:


 I'm trying to analyse some excel data in R.  The problem is that when i
 input the data with the first column as absolute values, everything works
 fine, can analyse as normal.  When I leave the first column unchanged to
 import negative numbers as well I get:

 Error in storage.mode(y) - double :
   invalid to change the storage mode of a factor
 In addition: Warning message:
 In model.response(mf, numeric) :
   using type=numeric with a factor response will be ignored




 You did not tell us anything how you got the data from Excel, so I have to
 guess. Try to re-arrange your Excel row so that the first (3 ? check the
 docs; which docs? your unknown function's) lines contain non-missing data.

 Dieter



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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062

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Re: [R] Searching within a ch. string

2009-05-11 Thread Duncan Murdoch

On 5/11/2009 10:07 AM, RON70 wrote:

Hi all, is there any function to find some words in a character-string? For
example suppose the string is : gdfsa-sdhchc-88, now I want to find
whether this string contains sdhch. Is there any R function to do that?


See ?grep.  There are several functions described there, with different 
outputs.


Duncan Murdoch

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Re: [R] Searching within a ch. string

2009-05-11 Thread Gabor Grothendieck
The following is TRUE if the indicated string is found:

regexpr(sdhch, gdfsa-sdhchc-88)  0

See ?regexpr

Either of the next two will break up a string into words. The
first needs to know the delimiters whereas the second
needs to know the contents:

 strsplit(gdfsa-sdhchc-88, \\W+)[[1]]
[1] gdfsa  sdhchc 88

 library(gsubfn)
 strapply(gdfsa-sdhchc-88, \\w+, c)[[1]]
[1] gdfsa  sdhchc 88

See ?strplit and ?regex and also the gsubfn home page at:
http://gsubfn.googlecode.com
where you can find info on strapply and also further resources
on regular expressions.

On Mon, May 11, 2009 at 10:07 AM, RON70 ron_michae...@yahoo.com wrote:

 Hi all, is there any function to find some words in a character-string? For
 example suppose the string is : gdfsa-sdhchc-88, now I want to find
 whether this string contains sdhch. Is there any R function to do that?

 Regards,
 --
 View this message in context: 
 http://www.nabble.com/Searching-within-a-ch.-string-tp23484010p23484010.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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Re: [R] Searching within a ch. string

2009-05-11 Thread Luc Villandre

Hi Ron,

Look up the grep() function.

Cheers,

--
*Luc Villandré*
/Biostatistician
McGill University Health Center -
Montreal Children's Hospital Research Institute/

RON70 wrote:

Hi all, is there any function to find some words in a character-string? For
example suppose the string is : gdfsa-sdhchc-88, now I want to find
whether this string contains sdhch. Is there any R function to do that?

Regards,
  

//

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[R] Inconsistency in representation of variables

2009-05-11 Thread Stavros Macrakis
In stats::D, I was wondering why variables are represented as symbols
in expressions, but as strings in lists of variables:

D(quote(x^2),x) = 2*x
D(quote(x^2),quote(x)) = error Variable must be a character string

Strings are not allowed in the expression to denote variables:

D(quote(x),quote(x)) == D(k,x) = NA  (why not an error?)

-s

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[R] Rcompression on Omegahat

2009-05-11 Thread Dieter Menne
I tried to install Rcompression (for reading Matlab files)

options(CRAN = c(getOption(CRAN), http://www.omegahat.org/R;))
install.packages(Rcompression)

Warning: unable to access index for repository
http://www.omegahat.org/R/bin/windows/contrib/2.9

Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
  package ‘Rcompression’ is not available

Manual inspection of the site showed that 2.7 is the last version.
Did I miss some change notes?

Dieter


R version 2.9.0 (2009-04-17) 
i386-pc-mingw32 

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

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

loaded via a namespace (and not attached):
[1] tools_2.9.0

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Re: [R] Citing R/Packages Question

2009-05-11 Thread stephen sefick
I have have a package that I wrote called StreamMetabolism; which I
use to calculate single station stream metabolism from diurnal oxygen
curves.  I would love to publish something about it (I am also an
entering PhD student and need publications); however, I am not sure
the applicability out side of a small subset of stream ecologists.
Also, JSS or Rnews may not be the proper forum.  We could publish a
bulletin or something with all of the packages as an official document
to site.  Just a half fleshed idea.
thanks

Stephen Sefick

On Mon, May 11, 2009 at 9:48 AM, Ravi Varadhan rvarad...@jhmi.edu wrote:
 It would be nice if each package went through a peer-review and had a
 related publication (either in R-news or J Stat Soft).  This publication can
 then be used as the official citation for the package.  However, this still
 would not address updates and versions of the package.

 Ravi.


 
 ---

 Ravi Varadhan, Ph.D.

 Assistant Professor, The Center on Aging and Health

 Division of Geriatric Medicine and Gerontology

 Johns Hopkins University

 Ph: (410) 502-2619

 Fax: (410) 614-9625

 Email: rvarad...@jhmi.edu

 Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html



 
 


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
 Behalf Of roger koenker
 Sent: Saturday, May 09, 2009 8:36 AM
 To: Derek Ogle
 Cc: r-help@r-project.org
 Subject: Re: [R] Citing R/Packages Question

 I've had an email exchange with the authors of a recent paper in Nature who
 also made a good faith effort to cite both R and the quantreg package, and
 were told that the Nature house style didn't allow such citations so they
 were dropped from the published paper and the supplementary material
 appearing on the Nature website.

 Since the CRAN website makes a special effort to make prior versions of
 packages available, it would seem to me to be much more useful to cite
 version numbers than access dates.  There  are serious questions about the
 ephemerality of url citations, not all of which are adequately resolved by
 the Wayback machine, and access dating, but it would be nice to have some
 better standards for such contingent citations rather than leave authors at
 the mercy of copy editors.  I would also be interested in suggestions by
 other contributors.


 url:    www.econ.uiuc.edu/~roger                Roger Koenker
 email   rkoen...@uiuc.edu                       Department of Economics
 vox:    217-333-4558                            University of Illinois
 fax:    217-244-6678                            Champaign, IL 61820


 On May 8, 2009, at 5:27 PM, Derek Ogle wrote:

 I used R and the quantreg package in a manuscript that is currently in
 the proofs stage.  I cited both R and quantreg as suggested by
 citation() and noted the version of R and quantreg that I used in the
 main text as



  All tests were computed with the R v2.9.0 statistical programming
 language (R Development Core 2008).  Quantile regressions were
 conducted with the quantreg v4.27 package (Koenker 2008) for R.



 The editor has asked me to also provide the date when the webpage was
 accessed for both R and quantreg.



 This does not seem like an appropriate request to me as both R and the
 quantreg package are versioned.  This request seems to me to be the
 same
 as asking someone when they purchased commercial package X version Y
 (which I don't think would be asked).



 Am I thinking about this correctly or has the editor made a valid
 request?



 I would be interested in any comments or opinions.



 Dr. Derek H. Ogle

 Associate Professor of Mathematical Sciences and Natural Resources

 Northland College

 1411 Ellis Avenue

 Box 112

 Ashland, WI

 715.682.1300

 www.ncfaculty.net/dogle/




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-- 
Stephen Sefick

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff 

[R] predict function the other way around

2009-05-11 Thread Gerrit Voigt

Dear List,
Consider the following example

x=c(1,2,3,4,5)
y=c(2,4,6,8,10)
linearmodel=lm(y~x)

To predict a y-value if you know the corresponding x value is very easy 
with the command predict.


predict(linearmodel, newdata=(x=1.5))

The other way around, to predict an x-value with a corresponding 
y-value, doesn't work unfortunately.
Is there another function that can do that, or do I need to solve that 
problem the classical way?


Thanks in advance

Gerrit Voigt

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Re: [R] Do you use R for data manipulation?

2009-05-11 Thread ronggui
2009/5/6 Emmanuel Charpentier charp...@bacbuc.dyndns.org:
 Le mercredi 06 mai 2009 à 00:22 -0400, Farrel Buchinsky a écrit :
 Is R an appropriate tool for data manipulation and data reshaping and data
 organizing?
 [ Large Snip ! ... ]

 Depends on what you have to do.

 I've done what can be more or less termed data management with almost
 uncountable tools (from Excel (sigh...) to R with SQL, APL, Pascal, C,
 Basic (in 1982 !), Fortran and even Lisp in passing...).


 SQL has strong points : join is, to my tastes, more easily expressed in
 SQL than in most languages, projection and aggregation are natural.

 However, in SQL, there is no natural ordering of row tables, which
 makes expressing algorithms using this order difficult. Try for example
 to express the differences of a time series ... (it can be done, but it
 is *not* a pretty sight).

 On the other hand, R has some unique expressive possibilities (reshape()
 comes to mind).

 So I tend to use a combination of tools : except for very small samples,
 I tend to manage my data in SQL and with associated tools (think data
 editing, for example ; a simple form in OpenOffice's Base is quite easy
 to create, can handle anything for which an ODBC driver exists, and
 won't crap out for more than a few hundreds line...). finer manipulation
 is usually done in R with  native tools and sqldf.

 But, at least in my trade, the ability to handle Excel files is a must
 (this is considered as a standard for data entry. Sigh ...). So the
 first task is usually a) import data in an SQL database, and b) prepare
 some routines to dump SQL tables / R dataframes in Excel tor returning
 back to the original data author...

I don't think Excel is  a standard tool for data entry. Epidata entry
is much more professional.


 HTH

                                        Emmanuel Charpentier

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-- 
HUANG Ronggui, Wincent
PhD Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html

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[R] Legend in plots

2009-05-11 Thread Mafalda Viana
Dear R users,

I have what I think it is a very simple question concerning plots in R. If
you could help me I would be very grateful.

How can I include a legend in my plot? Below I give an example of my data
and the code I am using.

What I wish to accomplish is a legend saying that the black line refers to
A, the red to B and the blue to C. Any suggestions?


# data
x-c(1,2,3,4,5)
y1-c(2,4,6,8,10)
y2-c(3,4,5,6,7)
y3-c(2,3,5,7,9)

# plot
plot(x, y1, main=plot, type=l, xlab=x-axis, ylab=y-axis,
col=black)
lines(x, y2, col=red)
lines(x, y3, col=blue)

Thanking in advance.

Best regards,
Mafalda

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Re: [R] Legend in plots

2009-05-11 Thread Jorge Ivan Velez
Dear Mafalda,
?legend is what you are looking for. Take a look at the help page. Here is
one way to do what you asked for:

legend('topleft', c('A','B','C'),lty=1, col=c(1,2,4))

HTH,

Jorge


On Mon, May 11, 2009 at 11:26 AM, Mafalda Viana via...@tcd.ie wrote:

 Dear R users,

 I have what I think it is a very simple question concerning plots in R. If
 you could help me I would be very grateful.

 How can I include a legend in my plot? Below I give an example of my data
 and the code I am using.

 What I wish to accomplish is a legend saying that the black line refers to
 A, the red to B and the blue to C. Any suggestions?


 # data
 x-c(1,2,3,4,5)
 y1-c(2,4,6,8,10)
 y2-c(3,4,5,6,7)
 y3-c(2,3,5,7,9)

 # plot
 plot(x, y1, main=plot, type=l, xlab=x-axis, ylab=y-axis,
 col=black)
 lines(x, y2, col=red)
 lines(x, y3, col=blue)

 Thanking in advance.

 Best regards,
 Mafalda

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[R] Removing any text beginning with...

2009-05-11 Thread Amélie Baud
Hi !

From an Ensembl annotation like ENSG /// ENSGy /// ENSG, I am 
trying to keep only the first part: ENSG. I wasn't able to find any 
helpful information about how to do it. Could you help me with that please ? 
Is the use of the equivalent to the Excel * (any text) a good way of doing it 
and how ? 
Your help will be very much appreciated.

Amelie



  
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[R] R and McAfee 8.5

2009-05-11 Thread Joyce, Warren
Hi,

I have been working with R for the last year and using the UKFSST package to 
look at satellite tag track data and SST information. Fpr those not familiar, 
the package uses the positions estimated by the satellite tags themselves and 
the associated SST data from servers (in this case, from the University of 
Hawaii and the NOAA Coastwatch website) for the time preiod to estimate a 
reasonable track the animal took while it was at liberty.  I'm currently having 
trouble downloading data from two different servers which I never had a problem 
before I recevied an upgraded computer with McAfee 8.5. When the package goes 
to the various servers for SST information to download, I continue to get a 
message in R:

trying URL 
'http://coastwatch.pfeg.noaa.gov/coastwatch/CWBrowserWW360.jsp?get=gridDatadataSet=TAGsstatimePeriod=3daycenteredTime=~2007-09-17T12%3A00%3A00maxLat=61.5minLon=295.06296maxLon=301.37891minLat=36filetype=.xyz'
Error in download.file(link, dest, mode = wb) : 
  cannot open URL 
'http://coastwatch.pfeg.noaa.gov/coastwatch/CWBrowserWW360.jsp?get=gridDatadataSet=TAGsstatimePeriod=3daycenteredTime=~2007-09-17T12%3A00%3A00maxLat=61.5minLon=295.06296maxLon=301.37891minLat=36filetype=.xyz'
In addition: Warning messages:
1: In dir.create(folder) :
  'C:\DOCUME~1\Joycew\LOCALS~1\Temp\Rtmp0DilA3\sst_temp' already exists
2: In download.file(link, dest, mode = wb) :
  cannot open: HTTP status was '404 Not Found'

I get a similar message when trying to access SST data from University of 
Hawaii 

trying URL 
'http://atlas.nmfs.hawaii.edu/cgi-bin/reynolds_extract.py?lon1=295.06296lon2=301.37891lat1=37lat2=64.5year1=2007day1=259year2=2007day2=355'
Error in download.file(link, dest, mode = wb) : 
  cannot open URL 
'http://atlas.nmfs.hawaii.edu/cgi-bin/reynolds_extract.py?lon1=295.06296lon2=301.37891lat1=37lat2=64.5year1=2007day1=259year2=2007day2=355'
In addition: Warning message:
In download.file(link, dest, mode = wb) :
  cannot open: HTTP status was '404 Not Found'

If I copy and paste the above url's 
http://coastwatch.pfeg.noaa.gov/coastwatch/CWBrowserWW360.jsp?get=gridDatadataSet=TAGsstatimePeriod=3daycenteredTime=~2007-09-17T12%3A00%3A00maxLat=61.5minLon=295.06296maxLon=301.37891minLat=36filetype=.xyz

Into my web browser, I get prompted with a file download window which quickly 
downloads an empty file. It looks like the data is available, I just can't seem 
to download it.

I've deduced that it must be a problem with the new firewall features of McAfee 
8.5 and was wondering if anyone else has run into a problem like this. I've 
tried using the exclusions options in 8.5 to exclude any R files but have come 
up empty handed. Still waiting on our informatics division to try and solve the 
problem as well but I thought I might try asking here as well.

Thanks and I appreciate any assistance!

Warren N. Joyce 
e-mail/courriel: warren.jo...@dfo-mpo.gc.ca  (new address)

Shark Technician/Technicien spécialiste du requin
ph/tél: 902-426-6382
laboratory: 902-426-1846
fax/téléc: 902-426-1506

Bedford Institute of Oceanography
1 Challenger Drive
P.O. Box 1006
Dartmouth, Nova Scotia 
Canada, B2Y 4A2

Institut Océanographique de Bedford
1, promenade Challenger
C.P. 1006
Dartmouth, Nouvelle-Écosse 
Canada, B2Y 4A2

Shark Website 
http://marinebiodiversity.ca/shark/
Otolith Website
http://marinebiodiversity.ca/otolith/
Skate Website
http://www.marinebiodiversity.ca/skatesandrays/index.htm



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[R] Plotting colors on a world map

2009-05-11 Thread dxc13

Hi useR's

I have created a simple map of the world using the following code:
m - map(xlim=c(-180,180), ylim=c(-90,90))
map.axes()

I then create a grid of dimension 36x72 using the code:
map.grid(m, nx=72, ny=36, labels=FALSE, col=black)

This gives 2592 grid cells.  In a separate data set of dimension 36x72, I
have 2592 temperature values (some missing values are present) ranging from
-20 degrees F to 90 degrees F.  

My question is, how can I create a reasonable color scheme (low temperatures
in light blue to higher temperatures in dark red) and plot the temperature
colors in their respective grid cells on the map?

Take this data matrix as some example data:
T - sample(-10:90, 2592, replace=TRUE)
mat - matrix(T, nrow=36, ncol=72)


Thanks in advance,
dxc13
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[R] Seewave options

2009-05-11 Thread Thiago de Carvalho



Good afternoon, 

My name is Thiago. I'm a graduate student and affiliated to an Anuran 
Bioacustic and Systematics Laboratory at UFU (Federal University of 
Uberlândia) in Brazil. The professor in charge of the lab (Dr. Ariovaldo 
Antonio Giaretta) and I have recently been making use of R, specially the 
packages Sound and Seewave so as to analize anuran recordings. This software is 
just fantastic, but I am writing to ask for help about the graphics generated 
by Seewave. A graphic and the script I used are attached to the e-mail, so that 
you are able to verify if it is possible to reduce the graphic scale of the 
Y-axis (see we need only until 10 KHz, even less than it). The fact of 
recording at 44100 Hz/16 bits resolution may be a possible cause for such a 
long scale, but that is the only way to record with the best quality... If 
you think it is easier to change my script and send me back, I would reallly 
appreciate it too.
I am looking forward to your reply.  
Thanks beforehand, 

__
Thiago Ribeiro de Carvalho
Laboratório de Comportamento, Ecologia e Sistemática de Anuros Neotropicais
Universidade Federal de Uberlândia
http://lattes.cnpq.br/0618169962335506


  Veja quais são os assuntos do momento no Yahoo! +Buscados
http://br.maisbuscados.yahoo.comsetWavPlayer(command=NULL)
x-loadSample(1.wav)
play(x)
spectro(x, wl = 256, wn = hanning, zp = 0,
ovlp = 0, plot = TRUE,
grid = FALSE, osc = TRUE, scale = TRUE, cont = FALSE,
collevels = seq(-30, 0, 1), palette = spectro.colors,
contlevels = seq(-30, 0, 10), colcont = black,
colbg = white, colgrid = black,
colaxis = black, collab=black,
plot.title = title(main = , xlab = Time (s),
ylab = Frequency (kHz)), scalelab = Amplitude\n(dB),
scalefontlab = 1, scalecexlab =0.75,
axisX = TRUE, axisY = TRUE, tlim = NULL, trel = TRUE,
flim = NULL, flimd = NULL, listen=TRUE)
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[R] Plot bivariate joint pdf

2009-05-11 Thread beetle2

For a homework question.
I was wondering if rcmdr has a function to plot a graph of a bivariate
function of X and Y.
I have a function with joint pdf

fX,Y(x,y) = x+y for 0x1  , 0y1

I've tried
 x - seq(0,1,.001) 
 y - seq(0,1,.001) 
 r = x+y
 plot(r)

but it seems to just add them together say .2+.2 .3+.3  not other
possibilities like

.9 + .1

Thanks



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Re: [R] working with groups of labels?

2009-05-11 Thread Phillip Porter
At 9:25 PM +1000 5/11/09, Jim Lemon wrote:
Hi Phillip,
I'm not exactly sure how you are positioning the labels. Is it 
possible to give us an example with some data that will produce a 
plot to show what you want? It shouldn't be too hard to do.

Jim


Data is something along the lines of:

variablescore
Real Bad Stuff  0
Real23
Bad 14
Stuff   17
Other Crazy Things  0
Other   18
Crazy   43
Things  13

The basic plot looks like this:

barplot(score, horiz=TRUE, axes=FALSE, las=2, space=1, 
names.arg=c(Real Bad Stuff, Real,Bad,Stuff,Other Crazy 
Things,Other,Crazy,Things))


I've been trying make it look the way I want it by doing this:

barplot(score, horiz=TRUE, axes=FALSE, las=1, space=1, 
names.arg=c(, Real,Bad,Stuff,,Other,Crazy,Things), 
cex.names=.5)
mtext(c(Real Bad Stuff,Other Crazy Things),at=c(8,15.6),side=2, 
line=1.5,adj=1, las=1, cex=.5, font=2)

The mtext is annoying to line up with the actual plot.  If there was 
a way to get it to line up with the variables that would solve my 
problems, but I've tried at=variable and at=score (putting in  for 
all of the variable names I don't want bolded and left justified) and 
neither way gets them to line up.

Thanks,
Phillip
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Re: [R] What does it mean by skip=2 and skip=7?

2009-05-11 Thread Richard . Cotton
 Can anyone tell me what is skip=2, skip =7

From ?read.csv:

skip: integer: the number of lines of the data file to skip before
  beginning to read data.

 and %in% mean here?

%in% matches values; see ?'%in%', and example('%in%')

Regards,
Richie.

Mathematical Sciences Unit
HSL



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Re: [R] Histogram frequencies with a normal pdf curve overlay

2009-05-11 Thread Greg Snow
You may also want to look at the rootogram function in the vcd package.

You could also use the updateusr function in the TeachingDemos package to reset 
the y-scale before plotting the curve (works well if you just want the shape of 
the curve, probably more work than the other suggestions if you want the areas 
to match).

Hope this helps,

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Jacques Wagnor
 Sent: Saturday, May 09, 2009 11:04 AM
 To: ted.hard...@manchester.ac.uk
 Cc: r-help@r-project.org
 Subject: Re: [R] Histogram frequencies with a normal pdf curve overlay
 
 Thank you!
 
 On Sat, May 9, 2009 at 11:31 AM, Ted Harding
 ted.hard...@manchester.ac.uk wrote:
  On 09-May-09 16:10:42, Jacques Wagnor wrote:
  Dear List,
  When I plot a histogram with 'freq=FALSE' and overlay the
  histogram with a normal pdf curve, everything looks as expected,
  as follows:
 
  x - rnorm(1000)
  hist(x, freq=FALSE)
  curve(dnorm(x), add=TRUE, col=blue)
 
  What do I need to do if I want to show the frequencies (freq=TRUE)
  with the same normal pdf overlay, so that the plot would still look
  the same?
 
  Regards,
  Jacques
 
  Think first about how you would convert the histogram densities
  (heights of the bars on the density scale) into histogram
 frequencies.
 
   Density * (bin width) * N = frequency
 
  where N = total number in sample. Then all you need to is multiply
  the Normal density by the same factor. To find out the bin width,
  take the difference between succesive values of the breaks
 component
  of the histogram. One way to do all this is
 
   N - 1000
   x - rnorm(N)
   H - hist(x, freq=TRUE)  ## This will plot the histogram as well
   dx - min(diff(H$breaks))
   curve(N*dx*dnorm(x), add=TRUE, col=blue)
 
  Ted.
 
  
  E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
  Fax-to-email: +44 (0)870 094 0861
  Date: 09-May-09                                       Time: 17:31:03
  -- XFMail --
 
 
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Re: [R] Reading large files quickly; resolved

2009-05-11 Thread Rob Steele
Rob Steele wrote:
 I'm finding that readLines() and read.fwf() take nearly two hours to
 work through a 3.5 GB file, even when reading in large (100 MB) chunks.
  The unix command wc by contrast processes the same file in three
 minutes.  Is there a faster way to read files in R?
 
 Thanks!
 

readChar() is fast.  I use strsplit(..., fixed = TRUE) to separate the
input data into lines and then use substr() to separate the lines into
fields.  I do a little light processing and write the result back out
with writeChar().  The whole thing takes thirty minutes where read.fwf()
took nearly two hours just to read the data.

Thanks for the help!

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[R] R errors when trying to read a file

2009-05-11 Thread Dennis Fisher

Colleagues,

I have encountered a problem in version 2.9 of R, running in both  
Vista and OS X.  My code is quite lengthy but the critical line is:

if (file.exists(FILENAME))  readLines(FILENAME))

This triggers the error:
Error in file(con, r) : cannot open the connection
Calls: readLines - file
Execution halted

The file that it is reading is being created on the fly, then  
eventually deleted.  The conditional was used to ensure that the  
readLines command would be executed only when the file existed.


Is it possible that the file is deleted AFTER the conditional is  
evaluated but before the readLines command is executed?  If so, is  
there any means to prevent an error condition?


Any help will be greatly appreciated.  Thanks.

Dennis

Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com

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Re: [R] need help with chisq

2009-05-11 Thread JC
Thank you Stephanie. This is perfect!

On May 10, 12:34 pm, Stephanie Kovalchik sko...@ucla.edu wrote:
 JC,

 If each row are the counts for a 2 x 2 contingency table - so for the  
 ith contingency table you have counts for row 1 c(Y08[i],Z08[i]) and  
 row 2 (Y09[i],Z09[i]) then you could use apply:

 X - cbind(vdata$Y08,vdata$X08-vdata$Y08,vdata$Y09,vdata$X09-vdata$Y98)

 f.chisq - function(x){
 m - matrix(x,2,2)
 chisq.test(m)$p.value

 }

 apply(X,1,f.chisq)

 Quoting JC jerome.co...@gmail.com:





  I am very new to R. I have some data from a CVS stored in vdata with 4
  columns labeled:
  X08, Y08, X09, Y09.

  I have created two new columns like so:

  Z08 - (vdata$X08-vdata$Y08)

  Z09 - (vdata$X09-vdata$Y09)

  I would like to use chisq.test for each row and output the p-value
  for each in a stored variable. I don't know how to do it. Can you
  help?

  so far I have done it for one row (but I want it done automatically
  for all my data):

  chidata=rbind(c(vdata$Y08[1],Z08[1]),c(vdata$Y09[1],Z09[1]))
  results - chisq.test(chidata)
  results$p.value

  I tried removing the [1] and the c() but that didn't work...  Any
  ideas?

  THANKS!

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Re: [R] Beyond double-precision?

2009-05-11 Thread Greg Snow
A large chunk of the function below could be replaced with a call to the Reduce 
function.  I don't know if it would be faster, slower, or depend on the 
situation, but it might make it a little more readable.

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


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Berwin A Turlach
 Sent: Saturday, May 09, 2009 10:18 AM
 To: spencerg
 Cc: r-help@r-project.org; joaks1
 Subject: Re: [R] Beyond double-precision?
 
 G'day all,
 
 On Sat, 09 May 2009 08:01:40 -0700
 spencerg spencer.gra...@prodsyse.com wrote:
 
The harmonic mean is exp(mean(logs)).  Therefore, log(harmonic
  mean) = mean(logs).
 
Does this make sense?
 
 I think you are talking here about the geometric mean and not the
 harmonic mean. :)
 
 The harmonic mean is a bit more complicated.  If x_i are positive
 values, then the harmonic mean is
 
 H= n / (1/x_1 + 1/x_2 + ... + 1/x_n)
 
 so
 
 log(H) = log(n) - log( 1/x_1 + 1/x_2 + ... + 1/x_n)
 
 now log(1/x_i) = -log(x_i) so if log(x_i) is available, the logarithm
 of the individual terms are easily calculated.  But we need to
 calculate the logarithm of a sum from the logarithms of the individual
 terms.
 
 At the C level R's API has the function logspace_add for such tasks, so
 it would be easy to do this at the C level.  But one could also
 implement the equivalent of the C routine using R commands.  The way to
 calculate log(x+y) from lx=log(x) and ly=log(y) according to
 logspace_add is:
 
   max(lx,ly) + log1p(exp(-abs(lx-ly)))
 
 So the following function may be helpful:
 
 logadd - function(x){
   logspace_add - function(lx, ly)
 max(lx, ly) + log1p(exp(-abs(lx-ly)))
 
   len_x - length(x)
if(len_x  1){
 res - logspace_add(x[1], x[2])
 if( len_x  2 ){
   for(i in 3:len_x)
 res - logspace_add(res, x[i])
 }
   }else{
 res - x
   }
   res
 }
 
 R set.seed(1)
 R x - runif(50)
 R lx - log(x)
 R log(1/mean(1/x))  ## logarithm of harmonic mean
 [1] -1.600885
 R log(length(x)) - logadd(-lx)
 [1] -1.600885
 
 Cheers,
 
   Berwin
 
 === Full address =
 Berwin A TurlachTel.: +65 6515 4416 (secr)
 Dept of Statistics and Applied Probability+65 6515 6650 (self)
 Faculty of Science  FAX : +65 6872 3919
 National University of Singapore
 6 Science Drive 2, Blk S16, Level 7  e-mail: sta...@nus.edu.sg
 Singapore 117546http://www.stat.nus.edu.sg/~statba
 
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Re: [R] Removing any text beginning with...

2009-05-11 Thread jim holtman
Is this what you want (using regular expressions):

 x - ENSG /// ENSGy /// ENSG
 sub(^([[:alpha:]]+).*, \\1 file://0.0.0.1/, x)
[1] ENSG




On Mon, May 11, 2009 at 9:01 AM, Amélie Baud ameli...@yahoo.fr wrote:

 Hi !

 From an Ensembl annotation like ENSG /// ENSGy /// ENSG, I am
 trying to keep only the first part: ENSG. I wasn't able to find any
 helpful information about how to do it. Could you help me with that please ?
 Is the use of the equivalent to the Excel * (any text) a good way of doing
 it and how ?
 Your help will be very much appreciated.

 Amelie




[[alternative HTML version deleted]]

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

What is the problem that you are trying to solve?

[[alternative HTML version deleted]]

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Re: [R] working with groups of labels?

2009-05-11 Thread Richard . Cotton
It seems that the structure of your data is that you have two groups 
(Real Bad Stuff and Other Crazy Things) which are then subdivided into 
further categories.  I'd be tempted to set your data up like this:

dfr - data.frame(
   score=c(23, 14, 17, 8, 43, 13),
   group=rep(c(Real Bad Stuff, Other Crazy Things), each=3),
   variable=c(Real,Bad,Stuff,Other,Crazy,Things))

Then you can use a lattice plot to emphasise the different groups, and 
avoid your mtext worries.

library(lattice)
barchart(
   variable~score|group, 
   data=dfr, 
   horizontal=TRUE,
   layout=c(1,2),
   scales=list(y=list(relation=free)),
   prepanel=function(x,y,...)
   {
  y - y[,drop=TRUE]
  prepanel.default.bwplot(x,y,...)
   }, 
   panel=function(x,y,...)
   {
  y - y[,drop=TRUE]
  panel.barchart(x,y,...)
   }
)

(The scales, prepanel and panel arguments are just to get rid of the 
variables that appear in one group but not the other.)  I realise it isn't 
exactly what you asked for, but I thought you might be interested in the 
alternate approach.

Regards,
Richie.

Mathematical Sciences Unit
HSL



ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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Re: [R] Problems to run SVM regression with e1071

2009-05-11 Thread Max Kuhn
Is the variable st character or a factor? What does str(train$st) show?

-- 

Max

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Re: [R] R errors when trying to read a file

2009-05-11 Thread Prof Brian Ripley

On Mon, 11 May 2009, Dennis Fisher wrote:


Colleagues,

I have encountered a problem in version 2.9 of R, running in both Vista and 
OS X.  My code is quite lengthy but the critical line is:

if (file.exists(FILENAME))  readLines(FILENAME))

This triggers the error:
Error in file(con, r) : cannot open the connection
Calls: readLines - file
Execution halted

The file that it is reading is being created on the fly, then eventually 
deleted.  The conditional was used to ensure that the readLines command would 
be executed only when the file existed.


Is it possible that the file is deleted AFTER the conditional is evaluated 
but before the readLines command is executed?  If so, is there any means to 
prevent an error condition?


It is possible.  But so are a lot of other things, and your pretest is 
just not enough -- for example another process might have the file 
open for exclusive access (quite common in the Windows world) in which 
case it exists but you cannot open it for reading.


There is only one sure way to find out if you can read a file, and 
that is to try to read it.  So I'd simply use try() or tryCatch() on 
the readLines call.




Any help will be greatly appreciated.  Thanks.

Dennis

Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com

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

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Re: [R] predict function the other way around

2009-05-11 Thread David Winsemius

Why not use the x=T and y=T paramemters and see if :

linearmodel=lm(y~x, x=T, y=T)linearmodel$model
(linearmodel$model)
   y x
1  2 1
2  4 2
3  6 3
4  8 4
5 10 5

 give you useful results. If you want to interpolate or  
extrapolate, then search for threads regarding those tasks.


--
David Winsemius
On May 11, 2009, at 11:20 AM, Gerrit Voigt wrote:


Dear List,
Consider the following example

x=c(1,2,3,4,5)
y=c(2,4,6,8,10)
linearmodel=lm(y~x)

To predict a y-value if you know the corresponding x value is very  
easy with the command predict.


predict(linearmodel, newdata=(x=1.5))

The other way around, to predict an x-value with a corresponding y- 
value, doesn't work unfortunately.
Is there another function that can do that, or do I need to solve  
that problem the classical way?


Thanks in advance

Gerrit Voigt

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] anyone know how to calculate chi square value from P value?

2009-05-11 Thread Anyuan Guo

Dear all,
   I have P value of a list of markers. Now I need the chi square value 
with degrees of freedom 2.
   I noticed there are several Chisquare functions (dchisq, pchisq, 
qchisq, and rchisq), but it seems all are not for my purpose.
   In microsoft excel, there is a function CHINV to do this, such as 
CHINV(0.184, 2) is 3.386, that means the chi square value for P value 
0.184, degree 2 is 3.386.

   Does the R package has some function to do this?

Thanks

Anyuan

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Re: [R] Plot bivariate joint pdf

2009-05-11 Thread David Winsemius


On May 11, 2009, at 9:46 AM, Ben Bolker wrote:





beetle2 wrote:


For a homework question.
I was wondering if rcmdr has a function to plot a graph of a  
bivariate

function of X and Y.
I have a function with joint pdf

fX,Y(x,y) = x+y for 0x1  , 0y1

I've tried

x - seq(0,1,.001)
y - seq(0,1,.001)
r = x+y
plot(r)


but it seems to just add them together say .2+.2 .3+.3  not other
possibilities like

.9 + .1




We don't do homework questions here, but a few hints:

(1) the question doesn't seem to be fully specified.  Are x and y  
supposed

to be uniform?

(2) you could get a reasonable representation of the solution  
(although

not necessarily one your prof would be happy with) by adding
runif(1,0,1) and runif(1,0,1)

(3) to do this analytically, think about convolutions and/or moment
generating functions.


(4) ?integrate
and search on double integration.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] anyone know how to calculate chi square value from P value?

2009-05-11 Thread Duncan Murdoch

On 5/11/2009 3:36 PM, Anyuan Guo wrote:

Dear all,
I have P value of a list of markers. Now I need the chi square value 
with degrees of freedom 2.
I noticed there are several Chisquare functions (dchisq, pchisq, 
qchisq, and rchisq), but it seems all are not for my purpose.
In microsoft excel, there is a function CHINV to do this, such as 
CHINV(0.184, 2) is 3.386, that means the chi square value for P value 
0.184, degree 2 is 3.386.

Does the R package has some function to do this?


qchisq(0.184, df=2, lower=FALSE)

Duncan Murdoch

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[R] Partial correlation function required

2009-05-11 Thread Moumita Das
-- Forwarded message --
From: r-help-boun...@r-project.org
Date: Mon, May 11, 2009 at 10:24 PM
Subject: The results of your email commands
To: das.moumita.onl...@gmail.com


The results of your email command are provided below. Attached is your
original message.

- Results:
   Ignoring non-text/plain MIME parts

- Unprocessed:
   What is the function for partial correlation.The function which i found
   here:-- http://www.yilab.gatech.edu/pcor.Rand used the recursive
   version, showed errors while computing partial correlation,where the
third
   variable i.e controlled variable for a pair is more than that .That is
when
   i had more than one third variable ,it always became an infinite loop,it
   else.
   # By using recursive formula
   pcor.rec - function(x,y,z,method=p,na.rm=T){
   #
   x - c(x)
   y - c(y)
   z - as.data.frame(z)
   if(dim(z)[2] == 0){
   stop(There should be given data\n)
   }
   data - data.frame(x,y,z)
   if(na.rm == T){
   data = na.omit(data)
   }

- Ignored:

   # recursive formula
   if(dim(z)[2] == 1){
   tdata - na.omit(data.frame(data[,1],data[,2]))
   rxy - cor(tdata[,1],tdata[,2],m=method)

   tdata - na.omit(data.frame(data[,1],data[,-c(1,2)]))
   rxz - cor(tdata[,1],tdata[,2],m=method)

   tdata - na.omit(data.frame(data[,2],data[,-c(1,2)]))
   ryz - cor(tdata[,1],tdata[,2],m=method)

   rxy.z - (rxy - rxz*ryz)/( sqrt(1-rxz^2)*sqrt(1-ryz^2) )

   return(rxy.z)
   }else{
   x - c(data[,1])
   y - c(data[,2])
   z0 - c(data[,3])
   zc - as.data.frame(data[,-c(1,2,3)])

   rxy.zc - pcor.rec(x,y,zc,method=method,na.rm=na.rm)
   rxz0.zc - pcor.rec(x,z0,zc,method=method,na.rm=na.rm)
   ryz0.zc - pcor.rec(y,z0,zc,method=method,na.rm=na.rm)

   rxy.z - (rxy.zc - rxz0.zc*ryz0.zc)/(
sqrt(1-rxz0.zc^2)*sqrt(1-ryz0.zc^2) )
   return(rxy.z)
   }
   }

   and this piece of code also i tried to use.

   Rinv - solve http://wiki.r-project.org/rwiki/rhelp.php?id=solve(R)
   D - diag http://wiki.r-project.org/rwiki/rhelp.php?id=diag(1 / sqrt
   http://wiki.r-project.org/rwiki/rhelp.php?id=sqrt(diag
   http://wiki.r-project.org/rwiki/rhelp.php?id=diag(Rinv)))
   P - -D http://wiki.r-project.org/rwiki/rhelp.php?id=D %*% Rinv %*%
   D http://wiki.r-project.org/rwiki/rhelp.php?id=D

   where R is teh correlation matrix.
   but NaN error at for
   sqrt http://wiki.r-project.org/rwiki/rhelp.php?id=sqrt(diag
   http://wiki.r-project.org/rwiki/rhelp.php?id=diag(Rinv) were
   generated.





http://wiki.r-project.org/rwiki/doku.php?id=tips:data-matrices:part_corrs=correlation


   Any convinient function for partial correlation for both one
   controlled variable or more.



   --
   Thanks
   Moumita

- Done.



-- Forwarded message --
From: Moumita Das das.moumita.onl...@gmail.com
To: r-help-requ...@r-project.org
Date: Mon, 11 May 2009 22:23:33 +0530
Subject: partial correlation function
Hi friends,
What is the function for partial correlation.The function which i found
here:-- http://www.yilab.gatech.edu/pcor.Rand used the recursive
version, showed errors while computing partial correlation,where the third
variable i.e controlled variable for a pair is more than that .That is when
i had more than one third variable ,it always became an infinite loop,it
else.

# By using recursive formula
pcor.rec - function(x,y,z,method=p,na.rm=T){
#

x - c(x)
y - c(y)
z - as.data.frame(z)

if(dim(z)[2] == 0){
stop(There should be given data\n)

}

data - data.frame(x,y,z)

if(na.rm == T){
data = na.omit(data)
}

# recursive formula
if(dim(z)[2] == 1){
tdata - na.omit(data.frame(data[,1],data[,2]))
rxy - cor(tdata[,1],tdata[,2],m=method)

tdata - na.omit(data.frame(data[,1],data[,-c(1,2)]))
rxz - cor(tdata[,1],tdata[,2],m=method)

tdata - na.omit(data.frame(data[,2],data[,-c(1,2)]))
ryz - cor(tdata[,1],tdata[,2],m=method)

rxy.z - (rxy - rxz*ryz)/( sqrt(1-rxz^2)*sqrt(1-ryz^2) )

return(rxy.z)
}else{
x - c(data[,1])
y - c(data[,2])
z0 - c(data[,3])
zc - as.data.frame(data[,-c(1,2,3)])

rxy.zc - pcor.rec(x,y,zc,method=method,na.rm=na.rm)
rxz0.zc - pcor.rec(x,z0,zc,method=method,na.rm=na.rm)
ryz0.zc - pcor.rec(y,z0,zc,method=method,na.rm=na.rm)

rxy.z - (rxy.zc - rxz0.zc*ryz0.zc)/( 
sqrt(1-rxz0.zc^2)*sqrt(1-ryz0.zc^2) )

return(rxy.z)
}   
}   


Re: [R] Plotting colors on a world map

2009-05-11 Thread Duncan Murdoch

On 5/11/2009 10:32 AM, dxc13 wrote:

Hi useR's

I have created a simple map of the world using the following code:
m - map(xlim=c(-180,180), ylim=c(-90,90))
map.axes()

I then create a grid of dimension 36x72 using the code:
map.grid(m, nx=72, ny=36, labels=FALSE, col=black)

This gives 2592 grid cells.  In a separate data set of dimension 36x72, I
have 2592 temperature values (some missing values are present) ranging from
-20 degrees F to 90 degrees F.  


My question is, how can I create a reasonable color scheme (low temperatures
in light blue to higher temperatures in dark red) and plot the temperature
colors in their respective grid cells on the map?

Take this data matrix as some example data:
T - sample(-10:90, 2592, replace=TRUE)
mat - matrix(T, nrow=36, ncol=72)



The colorspace package can make nice color sequences for this sort of 
thing, and image() can add them to a plot.  In colorspace, look at the 
examples for diverge_hcl.


Duncan Murdoch

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Re: [R] anyone know how to calculate chi square value from P value?

2009-05-11 Thread R. A. Bilonick
On Mon, 2009-05-11 at 12:36 -0700, Anyuan Guo wrote:
 Dear all,
 I have P value of a list of markers. Now I need the chi square value 
 with degrees of freedom 2.
 I noticed there are several Chisquare functions (dchisq, pchisq, 
 qchisq, and rchisq), but it seems all are not for my purpose.
 In microsoft excel, there is a function CHINV to do this, such as 
 CHINV(0.184, 2) is 3.386, that means the chi square value for P value 
 0.184, degree 2 is 3.386.
 Does the R package has some function to do this?
 
 Thanks
 
 Anyuan

It never hurts to read the documentation:

?qchisq

for example. Here q stands for quantile. You want the Chi-square
value that has 0.184 area to the left. To compute this:

 qchisq(1-0.184,2)
[1] 3.385639

You have to subtract 0.184 from 1, because qchisq finds the value with
the given area below it. To double check the result:

 1-pchisq(3.385639,2)
[1] 0.184


Rick B.

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Re: [R] Hoe to get RESIDUAL VARIANCE in logistic regression using lmer

2009-05-11 Thread tomal

Thank you very much Mr Onkelinx, I appreciated your answer. 
I'm sorry about my lack of knowledge on this subject, but I don't understand
why residual variance is fixed at 1 with the binomial family and I didn't
manage to find any exhaustive explanation for that.
May you kindly suggest me some reference I can consult in order to go into
depth on this matter?

Does anybody else can profide me furher information about this topic?

Thank you very much.

Best regards,   
  
Tommaso Alestra



ONKELINX, Thierry wrote:
 
 Dear Tommaso,
 
 The residuals variance is fixed at 1 with the binomial family.
 
 HTH,
 
 Thierry
 
 
 
 
 ir. Thierry Onkelinx
 Instituut voor natuur- en bosonderzoek / Research Institute for Nature
 and Forest
 Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
 methodology and quality assurance
 Gaverstraat 4
 9500 Geraardsbergen
 Belgium 
 tel. + 32 54/436 185
 thierry.onkel...@inbo.be 
 www.inbo.be 
 
 To call in the statistician after the experiment is done may be no more
 than asking him to perform a post-mortem examination: he may be able to
 say what the experiment died of.
 ~ Sir Ronald Aylmer Fisher
 
 The plural of anecdote is not data.
 ~ Roger Brinner
 
 The combination of some data and an aching desire for an answer does not
 ensure that a reasonable answer can be extracted from a given body of
 data.
 ~ John Tukey
 
 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens tomal
 Verzonden: donderdag 30 april 2009 12:10
 Aan: r-help@r-project.org
 Onderwerp: [R] Hoe to get RESIDUAL VARIANCE in logistic regression using
 lmer
 
 
 Hello everybody, 
 using the lmer function, I have fitted the following logistic mixed
 regression model on an experimental data set containing one fixed factor
 (Cond) and three random variables (Sito, Area, Trans): 
 
 
 model-lmer(Caul~Cond+(1|Sito)+(1|Area)+(1|Trans), data=dataset,
 family=binomial) 
 
 this is the output: 
 
 summary(model) 
 Generalized linear mixed model fit by the Laplace approximation 
 Formula: Caul ~ Cond + (1 | Sito) + (1 | Area) + (1 | Trans) 
Data: dataset 
AIC   BIC logLik deviance 
  548.7 573.7 -268.3536.7 
 Random effects: 
  Groups NameVariance  Std.Dev. 
  Trans  (Intercept) 3.2313398 1.797593 
  Area   (Intercept) 0.000 0.00 
  Sito   (Intercept) 0.0047151 0.068667 
 Number of obs: 480, groups: Trans, 48; Area, 12; Sito, 2 
 
 As you can see the residual variance is missing. Can anybody tell me
 why?
 Does anybody know how can I get it? 
 
 Thank you for your attention, I wish somebody can help me. 
 
 Have a nice day, best regards, 
   
 Tommaso Alestra 
 -- 
 View this message in context:
 http://www.nabble.com/Hoe-to-get-RESIDUAL-VARIANCE-in-logistic-regressio
 n-using-lmer-tp23313330p23313330.html
 Sent from the R help mailing list archive at Nabble.com.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
 weer 
 en binden het INBO onder geen enkel beding, zolang dit bericht niet
 bevestigd is
 door een geldig ondertekend document. The views expressed in  this message 
 and any annex are purely those of the writer and may not be regarded as
 stating 
 an official position of INBO, as long as the message is not confirmed by a
 duly 
 signed document.
 
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http://www.nabble.com/How-to-get-RESIDUAL-VARIANCE-in-logistic-regression-using-lmer-tp23313330p23487103.html
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Re: [R] Hoe to get RESIDUAL VARIANCE in logistic regression using lmer

2009-05-11 Thread tomal

Thank you very much Mr Onkelinx, I appreciated your answer. 
I'm sorry about my lack of knowledge on this subject, but I don't understand
why residual variance is fixed at 1 with the binomial family and I didn't
manage to find any exhaustive explanation for that.
May you kindly suggest me some reference I can consult in order to go into
depth on this matter?

Does anybody else can provide me furher information about this topic?

Thank you very much.

Best regards,   
  
Tommaso Alestra



ONKELINX, Thierry wrote:
 
 Dear Tommaso,
 
 The residuals variance is fixed at 1 with the binomial family.
 
 HTH,
 
 Thierry
 
 
 
 
 ir. Thierry Onkelinx
 Instituut voor natuur- en bosonderzoek / Research Institute for Nature
 and Forest
 Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
 methodology and quality assurance
 Gaverstraat 4
 9500 Geraardsbergen
 Belgium 
 tel. + 32 54/436 185
 thierry.onkel...@inbo.be 
 www.inbo.be 
 
 To call in the statistician after the experiment is done may be no more
 than asking him to perform a post-mortem examination: he may be able to
 say what the experiment died of.
 ~ Sir Ronald Aylmer Fisher
 
 The plural of anecdote is not data.
 ~ Roger Brinner
 
 The combination of some data and an aching desire for an answer does not
 ensure that a reasonable answer can be extracted from a given body of
 data.
 ~ John Tukey
 
 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 Namens tomal
 Verzonden: donderdag 30 april 2009 12:10
 Aan: r-help@r-project.org
 Onderwerp: [R] Hoe to get RESIDUAL VARIANCE in logistic regression using
 lmer
 
 
 Hello everybody, 
 using the lmer function, I have fitted the following logistic mixed
 regression model on an experimental data set containing one fixed factor
 (Cond) and three random variables (Sito, Area, Trans): 
 
 
 model-lmer(Caul~Cond+(1|Sito)+(1|Area)+(1|Trans), data=dataset,
 family=binomial) 
 
 this is the output: 
 
 summary(model) 
 Generalized linear mixed model fit by the Laplace approximation 
 Formula: Caul ~ Cond + (1 | Sito) + (1 | Area) + (1 | Trans) 
Data: dataset 
AIC   BIC logLik deviance 
  548.7 573.7 -268.3536.7 
 Random effects: 
  Groups NameVariance  Std.Dev. 
  Trans  (Intercept) 3.2313398 1.797593 
  Area   (Intercept) 0.000 0.00 
  Sito   (Intercept) 0.0047151 0.068667 
 Number of obs: 480, groups: Trans, 48; Area, 12; Sito, 2 
 
 As you can see the residual variance is missing. Can anybody tell me
 why?
 Does anybody know how can I get it? 
 
 Thank you for your attention, I wish somebody can help me. 
 
 Have a nice day, best regards, 
   
 Tommaso Alestra 
 -- 
 View this message in context:
 http://www.nabble.com/Hoe-to-get-RESIDUAL-VARIANCE-in-logistic-regressio
 n-using-lmer-tp23313330p23313330.html
 Sent from the R help mailing list archive at Nabble.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.
 
 Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
 weer 
 en binden het INBO onder geen enkel beding, zolang dit bericht niet
 bevestigd is
 door een geldig ondertekend document. The views expressed in  this message 
 and any annex are purely those of the writer and may not be regarded as
 stating 
 an official position of INBO, as long as the message is not confirmed by a
 duly 
 signed document.
 
 __
 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.
 
 

-- 
View this message in context: 
http://www.nabble.com/How-to-get-RESIDUAL-VARIANCE-in-logistic-regression-using-lmer-tp23313330p23487104.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] anyone know how to calculate chi square value from P val

2009-05-11 Thread Ted Harding
On 11-May-09 19:36:00, Anyuan Guo wrote:
 Dear all,
 I have P value of a list of markers. Now I need the chi square
 value with degrees of freedom 2.
 I noticed there are several Chisquare functions (dchisq, pchisq, 
 qchisq, and rchisq), but it seems all are not for my purpose.
 In microsoft excel, there is a function CHINV to do this, such as 
 CHINV(0.184, 2) is 3.386, that means the chi square value for P value 
 0.184, degree 2 is 3.386.
 Does the R package has some function to do this?
 
 Thanks
 Anyuan

Yes, and you already looked at it but apparently did not recognise it!

Either:

  qchisq(1-0.184, 2)  ## (note 1-0.184, since 0.184 is the upper tail)
# [1] 3.385639

Or:

  qchisq(0.184, 2, lower.tail=FALSE) ## Default for lower.tail is TRUE
# [1] 3.385639

Enter ?qchisq for more informatio0n on this and related function.
Ted.


E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 11-May-09   Time: 18:15:34
-- XFMail --

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[R] Error in NLME (nonlinear mixed effects model)

2009-05-11 Thread Lindsay Banin
Hi there,

I have been trying to fit an NLME to my data. My dataset has two category 
levels - one is a fixed effect (level1) and one is a random effect (level2), 
though so far I have only experimented with the highest level grouping (fixed, 
level1), with the following code:

mod1 - nlme(H ~ a*(1-exp(-b*D^c)),
data=sizes,
fixed=a+b+c~factor(Loc),
start=c(a=75,b=0.05,c=0.7))

This returns the error:
Error in getGroups.data.frame(dataMix, eval(parse(text = paste(~1, 
deparse(groups[[2]]),  : 
Invalid formula for groups


Other points that it may be useful to note in diagnosing the problem are that:
1) I have tried both specifying Loc=as.factor and not specifying this.
2) I have tried other configurations of writing fixed=list(...) or fixed=c(...) 
and that always generates an error
3) I have not specified groupedData
4) My groups do not have equal sample sizes (unbalanced?)

Look forward to hearing back!
Thanks in advance,
Lindsay

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[R] robustbase: cannot install

2009-05-11 Thread Agustin Lobo

This is really odd,
I've installed the binary of robustbase through synaptic on an ubuntu 
8.04 machine and get:

 library(robustbase)
Error in dyn.load(file, DLLpath = DLLpath, ...) :
  unable to load shared library 
'/usr/lib/R/site-library/robustbase/libs/robustbase.so':

  libRlapack.so: cannot open shared object file: No such file or directory
Error: package/namespace load failed for 'robustbase'


The package is where it is suposed to be:
/usr/lib/R/site-library/robustbase

is this not a problem at compiling?

The problem is that if I try to install the source with 
install.packages(robustbase), I get:
gcc -std=gnu99 -shared -o robustbase.so R-rng4ftn.o init.o lmrob.o mc.o 
qn_sn.o rf-common.o rffastmcd.o rfltsreg.o wgt_himed.o -L/usr/lib/R/lib 
-lRlapack -lblas -L/usr/lib/gcc/i486-linux-gnu/4.2 -lgfortran -lm 
-L/usr/lib/gcc/i486-linux-gnu/4.2 -lgfortran -lm -L/usr/lib/R/lib -lR

/usr/bin/ld: cannot find -lRlapack
collect2: ld returned 1 exit status

Despite the fact that I do have Rlapack:
~$ R CMD config LAPACK_LIBS
-L/usr/lib/R/lib -lRlapack

Am I doing something wrong?

Thanks

Agus
--
Dr. Agustin Lobo
Institut de Ciencies de la Terra Jaume Almera (CSIC)
LLuis Sole Sabaris s/n
08028 Barcelona
Spain
Tel. 34 934095410
Fax. 34 934110012
email: agustin.l...@ija.csic.es
http://www.ija.csic.es/gt/obster

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[R] maxLik package

2009-05-11 Thread amene kheradmandi
Hi all. 
Recently i have been used maxlik package for optimizing a function with 5 
parameters but i couldn't define gradient argument in function maxLik; 

How i can define a command to receive my goal? Whether I can change the core of 
this package?
 
Thanks for your attention and reply
A. kheradmandi


  
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Re: [R] Do you use R for data manipulation?

2009-05-11 Thread Dimitri Liakhovitski
I am not a statistician and not a computer scientist by education. I
consider myself an R novice and came to R - thanks to my boss - from
an SPSS background. I work for a market research company and the most
typical data files we deal with are not huge - up to several thousand
rows and up to a thousand variables.
I would say, on certain projects, most of what we do in R (if you look
at the number of lines in R we devote to a given task) is data
manipulation. The actual statistical method is frequently just a line
- all the rest is getting the data shaped right and then spitting out
the results of the analysis in a way that is usable (i.e.,
presentable).
I find R to be excellent in data manipulations that we perform. First
of all, it's great that you can always grab variables/cases you need
and ignore all the rest. In SPSS you just keep staring at all those
data and variables that you don't need - trying to find the one you
need.
Second - I find R to be incredibly fast (as opposed to SPSS or Excel)
with the amounts of data we are dealing with.
And third - nothing is written in stone and your original data is
always untouched - you can always read it in again and again. For
example, if I create a new variable and make a mistake, I can always
fix the code, rerun that piece of the code and that gives me the
corrected object that containes that new variable. I never touch the
original data and hence - never spoil it.

Dimitri



On Mon, May 11, 2009 at 11:20 AM, ronggui ronggui.hu...@gmail.com wrote:
 2009/5/6 Emmanuel Charpentier charp...@bacbuc.dyndns.org:
 Le mercredi 06 mai 2009 à 00:22 -0400, Farrel Buchinsky a écrit :
 Is R an appropriate tool for data manipulation and data reshaping and data
 organizing?
 [ Large Snip ! ... ]

 Depends on what you have to do.

 I've done what can be more or less termed data management with almost
 uncountable tools (from Excel (sigh...) to R with SQL, APL, Pascal, C,
 Basic (in 1982 !), Fortran and even Lisp in passing...).


 SQL has strong points : join is, to my tastes, more easily expressed in
 SQL than in most languages, projection and aggregation are natural.

 However, in SQL, there is no natural ordering of row tables, which
 makes expressing algorithms using this order difficult. Try for example
 to express the differences of a time series ... (it can be done, but it
 is *not* a pretty sight).

 On the other hand, R has some unique expressive possibilities (reshape()
 comes to mind).

 So I tend to use a combination of tools : except for very small samples,
 I tend to manage my data in SQL and with associated tools (think data
 editing, for example ; a simple form in OpenOffice's Base is quite easy
 to create, can handle anything for which an ODBC driver exists, and
 won't crap out for more than a few hundreds line...). finer manipulation
 is usually done in R with  native tools and sqldf.

 But, at least in my trade, the ability to handle Excel files is a must
 (this is considered as a standard for data entry. Sigh ...). So the
 first task is usually a) import data in an SQL database, and b) prepare
 some routines to dump SQL tables / R dataframes in Excel tor returning
 back to the original data author...

 I don't think Excel is  a standard tool for data entry. Epidata entry
 is much more professional.


 HTH

                                        Emmanuel Charpentier

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 --
 HUANG Ronggui, Wincent
 PhD Candidate
 Dept of Public and Social Administration
 City University of Hong Kong
 Home page: http://asrr.r-forge.r-project.org/rghuang.html

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-- 
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com

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[R] consultation

2009-05-11 Thread Nancy Tejerina
*Dear R Users;*

* *

*I´m writing to ask you how can I do Survivals Curves using Time-dependent
covariates? Which packages I need to Install?*
*Thanks for your help, I look foward you reply.*
**
*Sincerelly.*
**

*Nancy Tejerina.*

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Re: [R] Building US maps in R

2009-05-11 Thread stefan.d...@gmail.com
Hi DD,

if you have a .shp file (which in principal should exist of the kind
of map your describing) you can load into R using this packages:
http://r-spatial.sourceforge.net/

In addition, there is also the R-Geo list which is much more suited
for this kind of questions:
r-sig-...@stat.math.ethz.ch

And there is always the book when it comes to R and spatial modeling
(Applied Spatial Data Analysis With R):
http://www.asdar-book.org/

Best of luck,
Stefan

On Mon, May 11, 2009 at 3:55 AM, Dirty D dir...@overdetermined.net wrote:
 Hi,

 I'm trying to build some maps of the US by county that will have the
 following characteristics:

 Feature/Map
        Map 1
        Map2
        Both
 Broken out by county
        Yes
        Yes
        Yes
 Heatmaps of US Census Data for income by county
        Yes
        No
        Yes
 Heatmaps of US Census Data for race by county (recoded as white and
 non-white, with each county color coded based on the majority)
        No
        Yes
        No
 Polygon markers showing the coordinates of a list of restaurants that
 have geographical coordinates attached
        Yes
        Yes
        Yes


 I started learning R a year ago, but had to drop it for work reasons. Is
 there someone on the list who can help guide me through the process or
 refer me to /very, very/ detailed instructions and tutorials?

 Thanks,

 DD

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Re: [R] Error in NLME (nonlinear mixed effects model)

2009-05-11 Thread Ben Bolker



Lindsay Banin wrote:
 
 Hi there,
 
 I have been trying to fit an NLME to my data. My dataset has two category
 levels - one is a fixed effect (level1) and one is a random effect
 (level2), though so far I have only experimented with the highest level
 grouping (fixed, level1), with the following code:
 
 mod1 - nlme(H ~ a*(1-exp(-b*D^c)),
 data=sizes,
 fixed=a+b+c~factor(Loc),
 start=c(a=75,b=0.05,c=0.7))
 
 This returns the error:
 Error in getGroups.data.frame(dataMix, eval(parse(text = paste(~1,
 deparse(groups[[2]]),  : 
 Invalid formula for groups
 
 Other points that it may be useful to note in diagnosing the problem are
 that:
 1) I have tried both specifying Loc=as.factor and not specifying this.
 2) I have tried other configurations of writing fixed=list(...) or
 fixed=c(...) and that always generates an error
 3) I have not specified groupedData
 4) My groups do not have equal sample sizes (unbalanced?)
 
 

(1) You haven't given us a reproducible example ...
(2) You might try asking this on the r-sig-mixed-models list instead
(3) I suspect your problem might be that nlme can't fit models with no
random effect.  Have you tried the 
analogous model in gnls ?

  good luck
Ben Bolker

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Re: [R] Do you use R for data manipulation?

2009-05-11 Thread Emmanuel Charpentier
Le lundi 11 mai 2009 à 23:20 +0800, ronggui a écrit :

[ Snip... ]

  But, at least in my trade, the ability to handle Excel files is a must
  (this is considered as a standard for data entry. Sigh ...).

[ Re-snip... ]

 I don't think Excel is  a standard tool for data entry. Epidata entry
 is much more professional.

Irony squared ?

This *must* go in the fortunes file !
Emmanuel Charpentier

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[R] Warning trying to plot -log(log(survival))

2009-05-11 Thread John Sorkin
windows xp
R 2.8.1

I am trying to plot the -log(log(survival)) to visually test the proportional 
hazards assumption of a Cox regression. The plot, which should give two lines 
(one for each treatment) gives only one line and a warning message. I would 
appreciate help getting two lines, and an explanation of the warning message. 
My problem may the that I have very few events in one of my strata, but I don't 
know.
Thanks,
John

fit0-coxph(Surv(Time30,Died)~strata(Rx)+Age,data=GVHDdata)
plot(survfit(fit0),fun=cloglog)


WARNING: Warning in xy.coords (x, y, xlabel, ylabel, log)  :
 2 x values =0 omitted from logarithmic plot



 print(survfit(fit0),fun=cloglog)
Call: survfit.coxph(object = fit0)

  n events median 0.95LCL 0.95UCL
Rx=0 31  5Inf Inf Inf
Rx=1 32 15Inf1.17 Inf

John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)

Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}

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[R] explanation - what does R in the background?

2009-05-11 Thread Martin Batholdy
for an undergraduate course in working with R I would like to explain  
the R-environment and the associated file system. But I am not sure if  
I have understand all of that correctly myself.

Is there a document or a website out there which explains that in an  
understandable way? Things like; what files are loaded; what  
the .Rhistory does; what the different graphic devices are etc...
(perhaps with a nice graphic or table)


thanks für any help!
[[alternative HTML version deleted]]

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Re: [R] explanation - what does R in the background?

2009-05-11 Thread Bert Gunter
?Startup
 should help. 


Bert Gunter
Nonclinical Biostatistics
467-7374

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Martin Batholdy
Sent: Monday, May 11, 2009 12:53 PM
To: r-help@r-project.org
Subject: [R] explanation - what does R in the background?

for an undergraduate course in working with R I would like to explain  
the R-environment and the associated file system. But I am not sure if  
I have understand all of that correctly myself.

Is there a document or a website out there which explains that in an  
understandable way? Things like; what files are loaded; what  
the .Rhistory does; what the different graphic devices are etc...
(perhaps with a nice graphic or table)


thanks f|r any help!
[[alternative HTML version deleted]]

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Re: [R] Quantile Regression fixed effects model

2009-05-11 Thread Thuy T. Nguyen

 Dear R-users,
I am applying professor Koenker's code for fixed effect quantile regression.
However, I need to bootstrap and cluster the standard errors in my fitted
model. 
Assuming that I need to bootstrap then cluster the standard errors by s (the
strata indicator in Prof. Koenker's code), what do I have to do? 
I tried to bootstrap and cluster using the bootcov command in Design package
without any success. 
If you have used fixed effect quantile regression and were able to bootstrap
and cluster, please let me know how the process should be done! 
With much appreciation,

Thuy T. Nguyen



see: http://www.econ.uiuc.edu/~roger/research/panel/rq.fit.panel.R


url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820





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[R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread Mark Na
Hi R-helpers,

I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.

Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.

As an example:

Mark-read.csv(Mark.csv)

...but the code (or command) would do this automatically for every
.csv in the working directory without my specifying each file.

I'd appreciate any help or ideas you might have.

Thanks!

Mark Na

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


Re: [R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread Andreas Christoffersen
I really dont know hot to use it - but have a look at ?source

On Mon, May 11, 2009 at 10:37 PM, Mark Na mtb...@gmail.com wrote:

 Hi R-helpers,

 I would like to read into R all the .csv files that are in my working
 directory, without having to use a read.csv statement for each file.

 Each .csv would be read into a separate dataframe which would acquire
 the filename of the .csv.

 As an example:

 Mark-read.csv(Mark.csv)

 ...but the code (or command) would do this automatically for every
 .csv in the working directory without my specifying each file.

 I'd appreciate any help or ideas you might have.

 Thanks!

 Mark Na

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


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[R] Looking for a quick way to combine rows in a matrix

2009-05-11 Thread Crosby, Jacy R
I'm working with genotype data in a frequency table:

 a=matrix(1:16, nrow=4)
 rownames(a)=c(AA,AT,TA,TT)
 a
   [,1] [,2] [,3] [,4]
AA159   13
AT26   10   14
TA37   11   15
TT48   12   16

'AT' and 'TA' are essentially the same, and I'd like to combine (add) the rows 
to reflect this. The final matrix should be:

   [,1] [,2] [,3] [,4]
AA159   13
AT513   21   29
TT48   12   16

Is there a fast way to do this?

Thanks in advance!

Jacy Crosby
jacy.r.cro...@uth.tmc.edu


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


Re: [R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread Romain Francois


Hi,

Something like this perhaps:

files - dir( pattern  = \\.csv$ )
for( x in files){
   assign( sub( \\.csv$, , x ) , read.csv(x), envir = .GlobalEnv )
}

Romain

Mark Na wrote:

Hi R-helpers,

I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.

Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.

As an example:

Mark-read.csv(Mark.csv)

...but the code (or command) would do this automatically for every
.csv in the working directory without my specifying each file.

I'd appreciate any help or ideas you might have.

Thanks!

Mark Na
  



--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr

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


Re: [R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread Mose
Woo, global jinx!

On Mon, May 11, 2009 at 1:55 PM, Romain Francois romain.franc...@dbmail.com
 wrote:


 Hi,

 Something like this perhaps:

 files - dir( pattern  = \\.csv$ )
 for( x in files){
   assign( sub( \\.csv$, , x ) , read.csv(x), envir = .GlobalEnv )
 }

 Romain

 Mark Na wrote:

 Hi R-helpers,

 I would like to read into R all the .csv files that are in my working
 directory, without having to use a read.csv statement for each file.

 Each .csv would be read into a separate dataframe which would acquire
 the filename of the .csv.

 As an example:

 Mark-read.csv(Mark.csv)

 ...but the code (or command) would do this automatically for every
 .csv in the working directory without my specifying each file.

 I'd appreciate any help or ideas you might have.

 Thanks!

 Mark Na




 --
 Romain Francois
 Independent R Consultant
 +33(0) 6 28 91 30 30
 http://romainfrancois.blog.free.fr


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 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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Re: [R] Beyond double-precision?

2009-05-11 Thread Stavros Macrakis
On Sat, May 9, 2009 at 12:17 PM, Berwin A Turlach
ber...@maths.uwa.edu.au wrote:
 log(H) = log(n) - log( 1/x_1 + 1/x_2 + ... + 1/x_n)
...But we need to calculate the logarithm of a sum from the logarithms of the 
individual terms.

 ...The way to calculate log(x+y) from lx=log(x) and ly=log(y) ...
  max(lx,ly) + log1p(exp(-abs(lx-ly)))

Agreed completely so far. But instead of calculating the logsum
pairwise, you can do it all in one go, which is both more efficient
and more accurate.

Here are some timing and accuracy measurements of the one-shot logsum
compared to the loop and the Reduce versions. (Full code at the bottom
of this email.) The vector version is much faster and much more
accurate in general.  There must be cases where the log1p method
increases accuracy, but I couldn't find them.

-s

Large examples to test accuracy and speed

Test case: runif(1e+06)
  function. timeerror
1logsum 0.22 9.31e-16
2  logsum_s 0.15 9.31e-16
3  logsum_r 9.75 3.10e-13

Test case: rexp(1e+06)
  function.  time error
1logsum  0.21 -1.40e-15
2  logsum_s  0.15 -1.40e-15
3  logsum_r 10.13 -1.38e-14

Test case: abs(rnorm(1e+06))
  function.  time error
1logsum  0.24 -4.38e-16
2  logsum_s  0.14 -4.38e-16
3  logsum_r 10.01 -8.74e-14

Test case: rep(1, 1e+05)
  function. timeerror
1logsum 0.01 1.46e-16
2  logsum_s 0.01 1.46e-16
3  logsum_r 0.96 6.24e-14

Test case: rep(10^-(1:10), each = 1)
  function. time error
1logsum 0.02  6.14e-16
2  logsum_s 0.01  6.14e-16
3  logsum_r 0.95 -6.96e-12

More accurate even for small cases

Test case: 1:100
  function. time error
1logsum0 -3.60e-16
2  logsum_s0 -3.60e-16
3  logsum_r0  3.24e-15

Test case: abs(rnorm(100))
  function. time error
1logsum0 -3.48e-16
2  logsum_s0 -3.48e-16
3  logsum_r0 -2.09e-15


##
# Fast, accurate sum in log space
#
logsum - function(l) {
 maxi - which.max(l)
 maxl - l[maxi]
 maxl + log1p(sum(exp(l[-maxi]-maxl))) }
##

##
# Simpler, perhaps less accurate sum in log space
#
logsum_s - function(l) {
 maxl - max(l)
 maxl + log(sum(exp(l-maxl))) }
##



# Pairwise reduction
logsum_r - function(x) Reduce( function(lx, ly) max(lx, ly) +
log1p(exp(-abs(lx-ly))), x )


function_names - c(logsum,logsum_s,logsum_r)

logsum_test - function(l) {
  cat(\nTest case:,deparse(substitute(l)),\n)
  realsum - sum(l)
  logl - log(l)
  results - times - list()
  lapply( function_names, function(f) times[[f]] - system.time(
results[[f]] - getFunction(f)(logl))[1])
  data.frame(`function`=function_names,
 time=as.numeric(times),
 error=(exp(as.numeric(results))-realsum)/realsum )
}

set.seed(1)

cat(\n\nLarge examples to test accuracy and speed\n\n)
logsum_test(runif(100))
logsum_test(rexp(100))
logsum_test(abs(rnorm(100)))
logsum_test(rep(1,10))
logsum_test(rep(10^-(1:10),each=1))

cat(\n\nMore accurate even for small cases\n\n)
logsum_test(1:100)
logsum_test(abs(rnorm(100)))

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[R] stepAIC + cox.zph

2009-05-11 Thread Eric Rescorla
Hi,

I'm using stepAIC with a CoxPH model on some data. As
we expect, there are a bunch of predictors, and stepAIC
throws some out and then converges on others.

Anyway, when I run cox.zph on the final model, I get
some evidence of Beta(t) time dependence. Should I
be concerned about the impact of this on the entire
fit procedure, or is it just that I can't trust the final
hazard ratio that the Cox model spits out?

Thanks in advance,
-Ekr

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Re: [R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread baptiste auguie

Hi,

I once made this function (essentially the same as Romain),


assignFiles -
function (pattern = csv, strip = (_|.csv|-| ), ...)  # strip is  
any pattern you want to remove from the filenames

{
listFiles - list.files(pattern = pattern, all.files = FALSE,
full.names = FALSE, recursive = FALSE)
variables - gsub(strip, , listFiles)
for (ii in (1:length(listFiles))) {
name - listFiles[ii]
assign(variables[ii], read.csv(name, ...), env = .GlobalEnv)
}
}


HTH,

baptiste

On 11 May 2009, at 22:55, Romain Francois wrote:



Hi,

Something like this perhaps:

files - dir( pattern  = \\.csv$ )
for( x in files){
   assign( sub( \\.csv$, , x ) , read.csv(x), envir = .GlobalEnv )
}

Romain

Mark Na wrote:

Hi R-helpers,

I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.

Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.

As an example:

Mark-read.csv(Mark.csv)

...but the code (or command) would do this automatically for every
.csv in the working directory without my specifying each file.

I'd appreciate any help or ideas you might have.

Thanks!

Mark Na




--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr

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_

Baptiste Auguié

School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag

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Re: [R] Read many .csv files into an R session automatically, without specifying filenames

2009-05-11 Thread Rolf Turner


On 12/05/2009, at 8:41 AM, Andreas Christoffersen wrote:


I really dont know hot to use it - but have a look at ?source


No; ``source'' is irrelevant here.


On Mon, May 11, 2009 at 10:37 PM, Mark Na mtb...@gmail.com wrote:


Hi R-helpers,

I would like to read into R all the .csv files that are in my working
directory, without having to use a read.csv statement for each file.

Each .csv would be read into a separate dataframe which would acquire
the filename of the .csv.

As an example:

Mark-read.csv(Mark.csv)

...but the code (or command) would do this automatically for every
.csv in the working directory without my specifying each file.

I'd appreciate any help or ideas you might have.



datalist - list()
files - list.files(pattern=\\.csv$)
for(file in files) {
stem - gsub(\\.csv$,,file)
datalist[[stem]] - read.csv(file)
}

The foregoing does things the *right* way, putting the resulting data  
sets together
into a list whereby they can be accessed in an organized fashion.  If  
you wrong-headedly
*insist* on putting the data sets separately into your workspace then  
do:


files - list.files(pattern=\\.csv$)
for(file in files) {
stem - gsub(\\.csv$,,file)
assign(stem,read.csv(file))
}

cheers,

Rolf Turner

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

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[R] Add data count to plot?

2009-05-11 Thread MikSmith

Hi

I'm a relative newbie to R and had a query concerning plotting. I have
generated a par(mfrow = c(2, 2)) graphic with 10 rose diagrams using
circular. What I wanted to add to each individual plot was n = x for the
number of data observations in each dataset. How might I go about doing
this??

Thanks

Mike
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[R] inconsistent results for axis.POSIXct

2009-05-11 Thread Dan Kelley

Some time ago, I posted a note about what I considered to be a bug in
axis.POSIXt() for R 2.8.x, relating to whether timezones in the data are
obeyed on the axes.  A link to that note, and to a quick and helpful
response, is at the following URL 

http://www.nabble.com/patch-for-axis.POSIXct-%28related-to-timezones%29-td22338700.html#a22338700

Note that R 2.9.0 has been adjusted to help with this.  However, there are
still problems.  Test code is given below:

# test code (top panel, OK; bottom panel, incorrect times on x axis) 
par(mar=c(2,2,2,1)) 
par(mfrow=c(2,1)) 
start - as.POSIXct(2008-05-01 00:00:00, tz=UTC) 
end   - as.POSIXct(2008-05-01 00:10:00, tz=UTC) 
plot(c(start, end), c(0, 1), ylab=) 
mtext(paste(start, to, end), side=3) 

start - as.POSIXct(2008-05-01 00:00:00, tz=UTC) 
end   - as.POSIXct(2008-05-01 01:10:00, tz=UTC) 
plot(c(start, end), c(0, 1), ylab=) 
mtext(paste(start, to, end), side=3) 
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[R] Set working directory by dragging text file onto R shortcut? [WinXP, unfortunately]

2009-05-11 Thread Mark Na
Hi R-helpers,

I must use WinXP at work, and I'm missing a particular feature that's
available in R on my Mac at home...

In WinXP I would like to launch R by dragging and releasing a text
file on top of the R shortcut on my desktop. Then, I would like the
working directory to be automatically set to the location of that text
file.

That's what works in MacOS, but I can't figure out if it's possible in WinXP.

Any ideas? I'd appreciate it...

Thanks, Mark Na

PS What I do now is make a new R shortcut for each project, store that
shortcut in the directory containing that project's datafiles and R
programs (which are in text files), and set the Start in value to
that directory...but this has to be set up uniquely for each
project...

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Re: [R] R and McAfee 8.5

2009-05-11 Thread Bernardo Rangel Tura
On Mon, 2009-05-11 at 09:54 -0300, Joyce, Warren wrote:
 Hi,
 
 I have been working with R for the last year and using the UKFSST package to 
 look at satellite tag track data and SST information. Fpr those not familiar, 
 the package uses the positions estimated by the satellite tags themselves and 
 the associated SST data from servers (in this case, from the University of 
 Hawaii and the NOAA Coastwatch website) for the time preiod to estimate a 
 reasonable track the animal took while it was at liberty.  I'm currently 
 having trouble downloading data from two different servers which I never had 
 a problem before I recevied an upgraded computer with McAfee 8.5. When the 
 package goes to the various servers for SST information to download, I 
 continue to get a message in R:
 
(...)
 Into my web browser, I get prompted with a file download window which quickly 
 downloads an empty file. It looks like the data is available, I just can't 
 seem to download it.
 
 I've deduced that it must be a problem with the new firewall features of 
 McAfee 8.5 and was wondering if anyone else has run into a problem like this. 
 I've tried using the exclusions options in 8.5 to exclude any R files but 
 have come up empty handed. Still waiting on our informatics division to try 
 and solve the problem as well but I thought I might try asking here as well.
 
 Thanks and I appreciate any assistance!
 
 Warren N. Joyce 

Hi Joice,

In my opinion you need think about 2 things.

1- This is a problem of McAfee 8.5 not a R problem, so do you contact a
McAfee support for fix this problem ?

2- I presume  you use a windows but i do know your version (XP or Vista
or win 95 etc) neither your R configuration (do you use R --internet2
or not?)
-- 
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil

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Re: [R] Set working directory by dragging text file onto R shortcut? [WinXP, unfortunately]

2009-05-11 Thread Gabor Grothendieck
There are a several possibilities using batchfiles whose
home page is at:

http://batchfiles.googlecode.com

When on that home page note that there are different versions of
these utilities for XP and Vista.

There are two utilities that could be used.  Description is
for Vista so for XP you may have to vary it slightly.

1. RguiStart.bat is a batch file that starts R interpreting
its first argument as the folder in which to start R (or if its
an .Rdata file then it interprets it as the .Rdata file to start
R with).  The main reason for its existence is so that you can
place it in your SendTo folder.  On Vista do this:

  copy RguiStart.bat %APPDATA%\Microsoft\Windows\SendTo

Then, on Vista, when you are in Windows Explorer you can right click
on any folder in your user area and choose SendTo.  From the SendTo
menu choose RGuiStart.bat to start up R in that folder.  If you
right click on an .Rdata file rather than a folder then R will
start up with that file loaded (although in that case its probably
easier to just double click the .Rdata file provided you have
file type associations set up -- if you don't then the RguiStart.bat
will provide an alternative).

2. Rgui.bat is a batch file that starts up R.  Just shift right click
a folder in Windows
Explorer and choose Open Command Window Here and then when the command
window opens enter Rgui assuming you have placed Rgui.bat somewhere on
your path.

Note that Rgui.bat and RguiStart.bat are actually the same file just renamed.
When it runs it uses the name by which it was called to determine what to do.
The files have no dependencies and find R by looking in the registry so when
you install a new version of R they continue to work without any change
(because installing a new version of R updates the registry).

On Mon, May 11, 2009 at 6:30 PM, Mark Na mtb...@gmail.com wrote:
 Hi R-helpers,

 I must use WinXP at work, and I'm missing a particular feature that's
 available in R on my Mac at home...

 In WinXP I would like to launch R by dragging and releasing a text
 file on top of the R shortcut on my desktop. Then, I would like the
 working directory to be automatically set to the location of that text
 file.

 That's what works in MacOS, but I can't figure out if it's possible in WinXP.

 Any ideas? I'd appreciate it...

 Thanks, Mark Na

 PS What I do now is make a new R shortcut for each project, store that
 shortcut in the directory containing that project's datafiles and R
 programs (which are in text files), and set the Start in value to
 that directory...but this has to be set up uniquely for each
 project...

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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.


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