[R] Asymmetric colors for heatmap
Dear expeRts, Currently, my colors are as follows: mycol - c(blue1,blue2,blue3,blue4,black,yellow4,yellow3,yellow2,y ellow1) heatmap(snp, Rowv=NA, Colv=NA, col=mycol) However, I would like to have the following colors: bright blue - dark blue: for intensity range from 0 to 2 in steps of 0.5 (i.e. 4 grades of blue) black: for intensity 2 dark yellow - bright yellow: for intensity range from 2 to 8 in steps of 0.5 (i.e. 8 grades of yellow) You may realize that I want to display copy number data from SNP-chips as heatmap. Since copynumber = 2, is the default value, I want to display it in black, LOH in increasing blue, and amplifications in increasing yellow. Even though there may be higher amplification rates, a value of CN=8 should already display the brightest yellow. In Spotfire it is easy to achieve this, especially that CN=2 is always displayed as black, however, I do not know how to do it in R. Can you tell me how I have to create the colors to achieve this? (P.S.: Of course, I could do: snp[snp8] - 8, but this will not solve my problem with asymmetric colors) Thank you in advance Christian Stratowa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Asymmetric colors for heatmap
Dear Sean Thank you, however the heatmap() function from the stats package does not have this option. Browsing around I see that you mean heatmap.2() from package gplots, which we have not installed yet. Is there also another possibility besides heatmap.2() since I would also need this option for e.g. function image(). Best regards Christian -Original Message- From: Sean Davis [mailto:[EMAIL PROTECTED] Sent: Wednesday, July 27, 2005 13:16 To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Subject: Re: [R] Asymmetric colors for heatmap See the breaks argument to heatmap. Sean - Original Message - From: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] To: r-help@stat.math.ethz.ch mailto:r-help@stat.math.ethz.ch Sent: Wednesday, July 27, 2005 3:54 AM Subject: [R] Asymmetric colors for heatmap Dear expeRts, Currently, my colors are as follows: mycol - c(blue1,blue2,blue3,blue4,black,yellow4,yellow3,yellow2,y ellow1) heatmap(snp, Rowv=NA, Colv=NA, col=mycol) However, I would like to have the following colors: bright blue - dark blue: for intensity range from 0 to 2 in steps of 0.5 (i.e. 4 grades of blue) black: for intensity 2 dark yellow - bright yellow: for intensity range from 2 to 8 in steps of 0.5 (i.e. 8 grades of yellow) You may realize that I want to display copy number data from SNP-chips as heatmap. Since copynumber = 2, is the default value, I want to display it in black, LOH in increasing blue, and amplifications in increasing yellow. Even though there may be higher amplification rates, a value of CN=8 should already display the brightest yellow. In Spotfire it is easy to achieve this, especially that CN=2 is always displayed as black, however, I do not know how to do it in R. Can you tell me how I have to create the colors to achieve this? (P.S.: Of course, I could do: snp[snp8] - 8, but this will not solve my problem with asymmetric colors) Thank you in advance Christian Stratowa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html http://www.R-project.org/posting-guide.html [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Asymmetric colors for heatmap
Sorry, my mistake, I did not realize that image() has breaks and that heatmap() inherits from image(). However, I have the following problem, maybe I am doing something wrong. I have defined: mycol - c(blue1,blue2,blue3,blue4,black,yellow4,yellow3,yellow2,yel low1) breaks-c(0,0.5,1,1.5,2,2.5,3,4,5,6) When I call image() then the colors change slightly: image(t(tmp),col=mycol,axes=F) image(t(tmp),col=mycol,axes=F,breaks=breaks) However, when I call heatmap(): heatmap(tmp,Rowv=NA,Colv=NA,col=mycol) heatmap(tmp,Rowv=NA,Colv=NA,col=mycol,breaks=breaks) then breaks results in half of the heatmap drawn in white! Do you know what may be my mistake? Best regards Christian -Original Message- From: Stratowa,Dr.,Christian FEX BIG-AT-V Sent: Wednesday, July 27, 2005 14:07 To: 'Sean Davis'; r-help@stat.math.ethz.ch Subject: RE: [R] Asymmetric colors for heatmap Dear Sean Thank you, however the heatmap() function from the stats package does not have this option. Browsing around I see that you mean heatmap.2() from package gplots, which we have not installed yet. Is there also another possibility besides heatmap.2() since I would also need this option for e.g. function image(). Best regards Christian -Original Message- From: Sean Davis [mailto:[EMAIL PROTECTED] Sent: Wednesday, July 27, 2005 13:16 To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Subject: Re: [R] Asymmetric colors for heatmap See the breaks argument to heatmap. Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] RE: Comparison of correlation coefficients - Details
Dear Ioannis Thank you very much for pointing me to meta-analysis. Although it may not solve my problem with the normalization, it gives me some other options to display the different correlation coefficients. One possibility is the use of Funnel plots, which are even available in library(rmeta). Another possibility is the use of forest-plots, as implemented in rmeta as metaplot. Sorrowly, rmeta does not include the Rosenthal-Rubin method or the Hunter-Schmidt method, as described in Meta-Analysis of Correlations, see: http://www.sussex.ac.uk/Users/andyf/teaching/pg/meta.pdf Probably, the best solution for me may be to modify metaplot for the Hunter-Schmidt method. BTW, the manual to the program Meta-Analysis 5.3, is also very helpful, see: http://userpage.fu-berlin.de/~health/meta_e.htm Further suggestions in this direction are greatly appreciated. Best regards Christian Stratowa -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Wednesday, July 21, 2004 18:07 To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Re: [R] RE: Comparison of correlation coefficients - Details That sounds very close to a meta-analytic comparison of two statistics. As a matter of fact, the Rosenthal Rubin approach transforms all primary statistics into Pearson r and then to Fisher's z and then follows with comparisons. More, comparisons can take into account sample sizes, or the value of some other predictor variable. I believe there is a Rosenthal book on meta-analysis published by Sage publications, as well as a Brian Mullen book published by Lawrence Erlbaum. Brian Mullen's book comes (or used to come) with a meta.exe program to perform meta-analyses. Hope this helps, Ioannis Dear all I apologize for cross-posting, but first it is accepted custom to thank the repliers and give a summary, and second I have still the feeling that this problem might be a general statistical problem and not necessarily related to microarrays only, but I might be wrong. First, I want to thank Robert Gentleman, Mark Kimpel and Mark Reiners for their kind replies. Robert Gentleman kindly pointed me to the Bioconductor package MeasurementError.cor as alternative to cor.test. Mark Kimpel suggested that 2-way factorial Anova or the Bioconductor package limma, respectively, may be helpful. Mark Reiners suggested to use the p-value of cor.test to test the significance. Maybe, I miss the point, but being not a statistician I am still unsure if it is possible to compare correlation coefficients from different sample sets. Both, the p-values from cor.test and from compcorr, could be used as measure of the significance. However, is it possible to normalize correlation coefficients from different sample sets? Could an expression such as corr * (1 - pval) be used for normalization? Maybe, it is not possible to normalize correlation coefficients? Would a barplot comparing the correlation coefficients between two genes for different tissues be meaningful? (Alternatively, I have tried to use (1-pval) to calculate the gray-level of the bars.) Any further suggestions would be appreciated very much. Best regards Christian Stratowa -Original Message- From: Stratowa,Dr.,Christian FEX BIG-AT-V Sent: Monday, July 19, 2004 15:00 To: '[EMAIL PROTECTED]' Subject: Comparison of correlation coefficients - Details Dear all Maybe, my last mail did not explain my problem correctly: Since we are interested, which genes have similar expression profiles in a certain tissue or in different tissues, we have calculated the correlation coefficients between all 46,000 x 46,000 genes of the HG_U133A/B chipset for about 70 tissues, where the number of samples per tissue ranges from 10 to more than 200. While writing an R-function to display the correlation coefficients between gene A and B in the different tissues as bar-graph, I realized that it may not be correct to compare the different correlation coefficients directly, since the number of samples per tissue varyies between 10 and 200. Thus, the question is: Is there a way to compare different correlation coefficients and/or apply some kind of normalization? Assuming that this might be a well known statistical problem I was browsing statistics books and the web for more information, but could only find the function compcorr which gives a p-value how well you can trust the comparison of two correlation coefficients from different samples. Even though this might currently not be a direct Bioconductor question, it is certainly a microarray analysis related question. Any suggestions how to solve this problem would be greatly appreciated. Best regards Christian Stratowa -Original Message- From: Stratowa,Dr.,Christian FEX BIG-AT-V Sent: Tuesday, July 13, 2004 14:40 To: '[EMAIL PROTECTED]' Subject: Comparison of correlation coefficients Dear
[R] RE: Comparison of correlation coefficients - Details
Dear all
I apologize for cross-posting, but first it is accepted custom to
thank the repliers and give a summary, and second I have still
the feeling that this problem might be a general statistical problem
and not necessarily related to microarrays only, but I might be wrong.
First, I want to thank Robert Gentleman, Mark Kimpel and Mark Reiners
for their kind replies. Robert Gentleman kindly pointed me to the
Bioconductor package MeasurementError.cor as alternative to cor.test.
Mark Kimpel suggested that 2-way factorial Anova or the Bioconductor
package limma, respectively, may be helpful. Mark Reiners suggested
to use the p-value of cor.test to test the significance.
Maybe, I miss the point, but being not a statistician I am still unsure
if it is possible to compare correlation coefficients from different
sample sets. Both, the p-values from cor.test and from compcorr,
could be used as measure of the significance.
However, is it possible to normalize correlation coefficients from
different sample sets? Could an expression such as corr * (1 - pval)
be used for normalization? Maybe, it is not possible to normalize
correlation coefficients?
Would a barplot comparing the correlation coefficients between two
genes for different tissues be meaningful? (Alternatively, I have
tried to use (1-pval) to calculate the gray-level of the bars.)
Any further suggestions would be appreciated very much.
Best regards
Christian Stratowa
-Original Message-
From: Stratowa,Dr.,Christian FEX BIG-AT-V
Sent: Monday, July 19, 2004 15:00
To: '[EMAIL PROTECTED]'
Subject: Comparison of correlation coefficients - Details
Dear all
Maybe, my last mail did not explain my problem correctly:
Since we are interested, which genes have similar expression profiles in a
certain tissue or in different tissues, we have calculated the
correlation coefficients between all 46,000 x 46,000 genes of the
HG_U133A/B chipset for about 70 tissues, where the number of samples
per tissue ranges from 10 to more than 200.
While writing an R-function to display the correlation coefficients between
gene A and B in the different tissues as bar-graph, I realized that it may
not be correct to compare the different correlation coefficients directly,
since the number of samples per tissue varyies between 10 and 200.
Thus, the question is: Is there a way to compare different correlation
coefficients and/or apply some kind of normalization?
Assuming that this might be a well known statistical problem I was browsing
statistics books and the web for more information, but could only find the
function compcorr which gives a p-value how well you can trust the
comparison of two correlation coefficients from different samples.
Even though this might currently not be a direct Bioconductor question, it
is certainly a microarray analysis related question. Any suggestions how to
solve this problem would be greatly appreciated.
Best regards
Christian Stratowa
-Original Message-
From: Stratowa,Dr.,Christian FEX BIG-AT-V
Sent: Tuesday, July 13, 2004 14:40
To: '[EMAIL PROTECTED]'
Subject: Comparison of correlation coefficients
Dear Bioconductor expeRts
Is it possible to compare correlation coefficients or to normalize
different correlation coefficients?
Concretely, we have the following situation:
We have gene expression profiles for different tissues, where the
number of samples per tissue are different, ranging from 10 to 250.
We are able to determine the correlation between two genes A and B
for each tissue separately, using cor.test. However, the question
arises if the correlation coefficients between different tissues can
be compared or if they must somehow be normalized, since the
number of samples per tissue varyies.
Searching the web I found the function compcorr, see:
http://www.fon.hum.uva.nl/Service/Statistics/Two_Correlations.html
http://ftp.sas.com/techsup/download/stat/compcorr.html
and implemented it in R:
compcorr - function(n1, r1, n2, r2){
# compare two correlation coefficients
# return difference and p-value as list(diff, pval)
# Fisher Z-transform
zf1 - 0.5*log((1 + r1)/(1 - r1))
zf2 - 0.5*log((1 + r2)/(1 - r2))
# difference
dz - (zf1 - zf2)/sqrt(1/(n1 - 3) + (1/(n2 - 3)))
# p-value
pv - 2*(1 - pnorm(abs(dz)))
return(list(diff=dz, pval=pv))
}
Would it make sense to use the resultant p-value to normalize the
correlation coefficients, using: corr - corr * compcorr()$pval
Is there a better way or an alternative to normalize the correlation
coefficients obtained for different tissues?
Thank you in advance for your help.
Since in the company I am not subscribed to bioconductor-help, could you
please reply to me (in addition to bioconductor-help)
P.S.: I have posted this first at r-help and it was suggested to me to
post it here, too.
Best regards
Christian Stratowa
==
Christian Stratowa, PhD
[R] Comparison of correlation coefficients
Dear expeRts
Is it possible to compare correlation coefficients or to normalize
different correlation coefficients?
Concretely, we have the following situation:
We have gene expression profiles for different tissues, where the
number of samples per tissue are different, ranging from 10 to 250.
We are able to determine the correlation between two genes A and B
for each tissue separately, using cor.test. However, the question
arises if the correlation coefficients between different tissues
can be compared or if they must somehow be normalized, since the
number of samples per tissue varyies.
Searching the web I found the function compcorr, see:
http://www.fon.hum.uva.nl/Service/Statistics/Two_Correlations.html
http://ftp.sas.com/techsup/download/stat/compcorr.html
and implemented it in R:
compcorr - function(n1, r1, n2, r2){
# compare two correlation coefficients
# return difference and p-value as list(diff, pval)
# Fisher Z-transform
zf1 - 0.5*log((1 + r1)/(1 - r1))
zf2 - 0.5*log((1 + r2)/(1 - r2))
# difference
dz - (zf1 - zf2)/sqrt(1/(n1 - 3) + (1/(n2 - 3)))
# p-value
pv - 2*(1 - pnorm(abs(dz)))
return(list(diff=dz, pval=pv))
}
Would it make sense to use the resultant p-value to normalize
the correlation coefficients, using: corr - corr * compcorr()$pval
Is there a better way or an alternative to normalize the
correlation coefficients obtained for different tissues?
Thank you in advance for your help.
Since in the company I am not subscribed to r-help, could you
please reply to me (in addition to r-help)
Best regards
Christian Stratowa
==
Christian Stratowa, PhD
Boehringer Ingelheim Austria
Dept NCE Lead Discovery - Bioinformatics
Dr. Boehringergasse 5-11
A-1121 Vienna, Austria
Tel.: ++43-1-80105-2470
Fax: ++43-1-80105-2782
email: [EMAIL PROTECTED]
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] FW: [Fwd: Re: [S] Exact p-values]
Dear Spencer Thank you for this extensive explanation of the problem. I was just curious. Best regards Christian == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] -Original Message- From: Spencer Graves [SMTP:[EMAIL PROTECTED]] Sent: Friday, February 14, 2003 1:29 PM To: Stratowa,Dr,Christian FEX BIG-AT-V Cc: [EMAIL PROTECTED]; David Smith Subject: Re: [R] FW: [Fwd: Re: [S] Exact p-values] To understand the correct answer, you need to understand the following: pbinom(1, 2, .5) [1] 0.75 This is the binomial cumulative distribution function. *** pbinom(0, 2, .5) = 0.25 *** pbinom(1, 2, .5) = 0.75 = 0.25 + 0.5 *** pbinom(2, 2, .5) = 1 However, pbinom(1e15, 2e15, .5) is a computational challenge. Standard numerical algorithms often fail in situations like this. The code should test for such cases and use more numerically stable approximations in place of the exact algorithms. The standard deviation for a binomial is sqrt(p*(1-p)/n) = 0.5/sqrt(2e15), which is roughly 1e-8 in your case. I get the following from both S-Plus and R: pbinom(1e5+c(-1, 0, 1), 2e5, .5) [1] 0.4991079 0.5008921 0.5026762 For the problem you cite, the correct answer should be 0.5 to about 8 significant digits. Instead, I get 1 from R (as you did) and the following from S-Plus: pbinom(1e15,2e15,0.5) [1] 0.7411209 Both give wrong answers without warning, though in this case, S-Plus is closer. Answer the question? Spencer Graves # [EMAIL PROTECTED] wrote: Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] Original Message Subject: Re: [S] Exact p-values Date: Thu, 13 Feb 2003 18:31:38 +0100 From: Rau, Roland [EMAIL PROTECTED] To: 'Spencer Graves' [EMAIL PROTECTED], Jose María Fedriani Laffitte [EMAIL PROTECTED] CC: [EMAIL PROTECTED] Dear all, in relation to your question, the following working paper of Leo Knuesel, University of Munich, might be of interest: On the Accuracy of Statistical Distributions in S-Plus for Windows (1999) You can download the paper from (pdf-Format, 45k): http://www.stat.uni-muenchen.de/~knuesel/elv/accuracy.html Best, Roland -Original Message- From:Spencer Graves [SMTP:[EMAIL PROTECTED]] Sent:Thursday, February 13, 2003 6:12 PM To: Jose María Fedriani Laffitte Cc: [EMAIL PROTECTED] Subject: Re: [S] Exact p-values Try ( 1-pchisq(29.8, df=1)): With S-Plus 6.1, I got 4.78992e-008. By the way, the distribtion functions in R have more arguments. For example, pchisq(29.8, df=1, lower.tail=F) produces the same answer, and pchisq(29.8, df=1, lower.tail=F, log=T) produces its natural logarithm. Also, pchisq, dchisq, qchisq, and rchisq in R all have an ncp noncentrality parameter argument; only pchisq has such in S-Plus 6.1. Similarly, none of the Student's t functions in S-Plus have a non-centralitity parameter; in R, pt has an argument ncp, and from this one can easily program ncp for dt, qt and rt. Also, the distribution functions in the current release of S-Plus are known to have problems. For example, pt(-1, Inf) = 0.5 in S-Plus 6.1, but 0.159 in R; clearly, S-Plus gives a wrong answer without warning. Best Wishes, Spencer Graves Jose María Fedriani Laffitte wrote: Dear all, I want to get the exact p-values, on 1 degree of freedom, for an array of chi-square values. When my chi-square values are equal or lower than 29.7, I get the exact associated p-values. Thus, for instance: pchisq(29.7, df=1) [1] 0.999 However, when my chi-square values are greater or equal to 29.8 what I get is: pchisq(29.8, df=1) [1] 1 Could anyone tell me how to fix this trivial issue? Very grateful, Jose M. Fedriani Jose Mª Fedriani Laffitte Estacion Biologica de Donana (CSIC) Avda. Mª Luisa s/n 41013-Sevilla Spain Tel. +34-954232340 Fax +34-954621125 http://ebd.csic.es
[R] reading non-existent files
Dear R-experts I would like to read all files from a directory, the files have names myname0001.txt etc. I paste the directory plus file names and use read.delim(). My problem is that some file names are missing, so I get an error and my program stops. Is there a way to check for a null pointer analogous to C, so that I can simply skip non-existent filenames? Please do Reply to all since in the company I am not subscribed to the R-help list. Thank you in advance Best regards Christian Stratowa == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] reading non-existent files
Great, thank you for the fast reply Regards Christian Stratowa == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] -Original Message- From: Ben Bolker [SMTP:[EMAIL PROTECTED]] Sent: Tuesday, January 28, 2003 4:15 PM To: Stratowa,Dr,Christian FEX BIG-AT-V Cc: [EMAIL PROTECTED] Subject: Re: [R] reading non-existent files ?file.exists On Tue, 28 Jan 2003 [EMAIL PROTECTED] wrote: Dear R-experts I would like to read all files from a directory, the files have names myname0001.txt etc. I paste the directory plus file names and use read.delim(). My problem is that some file names are missing, so I get an error and my program stops. Is there a way to check for a null pointer analogous to C, so that I can simply skip non-existent filenames? Please do Reply to all since in the company I am not subscribed to the R-help list. Thank you in advance Best regards Christian Stratowa == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help -- 318 Carr Hall[EMAIL PROTECTED] Zoology Department, University of Floridahttp://www.zoo.ufl.edu/bolker Box 118525 (ph) 352-392-5697 Gainesville, FL 32611-8525 (fax) 352-392-3704 __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] reading non-existent files
Dear All Thank you all for your fast help, the solutions that I have received are: file.exists() try() allFiles-list.files(mydir) Best regards Christian Stratowa == Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead Discovery - Bioinformatics Dr. Boehringergasse 5-11 A-1121 Vienna, Austria Tel.: ++43-1-80105-2470 Fax: ++43-1-80105-2683 email: [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [SMTP:[EMAIL PROTECTED]] Sent: Tuesday, January 28, 2003 4:20 PM To: Stratowa,Dr,Christian FEX BIG-AT-V Cc: [EMAIL PROTECTED] Subject: Re: [R] reading non-existent files On Tue, 28 Jan 2003 [EMAIL PROTECTED] wrote: I would like to read all files from a directory, the files have names myname0001.txt etc. I paste the directory plus file names and use read.delim(). My problem is that some file names are missing, so I get an error and my program stops. How can some of `all files from a directory' be missing? Why not read them via e.g. list.files()? Is there a way to check for a null pointer analogous to C, so that I can simply skip non-existent filenames? ?file.exists -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
