aov() will handle multiple responses and that would be considerably more
efficient than running separate fits as you seem to be doing.
Your code is nigh unreadable: please use your spacebar and remove the
redundant semicolons: `Writing R Extensions' shows you how to tidy up
your code to make it presentable. But I think anova_[[1]] is really
coef(summary(aov_)) which is a lot more intelligible.
On Thu, 9 Aug 2007, Francoise PFIFFELMANN wrote:
Hi,
Im trying to replace some SAS statistical functions by R (batch calling).
But Ive seen that calling R in a batch mode (under Unix) takes about 2or 3
times more than SAS software. So its a great problem of performance for me.
Here is an extract of the calculation:
stoutput<-file("res_oneWayAnova.dat","w");
cat("Param|F|Prob",file=stoutput,"\n");
for (i in 1:n) {
p<-list_param[[i]]
aov_<-aov(A[,p]~ A[,"wafer"],data=A);
anova_<-summary(aov_);
if (!is.na(anova_[[1]][1,5]) & anova_[[1]][1,5]<=0.0001)
res_aov<-cbind(p,anova_[[1]][1,4],"<0.0001") else
res_aov<-cbind(p,anova_[[1]][1,4],anova_[[1]][1,5]);
cat(res_aov, file=stoutput, append = TRUE,sep = "|","\n");
};
close(stoutput);
A is a data.frame of about (400 lines and 1800 parameters).
Im a new user of R and I dont know if its a problem in my code or if
there are some tips that I can use to optimise my treatment.
Thanks a lot for your help.
Françoise Pfiffelmann
Engineering Data Analysis Group
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Tel: +33 438 92 29 84
Email: [EMAIL PROTECTED]
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
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