It worked. But this student of R is left wondering why or how. In particular the construction <foo>[.][.] was puzzling. I doubt that matters greatly, but I am using R in both a Mac (OSX 10.2 v 2.0.1) and WinXP (v 2.4.1.)
I have broken it into what I think are its pieces: > table(my.mat)[-1] my.mat 1 2 9 4 # a count of the non-zero elements > table(my.mat)[-1][my.mat] my.mat 1 1 1 1 1 1 1 1 1 2 2 2 2 9 9 9 9 9 9 9 9 9 4 4 4 4 I am having some trouble figuring out how the second index, the matrix, my.mat, is being handled by the table object ... if that is what is happening after the semantic parsing. I have searched for material on "subscripting" and found a rather small amount, but after looking the help file, I see that I should have been looking for "indexing" and "Extract". It appears that "[.]" is an operation that is represented internally by Extract(.) and Extract both extracts and replaces. Most of the discussions of "recursive indexing" I found were directed at the "[[" operation, but is this construct also recursive indexing? -- Sincerely; David Winsemius Gabor Grothendieck wrote: > Try this: > > my.mat[my.mat > 0] <- table(my.mat)[-1][my.mat] > > > On Nov 10, 2007 8:19 AM, Milton Cezar Ribeiro <[EMAIL PROTECTED]> wrote: > >> Hi R-gurus, >> >> I have a matrix which looks like >> >> 00000000000 >> 01110000220 >> 01110000220 >> 01110000000 >> 00000000000 >> >> As you can see we have non-zero levels 1 and 2. I would like to fill an >> other matrix with the frequency of non-zero levels in each cell. The results >> that I need is >> >> 00000000000 >> 09990000440 >> 09990000440 >> 09990000000 >> 00000000000 >> >> If I run the script below I can simulate the first matrix and count the >> cells for each non-zero levels. My question is how can I fill the second >> matrix in a easy way. >> >> my.mat<-matrix( >> c(0,0,0,0,0,0,0,0,0,0,0, >> 0,1,1,1,0,0,0,0,2,2,0, >> 0,1,1,1,0,0,0,0,2,2,0, >> 0,1,1,1,0,0,0,0,0,0,0, >> 0,0,0,0,0,0,0,0,0,0,0),nrow=5,byrow=T) >> my.mat.freq<-data.frame(table(my.mat)) >> my.mat.freq<-subset(my.mat.freq,my.mat.freq$my.mat!=0) >> my.mat.freq >> >> >> Any idea? >> >> Kind regards >> >> Miltinho >> >> >> >> para armazenamento! >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> >> > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ R-help@r-project.org mailing list 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.
