That's a very creative way to get the results! I haven't used the "table" call before, but that certainly made it much easier. Thanks!
Gabor Grothendieck wrote: > > The first line was missing. It should be: > > mat <- do.call(rbind, a) > sapply(apply(mat, 2, table), function(x) mean(as.numeric(names(x))[x > == length(a)])) > > > On Jan 11, 2008 7:23 PM, Gabor Grothendieck <[EMAIL PROTECTED]> > wrote: >> Assume that no component matrix can have repeated values. >> >> First we apply table to mat yielding a list of named counts. >> To each component of that we sapply the indicated function >> which picks out those that occur length(a) times (so they are >> in every column), converts the name to numeric and averages >> them. >> >> sapply(apply(mat, 2, table), function(x) mean(as.numeric(names(x))[x >> == length(a)])) >> >> By the way, note that the a you define is slightly different from the >> a displayed in your post. >> >> >> On Jan 11, 2008 6:09 PM, dxc13 <[EMAIL PROTECTED]> wrote: >> > >> > useR's, >> > >> > I want to match the real number elements of a list that has 3 matrices >> as >> > its elements and then average those numbers. I think I am close, but I >> > can't get it to quite work out. For example, >> > >> > > a <- list(matrix(c(10,NA,NA,12,11, >> > > 10,13,NA,14,12),ncol=2),matrix(c(10,12,15,13,11, >> > > 13,NA,NA,12,10),ncol=2),matrix(c(10,15,NA,13,NA, >> 13,12,NA,NA,10),ncol=2)) >> > > a >> > [[1]] >> > [,1] [,2] >> > [1,] 10 10 >> > [2,] NA 13 >> > [3,] NA NA >> > [4,] 13 14 >> > [5,] 11 12 >> > >> > [[2]] >> > [,1] [,2] >> > [1,] 10 13 >> > [2,] 12 NA >> > [3,] 15 NA >> > [4,] 13 12 >> > [5,] 11 10 >> > >> > [[3]] >> > [,1] [,2] >> > [1,] 10 13 >> > [2,] 15 12 >> > [3,] NA NA >> > [4,] 13 NA >> > [5,] NA 10 >> > >> > I want to average the non-NA numbers that simultaneously appear in each >> > column of all 3 matrices. For the first column, 10 and 13 are the only >> > numbers that appear in all 3 matrices. Thus, I want to take the >> average of >> > 10 and 13. Then repeat the same process for the second column. The >> data >> > here is only a sample, but with the data I work with, there are many >> more >> > columns in each matrix in the list. Does anyone know a efficient way >> to do >> > this? Maybe using lapply()? Thanks in advance. >> > >> > dxc13 >> > -- >> > View this message in context: >> http://www.nabble.com/matching-values-in-a-list-tp14767170p14767170.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. >> > >> > > ______________________________________________ > 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/matching-values-in-a-list-tp14767170p14768583.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.
