Dear Olivella, Assuming that x is your data, here is one possible way to do what you want:
t( apply(x, 1, function(x){ index <- x %in% c(1,2) x[index] <- cls[index] x[!index] <- NA x } ) ) How big is the data you are dealing with? HTH, Jorge On Wed, Jul 8, 2009 at 4:25 PM, Olivella <olive...@wustl.edu> wrote: > > Hello, > > I wish to perform a substitution of certain numerical values in a data > matrix with the corresponding column name. For instance, if I have a data > matrix > V1 V2 V3 > 2 0 1 > 0 1 2 > 1 5 0 > 5 0 0 > > I want to substitute the `1' and the `2' for the corresponding column name, > and make everything else `NA' like this > V1 V2 V3 > V1 NA V3 > NA V2 V3 > V1 NA NA > NA NA NA > > I have done this using an explicit `for' loop, but it takes a really long > time to finish. Is there any way I can do this using `apply' or some form > of > implicit looping? > > Thank you for your help, > > SO > -- > View this message in context: > http://www.nabble.com/Substituting-numerical-values-using-%60apply%27-tp24398687p24398687.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. > [[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.