lapply will do the trick, try something like lapply(1:length(dat1), function(x,dat1,dat2) cbind(dat1[[x]][,1:2], dat2[[x]]),dat1,dat2)
HTH Schalk On Mon, Mar 22, 2010 at 10:18 AM, Ron_M <ron_michae...@yahoo.com> wrote: > > Dear all, > > I have following two list object, both are basically collection of matrices > : > > dat1 <- matrix(rnorm(25*6), ncol=6) > dat1 <- split(dat1, seq(5,25,by=5)) > dat1 <- lapply(dat1, matrix, ncol=6) > > dat2 <- matrix(rnorm(25*4), ncol=4) > dat2 <- split(dat2, seq(5,25,by=5)) > dat2 <- lapply(dat2, matrix, ncol=4) > > > Now I want to replace last 4 columns of each matrix at "dat1" with the > corresponding matrix in "dat2". However I want to avoid the time consuming > loop to do that. Is there any way to do that without using loop? > > Thanks > -- > View this message in context: > http://n4.nabble.com/Replacing-elements-of-list-tp1677293p1677293.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.
