> identical(as.list(x), xz) [1] TRUE Bill Dunlap TIBCO Software wdunlap tibco.com
On Mon, Oct 26, 2015 at 6:31 PM, Erin Hodgess <erinm.hodg...@gmail.com> wrote: > Hello! > > The following (which is a toy example) works fine, but I wonder if there is > a better or more elegant way than to do the loop: > > xz <- vector("list",length=4) > x <- 6:9 > for(i in 1:4)xz[[i]] <- x[i] > xz > [[1]] > [1] 6 > > [[2]] > [1] 7 > > [[3]] > [1] 8 > > [[4]] > [1] 9 > > This does exactly what I want, but the "for" loop seems out of place. > Maybe not. > > Thanks, > Sincerely > Erin > > > -- > Erin Hodgess > Associate Professor > Department of Mathematical and Statistics > University of Houston - Downtown > mailto: erinm.hodg...@gmail.com > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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 -- To UNSUBSCRIBE and more, see 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.