will this do what you want: > f.rep <- function(x, times){ + # make sure the 'x' is long enough + x <- head(rep(x, length(times)), length(times)) + rep(x, times) + } > > f.rep(c(0,1), c(3,4,5,6,7)) [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 >
On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding <[EMAIL PROTECTED]> wrote: > Hi Folks, > I'm wondering if there's a compact way to achieve the > following. The "dream" is that, by analogy with > > rep(c(0,1),times=c(3,4)) > # [1] 0 0 0 1 1 1 1 > > one could write > > rep(c(0,1),times=c(3,4,5,6)) > > which would produce > > # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 > > in effect "recycling" x through 'times'. > > The objective is to produce a vector of alternating runs of > 0s and 1s, with the lengths of the runs supplied as a vector. > Indeed, more generally, something like > > rep(c(0,1,2), times=c(1,2,3,2,3,4)) > # [1] 0 1 1 2 2 2 0 0 1 1 1 2 2 2 2 > > Suggestions appreciated! With thanks, > Ted. > > -------------------------------------------------------------------- > E-Mail: (Ted Harding) <[EMAIL PROTECTED]> > Fax-to-email: +44 (0)870 094 0861 > Date: 20-Oct-08 Time: 21:57:15 > ------------------------------ XFMail ------------------------------ > > ______________________________________________ > 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. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? ______________________________________________ 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.