On 20-Oct-08 21:19:21, Stefan Evert wrote: > On 20 Oct 2008, at 22:57, (Ted Harding) wrote: >> I'm wondering if there's a compact way to achieve the >> following. The "dream" is that 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'. > > rep2 <- function (x, times) rep(rep(x, length.out=length(times)), > times) > > rep2(c(0,1),times=c(3,4,5,6)) > [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 > > Any prizes for shortest solution? ;-) > > Best, > Stefan
If ever we are both within reach of 'en øl', then yes. But Gabor came up with a shorter one. I tried to shorten Gabor's but failed. However, all competitors are entitled to a consolation prize! (And that includes me ... ) Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 20-Oct-08 Time: 22:59:16 ------------------------------ 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.
