Hi, first off, thanks for the suggestion. I managed to solve it by doing:
IND = rep(c(T,T,F,F), 5) X = rep(NA, 20) X[IND] = 1:10 X[!IND] = 1:10 which avoids any function -- I think mapply, apply etc call a for loop internally, which I'd rather avoid. BW F On 11 Nov 2013, at 12:35, andrija djurovic <djandr...@gmail.com> wrote: > Hi. Here are two approaches: > > c(mapply(function(x,y) rep(c(x,y), 2), (1:10)[c(T,F)], (1:10)[c(F,T)])) > > c(tapply(1:10, rep(1:(10/2), each=2), rep, 2), recursive=T) > > Andrija > > > > > > On Mon, Nov 11, 2013 at 1:11 PM, Federico Calboli <f.calb...@imperial.ac.uk> > wrote: > Hi All, > > I am trying to create an index that returns something like > > 1,2,1,2,3,4,3,4,5,6,5,6,7,8,7,8 > > and so on and so forth until a predetermined value (which is obviously even). > I am trying very hard to avoid for loops or for loops front ends. > > I'd be obliged if anybody could offer a suggestion. > > BW > > F > > ______________________________________________ > 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. > >
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______________________________________________ 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.
