[R] ? extended rep()

[jim holtman]

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
<Ted.Harding at manchester.ac.uk> 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) <Ted.Harding at manchester.ac.uk>
> Fax-to-email: +44 (0)870 094 0861
> Date: 20-Oct-08                                       Time: 21:57:15
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?