On Sat, Apr 12, 2008 at 01:30:13PM -0400, Vincent Goulet wrote:
> Le sam. 12 avr. à 12:47, carlos martinez a écrit :
> >> Looking for a simple, effective a minimum execution time solution.
> >>
> >> For a vector as:
> >>
> >> c(0,0,1,0,1,1,1,0,0,1,1,0,1,0,1,1,1,1,1,1)
> >>
> > To transform it to the following vector without using any loops:
> >
> >> (0,0,1,0,1,2,3,0,0,1,2,0,1,0,1,2,3,4,5,6)
> >>
> > Appreciate any suggetions.
>
> This does it -- but it is admittedly ugly:
>
> > x <- c(0,0,1,0,1,1,1,0,0,1,1,0,1,0,1,1,1,1,1,1)
> > ind <- which(x == 0)
> > unlist(lapply(mapply(seq, ind, c(tail(ind, -1) - 1, length(x))),
> function(y) cumsum(x[y])))
> [1] 0 0 1 0 1 2 3 0 0 1 2 0 1 0 1 2 3 4 5 6
>
> (The mapply() part is used to create the indexes of each sequence in x
> starting with a 0. The rest is then straightforward.)
Here's my effort. Maybe a bit easier to digest? Only one *apply so probably
more efficient.
function(x=c(0,0,1,0,1,1,1,0,0,1,1,0,1,0,1,1,1,1,1,1)) {
d <- diff(c(0,x,0))
starts <- which(d == 1)
ends <- which(d == -1)
x[x == 1] <- unlist(lapply(ends - starts, function(n) 1:n))
x
}
Dan
>
> HTH
>
> ---
> Vincent Goulet, Associate Professor
> École d'actuariat
> Université Laval, Québec
> [EMAIL PROTECTED] http://vgoulet.act.ulaval.ca
>
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