On 25-Mar-10 21:10:17, Jeff Brown wrote:
> I just had to solve this problem for myself, after not having
> luck with the code posted above. I'm posting in case others
> need a completely general function.
>
> riffle <- function (a,b) {
> # Interleave a & b, starting with a, without repeating.
> x <- NULL; count = 1;
> for (i in 1:max(length(a), length(b))) {
> if (i <= length(a)) {
> x[count] <- a[i];
> count = count+1;
> };
> if (i <= length(b)) {
> x[count] <- b[i];
> count = count+1;
> }
> };
> x
> };
> riffle( 1:10, 50:55 )
The following avoids the 'for' loop, though I'd like to simplify
the conditionals:
riffle2 <- function(x,y){
m0 <- length(x); n0 <- length(y)
m <- min(m0,n0); n <- max(m0,n0)
z <- numeric(m+n)
if(m0==n0){ z[(-1+2*(1:m))] <- x ; z[2*(1:m)] <- y } else
if(m0<n0){
z[(-1+2*(1:m))] <- x
z[2*(1:m)] <- y[1:m]
z[(2*m+1):(m+n)] <- y[(m+1):n]
} else {
z[(2*(1:m))] <- y
z[-1+2*(1:m)] <- x[1:m]
z[(2*m+1):(m+n)] <- x[(m+1):n]
}
z
}
riffle2((1:10),(50:55))
# [1] 1 50 2 51 3 52 4 53 5 54 6 55 7 8 9 10
riffle2((50:55),(1:10))
# [1] 50 1 51 2 52 3 53 4 54 5 55 6 7 8 9 10
riffle2((1:7),(8:14))
# [1] 1 8 2 9 3 10 4 11 5 12 6 13 7 14
Ted.
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E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 25-Mar-10 Time: 22:00:38
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