[R] yet another vectorization question

[Gabor Grothendieck]

On 1/30/06, Patricia J. Hawkins <phawkins at connact.com> wrote:
> >>>>> "AD" == Adrian Dusa <adi at roda.ro> writes:
>
> AD> set.seed(5)
> AD> aa <- matrix(sample(10, 15, replace=T), ncol=5)
> AD> bb <- matrix(NA, ncol=10, nrow=5)
> AD> for (i in 1:ncol(aa)) bb[i, aa[, i]] <- c(0, 1, 0)
>
> AD> Is there any possibility to vectorize this "for" loop?
> AD> (sometimes I have hundreds of columns in the "aa" matrix)
>
> Well, coming from ignorance of R, I came up with the below.  However,
> it means creating another vector that's the size of aa, so it's not
> clear that it's a win:
>
> #Problem:  Indexing bb correctly when vectorized
> #Solution:  Add the following matrix to aa:
> #     [,1] [,2] [,3] [,4] [,5]
> #[1,]    0   10   20   30   40
> #[2,]    0   10   20   30   40
> #[3,]    0   10   20   30   40
> #
> # or its vector equivalent:
> #
> # rep(0:(ncol(aa)-1)*ncol(bb), each=nrow(aa))
> # > [1]  0  0  0 10 10 10 20 20 20 30 30 30 40 40 40
>
> bb <- matrix(1:50, ncol=10, nrow=5, byrow=TRUE)
> bv <- as.vector(bb)
> ai <- as.vector(aa) + rep(0:4*10, each=3)
> bv[ai] <- c(0,1,0)
> bb <- matrix(bv, ncol=10, nrow=5, byrow=TRUE)
> bb
>
> #which generalizes to:
>
> bb <- matrix(1:50, ncol=10, nrow=5, byrow=TRUE)
> bv <- as.vector(bb)
> ai <- as.vector(aa) + rep((1:nrow(aa)-1)*10, each=3)
> bv[ai] <- c(0,1,0)
> bb <- matrix(bv, ncol=10, nrow=5, byrow=TRUE)
> bb

Try this:

bb <- matrix(NA, ncol=10, nrow=5)
bb[cbind(c(col(aa)), c(aa))] <- c(0,1,0)