Hi David,
Here are two suggestions, one that works for the example you provided and a
second one for a more general case.
# Your example
A <- matrix(c(1, 2, 3, 4, 5, 6), byrow=TRUE, ncol=3)
B <- matrix(c(7, 8, 9), byrow=TRUE, ncol=3)
t(abs(t(A) - as.vector(B)))
# More general case
B1 <- structure(c(7, 8, 10, 11, 8, 9, 11, 12, 9, 10, 12, 13), .Dim =
c(4L, 3L))
rA <- 1:nrow(A)
rB <- 1:nrow(B1)
grid <- as.matrix(expand.grid(rA, rB))
t(apply(grid, 1, function(x) abs(A[x[1], ] - B1[x[2], ])))
# The same using a function foo
foo <- function(A, B){
rA <- 1:nrow(A)
rB <- 1:nrow(B)
grid <- as.matrix(expand.grid(rA, rB))
t(apply(grid, 1, function(x) abs(A[x[1], ] - B1[x[2], ])))
}
foo(A, B)
foo(A, B1)
As usual, there might be better and faster ways to do this.
HTH,
Jorge
On Sat, May 22, 2010 at 1:35 PM, David Neu <> wrote:
> Hi,
>
> I have two (large) matrices A and B of dimensions (m,n) and (p,n)
> respectively.
>
> I'd like to see if the is a fast way to compute a new matrix C with
> dimension (m*p,n) in which each row in C is found by applying some
> function f to each pair of rows (x,y) where x is a row in A and y is a
> row in B.
>
> For example, if
>
> A <- matrix(c(1, 2, 3, 4, 5, 6), byrow=TRUE, ncol=3)
> B <- matrix(c(7, 8, 9), byrow=TRUE, ncol=3)
>
> and f <- function(x,y) abs(y-x)
>
> then
>
> C <- matrix(c(6, 6, 6, 3, 3, 3), byrow=TRUE, ncol=3)
>
> Many thanks!
>
> Cheers,
> David
>
> ______________________________________________
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