Hi:
Another alternative is
crossprod(A)
which is meant to produce an optimized A'A (not the vector cross-product
from intro physics :)
Example:
A <- matrix(rpois(9, 10), ncol = 3)
> A
[,1] [,2] [,3]
[1,] 6 10 14
[2,] 7 5 16
[3,] 12 16 10
> t(A) %*% A
[,1] [,2] [,3]
[1,] 229 287 316
[2,] 287 381 380
[3,] 316 380 552
> crossprod(A)
[,1] [,2] [,3]
[1,] 229 287 316
[2,] 287 381 380
[3,] 316 380 552
A little test:
f <- function(A) {
U <- crossprod(A)
diag(U) <- 0
U
}
g <- function(A) {
U <- t(A) %*% A
diag(U) <- 0
U
}
# matrix multiplication on a 100 x 10 matrix
> system.time(replicate(1000, g(matrix(rpois(1000, 10), ncol = 10))))
user system elapsed
0.24 0.00 0.23
# crossprod
> system.time(replicate(1000, f(matrix(rpois(1000, 10), ncol = 10))))
user system elapsed
0.20 0.00 0.21
# matrix multiplication on a 10 x 100 matrix
> system.time(replicate(1000, g(matrix(rpois(1000, 10), ncol = 100))))
user system elapsed
1.72 0.17 1.89
# crossprod
> system.time(replicate(1000, f(matrix(rpois(1000, 10), ncol = 100))))
user system elapsed
0.47 0.09 0.56
HTH,
Dennis
On Thu, Nov 11, 2010 at 5:19 PM, Michael Bedward
<michael.bedw...@gmail.com>wrote:
> On 12 November 2010 02:21, David Winsemius <dwinsem...@comcast.net> wrote:
> >
> >> The fastest and easiest solution is
> >>
> >> t(A) %*% A
> >
> > That is really elegant. (Wish I could remember my linear algebra lessons
> as
> > well from forty years ago.) I checked it against the specified output and
> > found that with one exception that the OP had planned for the diagonal to
> be
> > filled with zeroes. So that could be completed by a simple modification:
> >
> > temp <- t(A) %*% A
> > diag(temp) <- 0
> > temp
> >
>
> Excellent solution ! Small is beautiful :)
>
> Michael
>
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