[R] Is there a faster way to do this?
Peter Alspach
Peter.Alspach at plantandfood.co.nz
Fri Mar 26 02:38:13 CET 2010
Tena koe Marcio
Seems like you are simply multiplying transpose b by b and replacing the diagonal with 0. If this is correct, then use
a <- t(b) %*% b
diag(a) <- 0
If this is not a correct interpretation of what you are trying to do, could you show us with a small reproducible example.
HTH .....
Peter Alspach
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Márcio Resende
> Sent: Friday, 26 March 2010 2:15 p.m.
> To: r-help at r-project.org
> Subject: [R] Is there a faster way to do this?
>
>
> Hi guys, I am still learning R, and not well familiar with all the
> apply
> functions.
> I am trying to find faster alternatives to replace the for cycle.
> Is there a faster way to do the example below?
>
> nm <- 1000
> b <- matrix (rnorm (5000, 0, 1), nrow = 500, ncol = nm)
> a <- matrix (0, nm, nm)
> for (i in 1 : nm) {
> for (j in 1 : nm) {
> if ( j == i) {
> next }
> a[i, j] <- t (b [, i]) %*% b[, j]
> }
> }
>
> thanks
>
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> way-to-do-this-tp1691601p1691601.html
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>
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