Thanks to everyone who responded!
On Sun, May 23, 2010 at 1:32 AM, Jorge Ivan Velez
wrote:
> Yes, it should be. Thank you for pointing that out. Apologies for the noise.
>
> Regards,
> Jorge
>
>
> On Sun, May 23, 2010 at 1:07 AM, Berend Hasselman <> wrote:
>
>>
>>
>> Jorge Ivan Velez wrote:
>> >
Yes, it should be. Thank you for pointing that out. Apologies for the noise.
Regards,
Jorge
On Sun, May 23, 2010 at 1:07 AM, Berend Hasselman <> wrote:
>
>
> Jorge Ivan Velez wrote:
> >
> > # The same using a function foo
> > foo <- function(A, B){
> > rA <- 1:nrow(A)
> > rB <- 1:nrow(B)
>
Jorge Ivan Velez wrote:
>
> # 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 be
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 <- structur
One possibility:
n.A<-nrow(A)
n.B<-nrow(B)
abs( kronecker(A,rep(1,n.B)) - kronecker(rep(1,n.A),B) )
-tgs
On Sat, May 22, 2010 at 3:20 PM, Shi, Tao wrote:
> One way to do it:
>
> apply(B, 1, function(x) t(apply(A, 1, function(y) abs(y-x) )) )
>
>
>
>
>
> - Original Message
> > From: D
One way to do it:
apply(B, 1, function(x) t(apply(A, 1, function(y) abs(y-x) )) )
- Original Message
> From: David Neu
> To: r-help@r-project.org
> Sent: Sat, May 22, 2010 10:35:32 AM
> Subject: [R] Fast Matrix Computation
>
> Hi,
I have two (large) matrices A and B of dimensions
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