R help - May 2010 - Fast Matrix Computation

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

One way to do it:

apply(B, 1, function(x) t(apply(A, 1, function(y) abs(y-x) )) )





----- Original Message ----
> From: David Neu <david at davidneu.com>
> To: r-help at 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 (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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> href="mailto:R-help at r-project.org">R-help at r-project.org
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> href="https://stat.ethz.ch/mailman/listinfo/r-help" target=_blank
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and provide commented, 
> minimal, self-contained, reproducible code.

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
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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