Sara wins on memory use.
Rui wins on speed.
Bert wins on clarity.
library(microbenchmark)
N <- 1000
x <- matrix( runif( N*N ), ncol=N )
y <- seq.int( N )
microbenchmark( { t( y * t(x) ) }
, { x %*% diag( y ) }
, { sweep( x, 2, y, `*` ) }
)
Unit: milliseconds
expr min lq median uq max
neval
{ t(y * t(x)) } 6.659562 7.475414 7.871341 8.182623 47.01105
100
{ x %*% diag(y) } 9.859292 11.014021 11.281334 11.733825 48.79463
100
{ sweep(x, 2, y, `*`) } 16.535938 17.682175 18.283572 18.712342 55.47159
100
On Fri, 4 Nov 2016, Dimitri Liakhovitski wrote:
Nice!
Thanks a lot, everybody!
Dimitri
On Fri, Nov 4, 2016 at 10:35 AM, Bert Gunter <bgunter.4...@gmail.com> wrote:
My goodness!
x %*% diag(y)
[,1] [,2]
[1,] 2 12
[2,] 4 15
[3,] 6 18
will do.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Nov 3, 2016 at 2:33 PM, Sarah Goslee <sarah.gos...@gmail.com> wrote:
Like this?
sweep(x, 2, y, "*")
[,1] [,2]
[1,] 2 12
[2,] 4 15
[3,] 6 18
On Thu, Nov 3, 2016 at 5:05 PM, Dimitri Liakhovitski
<dimitri.liakhovit...@gmail.com> wrote:
Hello!
I have a matrix x and a vector y:
x <- matrix(1:6, ncol = 2)
y <- c(2,3)
I need to multiply the first column of x by 2 (y[1]) and the second
column of x by 3 (y[2]).
Of course, I could do this - but it's column by column:
x[,1] <- x[,1] * y[1]
x[,2] <- x[,2] * y[2]
x
Or I could repeat each element of y and multiply two matrices - that's better:
rep.row<-function(x,n){
matrix(rep(x,each=n),nrow=n)
}
y <- rep.row(y, nrow(x))
x * y
However, maybe there is a more elegant r-like way of doing it?
Thank you!
--
Dimitri Liakhovitski
--
Sarah Goslee
http://www.functionaldiversity.org
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--
Dimitri Liakhovitski
______________________________________________
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