# Re: [R] Matrix multiplication using apply() or lappy() ?

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Wed 06 Sep 2006 - 17:21:36 GMT

In terms of speed Toby's original idea was actually the fastest. Here they are decreasing order of the largest timing in each row of system.time. I also tried it with a 100x10 matrix and got almost the same order:

> library(reshape)
> system.time(for(i in 1:1000) iapply(a, 1, "/", a[1,]))

[1] 11.51 0.01 18.65 NA NA
> system.time(for(i in 1:1000) t(apply(a, 1, "/", a[1,])))
[1] 0.83 0.00 1.36 NA NA
> system.time(for(i in 1:1000) sweep(a, 2, a[1,], "/"))
[1] 0.27 0.00 0.39 NA NA
> system.time(for(i in 1:1000) a/outer(rep(1, nrow(a)), a[1,]))
[1] 0.23 0.00 0.39 NA NA
> system.time(for(i in 1:1000) a %*% diag(1/a[1,]))
[1] 0.25 0.00 0.38 NA NA
> system.time(for(i in 1:1000) a/rep(a[1,], each = nrow(a)))

[1] 0.09 0.00 0.16 NA NA
> system.time(for(i in 1:1000) t(t(a)/a[1,]))
[1] 0.10 0.00 0.13 NA NA
> system.time(for(i in 1:1000) a/matrix(a[1,], nrow(a), ncol(a), byrow = TRUE))
[1] 0.05 0.00 0.12 NA NA

> On 9/6/06, Rolf Turner <rolf@erdos.math.unb.ca> wrote:
> Prof. Brian Ripley wrote:
>
> > On Wed, 6 Sep 2006, Christos Hatzis wrote:
> >
> > > See ?sweep
> > >
> > > sweep(a, 2, a[1,],"/")
> >
> > That is less efficient than
> >
> > a/rep(a[1,], each=nrow(a))
>
> *My* first instinct was to use
>
> t(t(a)/a[1,])
>
> (which has not heretofore been suggested).
>
> This seems to be more efficient still (at least in respect of Prof.
> Grothendieck's toy example) by between 20 and 25 percent:
>
> > a <- matrix(1:24,4)

> > system.time(for(i in 1:1000) junk <- a / rep(a[1,], each = 4))

> [1] 0.690 0.080 1.051 0.000 0.000
> > system.time(for(i in 1:1000) junk <- t(t(a)/a[1,]))
> [1] 0.520 0.120 0.647 0.000 0.000
> > system.time(for(i in 1:10000) junk <- a / rep(a[1,], each = 4))
> [1] 7.08 0.99 10.08 0.00 0.00
> > system.time(for(i in 1:10000) junk <- t(t(a)/a[1,]))
> [1] 5.530 0.940 7.856 0.000 0.000
>
> cheers,
>
> Rolf Turner
>
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