[R] Matrix multiplication using apply() or lappy() ?
Prof Brian Ripley
ripley at stats.ox.ac.uk
Wed Sep 6 19:30:31 CEST 2006
What version of R was this?
In 2.4.0 alpha
> a <- matrix(1:24,4)
> system.time(for(i in 1:1000) junk <- a / rep(a[1,], each = 4))
[1] 0.014 0.000 0.014 0.000 0.000
> system.time(for(i in 1:1000) junk <- t(t(a)/a[1,]))
[1] 0.057 0.000 0.058 0.000 0.000
shows a large margin the other way, which increases with bigger matrices
> a <- matrix(pi*1:100, 100, 1000)
> system.time(for(i in 1:1000) junk <- t(t(a)/a[1,]))
[1] 18.329 2.238 20.595 0.000 0.000
> system.time(for(i in 1:1000) junk <- a / rep(a[1,], each = 4))
[1] 2.589 1.021 3.610 0.000 0.000
On Wed, 6 Sep 2006, Rolf Turner 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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--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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