[R] Speed up sum of outer products?
Doran, Harold
HDoran at air.org
Tue Mar 1 18:43:16 CET 2011
Isn't the following the canonical (R-ish) way of doing this:
X = matrix(rnorm(1000*50),1000,50)
system.time({C1 = t(X) %*% X}) # Phil's example
C2 <- crossprod(X) # use crossprod instead
> all.equal(C1,C2)
[1] TRUE
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
> Behalf Of Phil Spector
> Sent: Tuesday, March 01, 2011 12:31 PM
> To: AjayT
> Cc: r-help at r-project.org
> Subject: Re: [R] Speed up sum of outer products?
>
> What you're doing is breaking up the calculation of X'X
> into n steps. I'm not sure what you mean by "very slow":
>
> > X = matrix(rnorm(1000*50),1000,50)
> > n = 1000
> > system.time({C=matrix(0,50,50);for(i in 1:n)C = C + (X[i,] %o% X[i,])})
> user system elapsed
> 0.096 0.008 0.104
>
> Of course, you could just do the calculation directly:
>
> > system.time({C1 = t(X) %*% X})
> user system elapsed
> 0.008 0.000 0.007
> > all.equal(C,C1)
> [1] TRUE
>
>
> - Phil Spector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> spector at stat.berkeley.edu
>
>
>
> On Tue, 1 Mar 2011, AjayT wrote:
>
> > Hi, I'm new to R and stats, and I'm trying to speed up the following sum,
> >
> > for (i in 1:n){
> > C = C + (X[i,] %o% X[i,]) # the sum of outer products - this is very
> slow
> > according to Rprof()
> > }
> >
> > where X is a data matrix (nrows=1000 X ncols=50), and n=1000. The sum has to
> > be calculated over 10,000 times for different X.
> >
> > I think it is similar to estimating a co-variance matrix for demeaned data
> > X. I tried using cov, but got different answers, and it was'nt much quicker?
> >
> > Any help gratefully appreciated,
> >
> > --
> > View this message in context: http://r.789695.n4.nabble.com/Speed-up-sum-of-
> outer-products-tp3330160p3330160.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > ______________________________________________
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> >
>
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