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
spec...@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,
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