On 8/5/2005 12:43 PM, Uwe Ligges wrote: > Duncan Murdoch wrote: > >> On 8/5/2005 12:16 PM, Martin C. Martin wrote: >> >>>Hi, >>> >>>I have a 5x731 array A, and I want to compute the sums of the columns. >>>Currently I do: >>> >>>apply(A, 2, sum) >>> >>>But it turns out, this is slow: 70% of my CPU time is spent here, even >>>though there are many complicated steps in my computation. >>> >>>Is there a faster way? >> >> >> You'd probably do better with matrix multiplication: >> >> rep(1, nrow(A)) %*% A > > > No, better use colSums(), which has been optimized for this purpose: > > A <- matrix(seq(1, 10000000), ncol=10000) > system.time(colSums(A)) > # ~ 0.1 sec. > system.time(rep(1, nrow(A)) %*% A) > # ~ 0.5 sec.
I didn't claim my solution was the best, only better. :-) One point of interest: I think your example exaggerates the difference by using a matrix of integers. On my machine I get a ratio something like yours with the same example > A <- matrix(seq(1, 10000000), ncol=10000) > system.time(colSums(A)) [1] 0.08 0.00 0.08 NA NA > system.time(rep(1, nrow(A)) %*% A) [1] 0.25 0.01 0.23 NA NA but if I make A floating point, there's much less difference: > A <- matrix(as.numeric(seq(1, 10000000)), ncol=10000) > system.time(colSums(A)) [1] 0.09 0.00 0.09 NA NA > system.time(rep(1, nrow(A)) %*% A) [1] 0.11 0.00 0.12 NA NA Still, colSums is the winner in both cases. Duncan Murdoch ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html