Hello,
I work on large matrices and found something interesting. For multiplication of matrices, the order has a huge influence on computing time when one of them is a sparse matrix. In the below example, M is a full matrix and A is a sparse matrix in a regular matrix class. A %*% M takes much more time than M %*% A; moreover, t(t(M) %*% t(A)) is much faster than A %*% M with same result. I would like to know how it is possible. Even though I do not have an exact reason, this fact may be helpful to others. Thanks, Yongwan > n <- 1000 > M <- diag(n) - matrix(1/n,n,n) > A <- matrix(rnorm(n*n)>2,n,n) > system.time(M %*% A) [1] 0.10 0.03 0.12 NA NA > system.time(A %*% M) [1] 3.47 0.03 3.50 NA NA > system.time(t(t(M) %*% t(A))) [1] 0.23 0.00 0.23 NA NA ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
