On Tue, Dec 20, 2011 at 8:20 AM, Jean V Adams <jvad...@usgs.gov> wrote:
Hi Jean, > khai wrote on 12/19/2011 11:26:55 PM: > >> Hi, >> >> I'm very new to working with sparse matrices and would like to know how > I >> can column permute a sparse matrix. Here is a small example: >> >> > M1 <- >> > spMatrix(nrow=5,ncol=6,i=sample(5,15,replace=TRUE),j=sample(6, >> 15,replace=TRUE),x=round_any(rnorm(15,2),0.001)) >> > M1 >> 5 x 6 sparse Matrix of class "dgTMatrix" >> >> [1,] 2.983 . 1.656 5.003 . . >> [2,] . . 2.990 . . . >> [3,] . 0.592 5.349 1.115 . . >> [4,] 1.836 . 2.804 . . . >> [5,] . 6.961 . . . 1.077 >> >> I know I can permute entire columns this way >> >> > M1[,sample(6,6)] >> 5 x 6 sparse Matrix of class "dgTMatrix" >> >> [1,] 5.003 . . . 1.656 2.983 >> [2,] . . . . 2.990 . >> [3,] 1.115 0.592 . . 5.349 . >> [4,] . . . . 2.804 1.836 >> [5,] . 6.961 1.077 . . . >> >> But I would like the new sparse matrix to look like this...where only > the >> nonzero elements are permuted. >> >> [1,] 1.656 . 5.003 2.983 . . >> [2,] . . 2.990 . . . >> [3,] . 5.349 1.115 0.592 . . >> [4,] 2.804 . 1.836 . . . >> [5,] . 1.077 . . . 6.961 >> >> Thanks in advance for any advice! A peculiar request but you can do that by permuting the 'x' slot in your original matrix. > set.seed(1) > (M1 <- spMatrix(nrow=5,ncol=6,i=sample(5,15,replace=TRUE),j=sample(6, > 15,replace=TRUE),x=round(rnorm(15,2),3))) 5 x 6 sparse Matrix of class "dgTMatrix" [1,] . . 2.738 . 3.595 . [2,] . . . . . . [3,] 3.879 . 4.289 -0.215 1.374 . [4,] . 4.966 1.180 1.379 . 2.487 [5,] 3.512 . . . 2.330 . > (nnz <- nnzero(M1)) [1] 15 > M2 <- M1 > M2@x <- M2@x[sample(nnz, nnz)] > M2 5 x 6 sparse Matrix of class "dgTMatrix" [1,] . . 2.487 . 3.595 . [2,] . . . . . . [3,] 3.709 . 2.875 3.512 -0.215 . [4,] . 3.764 1.164 2.576 . 3.125 [5,] 2.184 . . . 2.738 . > > I don't have experience with sparse matrices, but I was able to get this > to work by converting the sparse matrix to a "base" matrix and back again. > > > library(Matrix) > > nonzero.cols <- !apply(M1==0, 2, all) > M2 <- as.matrix(M1) > reord <- sample(seq(dim(M1)[2])[nonzero.cols]) > M2[, nonzero.cols] <- as.matrix(M1[, reord]) > Matrix(M2, sparse=TRUE) This solution has the disadvantage of converting the sparse matrix to a dense matrix, which may end up producing a much larger object. ______________________________________________ R-help@r-project.org 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.