Transposing is fine! Thanks for that! On Sunday, 20 July 2014, Odd Andersen <odd.ander...@gmail.com> wrote:
> Sparse matrices in Julia are to my understanding stored as compressed > sparse columns. So it is very easy to get the nonzero elements for a given > column, but not so easy for rows. > > To get the indices of nonzeros rows for column 'c' in sparse matrix M, one > can use (at least in the current implementation): > > M.rowval[a.colptr[col] : M.colptr[col+1]-1] > > To do the same by rows would be more complicated (a quick-and-dirty > solution would of course be to transpose your matrix first). > I am however not a Julia expert, so perhaps there's a solution I am not > aware of. > >
