Robert Kern wrote: > On Wed, Apr 29, 2009 at 16:19, Dan Goodman <dg.gm...@thesamovar.net> wrote: >> Robert Kern wrote: >>> On Wed, Apr 29, 2009 at 08:03, Daniel Yarlett <daniel.yarl...@gmail.com> >>> wrote: >>> >>>> As you can see, Current is different in the two cases. Any ideas how I >>>> can recreate the behavior of the iterative process in a more numpy- >>>> friendly, vectorized (and hopefully quicker) way? >>> Use bincount(). >> Neat. Is there a memory efficient way of doing it if the indices are >> very large but there aren't many of them? e.g. if the indices were >> I=[10000, 20000] then bincount would create a gigantic array of 20000 >> elements for just two addition operations! > > indices -= indices.min() >
Ah OK, but bincount is still going to make an array of 10000 elements in that case... I came up with this trick, but I wonder whether it's overkill: Supposing you want to do y+=x[I] where x is big and indices in I are large but sparse (although potentially including repeats). J=unique(I) K=digitize(I,J)-1 b=bincount(K) y+=b*x[J] Well it seems to work, but surely there must be a nicer way? Dan _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
