2009/4/29 Dan Goodman <dg.gm...@thesamovar.net>: > 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?
Well, a few lines serve to rearrange the indices array so it contains each number only once, and then to add together (with bincount) all values with the same index: ix = np.unique1d(indices) reverse_index = np.searchsorted(ix, indices) r = np.bincount(reverse_index, weights=values) You can then do: y[ix] += r All this could be done away with if bincount supported input and output arguments. Then y[indices] += values could become np.bincount(indices, weights=values, input=y, output=y) implemented y the same nice tight loop one would use in C. (I'm not sure bincount is the right name for such a general function, though.) This question comes up fairly often on the mailing list, so it would be nice to have a single function to point to. Anne _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
