2010/3/5 Ian Mallett <geometr...@gmail.com>: > Cool--this works perfectly now :-)
:-) > Unfortunately, it's actually slower :P Most of the slowest part is in the > removing doubles section. Hmm. Let's see ... Can you tell me how I can test the time calls in a script take? I have no idea. > #takes 0.04 seconds > inner = np.inner(ns, v1s - some_point) I think I can do nothing about that at the moment. > #0.0840001106262 > sum_1 = sum.reshape((len(sum), 1)).repeat(len(sum), axis = 1) > > #0.0329999923706 > sum_2 = sum.reshape((1, len(sum))).repeat(len(sum), axis = 0) > > #0.0269999504089 > comparison_sum = (sum_1 == sum_2) We can leave out the repeat() calls and leave only the reshape() calls there. Numpy will substitute dimi == 1 dimensions with stride == 0, i.e., it will effectively repeat those dimension, just as we did it explicitly. > #0.0909998416901 > diff_1 = diff.reshape((len(diff), 1)).repeat(len(diff), axis = 1) > > #0.0340001583099 > diff_2 = diff.reshape((1, len(diff))).repeat(len(diff), axis = 0) > > #0.0269999504089 > comparison_diff = (diff_1 == diff_2) Same here. Delete the repeat() calls, but not the reshape() calls. > #0.0230000019073 > same_edges = comparison_sum * comparison_diff Hmm, maybe use numpy.logical_and(comparison_sum, comparison_diff)? I don't know, but I guess it is in some way optimised for such things. > #0.128999948502 > doublet_count = same_edges.sum(axis = 0) Maybe try axis = 1 instead. I wonder why this is so slow. Or maybe it's because he does the conversion to ints on-the-fly, so maybe try same_edges.astype(numpy.int8).sum(axis = 0). Hope this gives some improvement. I attach the modified version. Ah, one thing to mention, have you not accidentally timed also the printout functions? They should be pretty slow. Friedrich _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
