[Numpy-discussion] Iterative Matrix Multiplication
friedrichromstedt at gmail.com
Sat Mar 6 04:20:52 EST 2010
2010/3/5 Ian Mallett <geometrian at 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.
> sum_1 = sum.reshape((len(sum), 1)).repeat(len(sum), axis = 1)
> sum_2 = sum.reshape((1, len(sum))).repeat(len(sum), axis = 0)
> 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
> diff_1 = diff.reshape((len(diff), 1)).repeat(len(diff), axis = 1)
> diff_2 = diff.reshape((1, len(diff))).repeat(len(diff), axis = 0)
> comparison_diff = (diff_1 == diff_2)
Same here. Delete the repeat() calls, but not the reshape() calls.
> 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.
> 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.
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