[Numpy-discussion] Iterative Matrix Multiplication
Ian Mallett
geometrian at gmail.com
Sat Mar 6 13:45:36 EST 2010
Hi,
On Sat, Mar 6, 2010 at 1:20 AM, Friedrich Romstedt <
friedrichromstedt at gmail.com> wrote:
> Hmm. Let's see ... Can you tell me how I can test the time calls in a
> script take? I have no idea.
>
I've been doing:
t1 = time.time()
#code
t2 = time.time()
print "[code]:",t2-t1
> > #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.
>
Wow! Drops to immeasurably quick :D
> > #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.
>
Once again, drops to immeasurably quick :D
> > #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.
>
It's marginally faster.
> > #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).
>
Actually, it's marginally slower :S
> 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.
>
Nope--I've been timing as above.
> Friedrich
>
On Sat, Mar 6, 2010 at 1:42 AM, Friedrich Romstedt <
friedrichromstedt at gmail.com> wrote:
> 2010/3/5 Ian Mallett <geometrian at gmail.com>:
> > #takes 0.04 seconds
> > inner = np.inner(ns, v1s - some_point)
>
> Ok, I don't know why I was able to overlook this:
>
> dotprod = (ns * (v1s - some_point)).sum(axis = 1)
>
Much faster :D
So, the most costly lines:
comparison_sum = (sum_1 == sum_2) #0.024 sec
comparison_diff = (diff_1 == diff_2) #0.024 sec
same_edges = np.logical_and(comparison_sum, comparison_diff) #0.029 sec
doublet_count = same_edges.sum(axis = 0) #0.147
Thanks again,
Ian
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