[Numpy-discussion] Slicing slower than matrix multiplication?
Jasper van de Gronde
th.v.d.gronde@hccnet...
Mon Dec 14 11:27:08 CST 2009
Bruce Southey wrote:
>> So far this is the fastest code I've got:
>> ------------------------------------------------------------------------
>> import numpy as np
>>
>> nmax = 100
>>
>> def minover(Xi,S):
>> P,N = Xi.shape
>> SXi = Xi.copy()
>> for i in xrange(0,P):
>> SXi[i] *= S[i]
>> SXi2 = np.dot(SXi,SXi.T)
>> SXiSXi2divN = np.concatenate((SXi,SXi2),axis=1)/N
>> w = np.random.standard_normal((N))
>> E = np.dot(SXi,w)
>> wE = np.concatenate((w,E))
>> for s in xrange(0,nmax*P):
>> mu = wE[N:].argmin()
>> wE += SXiSXi2divN[mu]
>> # E' = dot(SXi,w')
>> # = dot(SXi,w + SXi[mu,:]/N)
>> # = dot(SXi,w) + dot(SXi,SXi[mu,:])/N
>> # = E + dot(SXi,SXi.T)[:,mu]/N
>> # = E + dot(SXi,SXi.T)[mu,:]/N
>> return wE[:N]
>> ------------------------------------------------------------------------
>>
>> I am particularly interested in cleaning up the initialization part, but
>> any suggestions for improving the overall performance are of course
>> appreciated.
>>
>>
> What is Xi and S?
> I think that your SXi is just:
> SXi=Xi*S
Sort of, it's actually (Xi.T*S).T, now that I think of it... I'll see if
that is any faster. And if there is a neater way of doing it I'd love to
hear about it.
> But really I do not understand what you are actually trying to do. As
> previously indicated, some times simplifying an algorithm can make it
> computationally slower.
It was hardly simplified, this was the original function body:
P,N = Xi.shape
SXi = Xi.copy()
for i in xrange(0,P):
SXi[i] *= S[i]
w = np.random.standard_normal((N))
for s in xrange(0,nmax*P):
E = np.dot(SXi,w)
mu = E.argmin()
w += SXi[mu]/N
return w
As you can see it's basically some basic linear algebra (which reduces
the time complexity from about O(n^3) to O(n^2)), plus some less nice
tweaks to avoid the high Python overhead.
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