[Numpy-discussion] Slicing slower than matrix multiplication?
Charles R Harris
charlesr.harris@gmail....
Mon Dec 14 14:46:17 CST 2009
On Mon, Dec 14, 2009 at 10:27 AM, Jasper van de Gronde <
th.v.d.gronde@hccnet.nl> wrote:
> 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.
>
>
Xi*S[:,newaxis]
Chuck
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