[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