>> idx=N.ravel(x)!=0 >> A[:,idx]*x[idx]
On Fri, 14 Dec 2007, dmitrey apparently wrote: > I expect <A,x> being vector of length n A and x change > every iter. A is not sparse. Still, whenever x has a lot of zeros, this should have a substantial payoff. That seems the only exploitable information you have offered in the problem. Cheers, Alan _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
