On Fri, 14 Dec 2007, dmitrey apparently wrote: > I guess it doesn't matter, but typical n are 1...1000. > However, I need to call the operation hundreds or thousands times (while > running NLP solver ralg, so 4..5 * nIter times). > Number of zeros can be 0...n-1
Do both A and x change every iteration? Anyway, if x is sparse I think you'll get some benefit by doing idx=N.ravel(x)!=0 A[:,idx]*x[idx] hth, Alan Isaac _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
