Thank you for the tip, however, I expect <A,x> being vector of length n A and x change every iter. A is not sparse.
Alan G Isaac wrote: > 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 > > > > _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
