Henrik Ronellenfitsch wrote: <snip> > Thanks very much for your solution, this is exactly what I needed! > If I'm not mistaken, though, you can achieve the same result with > > h = hamming(n) > ham2d = sqrt(outer(h,h)) > > which is a bit more compact. > > Regards, > Henrik
Yes, that's nicer. regards, Gary _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
