James Bergstra wrote: > In some cases a brute-force approach is also good.
true. > If r is a matrix of shape Nx2: > > (r*r).sum(axis=1) -2 * numpy.dot(r, r.T) + > (r*r).sum(axis=1).reshape((r.shape[0], 1)) < thresh**2 > > It's brute force, but it takes advantage of fast matrix multiplication. I'm more concerned about memory -- doesn't this use N^2 memory? Which could be an issue here. -Chris -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion