Rob Hetland wrote: > I think you want something like this: > > x1 = x1 * weights[np.newaxis,:] > x2 = x2 * weights[np.newaxis,:] > > x1 = x1[np.newaxis, :, :] > x2 = x2[:, np.newaxis, :] > distance = np.sqrt( ((x1 - x2)**2).sum(axis=-1) ) > > x1 and x2 are arrays with size of (npoints, ndimensions), and npoints > can be different for each array. I'm not sure I did your weights > right, but that part shouldn't be so difficult. > >
Weights seem not right but anyway here is the solution adapted from Bill Baxter's : def distance_matrix_final(data1,data2,weights): data1w = data1*weights dm = (data1w*data1).sum(1)[:,None]-2*N.dot(data1w,data2.T)+(data2*data2*weights).sum(1) dm[dm<0] = 0 return dm This solution is super-fast, stable and use little memory. It is based on the fact that: (x-y)^2*w = x*x*w - 2*x*y*w + y*y*w For size1=size2=dimensions=1000 requires ~0.6sec. to compute on my dual core duo. It is 2 order of magnitude faster than my previous solution, but 1-2 order of magnitude slower than using C with weave.inline. Definitely good enough for me. Emanuele _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
