On Wed, 2017-11-29 at 14:56 +0000, ZHUO QL (KDr2) wrote: > Hi, all > > suppose: > > - D, is the data matrix, its shape is M x N > - I, is the indices matrix, its shape is M x K, K<=N > > Is there a efficient way to get a Matrix R with the same shape of I > so that R[x,y] = D[x, I[x,y]] ? > > A nested for-loop or list-comprehension is too slow for me. >
Advanced indexing can do any odd thing you might want to do. I would not suggest to use the matrix class, but always use the array class in case you are doing that though. This should do the trick, I will refer the the documentation for how it works, except that it is basically: R[x,y] = D[I1[x, y], I2[x, y]] R = D[np.arange(I.shape[0])[:, np.newaxis], I] - Sebastian > Thanks. > > ---- > ZHUO QL (KDr2) http://kdr2.com > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion
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