On Sun, May 4, 2014 at 9:34 PM, srean <srean.l...@gmail.com> wrote: > Hi all, > > is there an efficient way to do the following without allocating A where > > A = np.repeat(x, [4, 2, 1, 3], axis=0) > c = A.dot(b) # b.shape >
If x is a 2D array you can call repeat **after** dot, not before, which will save you some memory and a few operations: >>> a = np.random.rand(4, 5) >>> b = np.random.rand(5, 6) >>> np.allclose(np.repeat(a, [4, 2, 1, 3], axis=0).dot(b), ... np.repeat(a.dot(b), [4, 2, 1, 3], axis=0)) True Similarly, if x is a 1D array, you can sum the corresponding items of b before calling dot: >>> a = np.random.rand(4) >>> b = np.random.rand(10) >>> idx = np.concatenate(([0], np.cumsum([4,2,1,3])[:-1])) >>> np.allclose(np.dot(np.repeat(a, [4,2,1,3] ,axis=0), b), ... np.dot(a, np.add.reduceat(b, idx))) ... ) True Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial.
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