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
Great ! thanks. I should have seen that.
Is there any way array multiplication (as opposed to matrix multiplication)
can be sped up without forming A and (A * b) explicitly.
A = np.repeat(x, [4, 2, 1, 3], axis = 0)# A.shape == 10,10
c = sum(b * A, axis = 1)# b.shape
If b is indeed big I don't see a problem with the python loop, elegance
aside; but Cython will not beat it on that front.
On Mon, May 5, 2014 at 9:34 AM, srean srean.l...@gmail.com wrote:
Great ! thanks. I should have seen that.
Is there any way array multiplication (as opposed to matrix
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
thanks
-- srean
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nope; its impossible to express A as a strided view on x, for the repeats
you have.
even if you had uniform repeats, it still would not work. that would make
it easy to add an extra axis to x without a new allocation; but
reshaping/merging that axis with axis=0 would again trigger a copy, as it
