On Sun, May 4, 2014 at 9:34 PM, srean 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 s
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
wo
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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Hi,
On behalf of the Scipy development team I'm pleased to announce the
availability of Scipy 0.14.0. This release contains new features (see
release notes below) and 8 months worth of maintenance work. 80 people
contributed to this release.
This is also the first release for which binary wheels
