[SciPy-User] Reshaping Question
David Warde-Farley
dwf@cs.toronto....
Wed Nov 4 19:08:35 CST 2009
Hi Skipper,
No, I don't believe so. The reason is that NumPy arrays have to obey constant stride along each dimension. Assuming
dtype is int32, the reshaping you describe (assuming you want to reshape c into d) would require the stride along dim
2 to be 4 bytes to get from 0 to 1, and then 12 bytes to get to 4, and then 4 bytes again to get to 5. This isn't
legal, you'd have to do a copy to construct this matrix.
David
On Wed, Nov 04, 2009 at 08:25:12PM -0500, Skipper Seabold wrote:
> My brain is failing me. Is there a clean way to reshape an array like
> the following?
>
> import numpy as np
>
> c = np.arange(16).reshape(4, 2, 2)
>
> In [209]: c
> Out[209]:
> array([[[ 0, 1],
> [ 2, 3]],
>
> [[ 4, 5],
> [ 6, 7]],
>
> [[ 8, 9],
> [10, 11]],
>
> [[12, 13],
> [14, 15]]])
>
> So that c == d where
>
> d = np.array(([0, 1, 4, 5], [2,3,6,7], [8,9,12,13], [10, 11, 14, 15]))
>
> In [211]: d
> Out[211]:
> array([[ 0, 1, 4, 5],
> [ 2, 3, 6, 7],
> [ 8, 9, 12, 13],
> [10, 11, 14, 15]])
>
> Cheers,
>
> Skipper
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