[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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