# [SciPy-User] Reshaping Question

Skipper Seabold jsseabold@gmail....
Wed Nov 4 21:56:10 CST 2009

```On Wed, Nov 4, 2009 at 9:05 PM, Anne Archibald
<peridot.faceted@gmail.com> wrote:
> 2009/11/4 David Warde-Farley <dwf@cs.toronto.edu>:
>> 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.

Ah ok.  This makes sense, and is kind of why I thought I couldn't do
what I wanted as easily, as I'd like.

>
> Reshape sometimes creates copies. It tries hard not to, and if you
> assign the shape attribute rather than calling reshape it won't ever
> make a copy, but if necessary reshape will copy the input array:
>
> In [42]: np.transpose(c.reshape(2,2,2,2),(0,2,1,3)).reshape(4,4)Out[42]:
> array([[ 0,  1,  4,  5],
>       [ 2,  3,  6,  7],
>       [ 8,  9, 12, 13],
>       [10, 11, 14, 15]])
>
> The trick is to use transpose to do an arbitrary permutation of the
> input axes, and also to rearrange the first axis with an additional
> reshape.
>
> Anne
>

This makes sense as well.  This is kind of what I was looking for I
just couldn't figure out the permutation.  I was trying to roll the
axes, though I guess this could still work if you add the extra axis.

I don't know if I'd use this in the end though, as it might sacrifice
too much readability in the code, but maybe that's just me...

What if I had the outermost container as a list?  Say,

c = [np.arange(4).reshape(2,2),np.arange(4,8).reshape(2,2),np.arange(8,12).reshape(2,2),np.arange(12,16).reshape(2,2)]

I seem to be running into much the same problems trying to use list
comprehension to end up with d.

It seems like I'm going to need a copy anyway, so maybe I'd be better
off just allocating a new array and filling it up transparently?

Skipper

>> 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
>>> _______________________________________________
>>> SciPy-User mailing list
>>> SciPy-User@scipy.org
>>> http://mail.scipy.org/mailman/listinfo/scipy-user
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```