alias of numpy array

Question

I've been thinking about this for a while now, and I just can't make sense of what is going on here.. hopefully it's something simple? In the output below I would expect a '41' in the first element of the second entry of 'c'.

>>> a = np.zeros(shape = (2,2))
>>> b = np.zeros(shape = (2,2))
>>> c = [np.array(x) for x in range(3)]
>>> c[1] = np.zeros(shape=(2,2,3))
>>> c[1][:,:,0] = a.view()
>>> a
array([[ 0.,  0.],
       [ 0.,  0.]])
>>> c
[array(0), array([[[ 0.,  0.,  0.],
                   [ 0.,  0.,  0.]],

                  [[ 0.,  0.,  0.],
                   [ 0.,  0.,  0.]]]), array(2)]
>>> a[0,0] = 41
>>> a
array([[ 41.,   0.],
       [  0.,   0.]])
>>> c
[array(0), array([[[ 0.,  0.,  0.],
                   [ 0.,  0.,  0.]],

                  [[ 0.,  0.,  0.],
                   [ 0.,  0.,  0.]]]), array(2)]

wim · Accepted Answer · 2013-02-01 00:10:27Z

3

Because you did:

c[1][:,:] = a.view()

Which makes a shallow copy.

If you did c[1] = a.view() instead you would see the behaviour you expected.

answered Feb 1, 2013 at 0:10

wim

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2 Comments

rokuingh Over a year ago

thanks, that makes sense because it's creating a slice right? (which is a copy).. soo, this situation is actually something more like this (updated the question), so I can't just do the direct indexing on 'c'..

codewarrior Over a year ago

I think wim is still right. What you are doing is assigning a.view() to a slice of the array c[1], modifying that array in-place. This copies the values from a.view() into c[1]. It's not possible to make a slice of one array act as a view into another array. numpy doesn't support this at all.

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