Numpy reshape on view

Question

I\'m confused about the results of numpy reshape operated on a view. In the following q.flags shows that it does not own the data, but q.base is neither x nor y, so what is

Answer 1

In short: you cannot always rely on the ndarray.flags['OWNDATA'].

>>> import numpy as np
>>> x = np.random.rand(2,2)
>>> y = x.T
>>> q = y.reshape(4)
>>> y[0,0]
0.86751629121019136
>>> y[0,0] = 1
>>> q
array([ 0.86751629,  0.87671107,  0.65239976,  0.41761267])
>>> x
array([[ 1.        ,  0.65239976],
       [ 0.87671107,  0.41761267]])
>>> y
array([[ 1.        ,  0.87671107],
       [ 0.65239976,  0.41761267]])
>>> y.flags['OWNDATA']
False
>>> x.flags['OWNDATA']
True
>>> q.flags['OWNDATA']
False
>>> np.may_share_memory(x,y)
True
>>> np.may_share_memory(x,q)
False

Because q didn't reflect the change in the first element, like x or y, it must somehow be the owner of the data (somehow is explained below).

There is more discussion about the OWNDATA flag over at the numpy-discussion mailinglist. In the How can I tell if NumPy creates a view or a copy? question, it is briefly mentioned that simply checking the flags.owndata of an ndarray sometimes seems to fail and that it seems unreliable, as you mention. That's because every ndarray also has a base attribute:

the base of an ndarray is a reference to another array if the memory originated elsewhere (otherwise, the base is None). The operation y.reshape(4) creates a copy, not a view, because the strides of y are (8,16). To get it reshaped (C-contiguous) to (4,), the memory pointer would have to jump 0->16->8->24, which is not doable with a single stride. Thus q.base points to the memory location generated by the forced-copy-operation y.reshape, which has the same shape as y, but copied elements and thus has normal strides again: (16, 8). q.base is thus not bound to by any other name as it was the result of the forced-copy operation y.reshape(4). Only now can the object q.base be viewed in a (4,) shape, because the strides allow this. q is then indeed a view on q.base.

For most people it would be confusing to see that q.flags.owndata is False, because, as shown above, it is not a view on y. However, it is a view on a copy of y. That copy, q.base, is the owner of the data however. Thus the flags are actually correct, if you inspect closely.