I took an example from view_as_blocks
, and tried your style of reshape:
A = np.arange(4*4).reshape(4,4)
B = view_as_blocks(A, block_shape=(2, 2))
print(A.__array_interface__)
print(B.__array_interface__)
C = B.reshape((2*2,2,2))
print(C.__array_interface__)
producing:
{'typestr': '<i4', 'data': (153226600, False), 'shape': (4, 4),
'descr': [('', '<i4')], 'version': 3, 'strides': None}
{'typestr': '<i4', 'data': (153226600, False), 'shape': (2, 2, 2, 2),
'descr': [('', '<i4')], 'version': 3, 'strides': (32, 8, 16, 4)}
{'typestr': '<i4', 'data': (150895960, False), 'shape': (4, 2, 2),
'descr': [('', '<i4')], 'version': 3, 'strides': None}
The data
pointer for A
and B
is the same; B
is a view on A
.
But the pointer for C
is different. It is a copy. That explains why it takes so long in your case.
Lets do that a little differently:
A = np.arange(4*4).reshape(4,4)
B = view_as_blocks(A, block_shape=(2, 2))
print(A.__array_interface__)
print(B.__array_interface__)
C = B.reshape((2*2,1,2,2))
print(C.__array_interface__)
D = as_strided(B, shape=(2*2,1,2,2))
print(D.__array_interface__)
print(B[1,1,:,:])
print(C[3,0,:,:])
print(D[3,0,:,:])
producing
1254:~/mypy$ python3 skshape.py
{'strides': None, 'typestr': '<i4', 'version': 3,
'data': (154278744, False), 'shape': (4, 4), 'descr': [('', '<i4')]}
{'strides': (32, 8, 16, 4), 'typestr': '<i4', 'version': 3,
'data': (154278744, False), 'shape': (2, 2, 2, 2), 'descr': [('', '<i4')]}
{'strides': None, 'typestr': '<i4', 'version': 3,
'data': (155705400, False), 'shape': (4, 1, 2, 2), 'descr': [('', '<i4')]}
{'strides': (32, 8, 16, 4), 'typestr': '<i4', 'version': 3,
'data': (154278744, False), 'shape': (4, 1, 2, 2), 'descr': [('', '<i4')]}
[[10 11]
[14 15]]
[[10 11]
[14 15]]
[[ 154561960 -1217783696]
[ 48 3905]]
Again the reshape creates a copy. The 2nd as_strides
returns a view, but the striding is screwed up. It is looking at memory outside the original data buffer (that's part of why playing with strides on your own is dangerous).
In my example, look at the first corner value of each block
print(B[:,:,0,0])
print(C[:,0,0,0])
[[ 0 2]
[ 8 10]]
[ 0 2 8 10]
For B
, the rows increase by 8, columns by 2; that's reflected in the (32,8)
(4*8,4*2) striding.
But in C
the steps are (2,6,2) - striding can't do that.
From this I conclude that the reshape is impossible without copy.