reshaping a view of a n-dimensional array without

问题:

tl;dr Can I reshape a view of a numpy array from 5x5x5x3x3x3 to 125x1x1x3x3x3 without using numpy.reshape?

I would like to perform a sliding window operation (with different strides) to a volume (size of MxMxM). The sliding window array can be generated with the use of numpy.lib.stride_tricks.as_strided, as previously suggested by Benjamin and Eickenberg, and demonstrated in the below code snippet, which uses a helper method from skimage that uses as_strided.

The output from this helper method gives me a shape of NxNxNxnxnxn, but I'd prefer the shape to be N^3x1xnxnxn. While I can use np.reshape to achieve this, np.reshape is slow if the volume gets large (> 100x100x100), which I'm not sure why. I thought I can use as_stride to reshape the output, but numpy crashes (code snippet below). Any ideas on how I can get a view of the output from the helper method as N**3x1xnxnxn without using np.reshape?

import numpy as np
import skimage
l = 15
s = 3
X = np.ones((l,l,l))      
print('actual shape',X.shape)
view = skimage.util.shape.view_as_blocks(X,(s,s,s))    
print('original view',view.shape)

new_shape = ((l/s)**3,1,1,s,s,s)
print('new view',new_shape)    

view_correct = view.reshape(new_shape)
print(view_correct.shape)
print('coord:','124,0,0,2,2,2','value:',view_correct[124,0,0,2,2,2])

view_incorrect = np.lib.stride_tricks.as_strided(view, shape=new_shape)
print(view_incorrect.shape)
print('coord:','124,0,0,2,2,2','value:',view_incorrect[124,0,0,2,2,2])

回答1:

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.