通过numpy数组优化迭代以进行邻居检查

Question

Is there a way to optimize the following iteration with neighbours checking:

for i in range(1, A.shape[0]):
    for j in range(1, A.shape[1]):
        v = (A[i, j], A[i-1,j-1], A[i-1, j], A[i, j-1])
        if v == something: print(v)

where A is a (very big) numpy array of 0's and 1's, v and something are tuples (eg (0, 1, 0, 0)).

Answer 1

Create the test data first:

import numpy as np
np.random.seed(1)
A = np.random.randint(0, 2, size=(10, 8)).astype(np.uint8)

A:

array([[1, 1, 0, 0, 1, 1, 1, 1],
       [1, 0, 0, 1, 0, 1, 1, 0],
       [0, 1, 0, 0, 0, 1, 0, 0],
       [1, 0, 0, 0, 1, 0, 0, 0],
       [1, 1, 1, 1, 1, 0, 0, 0],
       [1, 1, 1, 1, 1, 1, 0, 1],
       [1, 0, 0, 1, 0, 0, 1, 1],
       [1, 0, 1, 0, 0, 1, 1, 0],
       [1, 1, 1, 1, 0, 0, 1, 1],
       [0, 0, 0, 0, 1, 1, 1, 0]], dtype=uint8)

then create an array that use bits to represent neighbours value:

a = A[:-1, :-1] << 3
a |= A[:-1, 1:] << 2
a |= A[1:, :-1] << 1
a |= A[1:, 1:]

to check neighbours value of

11
00

you can use:

a == int("1100", 2)

the output:

array([[False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False],
       [False,  True, False, False,  True, False, False],
       [False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False],
       [ True,  True,  True, False, False, False, False]], dtype=bool)

Answer 2

All you need to do is generate the four arrays of boolean conditions. Roughly:

spot = A == v[0]
upleft = np.roll(A, (1,1), (0,1)) == v[1]
left = np.roll(A, 1, 1) == v[2]
up = np.roll(A, 1, 0) == v[3]

Then you can find the places where all four are true:

res = spot & upleft & left & up

This gives you a boolean array which is True wherever all the conditions are met. You should ignore the first row and column, as they cannot be computed:

res = res[1:,1:]
# or
res[0,:] = False
res[:,0] = False

Answer 3

Here is the convolution based method suggested by @Matt.St in the comments.

>>> import numpy as np
>>> from scipy import signal
>>> a = np.random.randint(0, 2, (n, n))
>>> t = np.random.randint(0, 2, (2, 2))
>>> 
>>> 
>>> tt = 2*t[::-1, ::-1] - 1
>>> ts = t.sum()
>>> 
>>> result = signal.convolve2d(a, tt, 'valid') == ts
>>> 
>>> a
array([[1, 0, 1, 0, 1, 0, 0, 1, 0, 1],
       [0, 1, 1, 1, 0, 1, 1, 1, 0, 0],
       [0, 1, 0, 1, 1, 1, 0, 1, 0, 0],
       [0, 1, 1, 1, 0, 0, 1, 1, 1, 1],
       [1, 1, 0, 1, 1, 1, 1, 0, 1, 0],
       [1, 1, 0, 0, 1, 0, 0, 1, 1, 1],
       [1, 1, 1, 0, 1, 0, 0, 1, 1, 1],
       [0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
       [0, 0, 0, 1, 1, 0, 0, 1, 0, 1],
       [1, 1, 0, 1, 0, 0, 1, 1, 1, 0]])
>>> t
array([[0, 1],
       [1, 1]])
>>> result
array([[False,  True, False, False, False, False,  True, False, False],
       [False, False, False, False,  True, False, False, False, False],
       [False, False,  True, False, False, False,  True, False, False],
       [ True, False, False, False, False,  True, False, False, False],
       [False, False, False, False, False, False, False,  True, False],
       [False, False, False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False, False, False],
       [False, False, False, False, False, False, False, False, False],
       [False, False, False, False, False, False,  True, False, False]], dtype=bool)