Python - 獲取矩陣中切片周圍的數組的方法

Question

我用隨機值（0,1）定義了一個矩陣 NxN。 我需要得到圍繞連續 1 的數字的總和。

例如：

100110001
101001000
100001001
000000000
000111001
000000100
.. ..

對於上面的 111，周圍的數字之和為 1。

有沒有辦法使用numpy或itertools或任何東西來獲得數字的總和或數組？ 請幫忙，加油

為了檢測隨機連續 1，我使用：

from itertools import groupby

def groups(l): 
    return [sum(g) for i, g in groupby(l) if i == 1] 

con += list(filter(lambda x: x > 1,groups(matrix[4])))

並獲得 1s 的索引：

idx+=[idx for idx, i in enumerate(matrix[4]) if i == 1]

Answer 1

這可能無法完全回答您的問題，但可能會為您指明正確的方向：

In [1]: import numpy as np

In [2]: from scipy import ndimage as nd

In [3]: mat = np.random.randint(0,2,(6,6))

In [4]: mat
Out[4]: 
array([[1, 1, 1, 0, 1, 1],
       [1, 1, 0, 0, 1, 0],
       [1, 0, 1, 1, 1, 1],
       [0, 0, 1, 1, 1, 0],
       [1, 0, 1, 0, 1, 0],
       [1, 0, 0, 1, 1, 1]])

In [5]: s = np.array([[0,0,0],[1,1,1],[0,0,0]])

In [6]: lab, nlab = nd.label(mat, s)

In [7]: lab
Out[7]: 
array([[ 1,  1,  1,  0,  2,  2],
       [ 3,  3,  0,  0,  4,  0],
       [ 5,  0,  6,  6,  6,  6],
       [ 0,  0,  7,  7,  7,  0],
       [ 8,  0,  9,  0, 10,  0],
       [11,  0,  0, 12, 12, 12]], dtype=int32)

In [8]: s2 = nd.generate_binary_structure(2,2)

In [9]: for l in range(1, nlab+1):
   ...:     is_l = lab==l
   ...:     if np.sum(is_l) ==1:
   ...:         continue
   ...:     is_l_padded = nd.binary_dilation(is_l, s2)
   ...:     border = np.logical_xor( is_l_padded, is_l)
   ...:     num_ones = mat[border].sum()
   ...:     print("label %d has %d ones around it" % (l, num_ones))
   ...: 
label 1 has 2 ones around it
label 2 has 1 ones around it
label 3 has 5 ones around it
label 6 has 5 ones around it
label 7 has 6 ones around it
label 12 has 2 ones around it