Question

让

X = [[2,3, 5, 6], 
     [4,5, 9, 10], 
     [6,1, 3, 9], 
     [3,7, 11, 12]]

如何输出子矩阵，使得X = [[A，B]，[C，D]]？

    A = [[2, 3], 
         [4,5]]
    B = [[5,6], 
        [9, 10]] 
    C = [[6, 1], 
         [3,7]]
    D = [[3,9], 
        [11, 12]]

Answer 1

没有numpy，你可以这样做....

X = [[2,3, 5, 6], 
         [4,5, 9, 10], 
         [6,1, 3, 9], 
         [3,7, 11, 12]]

for i in X:
   print([i[ :len(i)//2],i[len(i)//2:]])

Answer 2

你可以试试这个，假设X总是一个4X4矩阵：

X = [[2,3, 5, 6], 
 [4,5, 9, 10], 
 [6,1, 3, 9], 
 [3,7, 11, 12]]

new_matrix = [[X[i][:2], X[i+1][:2]] for i in range(0, len(X), 2)]
new_matrix.extend([[X[i][2:], X[i+1][2:]] for i in range(0, len(X), 2)]) 
print(new_matrix)

Answer 3

使用Numpy的解决方案如下所示：

A = X[0:2, 0:2]
B = X[0:2, 2:4]
C = X[2:4, 0:2]
D = X[2:4, 2:4]

Answer 4

对于第一个子矩阵，您可以使用

A = [row[:2] for row in X[:2]]

（和其他人一样）

Answer 5

这是一个不使用numpy的解决方案。

A = [X[i][:2] for i in range(2)]
B = [X[i][2:] for i in range(2)]
C = [X[i][:2] for i in range(2,4)]
D = [X[i][2:] for i in range(2,4)]

>>> A
[[2, 3], [4, 5]]
>>> B
[[5, 6], [9, 10]]
>>> C
[[6, 1], [3, 7]]
>>> D
[[3, 9], [11, 12]]

Answer 6

一个班轮版本，虽然它的可读性稍差

{{1}}

将列表的列表索引到块矩阵中X = [[A，B]，[C，D]]

6 个答案: