Python 将列表子索引为块矩阵X=[[a，B]，[C，D]]

让

如何输出子矩阵，使X=[[A，B]，[C，D]]

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

---

没有numpy，你可以这样做

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

---

假设X始终是4X4矩阵，您可以尝试此操作：

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:]])

---

对于第一个子矩阵，您可以使用

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)

---

（与其他类似）

使用Numpy的解决方案如下所示：

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

---

这里有一个不使用numpy的解决方案

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

---

一行程序版本，但可读性稍差

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]]

---

“到目前为止，您尝试了什么？”帕迪，这在Matlab中是微不足道的。对于Python，我尝试了A=x[：2][：2][：2]，但没有效果。最好将它们转换为Numpy数组。然后只需要给出开始索引。@sbimb非常确定X[:2]与X[0:2]相同。类似的问题：您有一个语法错误——一个额外的括号。另外，A的输出是

[[5,6]，[2,3]]

，这不是OP想要的。@rassar捕捉得好，谢谢

>>> (A,B),(C,D)= [([X[i][:2],X[i+1][:2]],[X[i][2:],X[i+1][2:]]) for i in range(0,len(X),2)]
>>>
>>? A
[[2, 3], [4, 5]]
>>? B
[[5, 6], [9, 10]]
>>? C
[[6, 1], [3, 7]]
>>? D
[[3, 9], [11, 12]]
>>?