Question

我的清单：

>>> l = [["A", "A1", 1, 2, 3], 
         ["B", "A2", 4, 5, 6], 
         ["C", "A3", 7, 8, 9], 
         ["D", "A4", 10, 11, 12]]

切片操作：

>>> [[n[0] for n in l], [u[1:] for u in l]]
[['A', 'B', 'C', 'D'], 
  [['A1', 1, 2, 3], 
  ['A2', 4, 5, 6], 
  ['A3', 7, 8, 9], 
  ['A4', 10, 11, 12]]]

有没有办法在没有额外方括号的情况下切割此列表？如下所示：

[['A', 'B', 'C', 'D'], 
 ['A1', 1, 2, 3], 
 ['A2', 4, 5, 6], 
 ['A3', 7, 8, 9], 
 ['A4', 10, 11, 12]]

Answer 1

您可以连接列表：

>>> [[n[0] for n in l]] + [u[1:] for u in l]
[['A', 'B', 'C', 'D'], ['A1', 1, 2, 3], ['A2', 4, 5, 6], ['A3', 7, 8, 9], ['A4', 10, 11, 12]]

Answer 2

你可以简单地创建另一个列表并加入像这样的结果

print [[n[0] for n in l]] + [u[1:] for u in l]

Answer 3

喜欢这个吗？

>>> mylist = [[n[0] for n in l]] + [u[1:] for u in l]
>>> mylist
[['A', 'B', 'C', 'D'], ['A1', 1, 2, 3], ['A2', 4, 5, 6], ['A3', 7, 8, 9], ['A4', 10, 11, 12]]

Python嵌套列表切片

3 个答案: