Question

请问我如何循环嵌套列表获取一个嵌套的元组列表，例如循环通过pot来获取rslt

pot = [[1,2,3,4],[5,6,7,8]]

我试过

b = []

for i in pot:
    for items in i:
        b = zip(pot[0][0:],pot[0][1:])

但没有得到所需的输出谢谢

期望结果=

rslt = [[(1,2),(3,4)],[(5,6),(7,8)]]

Answer 1

根据grouper recipe in the itertools documentation，您可以尝试这样的事情（假设您的子列表是您指定的长度）：

>>> def grouper(iterable, n):
    args = [iter(iterable)] * n  # creates a list of n references to the same iterator object (which is exhausted after one iteration)
    return zip(*args)

现在你可以测试一下：

>>> pot = [[1,2,3,4],[5,6,7,8]]
>>> rslt = []
>>> for sublist in pot:
    rslt.append(grouper(sublist, 2))
>>> rslt
[[(1, 2), (3, 4)], [(5, 6), (7, 8)]]

Answer 2

你也可以使用列表理解：

[[(a, b), (c, d)] for a, b, c, d in l]

将嵌套列表转换为嵌套元组

2 个答案: