迭代python嵌套列表

Question

我需要将列表'a'转换为列表'b'，怎么做？

a = [[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12,]]]

b = [[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12,]]]

Answer 1

您可以使用sum()的巧妙技巧来加入嵌套列表：

[[sum(l, [])] for l in a]
#[[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12]]]

---

为了更清楚地了解其工作原理，请考虑以下示例：

>>> sum([[1,2], [3,4]], [])
[1, 2, 3, 4]

---

或者您可以使用更有效的itertools.chain_from_iterable方法：

flatten = itertools.chain.from_iterable
[[list(flatten(l))] for l in a]
#[[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12]]]

Answer 2

我建议遵循以下列表：

In [1]: a = [[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12,]]]
   ...:

In [2]: [[[a for sub in nested for a in sub]] for nested in a]
Out[2]: [[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12]]]

它相当于以下嵌套的for循环：

result = []
for nested in a:
    _temp = []
    for sub in nested:
        for a in sub:
            _temp.append(a)
    result.append([_temp])

虽然，我会写它更像：

result = []
for nested in a:
    _temp = []
    for sub in nested:
        _temp.extend(sub)
    result.append([_temp])

Answer 3

您可以使用列表推导：这假设您在每个子列表中只有两个子子列表。由于您的输入和输出非常清晰，它可以满足您的需求

a = [[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12,]]]
b = [[c[0] + c[1]] for c in a ]
print (b)

产量

[[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12]]]

Answer 4

实现两个列表串联的另一种方法如下：

a = [[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12,]]]

b = list(map(lambda elem: [[*elem[0], *elem[1]]],a))
print(b)

输出：

[[[1, 2, 3, 4, 5, 6]], [[7, 8, 9, 10, 11, 12]]]

Answer 5

我建议使用循环方法。

a = [[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12,]]]
b = []
for i in a:
    new_j = []
    for j in i:
        new_j.extend(j)
    new_i = [new_j]
    b = b + [new_i]
b