将3D列表更改为2D列表

Question

我想将3D列表（列表_1）转换为2D列表（列表_2）。 例如：

list_1: [[[3], [4]], [[5], [6]], [[7], [8]]]
list_2: [[3, 4], [5, 6], [7, 8]]

这是我已经尝试过的：

[e for sl in lst for e in sl]

但是结果会有所不同：

[[3], [4], [5], [6], [7], [8]]

Answer 1

出于可读性考虑，我建议itertools.chain将此处的内部列表展平：

from itertools import chain
l = [[[3], [4]], [[5], [6]], [[7], [8]]]

[list(chain.from_iterable(i)) for i in l]
# [[3, 4], [5, 6], [7, 8]]

---

按照您的方法，您需要进行额外的循环以平化内部列表：

[[i for e in sl for i in e] for sl in l]
# [[3, 4], [5, 6], [7, 8]]

Answer 2

另一个选择：

[reduce(list.__add__, x)  for x in list_1]

Answer 3

您也可以这样：

l = [[[3], [4]], [[5], [6]], [[7], [8]]]

print ([sum(item, []) for item in l])

输出：

[[3, 4], [5, 6], [7, 8]]

Answer 4

您很亲密：

[[c for b in a for c in b] for a in lst]

Answer 5

如果所有嵌套列表都包含单个元素，则可以使用拆包：

d = [[[3], [4]], [[5], [6]], [[7], [8]]]
new_d = [[i for [i] in b] for b in d]

输出：

[[3, 4], [5, 6], [7, 8]]