Python函数从列表列表中返回二维列表

Question

这个问题已经有答案了:

How to get the Cartesian product of multiple lists (19 个回答)

已关闭 5 年前。

这里有一个Python问题。我想创建一个函数来从列表列表中返回二维列表。我在itertools中找不到这个函数。

可能的输入 = 如下所示列出 aa 或 ab(不同长度)

aa = [[230, 115, 90, 264], [12, 9, 5], [-1], ['on', 'off']]
ab = [[230, 115, 90, 264], ['on', 'off']]

ab 的输出应该是:

[[230, 'on'],
[230, 'off'],
[115, 'on'],
[115, 'off'],
[90, 'on'],
[90, 'off'],
[264, 'on'],
[264, 'off']]

aa 的输出应该是:

[[230, 12, -1, 'on'],
[230, 12, -1, 'off'],
[230, 9, -1, 'on'],
[230, 9, -1, 'off'],
....
[264, 5, -1, 'off']]

Answer 1

我想你要的是 cartesian product一些可迭代的，例如:

import itertools
import pprint

aa = [[230, 115, 90, 264], [12, 9, 5], [-1], ['on', 'off']]
ab = [[230, 115, 90, 264], ['on', 'off']]

for iterable in [aa, ab]:
    pprint.pprint(list(itertools.product(*aa)))

结果:

[(230, 12, -1, 'on'),
 (230, 12, -1, 'off'),
 (230, 9, -1, 'on'),
 (230, 9, -1, 'off'),
 (230, 5, -1, 'on'),
 (230, 5, -1, 'off'),
 (115, 12, -1, 'on'),
 (115, 12, -1, 'off'),
 (115, 9, -1, 'on'),
 (115, 9, -1, 'off'),
 (115, 5, -1, 'on'),
 (115, 5, -1, 'off'),
 (90, 12, -1, 'on'),
 (90, 12, -1, 'off'),
 (90, 9, -1, 'on'),
 (90, 9, -1, 'off'),
 (90, 5, -1, 'on'),
 (90, 5, -1, 'off'),
 (264, 12, -1, 'on'),
 (264, 12, -1, 'off'),
 (264, 9, -1, 'on'),
 (264, 9, -1, 'off'),
 (264, 5, -1, 'on'),
 (264, 5, -1, 'off')]
[(230, 12, -1, 'on'),
 (230, 12, -1, 'off'),
 (230, 9, -1, 'on'),
 (230, 9, -1, 'off'),
 (230, 5, -1, 'on'),
 (230, 5, -1, 'off'),
 (115, 12, -1, 'on'),
 (115, 12, -1, 'off'),
 (115, 9, -1, 'on'),
 (115, 9, -1, 'off'),
 (115, 5, -1, 'on'),
 (115, 5, -1, 'off'),
 (90, 12, -1, 'on'),
 (90, 12, -1, 'off'),
 (90, 9, -1, 'on'),
 (90, 9, -1, 'off'),
 (90, 5, -1, 'on'),
 (90, 5, -1, 'off'),
 (264, 12, -1, 'on'),
 (264, 12, -1, 'off'),
 (264, 9, -1, 'on'),
 (264, 9, -1, 'off'),
 (264, 5, -1, 'on'),
 (264, 5, -1, 'off')]