Question

假设我有一个嵌套的字符串列表

lst = [['a'], ['b', 'c'], ['d', ['e', 'f']]]

我想从嵌套列表中生成所有可能的组合，如下所示：

new_lst = [['a', 'b', 'd'],
           ['a', 'b', 'e', 'f'],
           ['a', 'c', 'd'],
           ['a', 'c', 'e', 'f']]

我发现了一些与我的问题有关的问题。 how to produce a nested list from two lists in python 但是，我的问题是更复杂的问题。

Answer 1

这就是诀窍 -

import itertools
lst = [['a'], ['b', 'c'], ['d', ['e', 'f']]]
outp = list(itertools.product(*lst))
out = []
for i in outp:
    temp = []
    for j in i:
        if isinstance(j, list):
            for k in j:
                temp.append(k)
        else:
            temp.append(j)
    out.append(temp)
print(out)

首先使用itertools.product形成输出材料，然后以嵌套列表展平的方式对其进行格式化。

输出

[['a', 'b', 'd'], ['a', 'b', 'e', 'f'], ['a', 'c', 'd'], ['a', 'c', 'e', 'f']]

Answer 2

与@VivekKalyanarangan类似，但有适当的扁平化：

>>> def flatten(nl):
...     for e in nl:
...         if isinstance(e, str):
...             yield e
...             continue
...         try:
...             yield from flatten(e)
...         except TypeError:
...             yield e
... 

>>> lst = [['a'], ['b', 'c'], ['d', ['e', 'f']]]
>>> 
>>> list(map(list, map(flatten, itertools.product(*lst))))
[['a', 'b', 'd'], ['a', 'b', 'e', 'f'], ['a', 'c', 'd'], ['a', 'c', 'e', 'f']]

Answer 3

使用列表理解的另一种方式

>>> ls = [['a'], ['b', 'c'], ['d', ['e', 'f']]]
>>> res = ['']
>>> for elem in ls:
...     res = [list(j) + list(e) for j in res for e in elem]
... 
>>> res
[['a', 'b', 'd'], ['a', 'b', 'e', 'f'], ['a', 'c', 'd'], ['a', 'c', 'e', 'f']]

Answer 4

您可以使用chain.from_iterable()来修饰结果：

from itertools import product, chain

lst = [['a'], ['b', 'c'], ['d', ['e', 'f']]]

[list(chain.from_iterable(i)) for i in product(*lst)]
# [['a', 'b', 'd'], ['a', 'b', 'e', 'f'], ['a', 'c', 'd'], ['a', 'c', 'e', 'f']]

Answer 5

这是你要找的吗？

from itertools import permutations
lst = [['a'], ['b', 'c'], ['d', ['e', 'f']]]
list(permutations(lst))

否则，试一试：

lst = ['a','b','c','d','e','f']
list(permutations(lst))   ##will return all possible combos

如何将itertools.product应用于Python

5 个答案: