如何从多个列表中获取所有组合？

Question

我不确定我的问题是否正确，但我不知道如何解释它。 所以我有一些清单，比如

a = ['11', '12']
b = ['21', '22']
c = ['31', '32']

我需要得到类似的东西：

result = [
    ['11', '21', '31'],
    ['11', '21', '32'],
    ['11', '22', '31'],
    ['11', '22', '32'],
    ['12', '21', '31'],
    ['12', '21', '32'],
    ['12', '22', '31'],
    ['12', '22', '32']
]

Answer 1

用户itertools ， 组合：

import itertools
a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
list(itertools.combinations(itertools.chain(a,b,c), 3))
[('11', '12', '21'), ('11', '12', '22'), ('11', '12', '31'), ('11', '12', '32'), ('11', '21', '22'), ('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('11', '31', '32'), ('12', '21', '22'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32'), ('12', '31', '32'), ('21', '22', '31'), ('21', '22', '32'), ('21', '31', '32'), ('22', '31', '32')]

或产品：

list(itertools.product(a,b,c))
[('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32')]

Answer 2

您需要itertools.product来返回输入迭代的笛卡尔积。

>>> a = ['11', '12']
>>> b = ['21', '22']
>>> c = ['31', '32']
>>>
>>> from itertools import product
>>>
>>> list(product(a,b,c))
[('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32')]

您可以使用列表推导将元组转换为列表：

>>> [list(i) for i in product(a,b,c)]
[['11', '21', '31'], ['11', '21', '32'], ['11', '22', '31'], ['11', '22', '32'], ['12', '21', '31'], ['12', '21', '32'], ['12', '22', '31'], ['12', '22', '32']]

Answer 3

import itertools
list(itertools.product(a,b,c))

或使用 numpy

import numpy
[list(x) for x in numpy.array(numpy.meshgrid(a,b,c)).T.reshape(-1,len(a))]

Answer 4

import numpy as np    
np.array(np.meshgrid(a, b, c)).T.reshape(-1,3)

编辑

import numpy as np 
len = 3 #combination array length
np.array(np.meshgrid(a, b, c)).T.reshape(-1,len)

Answer 5

不使用任何库的解决方案：

def subproblem(list_a, list_b):
    c = list()
    for i, a_i in enumerate(list_a):
        for j, b_j in enumerate(list_b):
            c.append(a_i + [b_j])
    return c


def multi_lists_find_all_combination(lists):
    com = [[]]
    if len(lists) == 1:
        return lists[0]
    for i in range(0, len(lists)):
        print(i)
        com = subproblem(com, lists[i])
    return com

a = ['11', '12']
b = ['21', '22']
c = ['31', '32']

ans = multi_lists_find_all_combination([a,b,c])