是否有任何pythonic方法可以在多个列表之间生成组合？ （类似于笛卡尔积，但更复杂）
示例：
a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]
# ...
# there are more than 3 lists
预期产出：
1. [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
2. [(1, 4, 8), (2, 5, 7), (3, 6, 9)]
3. [(1, 4, 9), (2, 5, 7), (3, 6, 8)]
4. [(1, 5, 7), (2, 4, 8), (3, 6, 9)]
5. ...
更新
感谢快速回复〜!!
澄清问题：
结果是列表a，b，c的笛卡尔乘积的所有非重复组合。
可以通过另一种丑陋的方法来完成：
1）生成笛卡尔积的整个列表
from itertools import product, combinations, chain
t = list(product(a, b, c))
2）使用组合生成所有可能的结果
p = list(combinations(t, 3))
3）过滤重复的条件
cnt = len(list(chain(a, b, c)))
f = [x for x in p if len(set(chain(*x))) == cnt]
UPDATE2：
丑陋方法产生的预期结果：
((1, 4, 7), (2, 5, 8), (3, 6, 9))
((1, 4, 7), (2, 5, 9), (3, 6, 8))
((1, 4, 7), (2, 6, 8), (3, 5, 9))
((1, 4, 7), (2, 6, 9), (3, 5, 8))
((1, 4, 8), (2, 5, 7), (3, 6, 9))
((1, 4, 8), (2, 5, 9), (3, 6, 7))
((1, 4, 8), (2, 6, 7), (3, 5, 9))
((1, 4, 8), (2, 6, 9), (3, 5, 7))
((1, 4, 9), (2, 5, 7), (3, 6, 8))
((1, 4, 9), (2, 5, 8), (3, 6, 7))
((1, 4, 9), (2, 6, 7), (3, 5, 8))
((1, 4, 9), (2, 6, 8), (3, 5, 7))
((1, 5, 7), (2, 4, 8), (3, 6, 9))
((1, 5, 7), (2, 4, 9), (3, 6, 8))
((1, 5, 7), (2, 6, 8), (3, 4, 9))
((1, 5, 7), (2, 6, 9), (3, 4, 8))
((1, 5, 8), (2, 4, 7), (3, 6, 9))
((1, 5, 8), (2, 4, 9), (3, 6, 7))
((1, 5, 8), (2, 6, 7), (3, 4, 9))
((1, 5, 8), (2, 6, 9), (3, 4, 7))
((1, 5, 9), (2, 4, 7), (3, 6, 8))
((1, 5, 9), (2, 4, 8), (3, 6, 7))
((1, 5, 9), (2, 6, 7), (3, 4, 8))
((1, 5, 9), (2, 6, 8), (3, 4, 7))
((1, 6, 7), (2, 4, 8), (3, 5, 9))
((1, 6, 7), (2, 4, 9), (3, 5, 8))
((1, 6, 7), (2, 5, 8), (3, 4, 9))
((1, 6, 7), (2, 5, 9), (3, 4, 8))
((1, 6, 8), (2, 4, 7), (3, 5, 9))
((1, 6, 8), (2, 4, 9), (3, 5, 7))
((1, 6, 8), (2, 5, 7), (3, 4, 9))
((1, 6, 8), (2, 5, 9), (3, 4, 7))
((1, 6, 9), (2, 4, 7), (3, 5, 8))
((1, 6, 9), (2, 4, 8), (3, 5, 7))
((1, 6, 9), (2, 5, 7), (3, 4, 8))
((1, 6, 9), (2, 5, 8), (3, 4, 7))
答案 0 :(得分：2)
你想要的不是组合，而是排列。 3个元素有6个排列，2个排列的笛卡尔乘积有36个。下午2Ring最初怀疑你想要所有这3个置换，因为你的问题没有说明。如果问题中的代码产生了所需的输出，则表示您希望b和c置换，而不是a。最初，我编写的代码计算了a，b和c的所有排列。但是，由于a不需要置换，我们只需将其包装在列表中即可。这使我们非常接近所需的输出：
import itertools as it
a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]
for i in it.product([tuple(a)], it.permutations(b), it.permutations(c)):
print(i)
输出为36行，以
开头((1, 2, 3), (4, 5, 6), (7, 8, 9))
((1, 2, 3), (4, 5, 6), (7, 9, 8))
((1, 2, 3), (4, 5, 6), (8, 7, 9))
它已经几乎与您想要的格式相同，但转换了索引，因此o[x][y]将匹配您所需输出的o[y][x]。我们使用一些zip magic来转置它们。作为一个加号，此函数现在适用于任意数量的参数：
import itertools as it
a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]
def funnyperms(first, *rest):
for i in it.product([first], *(it.permutations(j) for j in rest)):
yield tuple(zip(*i))
for i in funnyperms(a, b, c):
print(i)
输出
((1, 4, 7), (2, 5, 8), (3, 6, 9))
((1, 4, 7), (2, 5, 9), (3, 6, 8))
((1, 4, 8), (2, 5, 7), (3, 6, 9))
((1, 4, 8), (2, 5, 9), (3, 6, 7))
((1, 4, 9), (2, 5, 7), (3, 6, 8))
((1, 4, 9), (2, 5, 8), (3, 6, 7))
((1, 4, 7), (2, 6, 8), (3, 5, 9))
((1, 4, 7), (2, 6, 9), (3, 5, 8))
((1, 4, 8), (2, 6, 7), (3, 5, 9))
((1, 4, 8), (2, 6, 9), (3, 5, 7))
((1, 4, 9), (2, 6, 7), (3, 5, 8))
((1, 4, 9), (2, 6, 8), (3, 5, 7))
((1, 5, 7), (2, 4, 8), (3, 6, 9))
((1, 5, 7), (2, 4, 9), (3, 6, 8))
((1, 5, 8), (2, 4, 7), (3, 6, 9))
((1, 5, 8), (2, 4, 9), (3, 6, 7))
((1, 5, 9), (2, 4, 7), (3, 6, 8))
((1, 5, 9), (2, 4, 8), (3, 6, 7))
((1, 5, 7), (2, 6, 8), (3, 4, 9))
((1, 5, 7), (2, 6, 9), (3, 4, 8))
((1, 5, 8), (2, 6, 7), (3, 4, 9))
((1, 5, 8), (2, 6, 9), (3, 4, 7))
((1, 5, 9), (2, 6, 7), (3, 4, 8))
((1, 5, 9), (2, 6, 8), (3, 4, 7))
((1, 6, 7), (2, 4, 8), (3, 5, 9))
((1, 6, 7), (2, 4, 9), (3, 5, 8))
((1, 6, 8), (2, 4, 7), (3, 5, 9))
((1, 6, 8), (2, 4, 9), (3, 5, 7))
((1, 6, 9), (2, 4, 7), (3, 5, 8))
((1, 6, 9), (2, 4, 8), (3, 5, 7))
((1, 6, 7), (2, 5, 8), (3, 4, 9))
((1, 6, 7), (2, 5, 9), (3, 4, 8))
((1, 6, 8), (2, 5, 7), (3, 4, 9))
((1, 6, 8), (2, 5, 9), (3, 4, 7))
((1, 6, 9), (2, 5, 7), (3, 4, 8))
((1, 6, 9), (2, 5, 8), (3, 4, 7))
将这些存储到一个集合中并与您的方法生成的值进行比较，证明它们具有相同的输出：
print(set(funnyperms(a, b, c)) == set(f))
打印True，Q.E.D。
答案 1 :(得分：0)
您可以使用itertools的产品进行所有组合
>>> from itertools import product
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> c = [7, 8, 9]
>>> A = [a,b,c]
>>> prod = list(product(*A))
>>> print(prod)
预期产出：
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9)]