有没有一种Pythonic的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)
例如:
例如:
a = [1, 2, 3]
b = [4, 5, 6]
c = [7, 8, 9]
# ...
# there are more than 3 lists
预期输出:
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:
谢谢您的快速回复~!!
为了澄清问题:
结果是列表a、b、c的笛卡尔积的所有非重复组合。
还可以用另一种不太好看的方法完成:
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]
更新2:
通过不太优美的方法生成的预期结果:
((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))
[(1, 4, 8), (2, 5, 9), (3, 6, 7)]
? - Moses Koledoye