分组间无重复的组合/排列

from itertools import product
def function1(*seqs):
  return (x for x in product(*seqs) if len(x) == len(set(x)))

>>> list(function1([0,1,2], [0,3,4], [0,2,4]))
[(0, 3, 2), (0, 3, 4), (0, 4, 2), (1, 0, 2), (1, 0, 4), (1, 3, 0), (1, 3, 2), (1, 3, 4), (1, 4, 0), (1, 4, 2), (2, 0, 4), (2, 3, 0), (2, 3, 4), (2, 4, 0)]

P¤r Wieslander为函数1给出了一个很好的通用解决方案。下面是函数2的一般解决方案

from itertools import product
def function2(*args):
    return [i for i in product(*args) if (lambda x: len(x) == len(set(x)))([k for j in i for k in j])]

当然,如果生成器表达式更符合您的目的,您可以返回它。
例如:

def function2(*args):
    return (i for i in product(*args) if (lambda x: len(x) == len(set(x)))([k for j in i for k in j]))

>>> [(x,y,z) for x in list1 for y in list2 for z in list3 if (x != y and y != z
and x != z)]
[(0, 3, 2), (0, 3, 4), (0, 4, 2), (1, 0, 2), (1, 0, 4), (1, 3, 0), (1, 3, 2), (1
, 3, 4), (1, 4, 0), (1, 4, 2), (2, 0, 4), (2, 3, 0), (2, 3, 4), (2, 4, 0)]

对于第一个

def f2(first, second):
    for a in first:
        for b in second:
            if len(set(a + b)) == 4:
                yield (a, b)

对于任何想知道的人来说,这里是最终的实现:

def function2(arg1, arg2):
    return [i for i in product(*arg1) if (lambda x: len(x) == len(set(x)))
            ([k for j in i for k in j] + arg2)]

gnibbler卓越解决方案的两个变化:

2) arg2是我要从任何组合中排除的内容的列表。将它们包含在lambda函数的参数中,可以进一步约束乘积(*arg1)的结果,而无需引入组合。

使用命名变量向您展示我对其所做的操作,下面是确切的代码:

def makeMyMenu(allmeals, dont_eat):
    return [menu for menu in product(*allmeals) if (lambda x: len(x) == len(set(x)))
            ([ingredient for meal in menu for ingredient in meal] + dont_eat)]