米琪卡哇伊
OP 保持 list1 的顺序，并用 list2 中的元素屏蔽它。我创建了 list2 的幂集，并使用 None 对其进行扩展以匹配 list1 （= helper）的长度。然后，我迭代 helper 的唯一排列，用另一个数组掩盖一个数组，以隐藏 list1 中掩码包含非 None 值的部分。from itertools import chain, combinations, permutationsdef powerset(iterable): "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)" s = list(iterable) return chain.from_iterable(combinations(s, r) for r in range(len(s) + 1))list1 = [1, 2, 3]list2 = [4, 5]solutions = []for s in powerset(list2): # for every possibility of list2 helper = [None] * (len(list1) - len(s)) helper.extend(s) # create something like [None, None, 4] for perm in set(permutations(helper, len(helper))): # for each permutation of the helper, mask out nones solution = [] for listchar, helperchar in zip(list1, perm): if helperchar != None: solution.append(helperchar) else: solution.append(listchar) solutions.append(solution)print(solutions)# [[1, 2, 3], [4, 2, 3], [1, 4, 3], [1, 2, 4], [1, 2, 5], [5, 2, 3], [1, 5, 3], [1, 5, 4], [4, 2, 5], [5, 2, 4], [5, 4, 3], [1, 4, 5], [4, 5, 3]]