通过连接初始列表的排列形成的唯一元素列表

Question

我想合并几个列表，每个列表都应该保留一个排列。

这是一个例子：

我想合并这些列表

[[0, 7], [2, 4], [0, 1, 7], [0, 1, 4, 7]]

我想要获得的输出是例如此列表

 [2, 4, 0, 7, 1]

或者正如 Sembei Norimaki 对任务的描述：

the result must be a list of unique elements formed by concatenating permutations of the initial lists.

解决方案不是唯一的，可能并不总是有可能的解决方案

Answer 1

第三次幸运。这有点俗气 - 它检查源列表元素的每个排列以查看哪些是有效的：

from itertools import permutations


def check_sublist(sublist, candidate):
    # a permutation of sublist must exist within the candidate list
    sublist = set(sublist)
    # check each len(sublist) portion of candidate
    for i in range(1 + len(candidate) - len(sublist)):
        if sublist == set(candidate[i : i + len(sublist)]):
            return True
    return False


def check_list(input_list, candidate):
    for sublist in input_list:
        if not check_sublist(sublist, candidate):
            return False
    return True


def find_candidate(input_list):
    # flatten input_list and make set of unique values
    values = {x for sublist in input_list for x in sublist}
    for per in permutations(values):
        if check_list(input_list, per):
            print(per)


find_candidate([[0, 7], [2, 4], [0, 1, 7], [0, 1, 4, 7]])
# (0, 7, 1, 4, 2)
# (1, 0, 7, 4, 2)
# (1, 7, 0, 4, 2)
# (2, 4, 0, 7, 1)
# (2, 4, 1, 0, 7)
# (2, 4, 1, 7, 0)
# (2, 4, 7, 0, 1)
# (7, 0, 1, 4, 2)

你肯定会更好地应用图论知识并使用图形库，但这超出了我目前的驾驶室！

通过连接初始列表的排列形成的唯一元素列表

list of unique elements formed by concatenating permutations of the initial lists

python

list

concatenation

permutation