@JohnGordon 你说得完全正确。实际的列表是 100 个项目，我只是快速将其放在那里，更新了，谢谢！

@user24714692 我希望输出返回在列表列表中找到的最常见组合，然后返回找到的次数。这似乎很简单，但出于某种原因，有点难。

@user24714692 不行，因为组合模式应该只包含唯一字符，不能有重复。模式长度最多为 5。

首先将 itertools 生成的组合存储在一个集合中，以便快速进行成员资格测试。

   combinations_of_5 = set(combinations(lst, 5))

接下来将您的列表过滤为 collections.Counter ：

   data = map(tuple, lst_of_lsts)
   c = Counter(comb for comb in data if comb in combinations_of_5)
   print(c.most_common(10))

话虽如此，我认为您实际上可能正在寻找排列而不是组合。

对于这个顺序并不重要，我只是想计算某个模式在列表列表中出现的次数。这是针对几百万个列表运行的，所以我需要它尽可能高效。每个字母代表一个动作，我试图找到最常见的动作序列。

您可以简单地使用计数器，如下所示 [Counter(l) for l in lst_of_lsts] ：

代码 1

from collections import Counter

lst_of_lsts = [
    ["e", "d", "c", "b", "a", "c", "d", "b"],
    ["e", "b", "c", "a", "b", "e", "c"],
    ["a", "b", "a", "d", "b", "a", "a"],
    ["a", "b", "c", "d", "e", "c", "d"],
    ["e", "d", "c", "b", "a"]
]

counters = [Counter(l) for l in lst_of_lsts]
res = [counter.most_common(1)[0] for counter in counters]
print(counters)
print(res)

印刷

[Counter({'d': 2, 'c': 2, 'b': 2, 'e': 1, 'a': 1}), Counter({'e': 2, 'b': 2, 'c': 2, 'a': 1}), Counter({'a': 4, 'b': 2, 'd': 1}), Counter({'c': 2, 'd': 2, 'a': 1, 'b': 1, 'e': 1}), Counter({'e': 1, 'd': 1, 'c': 1, 'b': 1, 'a': 1})]
[('d', 2), ('e', 2), ('a', 4), ('c', 2), ('e', 1)]

代码 2

from collections import Counter

lst_of_lsts = [
    ["e", "d", "c", "b", "a", "c", "d", "b"],
    ["e", "b", "c", "a", "b", "e", "c"],
    ["a", "b", "a", "d", "b", "a", "a"],
    ["a", "b", "c", "d", "e", "c", "d"],
    ["e", "d", "c", "b", "a"]
]

counter = Counter([tuple(sorted(l)) for l in lst_of_lsts])
print(counter.most_common(1)[0])

印刷

(('a', 'b', 'b', 'c', 'c', 'd', 'd', 'e'), 1)

代码 3

from collections import Counter

lst_of_lsts = [
    ["e", "d", "c", "b", "a", "c", "d", "b"],
    ["e", "b", "c", "a", "b", "e", "c"],
    ["a", "b", "a", "d", "b", "a", "a"],
    ["a", "b", "c", "d", "e", "c", "d"],
    ["e", "d", "c", "b", "a"]
]

counter = Counter([tuple(sorted(set(l))) for l in lst_of_lsts])
print(counter)
print(counter.most_common(1)[0])

印刷

Counter({('a', 'b', 'c', 'd', 'e'): 3, ('a', 'b', 'c', 'e'): 1, ('a', 'b', 'd'): 1})
(('a', 'b', 'c', 'd', 'e'), 3)

代码4：

import string
from collections import Counter
from itertools import combinations

lst = string.ascii_lowercase[0:6]
combinations_of_5 = tuple(combinations(lst, 5))
# print(combinations_of_5)

lst_of_lsts = [
    ["e", "d", "c", "b", "a", "c", "d", "b"],
    ["e", "b", "c", "a", "b", "e", "c"],
    ["a", "b", "a", "d", "b", "a", "a"],
    ["a", "b", "c", "d", "e", "c", "d"],
    ["e", "d", "c", "b", "a"]
]

res = Counter([tuple(sorted(l)) for l in lst_of_lsts if len(l) == len(set(l)) and len(l) == 5])
print(res)

印刷

Counter({('a', 'b', 'c', 'd', 'e'): 1})

如果我想使用排列而不是组合，代码 4 会有很大不同吗？或者我只需从 itertools 中将组合换成排列？

@joelion2 不，你只需将其替换为：tuple(permutations(lst, 5))。问题是我真的不知道你想做什么。lst_of_lsts 由你手动定义，这几乎是 Counter 的核心部分：Counter([tuple(sorted(l)) for l in lst_of_lsts if len(l) == len(set(l)) and len(l) == 5])。只需提出新问题，并使用这些代码明确定义预期输出。