Python:在列表列表中查找最频繁出现的任意长度的组合

Question

如何在列表列表中找到最常出现的组合。组合长度可以是任意的。
所以，样本数据:

l = [['action','mystery','horror','thriller'],
 ['drama','romance'],
 ['comedy','drama','romance'],
 ['scifi','mystery','horror','thriller'],
 ['horror','mystery','thriller']]

预期输出:

'mystery','horror','thriller' - 3 times
'drama','romance' - 2 times

在 this post 的帮助下，我能够找出大多数出现的对(2的组合)，但是如何扩展它可以找到任何长度的组合。
编辑:根据@CrazyChucky 的评论:
样本输入:

l = [['action','mystery','horror','thriller'],
     ['drama','romance'],
     ['comedy','drama','romance'],
     ['scifi','mystery','horror','thriller'],
     ['horror','mystery','thriller'],
     ['mystery','horror']]

预期输出:

'mystery','horror' - 4 times
'mystery','horror','thriller' - 3 times
'drama','romance' - 2 times

Answer 1

您可以修改该问题中的代码以迭代每个子列表中每个可能大小的所有可能组合:

from collections import Counter
from itertools import combinations

l = [['action','mystery','horror','thriller'],
 ['drama','romance'],
 ['comedy','drama','romance'],
 ['scifi','mystery','horror','thriller'],
 ['horror','mystery','thriller']]
d  = Counter()
for sub in l:
    if len(sub) < 2:
        continue
    sub.sort()
    for sz in range(2, len(sub)+1):
        for comb in combinations(sub, sz):
            d[comb] += 1

print(d.most_common())

输出:

[
 (('horror', 'mystery'), 3),
 (('horror', 'thriller'), 3),
 (('mystery', 'thriller'), 3),
 (('horror', 'mystery', 'thriller'), 3),
 (('drama', 'romance'), 2),
 (('action', 'horror'), 1),
 (('action', 'mystery'), 1),
 (('action', 'thriller'), 1),
 (('action', 'horror', 'mystery'), 1),
 (('action', 'horror', 'thriller'), 1),
 (('action', 'mystery', 'thriller'), 1),
 (('action', 'horror', 'mystery', 'thriller'), 1),
 (('comedy', 'drama'), 1),
 (('comedy', 'romance'), 1),
 (('comedy', 'drama', 'romance'), 1),
 (('horror', 'scifi'), 1),
 (('mystery', 'scifi'), 1),
 (('scifi', 'thriller'), 1),
 (('horror', 'mystery', 'scifi'), 1),
 (('horror', 'scifi', 'thriller'), 1),
 (('mystery', 'scifi', 'thriller'), 1),
 (('horror', 'mystery', 'scifi', 'thriller'), 1)
]

要获得计数最高的流派，您可以在计数器上进行迭代:

most_frequent = [g for g, cnt in d.items() if cnt == d.most_common(1)[0][1]]