如何基于python嵌套列表中的公共元素计算子列表的数量？

Question

I have a list of lists like this: [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3], [3, 4]] . How can I count the lists which are sublists of more than two lists? For example, here [2, 3] and [3, 4] would be the lists that are sublists of first 3 lists. I want to get rid of them.

Answer 1

This comprehension should do it:

data = [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3], [3, 4]]
solution = [i for i in data if sum([1 for j in data if set(i).issubset(set(j))]) < 3]

Answer 2

set_list = [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3], [3, 4]]
check_list = [[2, 3], [3, 4]]
sublist_to_list = {}

for set in set_list:
    for i, sublist in enumerate(check_list):
        count = 0
        for element in sublist:
            if element in set:
                count += 1

        if count == len(sublist):
            if i not in sublist_to_list:
                sublist_to_list[i] = [set]
            else:
                sublist_to_list[i].append(set)

print(sublist_to_list)

Output: {0: [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3]], 1: [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [3, 4]]}

• which means [2, 3] is subset of [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3]]
• and [3, 4] is subset of [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [3, 4]]

Answer 3

You can first make a function that gets sub lists of a list:

def sublists(lst):
    length = len(lst)

    for size in range(1, length + 1):
        for start in range(length - size + 1):
            yield lst[start:start+size]

Which works as follows:

>>> list(sublists([1, 2, 3, 4, 5]))
[[1], [2], [3], [4], [5], [1, 2], [2, 3], [3, 4], [4, 5], [1, 2, 3], [2, 3, 4], [3, 4, 5], [1, 2, 3, 4], [2, 3, 4, 5], [1, 2, 3, 4, 5]]

Then you can use this to collect all the sublists list indices into a collections.defaultdict :

from collections import defaultdict

lsts = [[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6, 7], [2, 3], [3, 4]]

d = defaultdict(list)
for i, lst in enumerate(lsts):
    subs = sublists(lst)
    while True:
        try:
            curr = tuple(next(subs))
            d[curr].append(i)
        except StopIteration:
            break

Which will have tuple keys for the sublists, and the list indices as the values.

Then to determine sub lists that occur more than twice in all the lists, you can check if the set of all the indices has a length of more than two:

print([list(k) for k, v in d.items() if len(set(v)) > 2])

Which will give the following sublists:

[[2], [3], [4], [5], [2, 3], [3, 4], [4, 5], [2, 3, 4], [3, 4, 5], [2, 3, 4, 5]]