问Python子列表在list2中没有元素，在list3中有一个元素

EN

def subgroup_join(data, exclude, include):
    exclude = set(exclude)  # turn into set for faster lookup/compare
    include = set(include)  # turn into set for faster lookup/compare
    data = [set(element) - exclude for element in data]  # remove excluded elements
    results = [set()]  # init with an empty set
    for element in data:  # loop through our remaining elements
        groups = []  # store elements / current results filtered by exclude list
        ignore_element = False  # flag if we should add the element as a standalone
        for result in results:  # go through each subgroup in the results
            if element & result:  # if the current element has common items with the result
                result |= element  # ... concatenate both into a subgroup
                ignore_element = True
            groups.append(result)  # add the current result subgroup
        if not ignore_element:  # add only if the element wasn't concatenated
            groups.append(element)  # add the current element
        results = groups  # our element store becomes our new results set
    return sorted([sorted(res) for res in results if result & include])  # sort & return

list1 = [[1, 4], [4, 5], [5, 7], [6, 7], [7, 8], [9, 7], [10, 9], [8, 10], [8, 11], [8, 13], [13, 15]]
list2 = [7, 8]
list3 = [1, 6, 11, 13]

print(subgroup_join(list1, list2, list3))
# prints: [[1, 4, 5], [6], [11], [13, 15]]

list1 = [[1, 4], [4, 5], [5, 7], [6, 7], [9, 7], [10, 9], [8, 10], [8, 11], [8, 13], [6, 8], [20, 2], [20, 3], [11, 14], [14, 16], [13, 15]]
list2 = [7, 8, 20]
list3 = [1, 2, 3, 16, 15]

print(subgroup_join(list1, list2, list3))
# prints: [[1, 4, 5], [2], [3], [11, 14, 16], [13, 15]]

zwer_join - 100,000 loops: 2.849 s; per loop: 28.399 µs
kuhn_join - 100,000 loops: 3.071 s; per loop: 30.706 µs
nuag_join -   1,000 loops: 15.82 s; per loop: 15.819 ms (had to reduce the number of loops)

list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[6,8],[20,2],[20,3],[11,14],[14,16],[13,15]]
list2 = [7,8,20]
list3 = [1,2,3,16,15]

list1a = [set(l) for l in list1]
#removing elements from list1 that contain numbers of list2:
for x in list2:
    for l in list(list1a):
        if x in l:
            l.remove(x)

#joining sub-lists in list1:
list1b = [set(l) for l in list1a]
list1c = []
while list1b:
    s1 = list1b.pop(0)
    for s2 in list(list1b):
        if s1 & s2:
            s1 |= s2
            list1b.remove(s2)
    list1c.append(s1)

#generating final list with only sub-lists that contain elements of list2
list1_new = sorted([sorted(list(s)) for s in list1c if s & set(list3)])

[[1, 4, 5], [6], [11], [13, 15]]

[[1, 4, 5], [2], [3], [11, 14, 16], [13, 15]]

def join(list1, list2):
    l = []
    for ee in list1:
        # We consider here that list1 only have pairs
        if ee[0] not in list2 and ee[1] not in list2:
            flat_l = [x for e in l for x in e]
            if ee[0] in flat_l or ee[1] in flat_l:
                for i, e in enumerate(l):
                    if ee[0] in e:
                        l[i].append(ee[1])
                    if ee[1] in e:
                        l[i].append(ee[0])
            else:
                l.append(ee)
    return l

def f(list1,list2,list3):
    l = [[e] for e in list3]
    list1 = join(list1, list2)
    for ee in list1:
        flat_l = [x for e in l for x in e]
        for e in ee:
            if e in flat_l:
                for i in range(len(l)):
                    if e in l[i]:
                        l[i] = list(set(l[i]+ee))
    print(l)

list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[13,15]]
list2 = [7,8]
list3 = [1,6,11,13]

f(list1,list2,list3)
# [[1, 4, 5], [6], [11], [13, 15]]

list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[6,8],[20,2],[20,3],[11,14],[14,16],[13,15]]
list2 = [7,8,20]
list3 = [1,2,3,16,15]

f(list1,list2,list3)
# [[1, 4, 5], [2], [3], [16, 11, 14], [13, 15]]