Question

以下问题是关于python 3.6的。假设我有集合列表，例如

L1 = [{2,7},{2,7,8},{2,3,6,7},{1,2,4,5,7}]      
L2 = [{3,6},{1,3,4,6,7},{2,3,5,6,8}]      
L3 = [{2,5,7,8},{1,2,3,5,7,8}, {2,4,5,6,7,8}]

我需要找到L1，L2和L3的每个元素之间的所有交集。例如：

    {2,7}.intersection({3,6}).intersection({2,5,7,8})= empty  
    {2,7}.intersection({3,6}).intersection({1,2,3,5,7,8})= empty  
    {2,7}.intersection({3,6}).intersection({2,4,5,6,7,8})= empty  
    {2,7}.intersection({1,3,4,6,7}).intersection({2,5,7,8})= {7}  
    {2,7}.intersection({1,3,4,6,7}).intersection({1,2,3,5,7,8})= {7}  
    {2,7}.intersection({1,3,4,6,7}).intersection({2,4,5,6,7,8})= {7}

................................

如果我们继续这样做，我们最终得到以下集合：

{{空}，{2}，{3}，{6}，{7}，{2,3}，{2,5}，{2,6}，{2,8}，{3 ，7}，{4,7}，{6,7}}

假设：
- 我有很多列表L1，L2，L3，... Ln。我不知道我有多少名单 - 每个列表L1，L2，L3..Ln都很大，所以我无法将所有列表加载到内存中。

我的问题是：有没有办法计算顺序集合，例如，计算L1和L2之间，然后使用结果来计算L3，依此类推......

Answer 1

您可以先计算L1和L2之间所有可能的交点，然后计算该组与L3之间的交点，依此类推。

list_generator = iter([  # some generator that produces your lists 
    [{2,7}, {2,7,8}, {2,3,6,7}, {1,2,4,5,7}],      
    [{3,6}, {1,3,4,6,7}, {2,3,5,6,8}],      
    [{2,5,7,8}, {1,2,3,5,7,8}, {2,4,5,6,7,8}], 
])
# for example, you can read from a file:
# (adapt the format to your needs)
def list_generator_from_file(filename):
    with open(filename) as f:
        for line in f:
            yield list(map(lambda x: set(x.split(',')), line.strip().split('|')))
# list_generator would be then list_generator_from_file('myfile.dat')

intersections = next(list_generator)  # get first list
new_intersections = set()

for list_ in list_generator:
    for old in intersections:
        for new in list_:
            new_intersections.add(frozenset(old.intersection(new)))
    # at this point we don't need the current list any more
    intersections, new_intersections = new_intersections, set()

print(intersections)

输出看起来像{frozenset({7}), frozenset({3, 7}), frozenset({3}), frozenset({6}), frozenset({2, 6}), frozenset({6, 7}), frozenset(), frozenset({8, 2}), frozenset({2, 3}), frozenset({1, 7}), frozenset({4, 7}), frozenset({2, 5}), frozenset({2})}，与你错过的{1,7}套装除外。

Answer 2

您可以使用functools.reduce(set.intersection, sets)来处理变量输入。并itertools.product(nested_list_of_sets)与几个序列中的每一个生成一个元素的组合。

通过使用生成器函数（yield）和延迟迭代器（如itertools.product），可以减少数量级的内存使用量。

import itertools
import functools

nested_list_of_sets = [
    [{2,7}, {2,7,8}, {2,3,6,7}, {1,2,4,5,7}], 
    [{3,6}, {1,3,4,6,7}, {2,3,5,6,8}],
    [{2,5,7,8}, {1,2,3,5,7,8}, {2,4,5,6,7,8}],
]

def find_intersections(sets):
    """Take a nested sequence of sets and generate intersections"""
    for combo in itertools.product(*sets):
        yield (combo, functools.reduce(set.intersection, combo))

for input_sets, output_set in find_intersections(nested_list_of_sets):
    print('{:<55}  ->   {}'.format(repr(input_sets), output_set))

输出

({2, 7}, {3, 6}, {8, 2, 5, 7})                           ->   set()
({2, 7}, {3, 6}, {1, 2, 3, 5, 7, 8})                     ->   set()
({2, 7}, {3, 6}, {2, 4, 5, 6, 7, 8})                     ->   set()
({2, 7}, {1, 3, 4, 6, 7}, {8, 2, 5, 7})                  ->   {7}
({2, 7}, {1, 3, 4, 6, 7}, {1, 2, 3, 5, 7, 8})            ->   {7}
({2, 7}, {1, 3, 4, 6, 7}, {2, 4, 5, 6, 7, 8})            ->   {7}
({2, 7}, {2, 3, 5, 6, 8}, {8, 2, 5, 7})                  ->   {2}
({2, 7}, {2, 3, 5, 6, 8}, {1, 2, 3, 5, 7, 8})            ->   {2}
# ... etc

Online demo on repl.it

Answer 3

这可能是您正在寻找的：

res = {frozenset(frozenset(x) for x in (i, j, k)): i & j & k \
       for i in L1 for j in L2 for k in L3}

<强>解释

frozenset是必需的，因为set不可播放。字典键必须是可清除的。
循环浏览L1，L2，L3中每个长度为3的项目组合。
通过&操作计算交点，相当于set.intersection。

在列表集之间查找交集

3 个答案: