查找集列表之间的交集

Question

下面的问题是python3.6。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，我们最终会得到以下集合：

{{empty},{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。而且我不知道我有多少个列表。
- 每个列表 L1、L2、L3..Ln 都很大，所以我无法将它们全部加载到内存中。

我的问题是：有什么方法可以顺序计算那个集合，例如，在L1和L2之间计算，然后用结果计算L3，等等。 .

Answer 1

你可以先计算出L1和L2之间所有可能的交集，然后计算那个集合和L3之间的交集等等。

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})}，除了您错过的 {1,7} 集外，它与您现有的相匹配。

Answer 2

这可能是您要找的：

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}

说明

frozenset 是必需的，因为 set 不可散列。字典键必须是可哈希的。
循环遍历 L1、L2、L3 中每个长度为 3 的项目组合。
通过&运算计算交集，相当于set.intersection.

Answer 3

您可以使用 functools.reduce(set.intersection, sets) to handle variable inputs. And itertools.product(nested_list_of_sets) 从多个序列中的每个序列生成一个元素的组合。

通过使用生成器函数 (yield) 和 itertools.product 等惰性迭代器，您可以将内存使用量减少几个数量级。

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))

输出为

({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

查找集列表之间的交集

Find intersection sets between list of sets

python

set-intersection

python-3.x