计算列表中所有集合组合的交集数量

Question

我有一组集合，我想找出每个集合组合中，只有在交集中出现的项目数量。简单来说，我想做的事情就像在维恩图中计算数字一样。

一个简单的例子可能会让这个问题更清楚。

a = set(1,2,5,10,12)
b = set(1,2,6,9,12,15)
c = set(1,2,7,8,15)

我最终应该得到的结果是，只有在以下情况下找到的项目数量：

• a
• b
• c
• a和b的交集
• a和c的交集
• b和c的交集
• a、b和c的交集

一种不太灵活的做法是

num_a = len(a - b - c)  # len(set([5,10])) -> 2
num_b = len(b - a - c)  # len(set([6,9])) -> 2
num_c = len(c - a - b)  # len(set([7,8])) -> 2

num_ab = len((a & b) - c)  # 1
num_ac = len((a & c) - b)  # 0
num_bc = len((b & c) - a)  # 1

num_abc = len(a & b & c)  # 2

虽然这种方法适用于3个集合，但我的集合是动态变化的。

Answer 1

你可以使用 itertools.combinations 来获取所有可能的组合。http://docs.python.org/2/library/itertools.html

Answer 2

我建议你试试位掩码：

sets = [
    set([1,2,5,10,12]),
    set([1,2,6,9,12,15]),
    set([1,2,7,8,15]),
]

d = {}

for n, s in enumerate(sets):
    for i in s:
        d[i] = d.get(i, 0) | (1 << n)

for mask in range(1, 2**len(sets)):
    cnt = sum(1 for x in d.values() if x & mask == mask)
    num = ','.join(str(j) for j in range(len(sets)) if mask & (1 << j))
    print 'number of items in set(s) %s = %d' % (num, cnt)

你输入的结果是：

number of items in set(s) 0 = 5
number of items in set(s) 1 = 6
number of items in set(s) 0,1 = 3
number of items in set(s) 2 = 5
number of items in set(s) 0,2 = 2
number of items in set(s) 1,2 = 3
number of items in set(s) 0,1,2 = 2

Answer 3

如果我理解正确的话，像这样的代码应该可以正常运行：

from itertools import combinations

def venn_count(named_sets):
    names = set(named_sets)
    for i in range(1, len(named_sets)+1):
        for to_intersect in combinations(sorted(named_sets), i):
            others = names.difference(to_intersect)
            intersected = set.intersection(*(named_sets[k] for k in to_intersect))
            unioned = set.union(*(named_sets[k] for k in others)) if others else set()
            yield to_intersect, others, len(intersected - unioned)


ns = {"a": {1,2,5,10,12}, "b": {1,2,6,9,12,15}, "c": {1,2,7,8,15}}
for intersected, unioned, count in venn_count(ns):
    print 'len({}{}) = {}'.format(' & '.join(sorted(intersected)),
                                  ' - ' + ' - '.join(sorted(unioned)) if unioned else '',
                                  count)

执行后会得到：

len(a - b - c) = 2
len(b - a - c) = 2
len(c - a - b) = 2
len(a & b - c) = 1
len(a & c - b) = 0
len(b & c - a) = 1
len(a & b & c) = 2