Question

我有一个包含数千个与此类似的集合的列表：

set_list = [a, b, c, d]

列表中的每个设置如下所示：

a = set([1, 2, 3, 4, 5])
b = set([4, 5, 6, 7, 7, 9])
c = set([1, 2, 6, 8, 10, 12, 45])
d = set([11, 3, 23, 3, 4, 44])

我想进行设置操作：例如，列表中的每个设置都使用X-（YUZUAUB ......等），如下所示：在set_list中的所有元素上应用此操作后，新元素将如下所示：

a = a.difference(b.union(c, d))
b = b.difference(c.union(a, d))
c = c.difference(d.union(b, a))
d = d.difference(a.union(c, b))

我如何做到这一点？

Answer 1

一种可能性是利用multiset module来预先计算fetchData(){ return this.http.get('assets/ninjas.json'); }中所有元素的多集并集，就像这样：

set_list

在这里，{{ 1}}用您的符号计算from multiset import Multiset union = sum(set_list, Multiset()) set_list = [s - (union - s) for s in set_list]。

有关仅使用标准库实现的方法（更详细地说，请参见Aran-Fey's answer）。

Answer 2

如果我理解正确，那么您希望每个集合都具有差异，而其余集合则需要并集。我将使用循环以及functools.reduce和operator.or_：

设置

import functools
import operator

a = set([1, 2, 3, 4, 5])
b = set([4, 5, 6, 7, 7, 9])
c = set([1, 2, 6, 8, 10, 12, 45])
d = set([11, 3, 23, 3, 4, 44])
set_list = [a, b, c, d]

循环并保存结果

# I don't know what you want to do with the results so
# I'll save them in a list...
results = [] 
for i in set_list:
    list_copy = list(set_list)
    list_copy.remove(i)
    r = i - functools.reduce(operator.or_, list_copy)
    results.append(r)

print(results)
# prints [set(), {9, 7}, {8, 10, 12, 45}, {11, 44, 23}]

Answer 3

[value - {item for subset in set_list[0:index] + set_list[index + 1:] for item in subset} for index, value in enumerate(set_list)]

这意味着：

result = []
for index, value in enumerate(set_list):
    union = {
        item
        for subset in set_list[0:index] + set_list[index + 1:]
        for item in subset
    }
    result.append(value - union)

print(result)

输出：

[set(), {9, 7}, {8, 10, 12, 45}, {11, 44, 23}]

Answer 4

这是使用标准库中的NPE's answer对collections.Counter的重新实现：

from collections import Counter

def mutual_difference(set_list):
    # create a multiset out of the union of all sets
    union = Counter()
    for s in set_list:
        union.update(s)

    new_set_list = []
    for s in set_list:
        # subtract s from the union and discard 0-count elements
        union.subtract(s)
        union += {}

        # take the difference
        new_set_list.append(s.difference(union))

        # add s to the union again
        union.update(s)

    return new_set_list

示例：

>>> mutual_difference([{1,2}, {2,3}, {1,4,5}])
[set(), {3}, {4, 5}]

在元素列表上设置操作

4 个答案: