如何提取集合列表中的唯一集合并统计其出现次数？

给定初始字符串列表：

['abc', 'acb', 'bac', 'foo', 'bca', 'cab', 'cba']

通过list(map(set, x))转换得到集合列表：

[{'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'f', 'o'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}]

需求一：提取唯一集合 #

由于普通集合是不可哈希类型，无法直接用set()去重，需先转为可哈希的frozenset处理，再转回普通集合：

original_sets = [{'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'f', 'o'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}]
unique_sets = [set(s) for s in set(frozenset(item) for item in original_sets)]
print(unique_sets)

输出结果：

[{'a', 'b', 'c'}, {'f', 'o'}]

需求二：统计各集合出现次数 #

同样借助frozenset作为字典键（字典键需为可哈希类型），结合defaultdict完成频次统计，最后可将键转回普通集合：

from collections import defaultdict

original_sets = [{'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'f', 'o'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}, {'a', 'b', 'c'}]
count_dict = defaultdict(int)

for s in original_sets:
    count_dict[frozenset(s)] += 1

# 转换为以普通集合为键的字典
result_dict = {set(k): v for k, v in count_dict.items()}
print(result_dict)

输出结果：

{{'a', 'b', 'c'}: 6, {'f', 'o'}: 1}

内容的提问来源于stack exchange，提问作者Ashtart