Python 在多个集合中查找子集的频率

我的技能清单如下：

skills = ['Listening', 'Written_Expression','Clerical',
         'Night_Vision', 'Accounting']

skill_pairs = [{'Listening', 'Written_Expression', 2},
              {'Listening', 'Clerical', 2},
              .
              .
              {'Night_Vision', 'Accounting', 0}]

我有一个单独的集合列表，每个集合都包含与特定工作相关的技能：

job_skills =  
     [{'Listening','Written_Expression','Clerical','Night_Vision'},
     {'Chemistry','Written_Expression','Clerical','Listening'},
     .
     .
     ]

我想计算两种独特技能的每种组合作为job_skills集合子集的频率，并返回列表/集合列表，其组合和频率如下：

skills = ['Listening', 'Written_Expression','Clerical',
         'Night_Vision', 'Accounting']

skill_pairs = [{'Listening', 'Written_Expression', 2},
              {'Listening', 'Clerical', 2},
              .
              .
              {'Night_Vision', 'Accounting', 0}]

目前，我正在做以下工作：

skill_combos = []
for idx, i in enumerate(skills):
    for jdx, j in enumerate(skills[idx+1:]):
        temp = []
        for job in range(len(job_skills)):
            temp.append(set([i,j]).issubset(job_skills[job])
        skill_combos.append([i,j,sum(temp)])

这项工作完成了，但速度很慢，因为我有大约50万个技能组合。有没有更快的方法？理想情况下不使用3个循环

---

谢谢

您只需计算存在的组合，其余为零，例如：

from collections import Counter
from itertools import combinations

job_skills = [{'Listening', 'Written_Expression', 'Clerical', 'Night_Vision'},
              {'Chemistry', 'Written_Expression', 'Clerical', 'Listening'}]


counts = Counter(combo for skill_set in job_skills for combo in combinations(skill_set, 2))

for key, value in counts.items():
    print(key, value)

输出

('Clerical', 'Written_Expression') 2
('Clerical', 'Listening') 2
('Clerical', 'Night_Vision') 1
('Written_Expression', 'Listening') 2
('Written_Expression', 'Night_Vision') 1
('Listening', 'Night_Vision') 1
('Clerical', 'Chemistry') 1
('Written_Expression', 'Chemistry') 1
('Listening', 'Chemistry') 1

0

见和。如果希望字典对缺少的字典返回0，请使用以下符号换行

counts

：

输出

('Clerical', 'Written_Expression') 2
('Clerical', 'Listening') 2
('Clerical', 'Night_Vision') 1
('Written_Expression', 'Listening') 2
('Written_Expression', 'Night_Vision') 1
('Listening', 'Night_Vision') 1
('Clerical', 'Chemistry') 1
('Written_Expression', 'Chemistry') 1
('Listening', 'Chemistry') 1

0

---

不确定它是否更快，但您可以使用带计数器的组合。我的解决方案只计算一次组合。然后它使用

issubset

符号

from itertools import combinations
from collections import Counter

skills = ['Listening', 'Written_Expression','Clerical',
         'Night_Vision', 'Accounting']
job_skills = [{'Listening','Written_Expression','Clerical','Night_Vision'}, {'Chemistry','Written_Expression','Clerical','Listening'}]

pairs = {frozenset(x) for x in combinations(skills, 2)}
c = Counter(pair for pair in pairs for job in job_skills if pair.issubset(job))

for pair in pairs: # Adding the pairs that had no matches
    if pair not in c:
        c[pair] = 0

for key, count in c.items():
    print(key, count)

输出：

frozenset({'Written_Expression', 'Clerical'}) 2
frozenset({'Listening', 'Clerical'}) 2
frozenset({'Written_Expression', 'Listening'}) 2
frozenset({'Written_Expression', 'Night_Vision'}) 1
frozenset({'Listening', 'Night_Vision'}) 1
frozenset({'Clerical', 'Night_Vision'}) 1
frozenset({'Written_Expression', 'Accounting'}) 0
frozenset({'Clerical', 'Accounting'}) 0
frozenset({'Listening', 'Accounting'}) 0
frozenset({'Night_Vision', 'Accounting'}) 0

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这可能是codereview.stackexchange.com的一个很好的候选者