Question

我的数据集表示许多技术报告的作者列表。每份报告都可以由一个或多个人撰写：

a = [
['John', 'Mark', 'Jennifer'],
['John'],
['Joe', 'Mark'],
['John', 'Anna', 'Jennifer'],
['Jennifer', 'John', 'Mark']
]

我发现最常见的一对，即过去合作最多的人：

['John', 'Jennifer'] - 3 times
['John', 'Mark'] - 2 times
['Mark', 'Jennifer'] - 2 times
etc...

如何在Python中执行此操作？

Answer 1

collections.Counter使用itertools.combinations dict：

from collections import Counter
from itertools import combinations

d  = Counter()
for sub in a:
    if len(a) < 2:
        continue
    sub.sort()
    for comb in combinations(sub,2):
        d[comb] += 1

print(d.most_common())
[(('Jennifer', 'John'), 3), (('John', 'Mark'), 2), (('Jennifer', 'Mark'), 2), (('Anna', 'John'), 1), (('Joe', 'Mark'), 1), (('Anna', 'Jennifer'), 1)]

most_common()会按照最常见的顺序返回配对，您希望第一个n最常见的传递n d.most_common(n)

Answer 2

import collections
import itertools

a = [
['John', 'Mark', 'Jennifer'],
['John'],
['Joe', 'Mark'],
['John', 'Anna', 'Jennifer'],
['Jennifer', 'John', 'Mark']
]


counts = collections.defaultdict(int)
for collab in a:
    collab.sort()
    for pair in itertools.combinations(collab, 2):
        counts[pair] += 1

for pair, freq in counts.items():
    print(pair, freq)

输出：

('John', 'Mark') 2
('Jennifer', 'Mark') 2
('Anna', 'John') 1
('Jennifer', 'John') 3
('Anna', 'Jennifer') 1
('Joe', 'Mark') 1

Answer 3

您可以使用集合理解来创建一组所有数字，然后使用列表推导来计算子列表中对名称的出现次数：

>>> from itertools import combinations as comb
>>> all_nam={j for i in a for j in i}
>>> [[(i,j),sum({i,j}.issubset(t) for t in a)] for i,j in comb(all_nam,2)]

[[('Jennifer', 'John'), 3], 
 [('Jennifer', 'Joe'), 0], 
 [('Jennifer', 'Anna'), 1], 
 [('Jennifer', 'Mark'), 2], 
 [('John', 'Joe'), 0], 
 [('John', 'Anna'), 1], 
 [('John', 'Mark'), 2], 
 [('Joe', 'Anna'), 0], 
 [('Joe', 'Mark'), 1], 
 [('Anna', 'Mark'), 0]]

在列表列表中查找最常出现的对

3 个答案: