使用Pandas的.isin()进行布尔索引未返回预期结果的问题求助

嗨，我来帮你捋清楚这个问题～

首先得说清楚你原来的代码为什么没得到预期结果：
你写的couples['man'][~couples['man'].isin(couples['woman'])]里，isin()的逻辑是检查整个man的爱好列表是否存在于couples['woman']这一列的所有列表中，而不是拿当前行的man爱好和同一行的woman爱好做元素级别的对比。比如第一行的man列表['fishing', 'biking', 'reading']并不存在于woman列的任何一个列表里，所以~isin()返回True，直接把整个man列表保留下来了，这显然不是你要的“找出单个爱好差异”的效果。

要实现“每行独立对比，找出一方有而另一方没有的爱好”，咱们可以用Pandas的apply()方法逐行处理，结合集合的差集操作来做——集合求差集正好就是找“我有你没有”的元素，非常顺手。

解决代码示例 #

import numpy as np
import pandas as pd

couples = pd.DataFrame({
    'man': [
        ['fishing', 'biking', 'reading'],
        ['hunting', 'mudding', 'fishing'],
        ['reading', 'movies', 'running'],
        ['running', 'reading', 'biking', 'mudding'],
        ['movies', 'reading', 'yodeling']
    ],
    'woman': [
        ['biking', 'reading', 'movies'],
        ['fishing', 'drinking'],
        ['knitting', 'reading'],
        ['running', 'biking', 'fishing', 'movies'],
        ['movies']
    ]
})

# 计算男人有但妻子没有的爱好
couples['man_only_hobbies'] = couples.apply(
    lambda row: list(set(row['man']) - set(row['woman'])),
    axis=1
)

# 计算女人有但丈夫没有的爱好
couples['woman_only_hobbies'] = couples.apply(
    lambda row: list(set(row['woman']) - set(row['man'])),
    axis=1
)

print(couples)

运行结果 #

man                               woman man_only_hobbies woman_only_hobbies
0           [fishing, biking, reading]           [biking, reading, movies]       [fishing]           [movies]
1          [hunting, mudding, fishing]                 [fishing, drinking]  [hunting, mudding]           [drinking]
2           [reading, movies, running]                 [knitting, reading]  [movies, running]           [knitting]
3  [running, reading, biking, mudding]  [running, biking, fishing, movies]  [reading, mudding]           [fishing, movies]
4          [movies, reading, yodeling]                            [movies]  [reading, yodeling]                  []

补充说明 #

• 用set()转换是因为集合支持快速的差集运算，效率比手动遍历列表高很多；
• 如果需要保留原爱好列表的顺序，可以不用集合，改用列表推导式筛选：

couples['man_only_hobbies'] = couples.apply(
    lambda row: [hobby for hobby in row['man'] if hobby not in row['woman']],
    axis=1
)

这样结果会严格遵循原列表里的爱好顺序，比如第一行还是['fishing']，和集合方法的结果一致，只是顺序保留了原始排列。

备注：内容来源于stack exchange，提问作者katjacodes