如何按列表条件过滤Pandas DataFrame并替换临时员工的Job值？

现有如下Pandas DataFrame：

ID  Type                Job
1   Employee            Doctor
2   Contingent Worker   Doctor
3   Employee            Employee
4   Employee            Employee
5   Contingent Worker   Employee
6   Contingent Worker   Consultant
7   Contingent Worker   Trainee
8   Contingent Worker   SSS
9   Contingent Worker   Agency Worker
10  Contingent Worker   

针对Type为Contingent Worker的行，设定可接受的Job值列表：

valid_jobs = ['Agency Worker', 'Consultant']

需求：检查所有Type为Contingent Worker的行的Job值，若不在上述列表中或为空值，则将其替换为"Consultant"，得到目标DataFrame。

使用Pandas的loc索引器结合条件判断是最直观且高效的实现方式，代码如下：

import pandas as pd

# 定义合法Job列表（避免用内置关键字`list`作为变量名）
valid_jobs = ['Agency Worker', 'Consultant']

# 假设原DataFrame名为df，执行替换操作
df.loc[
    (df['Type'] == 'Contingent Worker') & 
    (~df['Job'].isin(valid_jobs) | df['Job'].isna()),
    'Job'
] = 'Consultant'

代码逻辑说明 #

• (df['Type'] == 'Contingent Worker')：精准筛选出所有临时员工的行
• ~df['Job'].isin(valid_jobs)：筛选出Job不在合法列表中的行
• df['Job'].isna()：筛选出Job为空（包括NaN或空字符串）的行
• &和|分别代表逻辑“与”和“或”，通过括号明确条件优先级，确保仅对符合要求的临时员工行进行处理
• loc直接定位目标行和列，赋值操作是Pandas中效率最高的批量修改方式之一

处理后的目标DataFrame #

ID  Type                Job
1   Employee            Doctor
2   Contingent Worker   Consultant
3   Employee            Employee
4   Employee            Employee
5   Contingent Worker   Consultant
6   Contingent Worker   Consultant
7   Contingent Worker   Consultant
8   Contingent Worker   Consultant
9   Contingent Worker   Agency Worker
10  Contingent Worker   Consultant

内容的提问来源于stack exchange，提问作者Paulo Cortez