在 Pandas 中，如何从基于另一个数据帧的数据帧中删除行? [英] In Pandas, how to delete rows from a Data Frame based on another Data Frame?

问题描述

我有 2 个数据帧，一个名为 USERS，另一个名为 EXCLUDE.他们都有一个名为email"的字段.

I have 2 Data Frames, one named USERS and another named EXCLUDE. Both of them have a field named "email".

基本上，我想删除 USERS 中包含在 EXCLUDE 中的电子邮件的每一行.

Basically, I want to remove every row in USERS that has an email contained in EXCLUDE.

我该怎么做?

推荐答案

您可以使用 boolean 索引 和条件 isin，反转布尔值 Series 是由 ~:

You can use boolean indexing and condition with isin, inverting boolean Series is by ~:

import pandas as pd

USERS = pd.DataFrame({'email':['[e-mail]','[e-mail]','[e-mail]','[e-mail]','[e-mail]']})
print (USERS)
     email
0  [e-mail]
1  [e-mail]
2  [e-mail]
3  [e-mail]
4  [e-mail]

EXCLUDE = pd.DataFrame({'email':['[e-mail]','[e-mail]']})
print (EXCLUDE)
     email
0  [e-mail]
1  [e-mail]

print (USERS.email.isin(EXCLUDE.email))
0     True
1    False
2    False
3    False
4     True
Name: email, dtype: bool

print (~USERS.email.isin(EXCLUDE.email))
0    False
1     True
2     True
3     True
4    False
Name: email, dtype: bool

print (USERS[~USERS.email.isin(EXCLUDE.email)])
     email
1  [e-mail]
2  [e-mail]
3  [e-mail]

<小时>

merge:

df = pd.merge(USERS, EXCLUDE, how='outer', indicator=True)
print (df)
     email     _merge
0  [e-mail]       both
1  [e-mail]  left_only
2  [e-mail]  left_only
3  [e-mail]  left_only
4  [e-mail]       both

print (df.loc[df._merge == 'left_only', ['email']])
     email
1  [e-mail]
2  [e-mail]
3  [e-mail]

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