我有此dataframe：
user_id status_id date_created
1 1 2018-02-14 11:49:07.429000-02:00
1 4 2018-02-19 12:51:43.622000-03:00
1 3 2018-02-15 09:21:23.116000-02:00
2 3 2018-02-19 12:52:08.646000-03:00
3 3 2016-08-29 11:02:39.449000-03:00
4 4 2016-08-29 11:18:31.742000-03:00
4 2 2018-02-21 10:43:45.747000-03:00
5 3 2018-02-15 09:34:57.478000-02:00
5 2 2018-02-19 11:52:16.629000-03:00
我只想返回具有特定status_id且仅此特定状态的用户，因此，对于status_id=3，它应该返回以下内容：
user_id status_id date_created
2 3 2018-02-19 12:52:08.646000-03:00
3 3 2016-08-29 11:02:39.449000-03:00
我尝试过滤所有具有我需要的status_id的用户，但它还返回具有多个status_id的用户：
> df.loc[df.user_id.isin(df.user_id.loc[df.status_id == 3])]
user_id status_id date_created
1 1 2018-02-14 11:49:07.429000-02:00
1 4 2018-02-19 12:51:43.622000-03:00
1 3 2018-02-15 09:21:23.116000-02:00
2 3 2018-02-19 12:52:08.646000-03:00
3 3 2016-08-29 11:02:39.449000-03:00
5 3 2018-02-15 09:34:57.478000-02:00
5 2 2018-02-19 11:52:16.629000-03:00
通过使用transform nunique
df[df.groupby('user_id').status_id.transform('nunique').eq(1)].loc[lambda x :x['status_id']==3,:]
更多信息
df.groupby('user_id').status_id.transform('nunique') # get the number of unique value within each group, after this we just need to select the group only contain one value , which is index 3,4
Out[426]:
0 3
1 3
2 3
3 1
4 1
5 2
6 2
7 2
8 2
Name: status_id, dtype: int64
您可以使用df.loc[df['status_id'] == 3]如下所述
带有相关输入的Python文件
示例