python - 如何返回涉及 Pandas 中两列的特定条件的行？

Question

我有这个数据框:

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，例如，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

我尝试过滤具有所需 status_id 的所有用户，但它也会返回具有多个 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

Answer 1

通过使用transform +独特

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