I have a dataframe like this one and would like to make sure to sort it by date and have the same ordering for side for status emtries, I would like to see B entries on top before S for each date. But also keep the dates where there is only one entry.
tempDF = pd.DataFrame({ 'id': [12,12,12,12,45,45,45,51,51,51,51,51,51,76,76,76],
'date': ['2015-05-01','2015-05-22','2015-05-14','2015-05-06','2015-05-03','2015-05-12','2015-05-02','2015-05-05','2015-05-01','2015-05-23','2015-05-17','2015-05-03','2015-05-05','2015-05-04','2015-05-22','2015-05-08'],
'status': ['B','S','B','S','B','B','S','B','S','B','B','S','S','B','B','S']})
tempDF['date'] = pd.to_datetime(tempDF['date'])
I tried
tempDF.sort_values(['status']).groupby(tempDF['date']).head(2)
but this ends up the entries sorted before being grouped
If I understand correctly, first sort_values then grouby.head with sort=False:
(tempDF.sort_values(by=['date', 'status'], ascending=True)
.groupby('date', sort=False).head(2)
)
Output:
id date status
0 12 2015-05-01 B
8 51 2015-05-01 S
6 45 2015-05-02 S
4 45 2015-05-03 B
11 51 2015-05-03 S
13 76 2015-05-04 B
7 51 2015-05-05 B
12 51 2015-05-05 S
3 12 2015-05-06 S
15 76 2015-05-08 S
5 45 2015-05-12 B
2 12 2015-05-14 B
10 51 2015-05-17 B
14 76 2015-05-22 B
1 12 2015-05-22 S
9 51 2015-05-23 B
Similarly, if you want all Bs before all Ss:
tempDF.sort_values(by=['status', 'date'], ascending=True)
.groupby('date', sort=False).head(2)
)
Output:
id date status
0 12 2015-05-01 B
4 45 2015-05-03 B
13 76 2015-05-04 B
7 51 2015-05-05 B
5 45 2015-05-12 B
2 12 2015-05-14 B
10 51 2015-05-17 B
14 76 2015-05-22 B
9 51 2015-05-23 B
8 51 2015-05-01 S
6 45 2015-05-02 S
11 51 2015-05-03 S
12 51 2015-05-05 S
3 12 2015-05-06 S
15 76 2015-05-08 S
1 12 2015-05-22 S