Pandas aggregate count distinct

Use the below code:

>>> df
date duration user_id
0 2013-04-01 30 0001
1 2013-04-01 15 0001
2 2013-04-01 20 0002
3 2013-04-02 15 0002
4 2013-04-02 30 0002
>>> df.groupby("date").agg({"duration": np.sum, "user_id": pd.Series.nunique})
duration user_id
date
2013-04-01 65 2
2013-04-02 45 1
>>> df.groupby("date").agg({"duration": np.sum, "user_id": lambda x: x.nunique()})
duration user_id
date
2013-04-01 65 2
2013-04-02 45 1

The function pd.Series.nunique is much quicker:

%time _=g.agg({"id": lambda x: x.nunique()})
CPU times: user 3min 3s, sys: 2.94 s, total: 3min 6s
Wall time: 3min 20s
 
%time _=g.agg({"id": pd.Series.nunique})
CPU times: user 3min 2s, sys: 2.44 s, total: 3min 4s
Wall time: 3min 18s
 
%time _=g.agg({"id": 'nunique'})
CPU times: user 14 s, sys: 4.76 s, total: 18.8 s
Wall time: 24.4 s