Issue
Let's say I have a log of user activity and I want to generate a report of the total duration and the number of unique users per day.
import numpy as np
import pandas as pd
df = pd.DataFrame({'date': ['2013-04-01','2013-04-01','2013-04-01','2013-04-02', '2013-04-02'],
'user_id': ['0001', '0001', '0002', '0002', '0002'],
'duration': [30, 15, 20, 15, 30]})
Aggregating duration is pretty straightforward:
group = df.groupby('date')
agg = group.aggregate({'duration': np.sum})
agg
duration
date
2013-04-01 65
2013-04-02 45
What I'd like to do is sum the duration and count distincts at the same time, but I can't seem to find an equivalent for count_distinct:
agg = group.aggregate({ 'duration': np.sum, 'user_id': count_distinct})
This works, but surely there's a better way, no?
group = df.groupby('date')
agg = group.aggregate({'duration': np.sum})
agg['uv'] = df.groupby('date').user_id.nunique()
agg
duration uv
date
2013-04-01 65 2
2013-04-02 45 1
I'm thinking I just need to provide a function that returns the count of distinct items of a Series object to the aggregate function, but I don't have a lot of exposure to the various libraries at my disposal. Also, it seems that the groupby object already knows this information, so wouldn't I just be duplicating the effort?
Solution
How about either of:
>>> 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
Answered By - DSM
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