Pandas aggregate count distinct

Question

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?

Groot

Answer 1

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

Dryclean

That's it. pd.Series.nunique is what I couldn't find, well, couldn't get to work correctly. Pretty obvious in hindsight. Thanks! – Groot

This answer is outdated. You can now use nunique directly. See @Blodwyn Pig's solution below – Greenfield

Hey do you know how to get non-duplicate count? – Cathcart

Answer 2

'nunique' is an option for .agg() since pandas 0.20.0, so:

df.groupby('date').agg({'duration': 'sum', 'user_id': 'nunique'})

Kong

Is it possible to agg and get the unique values? something like duration: np.unique – Meingolda

@Meingolda Try df.groupby('date').agg({'user_id': lambda s: s.unique().reset_index(drop=True)}) – Aponeurosis

How do we get the output? – Horizon

Answer 3

Just adding to the answers already given, the solution using the string "nunique" seems much faster, tested here on ~21M rows dataframe, then grouped to ~2M

%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

Impacted

Nice catch! I guess it's b/c in a "lambda"/"other function" case it is applied sequentially, while "known" functions are applied to the whole column in a vectorized fashion. – Isentropic

which solution is from @Blodwyn Pig? – Tound

@Chogg, the fastest one! – Finzer

@Tound - sorry I changed my username. It was me. – Kong

Answer 4

If you want to get only a number of distinct values per group you can use the method nunique directly with the DataFrameGroupBy object:

df.groupby('date')['user_id'].nunique()

Genniegennifer

Answer 5

You can find it for all columns at once with the aggregate method,

df.aggregate(func=pd.Series.nunique, axis=0)
# or
df.aggregate(func='nunique', axis=0)

See aggregate | Pandas Docs

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