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?

score 218 · Accepted Answer · answered Sep 01 '13 at 03:31

218

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 Sep 01 '13 at 03:31

DSM

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That's it. pd.Series.nunique is what I couldn't find, well, couldn't get to work correctly. Pretty obvious in hindsight. Thanks! – dave Sep 01 '13 at 03:36
13

This answer is outdated. You can now use `nunique` directly. See @Blodwyn Pig's solution below – Ted Petrou Nov 05 '17 at 20:07
1

Hey do you know how to get non-duplicate count? – haneulkim Feb 25 '20 at 04:17

score 109 · Answer 2 · edited Oct 08 '19 at 11:40

109

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

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

edited Oct 08 '19 at 11:40

thorbjornwolf

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answered Jul 11 '17 at 21:27

Ricky McMaster

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Is it possible to agg and get the unique values? something like `duration: np.unique` – guy Sep 26 '17 at 20:31
@guy Try `df.groupby('date').agg({'user_id': lambda s: s.unique().reset_index(drop=True)})` – BallpointBen Aug 14 '18 at 19:01
How do we get the output? – Oct 15 '18 at 21:09

score 37 · Answer 3 · edited Oct 08 '19 at 13:55

37

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

edited Oct 08 '19 at 13:55

thorbjornwolf

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answered Sep 27 '17 at 15:53

user6903745

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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. – Ufos Dec 09 '17 at 20:16
which solution is from @Blodwyn Pig? – Chogg Mar 30 '18 at 19:26
@Chogg, the fastest one! – m-dz Apr 23 '18 at 14:11
@Chogg - sorry I changed my username. It was me. – Ricky McMaster Oct 16 '18 at 07:49

score 0 · Answer 4 · answered May 04 '22 at 07:12

0

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()

answered May 04 '22 at 07:12

Mykola Zotko

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score 0 · Answer 5 · answered Jan 05 '23 at 20:01

0

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

answered Jan 05 '23 at 20:01

HT.

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Pandas aggregate count distinct

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