Get statistics for each group (such as count, mean, etc) using pandas GroupBy?

Question

I have a dataframe df and I use several columns from it to groupby:

df['col1','col2','col3','col4'].groupby(['col1','col2']).mean()

In the above way, I almost get the table (dataframe) that I need. What is missing is an additional column that contains number of rows in each group. In other words, I have mean but I also would like to know how many were used to get these means. For example in the first group there are 8 values and in the second one 10 and so on.

In short: How do I get group-wise statistics for a dataframe?

Blowfish

Answer 1

On groupby object, the agg function can take a list to apply several aggregation methods at once. This should give you the result you need:

df[['col1', 'col2', 'col3', 'col4']].groupby(['col1', 'col2']).agg(['mean', 'count'])

Weathers

This creates four count columns, but how to get only one? (The question asks for "an additional column" and that's what I would like too.) – Taliesin

Please see my answer if you want to get only one count column per group. – Unhallowed

What if I have a separate called Counts and instead of count the rows of the grouped type, I need to add along the column Counts. – Potentiometer

@Taliesin result = df['col1','col2','col3','col4'].groupby(['col1', 'col2']).mean() ; counts = times.groupby(['col1', 'col2']).size() ; result['count'] = counts – Deservedly

this is not working if the group by is a date and formated as datetime. – Bullhead

Answer 2

Quick Answer:

The simplest way to get row counts per group is by calling .size(), which returns a Series:

df.groupby(['col1','col2']).size()

Usually you want this result as a DataFrame (instead of a Series) so you can do:

df.groupby(['col1', 'col2']).size().reset_index(name='counts')

If you want to find out how to calculate the row counts and other statistics for each group continue reading below.

---

Detailed example:

Consider the following example dataframe:

In [2]: df
Out[2]: 
  col1 col2  col3  col4  col5  col6
0    A    B  0.20 -0.61 -0.49  1.49
1    A    B -1.53 -1.01 -0.39  1.82
2    A    B -0.44  0.27  0.72  0.11
3    A    B  0.28 -1.32  0.38  0.18
4    C    D  0.12  0.59  0.81  0.66
5    C    D -0.13 -1.65 -1.64  0.50
6    C    D -1.42 -0.11 -0.18 -0.44
7    E    F -0.00  1.42 -0.26  1.17
8    E    F  0.91 -0.47  1.35 -0.34
9    G    H  1.48 -0.63 -1.14  0.17

First let's use .size() to get the row counts:

In [3]: df.groupby(['col1', 'col2']).size()
Out[3]: 
col1  col2
A     B       4
C     D       3
E     F       2
G     H       1
dtype: int64

Then let's use .size().reset_index(name='counts') to get the row counts:

In [4]: df.groupby(['col1', 'col2']).size().reset_index(name='counts')
Out[4]: 
  col1 col2  counts
0    A    B       4
1    C    D       3
2    E    F       2
3    G    H       1

Including results for more statistics

When you want to calculate statistics on grouped data, it usually looks like this:

In [5]: (df
   ...: .groupby(['col1', 'col2'])
   ...: .agg({
   ...:     'col3': ['mean', 'count'], 
   ...:     'col4': ['median', 'min', 'count']
   ...: }))
Out[5]: 
            col4                  col3      
          median   min count      mean count
col1 col2                                   
A    B    -0.810 -1.32     4 -0.372500     4
C    D    -0.110 -1.65     3 -0.476667     3
E    F     0.475 -0.47     2  0.455000     2
G    H    -0.630 -0.63     1  1.480000     1

The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.

To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:

In [6]: gb = df.groupby(['col1', 'col2'])
   ...: counts = gb.size().to_frame(name='counts')
   ...: (counts
   ...:  .join(gb.agg({'col3': 'mean'}).rename(columns={'col3': 'col3_mean'}))
   ...:  .join(gb.agg({'col4': 'median'}).rename(columns={'col4': 'col4_median'}))
   ...:  .join(gb.agg({'col4': 'min'}).rename(columns={'col4': 'col4_min'}))
   ...:  .reset_index()
   ...: )
   ...: 
Out[6]: 
  col1 col2  counts  col3_mean  col4_median  col4_min
0    A    B       4  -0.372500       -0.810     -1.32
1    C    D       3  -0.476667       -0.110     -1.65
2    E    F       2   0.455000        0.475     -0.47
3    G    H       1   1.480000       -0.630     -0.63

---

Footnotes

The code used to generate the test data is shown below:

In [1]: import numpy as np
   ...: import pandas as pd 
   ...: 
   ...: keys = np.array([
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['C', 'D'],
   ...:         ['C', 'D'],
   ...:         ['C', 'D'],
   ...:         ['E', 'F'],
   ...:         ['E', 'F'],
   ...:         ['G', 'H'] 
   ...:         ])
   ...: 
   ...: df = pd.DataFrame(
   ...:     np.hstack([keys,np.random.randn(10,4).round(2)]), 
   ...:     columns = ['col1', 'col2', 'col3', 'col4', 'col5', 'col6']
   ...: )
   ...: 
   ...: df[['col3', 'col4', 'col5', 'col6']] = \
   ...:     df[['col3', 'col4', 'col5', 'col6']].astype(float)
   ...: 

Disclaimer:

If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.

Unhallowed

Hey, I really like your solution, particularly the last, where you use method chaining. However, since it is often necessary, to apply different aggregation functions to different columns, one could also concat the resulting data frames using pd.concat. This maybe easier to read than subsqeuent chaining – Pterous

Thanks for the "Including results for more statistics" bit! Since my next search was about flattening the resulting multiindex on columns, I'll link to the answer here: https://mcmap.net/q/55210/-how-to-flatten-a-hierarchical-index-in-columns – Langton

Great! Could you please give me a hint how to add isnull to this query to have it in one column as well? 'col4': ['median', 'min', 'count', 'isnull'] – Hetaerism

How do you broadcast the count? (I can't make transform work when using groupby(...).size – Jacktar

Instead of reset_index, another way to get a DataFrame is to use the as_index parameter: df.groupby(['col1','col2'], as_index=False).size(). I use the as_index=False as a habit for all my groupbys. – Reign

If you wanted to sort descendently by a column? (counts) – Groundsel

The use of join intrigued me, but don't you think the following gives you the equivalent thing in an easier way? Abbreviated code because in a comment: gb.agg(counts=('col3', "size"), col3_mean=('col3', 'mean'), col4_median=('col4', 'median')).reset_index() – Marielamariele

@Hetaerism isnull is not an aggregation, it's a scalar to scalar function. Compare sum(1, 2, 3) produces one value (6). Isnull produces a value per input. – Marielamariele

Answer 3

Swiss Army Knife: GroupBy.describe

Returns count, mean, std, and other useful statistics per-group.

df.groupby(['A', 'B'])['C'].describe()

           count  mean   std   min   25%   50%   75%   max
A   B                                                     
bar one      1.0  0.40   NaN  0.40  0.40  0.40  0.40  0.40
    three    1.0  2.24   NaN  2.24  2.24  2.24  2.24  2.24
    two      1.0 -0.98   NaN -0.98 -0.98 -0.98 -0.98 -0.98
foo one      2.0  1.36  0.58  0.95  1.15  1.36  1.56  1.76
    three    1.0 -0.15   NaN -0.15 -0.15 -0.15 -0.15 -0.15
    two      2.0  1.42  0.63  0.98  1.20  1.42  1.65  1.87

To get specific statistics, just select them,

df.groupby(['A', 'B'])['C'].describe()[['count', 'mean']]

           count      mean
A   B                     
bar one      1.0  0.400157
    three    1.0  2.240893
    two      1.0 -0.977278
foo one      2.0  1.357070
    three    1.0 -0.151357
    two      2.0  1.423148

_{Note: if you only need to compute 1 or 2 stats then it might be faster to use groupby.agg and just compute those columns otherwise you are performing wasteful computation.}

describe works for multiple columns (change ['C'] to ['C', 'D']—or remove it altogether—and see what happens, the result is a MultiIndexed columned dataframe).

You also get different statistics for string data. Here's an example,

df2 = df.assign(D=list('aaabbccc')).sample(n=100, replace=True)

with pd.option_context('precision', 2):
    display(df2.groupby(['A', 'B'])
               .describe(include='all')
               .dropna(how='all', axis=1))

              C                                                   D                
          count  mean       std   min   25%   50%   75%   max count unique top freq
A   B                                                                              
bar one    14.0  0.40  5.76e-17  0.40  0.40  0.40  0.40  0.40    14      1   a   14
    three  14.0  2.24  4.61e-16  2.24  2.24  2.24  2.24  2.24    14      1   b   14
    two     9.0 -0.98  0.00e+00 -0.98 -0.98 -0.98 -0.98 -0.98     9      1   c    9
foo one    22.0  1.43  4.10e-01  0.95  0.95  1.76  1.76  1.76    22      2   a   13
    three  15.0 -0.15  0.00e+00 -0.15 -0.15 -0.15 -0.15 -0.15    15      1   c   15
    two    26.0  1.49  4.48e-01  0.98  0.98  1.87  1.87  1.87    26      2   b   15

For more information, see the documentation.

---

pandas >= 1.1: DataFrame.value_counts

This is available from pandas 1.1 if you just want to capture the size of every group, this cuts out the GroupBy and is faster.

df.value_counts(subset=['col1', 'col2'])

Minimal Example

# Setup
np.random.seed(0)
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
                          'foo', 'bar', 'foo', 'foo'],
                   'B' : ['one', 'one', 'two', 'three',
                          'two', 'two', 'one', 'three'],
                   'C' : np.random.randn(8),
                   'D' : np.random.randn(8)})

df.value_counts(['A', 'B']) 

A    B    
foo  two      2
     one      2
     three    1
bar  two      1
     three    1
     one      1
dtype: int64

---

Other Statistical Analysis Tools

If you didn't find what you were looking for above, the User Guide has a comprehensive listing of supported statical analysis, correlation, and regression tools.

Brandebrandea

By doing .describe()[['count', 'mean']] you compute statistics that you would drop afterwards. Using .agg(['count', 'mean'] is a better option, about 7 times faster, as you only compute the ones actually needed – Paton

Answer 4

To get multiple stats, collapse the index, and retain column names:

df = df.groupby(['col1','col2']).agg(['mean', 'count'])
df.columns = [ ' '.join(str(i) for i in col) for col in df.columns]
df.reset_index(inplace=True)
df

Produces:

Tabithatablature

New users: warning this inserts spaces in column names. I think that should be avoided, generally speaking, so replace line 2 df.columns = [ '_'.join(str(i) for i in col) for col in df.columns] . – Dastardly

Answer 5

We can easily do it by using groupby and count. But, we should remember to use reset_index().

df[['col1','col2','col3','col4']].groupby(['col1','col2']).count().\
reset_index()

Cacodemon

Answer 6

Please try this code

new_column=df[['col1', 'col2', 'col3', 'col4']].groupby(['col1', 'col2']).count()
df['count_it']=new_column
df

I think that code will add a column called 'count it' which count of each group

Amaliaamalie

Answer 7

Create a group object and call methods like below example:

grp = df.groupby(['col1',  'col2',  'col3']) 

grp.max() 
grp.mean() 
grp.describe() 

Neral

Answer 8

If you are familiar with tidyverse R packages, here is a way to do it in python:

from datar.all import tibble, rnorm, f, group_by, summarise, mean, n, rep

df = tibble(
  col1=rep(['A', 'B'], 5), 
  col2=rep(['C', 'D'], each=5), 
  col3=rnorm(10), 
  col4=rnorm(10)
)
df >> group_by(f.col1, f.col2) >> summarise(
  count=n(),
  col3_mean=mean(f.col3), 
  col4_mean=mean(f.col4)
)

  col1 col2  n  mean_col3  mean_col4
0    A    C  3  -0.516402   0.468454
1    A    D  2  -0.248848   0.979655
2    B    C  2   0.545518  -0.966536
3    B    D  3  -0.349836  -0.915293
[Groups: ['col1'] (n=2)]

I am the author of the datar package. Please feel free to submit issues if you have any questions about using it.

Smitt

Answer 9

Another alternative:

import pandas as pd
import numpy as np

df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
                          'foo', 'bar', 'foo', 'foo'],
                   'B' : ['one', 'one', 'two', 'three',
                          'two', 'two', 'one', 'three'],
                   'C' : np.random.randn(8),
                   'D' : np.random.randn(8)})
df

    A   B       C           D
0   foo one   0.808197   2.057923
1   bar one   0.330835  -0.815545
2   foo two  -1.664960  -2.372025
3   bar three 0.034224   0.825633
4   foo two   1.131271  -0.984838
5   bar two   2.961694  -1.122788
6   foo one   -0.054695  0.503555
7   foo three 0.018052  -0.746912

---

pd.crosstab(df.A, df.B).stack().reset_index(name='count')

Output:

    A   B     count
0   bar one     1
1   bar three   1
2   bar two     1
3   foo one     2
4   foo three   1
5   foo two     2

Michaelamichaele

Answer 10

pivot_table with specific aggfuncs

For a dataframe of aggregate statistics, pivot_table can be used as well. It produces a table not too dissimilar from Excel pivot table. The basic idea is to pass in the columns to be aggregated as values= and grouper columns as index= and whatever aggregator functions as aggfunc= (all of the optimized functions that are admissible for groupby.agg are OK).

One advantage of pivot_table over groupby.agg is that for multiple columns it produces a single size column whereas groupby.agg which creates a size column for each column (all except one are redundant).

agg_df = df.pivot_table(
    values=['col3', 'col4', 'col5'], 
    index=['col1', 'col2'], 
    aggfunc=['size', 'mean', 'median']
).reset_index()
# flatten the MultiIndex column (should be omitted if MultiIndex is preferred)
agg_df.columns = [i if not j else f"{j}_{i}" for i,j in agg_df.columns]

Use named aggregation for custom column names

For custom column names, instead of multiple rename calls, use named aggregation from the beginning.

From the docs:

To support column-specific aggregation with control over the output column names, pandas accepts the special syntax in GroupBy.agg(), known as “named aggregation”, where

• The keywords are the output column names
• The values are tuples whose first element is the column to select and the second element is the aggregation to apply to that column. pandas provides the pandas.NamedAgg namedtuple with the fields ['column', 'aggfunc'] to make it clearer what the arguments are. As usual, the aggregation can be a callable or a string alias.

As an example, to produce aggregate dataframe where each of col3, col4 and col5 has its mean and count computed, the following code could be used. Note that it does the renaming columns step as part of groupby.agg.

aggfuncs = {f'{c}_{f}': (c, f) for c in ['col3', 'col4', 'col5'] for f in ['mean', 'count']}
agg_df = df.groupby(['col1', 'col2'], as_index=False).agg(**aggfuncs)

Another use case of named aggregation is if each column needs a different aggregator function. For example, if only the mean of col3, median of col4 and min of col5 are needed with custom column names, it can be done using the following code.

agg_df = df.groupby(['col1', 'col2'], as_index=False).agg(col3_mean=('col3', 'mean'), col4_median=('col4', 'median'), col5_min=('col5', 'min'))
# or equivalently,
agg_df = df.groupby(['col1', 'col2'], as_index=False).agg(**{'_'.join(p): p for p in [('col3', 'mean'), ('col4', 'median'), ('col5', 'min')]})

Shaduf

Answer 11

df_group = (df.groupby(['city','gender'])[['age',"grade"]]
     .agg([('average','mean'),('freq','count')])
     .reset_index())

city and gender are the groups columns.

age and grade are the the depend columns (which I calculate mean on them).

the name of the mean column would be average the name of the count column would be freq

Deprave