Multiple aggregations of the same column using pandas GroupBy.agg()

PythonPandasDataframeAggregatePandas Groupby

Python Problem Overview


Is there a pandas built-in way to apply two different aggregating functions f1, f2 to the same column df["returns"], without having to call agg() multiple times?

Example dataframe:

import pandas as pd
import datetime as dt
import numpy as np

pd.np.random.seed(0)
df = pd.DataFrame({
         "date"    :  [dt.date(2012, x, 1) for x in range(1, 11)], 
         "returns" :  0.05 * np.random.randn(10), 
         "dummy"   :  np.repeat(1, 10)
}) 

The syntactically wrong, but intuitively right, way to do it would be:

# Assume `f1` and `f2` are defined for aggregating.
df.groupby("dummy").agg({"returns": f1, "returns": f2})

Obviously, Python doesn't allow duplicate keys. Is there any other manner for expressing the input to agg()? Perhaps a list of tuples [(column, function)] would work better, to allow multiple functions applied to the same column? But agg() seems like it only accepts a dictionary.

Is there a workaround for this besides defining an auxiliary function that just applies both of the functions inside of it? (How would this work with aggregation anyway?)

Python Solutions


Solution 1 - Python

You can simply pass the functions as a list:

In [20]: df.groupby("dummy").agg({"returns": [np.mean, np.sum]})
Out[20]:         
           mean       sum
dummy                    
1      0.036901  0.369012

or as a dictionary:

In [21]: df.groupby('dummy').agg({'returns':
                                  {'Mean': np.mean, 'Sum': np.sum}})
Out[21]: 
        returns          
           Mean       Sum
dummy                    
1      0.036901  0.369012

To avoid deprecation warining:

df.groupby('dummy').agg(Mean=('returns', np.mean),
                        Sum=('returns', np.sum))

Solution 2 - Python

TLDR; Pandas groupby.agg has a new, easier syntax for specifying (1) aggregations on multiple columns, and (2) multiple aggregations on a column. So, to do this for pandas >= 0.25, use

df.groupby('dummy').agg(Mean=('returns', 'mean'), Sum=('returns', 'sum'))

           Mean       Sum
dummy                    
1      0.036901  0.369012

OR

df.groupby('dummy')['returns'].agg(Mean='mean', Sum='sum')

           Mean       Sum
dummy                    
1      0.036901  0.369012

#Pandas >= 0.25: Named Aggregation

Pandas has changed the behavior of GroupBy.agg in favour of a more intuitive syntax for specifying named aggregations. See the 0.25 docs section on Enhancements as well as relevant GitHub issues GH18366 and GH26512.

From the documentation, > 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.

You can now pass a tuple via keyword arguments. The tuples follow the format of (<colName>, <aggFunc>).

import pandas as pd

pd.__version__                                                                                                                            
# '0.25.0.dev0+840.g989f912ee'

# Setup
df = pd.DataFrame({'kind': ['cat', 'dog', 'cat', 'dog'],
                   'height': [9.1, 6.0, 9.5, 34.0],
                   'weight': [7.9, 7.5, 9.9, 198.0]
})

df.groupby('kind').agg(
    max_height=('height', 'max'), min_weight=('weight', 'min'),)
              
      max_height  min_weight
kind                        
cat          9.5         7.9
dog         34.0         7.5

Alternatively, you can use pd.NamedAgg (essentially a namedtuple) which makes things more explicit.

df.groupby('kind').agg(
    max_height=pd.NamedAgg(column='height', aggfunc='max'), 
    min_weight=pd.NamedAgg(column='weight', aggfunc='min')
)

      max_height  min_weight
kind                        
cat          9.5         7.9
dog         34.0         7.5
    

It is even simpler for Series, just pass the aggfunc to a keyword argument.

df.groupby('kind')['height'].agg(max_height='max', min_height='min')    

      max_height  min_height
kind                        
cat          9.5         9.1
dog         34.0         6.0       

Lastly, if your column names aren't valid python identifiers, use a dictionary with unpacking:

df.groupby('kind')['height'].agg(**{'max height': 'max', ...})

                                                                                                           

Pandas < 0.25

In more recent versions of pandas leading upto 0.24, if using a dictionary for specifying column names for the aggregation output, you will get a FutureWarning:

df.groupby('dummy').agg({'returns': {'Mean': 'mean', 'Sum': 'sum'}})
# FutureWarning: using a dict with renaming is deprecated and will be removed 
# in a future version

Using a dictionary for renaming columns is deprecated in v0.20. On more recent versions of pandas, this can be specified more simply by passing a list of tuples. If specifying the functions this way, all functions for that column need to be specified as tuples of (name, function) pairs.

df.groupby("dummy").agg({'returns': [('op1', 'sum'), ('op2', 'mean')]})

        returns          
            op1       op2
dummy                    
1      0.328953  0.032895

Or,

df.groupby("dummy")['returns'].agg([('op1', 'sum'), ('op2', 'mean')])

            op1       op2
dummy                    
1      0.328953  0.032895

Solution 3 - Python

Would something like this work:

In [7]: df.groupby('dummy').returns.agg({'func1' : lambda x: x.sum(), 'func2' : lambda x: x.prod()})
Out[7]: 
              func2     func1
dummy                        
1     -4.263768e-16 -0.188565

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionelyView Question on Stackoverflow
Solution 1 - PythonbmuView Answer on Stackoverflow
Solution 2 - Pythoncs95View Answer on Stackoverflow
Solution 3 - PythonChang SheView Answer on Stackoverflow