Implement different functions for different groupby objects pandas

Question

After doing some research, I found the following (Apply different functions to different items in group object: Python pandas). This is perhaps the exact same thing that I want, but I am unable to make sense of the answers that are being proposed. Let me try and explain with a simple example what I want:

import pandas as pd
import numpy as np

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

Let us say we have the simple data set built from the above that looks like this:

       B         C         D
0    one -1.758565 -1.544788
1    one -0.309472  2.289912
2    two -1.885911  0.384215
3  three  0.444186  0.551217
4    two -0.502636  2.125921
5    two -2.247551 -0.188705
6    one -0.575756  1.473056
7  three  0.640316 -0.410318

Upon grouping them on column 'B', there were 3 groups created

one
two
three

Now, how can I apply different functions on these groups, but still have them as part of the same data frame. For e.g. if I wanted to check if elements were < 0.5 in group 1, divisible by 2 in group 2 and -ve in group 3. These functions are for illustrative purposes only, the point I want to stress on is that they should be different custom functions that should be applied on each group, but the result should be something we can look at in one data frame. Any advice is appreciated.

Can you show what you mean exactly? Run the functions manually on the groups? — Mad Physicist
– Mad Physicist, Commented Aug 4, 2020 at 22:43
@MadPhysicist, I don't want to implement them manually. I just want different functions applied on different groups and then have the result in one data frame rather than handling each group separately as its own data frame — UGuntupalli
– UGuntupalli, Commented Aug 4, 2020 at 22:45

David Erickson · Accepted Answer · 2020-08-04 23:56:32Z

3

You can use np.where to define whatever logic you want:

df['Flag'] = np.where((df['B'] == 'one') & (df['C'] < 0.5), True, False)
df['Flag'] = np.where((df['B'] == 'two') & (df['C'] >= 0.5), True, df['Flag'])
df['Flag'] = np.where((df['B'] == 'three') & (df['C'] < 0.5), True, df['Flag'])

Out[85]: 
       B         C         D   Flag
0    one -1.758565 -1.544788   True
1    one -0.309472  2.289912   True
2    two -1.885911  0.384215  False
3  three  0.444186  0.551217   True
4    two -0.502636  2.125921  False
5    two -2.247551 -0.188705  False
6    one -0.575756  1.473056   True
7  three  0.640316 -0.410318  False

From there, let's say you then want to groupby the total that are True:

df = df.groupby('B')['Flag'].sum().reset_index()

       B    Flag
0    one     3.0
1  three     1.0
2    two     0.0

To implement as an adjustable custom function (per comment), you can do:

def flag(one, two, three):
    df['Flag'] = np.where((df['B'] == 'one') & (one), True, False)
    df['Flag'] = np.where((df['B'] == 'two') & (two), True, df['Flag'])
    df['Flag'] = np.where((df['B'] == 'three') & (three), True, df['Flag'])


flag(one=df['C'] < 0.5, two=df['C'] >= 0.5, three=df['C'] < 0.5)
df

B         C         D   Flag
0    one -1.758565 -1.544788   True
1    one -0.309472  2.289912   True
2    two -1.885911  0.384215  False
3  three  0.444186  0.551217   True
4    two -0.502636  2.125921  False
5    two -2.247551 -0.188705  False
6    one -0.575756  1.473056   True
7  three  0.640316 -0.410318  False

edited Aug 4, 2020 at 23:56

answered Aug 4, 2020 at 22:57

David Erickson

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2 Comments

UGuntupalli Over a year ago

David Erickson, Can the same approach be applied if a custom function needs to be applied instead of a simple logical check as you have shown in np.where ?

David Erickson Over a year ago

@UGuntupalli see my revised answer

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