Create Column with ELIF in Pandas

Question

I am having trouble figuring out how to create new DataFrame column based on the values in two other columns. I need to use if/elif/else logic. But all of the documentation and examples I have found only show if/else logic. Here is a sample of what I am trying to do:

Code

df['combo'] = 'mobile' if (df['mobile'] == 'mobile') elif (df['tablet'] =='tablet') 'tablet' else 'other')

I am open to using where() also. Just having trouble finding the right syntax.

Answers:

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Method 1

In cases where you have multiple branching statements it’s best to create a function that accepts a row and then apply it along the axis=1. This is usually much faster then iteration through rows.

def func(row):
    if row['mobile'] == 'mobile':
        return 'mobile'
    elif row['tablet'] =='tablet':
        return 'tablet' 
    else:
        return 'other'

df['combo'] = df.apply(func, axis=1)

Method 2

I tried the following and the result was much faster. Hope it’s helpful for others.

df['combo'] = 'other'
df.loc[df['mobile'] == 'mobile', 'combo'] = 'mobile'
df.loc[df['tablet'] == 'tablet', 'combo'] = 'tablet'

Method 3

ELIF logic can be implemented with np.select or nested np.where:

import numpy as np

df['combo'] = np.select([df.mobile == 'mobile', df.tablet == 'tablet'], 
                        ['mobile', 'tablet'], 
                        default='other')
# or 
df['combo'] = np.where(df.mobile == 'mobile', 'mobile', 
                       np.where(df.tablet == 'tablet', 'tablet', 'other'))

Sample Data + Output:

   mobile  tablet   combo
0  mobile     bar  mobile
1     foo  tablet  tablet
2     foo     nan   other
3  mobile  tablet  mobile
4  mobile     nan  mobile
5     foo  tablet  tablet
6  mobile     bar  mobile
7  mobile  tablet  mobile
8  mobile     bar  mobile
9  mobile     nan  mobile

Method 4

Adding to np.where solution :

df['col1']= np.where(df['col'] < 3, 1,np.where( (df['col'] >3 )& (df['col'] <5),2,3))

Overall Logic is :

np.where(Condition, 'true block','false block').

With each true/false block can in turn again be nested.

Also, Notice the & for ANDing! (not 'and')

Method 5

Adding to apply lambda solution :

 df['combo'] = df.apply(lambda x: 'mobile' if x['mobile'] == 'mobile' else 
                                 ('tablet' if x[''mobile']== 'tablet' else 'other'), axis=1)

Structure :

df['column_name'] = df.apply(lambda x: 'value if true 1' if x['column_name_check_1'] == 'condition_1' else 
                                      ('value if true 2' if x['column_name_check_2'] == 'condition_2' else 
                                      ('value if true 3' if x['column_name_check_3'] == 'condition_3' else 'default_value')),axis=1)

Note :
axis 1 for format value as column / series

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0