Create Column with ELIF in Pandas

Question

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.

score 58 · Accepted Answer · answered Aug 12 '13 at 18:50

58

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)

answered Aug 12 '13 at 18:50

Viktor Kerkez

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`This is usually much faster then iteration through rows.` -- doesn't this approach iterate through rows as well? – Dudelstein Apr 08 '23 at 08:29

score 47 · Answer 2 · edited Sep 12 '17 at 08:27

47

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'

edited Sep 12 '17 at 08:27

Eric O. Lebigot

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answered Dec 18 '15 at 04:02

tbk

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This is the method recommended in the pandas cookbook. http://pandas.pydata.org/pandas-docs/stable/cookbook.html#idioms – Eric Ness Mar 02 '17 at 14:20
1

`.ix()` is officially deprecated. I am updating this answer with the current cookbook solution. – Eric O. Lebigot Sep 12 '17 at 08:25
FYI for others. In my own specific example of using `df.apply()` `242ms +- 1.98ms` vs. `.loc` `213ms +- 1.3ms` on about `3900` rows. Done using `%%timeit`. – kevin_theinfinityfund Sep 18 '20 at 21:34

ALollz · Answer 3 · 2021-03-18T16:36:24.140

27

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

edited Mar 18 '21 at 16:36

answered Feb 20 '19 at 21:37

ALollz

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this really should be the accepted solution; it's scalable, and it doesn't even require an additional import: `pd.np.select`. – circld Mar 22 '19 at 23:33
ageed - I can't get any of these calling functions to work. – GenDemo Jun 24 '21 at 00:54
how would you do an 'in' statement: example `df['combo'] = np.where(df.device in ('moblile, 'Mobile', 'Phone'), 'mobile' , np.where(df.device like '%tablet%', 'tablet', 'other'))` – GenDemo Jun 24 '21 at 00:56
1

@GenDemo you can use `isin` for checking if something is in a list: https://stackoverflow.com/questions/19960077/how-to-filter-pandas-dataframe-using-in-and-not-in-like-in-sql. IIU the other logic, for the like you can use `Series.str.contains('tablet', case=False)` – ALollz Jun 24 '21 at 01:06
@ALollz thank you. I came right with `sale_method = pd.DataFrame(model_data['Sale Method'].str.upper()) sale_method['sale_classification'] = np.where(sale_method['Sale Method'].isin (['PRIVATE']), 'private' ,np.where(sale_method['Sale Method'].str.contains('AUCTION'), 'auction' , 'other'))` – GenDemo Jun 24 '21 at 01:41

score 3 · Answer 4 · edited Jun 10 '19 at 09:35

3

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')

edited Jun 10 '19 at 09:35

Sfili_81

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answered Jun 10 '19 at 09:08

Abhijeet Kelkar

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Fariliana Eri · Answer 5 · 2021-09-02T06:36:54.893

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

Create Column with ELIF in Pandas

5 Answers5

Sample Data + Output:

Linked