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Applying Lambda functions to Pandas Dataframe

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  • Difficulty Level : Easy
  • Last Updated : 20 Apr, 2022

In Pandas, we have the freedom to add different functions whenever needed like lambda function, sort function, etc. We can apply a lambda function to both the columns and rows of the Pandas data frame.

Syntax: lambda arguments: expression

An anonymous function which we can pass in instantly without defining a name or any thing like a full traditional function.

Example 1: Applying lambda function to single column using Dataframe.assign()

Python3




# importing pandas library
import pandas as pd
  
# creating and initializing a list
values= [['Rohan',455],['Elvish',250],['Deepak',495],
         ['Soni',400],['Radhika',350],['Vansh',450]]
 
# creating a pandas dataframe
df = pd.DataFrame(values,columns=['Name','Total_Marks'])
 
# Applying lambda function to find
# percentage of 'Total_Marks' column
# using df.assign()
df = df.assign(Percentage = lambda x: (x['Total_Marks'] /500 * 100))
 
# displaying the data frame
df

Output :

In the above example, the lambda function is applied to the ‘Total_Marks’ column and a new column ‘Percentage’ is formed with the help of it.

Example 2: Applying lambda function to multiple columns using Dataframe.assign()

Python3




# importing pandas library
import pandas as pd
 
# creating and initializing a nested list
values_list = [[15, 2.5, 100], [20, 4.5, 50], [25, 5.2, 80],
               [45, 5.8, 48], [40, 6.3, 70], [41, 6.4, 90],
               [51, 2.3, 111]]
 
# creating a pandas dataframe
df = pd.DataFrame(values_list, columns=['Field_1', 'Field_2', 'Field_3'])
 
# Applying lambda function to find
# the product of 3 columns using
# df.assign()
df = df.assign(Product=lambda x: (x['Field_1'] * x['Field_2'] * x['Field_3']))
 
# printing dataframe
df

Output :

In the above example, lambda function is applied to 3 columns i.e ‘Field_1’, ‘Field_2’, and ‘Field_3’.

Example 3: Applying lambda function to single row using Dataframe.apply()

Python3




# importing pandas and numpy libraries
import pandas as pd
import numpy as np
 
# creating and initializing a nested list
values_list = [[15, 2.5, 100], [20, 4.5, 50], [25, 5.2, 80],
               [45, 5.8, 48], [40, 6.3, 70], [41, 6.4, 90],
               [51, 2.3, 111]]
 
# creating a pandas dataframe
df = pd.DataFrame(values_list, columns=['Field_1', 'Field_2', 'Field_3'],
                  index=['a', 'b', 'c', 'd', 'e', 'f', 'g'])
 
 
# Apply function numpy.square() to square
# the values of one row only i.e. row
# with index name 'd'
df = df.apply(lambda x: np.square(x) if x.name == 'd' else x, axis=1)
 
 
# printing dataframe
df

 
 

Output :

 

 

In the above example, a lambda function is applied to row starting with ‘d’ and hence square all values corresponds to it.

 

Example 4: Applying lambda function to multiple rows using Dataframe.apply()

 

Python3




# importing pandas and numpylibraries
import pandas as pd
import numpy as np
 
# creating and initializing a nested list
values_list = [[15, 2.5, 100], [20, 4.5, 50], [25, 5.2, 80],
               [45, 5.8, 48], [40, 6.3, 70], [41, 6.4, 90],
               [51, 2.3, 111]]
 
# creating a pandas dataframe
df = pd.DataFrame(values_list, columns=['Field_1', 'Field_2', 'Field_3'],
                  index=['a', 'b', 'c', 'd', 'e', 'f', 'g'])
 
 
# Apply function numpy.square() to square
# the values of 3 rows only i.e. with row
# index name 'a', 'e' and 'g' only
df = df.apply(lambda x: np.square(x) if x.name in [
              'a', 'e', 'g'] else x, axis=1)
 
# printing dataframe
df

 
 

Output :

 

 

In the above example, a lambda function is applied to 3 rows starting with ‘a’, ‘e’, and ‘g’.

 

Example 5: Applying the lambda function simultaneously to multiple columns and rows

 

Python3




# importing pandas and numpylibraries
import pandas as pd
import numpy as np
 
# creating and initializing a nested list
values_list = [[1.5, 2.5, 10.0], [2.0, 4.5, 5.0], [2.5, 5.2, 8.0],
               [4.5, 5.8, 4.8], [4.0, 6.3, 70], [4.1, 6.4, 9.0],
               [5.1, 2.3, 11.1]]
 
# creating a pandas dataframe
df = pd.DataFrame(values_list, columns=['Field_1', 'Field_2', 'Field_3'],
                  index=['a', 'b', 'c', 'd', 'e', 'f', 'g'])
 
 
# Apply function numpy.square() to square
# the values of 2 rows only i.e. with row
# index name 'b' and 'f' only
df = df.apply(lambda x: np.square(x) if x.name in ['b', 'f'] else x, axis=1)
 
# Applying lambda function to find product of 3 columns
# i.e 'Field_1', 'Field_2' and 'Field_3'
df = df.assign(Product=lambda x: (x['Field_1'] * x['Field_2'] * x['Field_3']))
 
 
# printing dataframe
df

Output :

In this example, a lambda function is applied to two rows and three columns. 


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