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Python – Sort Nested keys by Value

Last Updated : 10 May, 2023
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Sometimes, while working with data records, we can have a problem in which we need to perform the sorting of nested keys of dictionary by the value of occurrence. This can have applications in arranging scores, prices etc. Lets discuss a way in which this task can be performed.

Method #1 : Using sorted() + lambda + generator expression 

The combination of above methods can be used to perform this task. In this, we perform the task of sorting using sorted() and lambda and generator expression are used to bind and extracting values of dictionaries. 

Python3




# Python3 code to demonstrate working of
# Sort Nested keys by Value
# Using sorted() + generator expression + lambda
 
# initializing dictionary
test_dict = {'Nikhil' : {'English' : 5, 'Maths' :  2, 'Science' : 14},
             'Akash' : {'English' : 15, 'Maths' :  7, 'Science' : 2},
             'Akshat' : {'English' : 5, 'Maths' :  50, 'Science' : 20}}
 
# printing original dictionary
print("The original dictionary : " + str(test_dict))
 
# Sort Nested keys by Value
# Using sorted() + generator expression + lambda
res = {key : dict(sorted(val.items(), key = lambda ele: ele[1]))
       for key, val in test_dict.items()}
     
# printing result
print("The sorted dictionary : " + str(res)) 






                        










Output : 




The original dictionary : {‘Nikhil’: {‘English’: 5, ‘Maths’: 2, ‘Science’: 14}, ‘Akash’: {‘English’: 15, ‘Maths’: 7, ‘Science’: 2}, ‘Akshat’: {‘English’: 5, ‘Maths’: 50, ‘Science’: 20}} The sorted dictionary : {‘Nikhil’: {‘Maths’: 2, ‘English’: 5, ‘Science’: 14}, ‘Akash’: {‘Science’: 2, ‘Maths’: 7, ‘English’: 15}, ‘Akshat’: {‘English’: 5, ‘Science’: 20, ‘Maths’: 50}}






Time Complexity: O(n*nlogn), where n is the length of the list test_list 
Auxiliary Space: O(n), where n is the number of elements in the res list 


Method #2: Using dictionary comprehension and itemgetter


    

  • Import the “itemgetter” module from the “operator” module.
    • 

  • Initialize an empty dictionary “res”.
    • 

  • Use a nested dictionary comprehension to iterate over the key-value pairs of “test_dict”.
    • 

  • For each key-value pair, sort the nested dictionary by values using the “sorted” function and passing the key and the “itemgetter” function as arguments.Create a new dictionary with the 
    • 

  • sorted nested dictionary and add it to the “res” dictionary.
    • 

  • Return the “res” dictionary.
    • 





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from operator import itemgetter


 


# initializing dictionary


test_dict = {'Nikhil' : {'English' : 5, 'Maths' :  2, 'Science' : 14},


             'Akash' : {'English' : 15, 'Maths' :  7, 'Science' : 2},


             'Akshat' : {'English' : 5, 'Maths' :  50, 'Science' : 20}}


 


# printing original dictionary


print("The original dictionary : " + str(test_dict))


 


# Sort Nested keys by Value


# Using dictionary comprehension and itemgetter


res = {key : dict(sorted(val.items(), key=itemgetter(1)))


       for key, val in test_dict.items()}


 


# printing result 


print("The sorted dictionary : " + str(res))



















                        







                        










Output


The original dictionary : {'Nikhil': {'English': 5, 'Maths': 2, 'Science': 14}, 'Akash': {'English': 15, 'Maths': 7, 'Science': 2}, 'Akshat': {'English': 5, 'Maths': 50, 'Science': 20}}
The sorted dictionary : {'Nikhil': {'Maths': 2, 'English': 5, 'Science': 14}, 'Akash': {'Science': 2, 'Maths': 7, 'English': 15}, 'Akshat': {'English': 5, 'Science': 20, 'Maths': 50}}




Time complexity: O(n log n), where n is the number of elements in the dictionary. 
Auxiliary space: O(n), where n is the number of elements in the dictionary. 




                                    


                                                                        

                                        

                                            

                                                

                                                    
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                                                    manjeet_04                                                

                                            

                                            

            
                                        

                                        

                                            
    

        
        
    
                                            

                                                
                                                
                                            

                                            

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