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Python – Create Nested Dictionary using given List

Last Updated : 14 Mar, 2023
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Given a list and dictionary, map each element of list with each item of dictionary, forming nested dictionary as value.

Input : test_dict = {‘Gfg’ : 4, ‘best’ : 9}, test_list = [8, 2] Output : {8: {‘Gfg’: 4}, 2: {‘best’: 9}} Explanation : Index-wise key-value pairing from list [8] to dict {‘Gfg’ : 4} and so on. Input : test_dict = {‘Gfg’ : 4}, test_list = [8] Output : {8: {‘Gfg’: 4}} Explanation : Index-wise key-value pairing from list [8] to dict {‘Gfg’ : 4}.

Method #1 : Using loop + zip() 

This is one of the ways in which this task can be performed. In this, we combine both the lists using zip() and loop is used to do iteration of zipped keys and dictionary construction.

Python3




# Python3 code to demonstrate working of
# Nested Dictionary with List
# Using loop + zip()
 
# initializing dictionary and list
test_dict = {'Gfg' : 4, 'is' : 5, 'best' : 9}
test_list = [8, 3, 2]
 
# printing original dictionary and list
print("The original dictionary is : " + str(test_dict))
print("The original list is : " + str(test_list))
 
# using zip() and loop to perform
# combining and assignment respectively.
res = {}
for key, ele in zip(test_list, test_dict.items()):
    res[key] = dict([ele])
         
# printing result
print("The mapped dictionary : " + str(res)) 






                        










Output


The original dictionary is : {'Gfg': 4, 'is': 5, 'best': 9}
The original list is : [8, 3, 2]
The mapped dictionary : {8: {'Gfg': 4}, 3: {'is': 5}, 2: {'best': 9}}




Time complexity: O(n*n), where n is the length of the test_list. The zip() + loop takes O(n*n) time
Auxiliary Space: O(n), extra space of size n is required


Method #2 : Using dictionary comprehension + zip()


This is yet another way in which this task can be performed. In this, we perform similar task as above method, but in one liner using dictionary comprehension




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# Python3 code to demonstrate working of 


# Nested Dictionary with List


# Using dictionary comprehension + zip()


 


# initializing dictionary and list


test_dict = {'Gfg' : 4, 'is' : 5, 'best' : 9} 


test_list = [8, 3, 2]


 


# printing original dictionary and list


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


print("The original list is : " + str(test_list))


 


# zip() and dictionary comprehension mapped in one liner to solve


res = {idx: {key : test_dict[key]} for idx, key in zip(test_list, test_dict)}


         


# printing result 


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



















                        







                        










Output


The original dictionary is : {'Gfg': 4, 'is': 5, 'best': 9}
The original list is : [8, 3, 2]
The mapped dictionary : {8: {'Gfg': 4}, 3: {'is': 5}, 2: {'best': 9}}






                                    


                                                                        

                                        

                                            

                                                

                                                    
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                                                    manjeet_04                                                

                                            

                                            

            
                                        

                                        

                                            
    

        
        
    
                                            

                                                
                                                
                                            

                                            

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