Python - How to Sort a Dictionary by Kth Index Value

While working with Python, one might come to a problem in which one needs to perform a sort on dictionary based on Kth index of value list. This can be typically in case of scoring or web development. Let’s discuss a method by which this task can be performed.

Input : test_dict = {‘gfg’ : [5, 6, 7], ‘is’ : [1, 4, 7], ‘best’ : [8, 3, 1]}, K = 2
Output : [(‘best’, [8, 3, 1]), (‘gfg’, [5, 6, 7]), (‘is’, [1, 4, 7])]

Input : test_dict = {‘gfg’ : [5, 6, 7], ‘is’ : [1, 4, 7], ‘best’ : [8, 3, 1]}, K = 0
Output : [(‘is’, [1, 4, 7]), (‘gfg’, [5, 6, 7]), (‘best’, [8, 3, 1])]

Method : Using sorted() + lambda
The combination of above functions can be used to solve this problem. In this, we perform sort using sorted() and lambda function is used to drive Kth index logic.

# Python3 code to demonstrate working of  
# Sort Dictionary by Kth Index Value 
# Using sorted() + lambda 
  
# initializing dictionary 
test_dict = {'gfg' : [5, 6, 7], 
             'is' : [1, 4, 7], 
             'best' : [8, 3, 1]} 
  
# printing original dictionary 
print("The original dictionary is : " + str(test_dict)) 
  
# initializing K 
K = 1
  
# Sort Dictionary by Kth Index Value 
# Using sorted() + lambda 
res = sorted(test_dict.items(), key = lambda key: key[1][K]) 
  
# printing result  
print("The sorted dictionary : " + str(res))  

Output :

The original dictionary is : {'gfg': [5, 6, 7], 'is': [1, 4, 7], 'best': [8, 3, 1]}
The sorted dictionary : [('best', [8, 3, 1]), ('is', [1, 4, 7]), ('gfg', [5, 6, 7])]

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