Python – Sort Dictionary by key-value Summation

Given a Dictionary, sort by summation of key and value. 

Input : test_dict = {3:5, 1:3, 4:6, 2:7, 8:1} 
Output : {1: 3, 3: 5, 2: 7, 8: 1, 4: 6} 
Explanation : 4 < 8 < 9 = 9 < 10 are increasing summation of keys and values.

Input : test_dict = {3:5, 1:3, 4:6, 2:7} 
Output : {1: 3, 3: 5, 2: 7, 4: 6} 
Explanation : 4 < 8 < 9 < 10 are increasing summation of keys and values. 

Method 1: Using sorted() + lambda + items()

In this sort operation is performed using sorted(), lambda function is used to provide addition logic. The items() is used to get both keys and values.

Python3

# Python3 code to demonstrate working of
# Sort Dictionary by key-value Summation
# Using sorted() + lambda + items()

# initializing dictionary
test_dict = {3: 5, 1: 3, 4: 6, 2: 7, 8: 1}

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

# sorted() to sort, lambda provides key-value addition
res = sorted(test_dict.items(), key=lambda sub: sub[0] + sub[1])

# converting to dictionary
res = {sub[0]: sub[1] for sub in res}

# printing result
print("The sorted result : " + str(res))

Output:

The original dictionary is : {3: 5, 1: 3, 4: 6, 2: 7, 8: 1}
The sorted result : {1: 3, 3: 5, 2: 7, 8: 1, 4: 6}

Method 2: Using dictionary comprehension and sorted()

Use dictionary comprehension to create a new dictionary with sorted key-value pairs. Sort the dictionary items using the lambda function which returns the sum of key-value pairs. Finally, use sorted() method to sort the dictionary items based on the lambda function result.

Python3

# Python3 code to demonstrate working of
# Sort Dictionary by key-value Summation
# Using dictionary comprehension and sorted()

# initializing dictionary
test_dict = {3: 5, 1: 3, 4: 6, 2: 7, 8: 1}

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

# sorting dictionary by key-value summation
res = {k: v for k, v in sorted(test_dict.items(), key=lambda x: x[0]+x[1])}

# printing result
print("The sorted result : " + str(res))

Output

The original dictionary is : {3: 5, 1: 3, 4: 6, 2: 7, 8: 1}
The sorted result : {1: 3, 3: 5, 2: 7, 8: 1, 4: 6}

Time complexity: O(n log n), where n is the number of items in the dictionary. This is because sorting the dictionary items takes O(n log n) time complexity.
Auxiliary space: O(n), where n is the number of items in the dictionary. This is because we create a new dictionary with the same number of items as the original dictionary.