Sort a dictionary by key and value difference in Python

Given dictionary, sort its items using key-value difference.

Input : test_dict = {3 : 9, 1 : 4, 9 : 13, 1 : 10, 8 : 9 }
Output : {8: 9, 9: 13, 3: 9, 1: 10}
Explanation : 1 < 4 < 6 < 9, is increasing difference.

Input : test_dict = {3 : 9, 1 : 4, 9 : 13, 1 : 9, 8 : 9 }
Output : {8: 9, 9: 13, 3: 9, 1: 9}
Explanation : 1 < 4 < 6 < 8, is increasing difference.

Method #1 : Using abs() + sort() + dictionary comprehension

In this, task of sorting is done using sort(), items() is used to get key-value pairs of dictionary as tuple list. Then, abs() is used to compute absolute difference and passed as key to sort. At last, dictionary comprehension is used to reconvert result to dictionary.

# Python3 code to demonstrate working of 
# Sort by dictionary key-value difference
# Using abs() + sort() + dictionary comprehension

# getting abs diff 
def abs_diff(ele):
    return abs(ele[0] - ele[1])

# initializing dictionary
test_dict = {3 : 9, 1 : 4, 9 : 13, 1 : 10, 8 : 9 }

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

# sorting using sort 
test_dict_list = list(test_dict.items())
test_dict_list.sort(key = abs_diff)

# reordering to dictionary
res = {ele[0] : ele[1]  for ele in test_dict_list}

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

The original dictionary is : {3: 9, 1: 10, 9: 13, 8: 9}
The sorted dictionary : {8: 9, 9: 13, 3: 9, 1: 10}

Method #2 : Using sorted() + lambda + abs() + dictionary comprehension

In this, we perform task of sorting and using sorted() and lambda function is used to input logic of getting absolute difference using abs().

# Python3 code to demonstrate working of 
# Sort by dictionary key-value difference
# Using sorted() + lambda + abs() + dictionary comprehension

# initializing dictionary
test_dict = {3 : 9, 1 : 4, 9 : 13, 1 : 10, 8 : 9 }

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

# sorting using sorted()
# lambda function to render logic 
test_dict_list = sorted(list(test_dict.items()), key = lambda ele : abs(ele[0] - ele[1]))

# reordering to dictionary
res = {ele[0] : ele[1]  for ele in test_dict_list}

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

The original dictionary is : {3: 9, 1: 10, 9: 13, 8: 9}
The sorted dictionary : {8: 9, 9: 13, 3: 9, 1: 10}