Sometimes, while working with Python dictionaries, we can have problem in which in which we need to perform sorting of items on basis of various factors. One such can be on basis of absolute difference of dual value list. This can occur in Python > 3.6, as dictionaries are ordered. This kind of problem can come in data domain. Let’s discuss a way in which this problem can be solved.
Method : Using sorted() + lambda + abs() + dictionary comprehension
The combination of above functions can be used to solve this problem. In this, we perform the task of sorting using sorted(), lambda function is used to provide the logic and abs() function is used to compute the absolute difference.
Python3
# Python3 code to demonstrate working of # Sort Dictionary by Value Difference# Using sorted() + lambda + abs() + dictionary comprehension# initializing dictionarytest_dict = {'gfg' : [34, 87], 'is' : [10, 13], 'best' : [19, 27], 'for' : [10, 50], 'neveropen' : [15, 45]}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# Sort Dictionary by Value Difference# Using sorted() + lambda + abs() + dictionary comprehensionres = dict(sorted(test_dict.items(), key = lambda sub: abs(sub[1][0] - sub[1][1])))# printing result print("The sorted dictionary : " + str(res)) |
The original dictionary is : {‘gfg’: [34, 87], ‘is’: [10, 13], ‘best’: [19, 27], ‘for’: [10, 50], ‘neveropen’: [15, 45]} The sorted dictionary : {‘is’: [10, 13], ‘best’: [19, 27], ‘neveropen’: [15, 45], ‘for’: [10, 50], ‘gfg’: [34, 87]}
Time Complexity: O(nlogn), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
Method 2 : using the items() method
Explanation:
We define the sort_by_difference function that takes an item (key-value pair) of the dictionary and returns the absolute difference between the two values in the list.
We use the sorted function with the key parameter set to the sort_by_difference function to sort the dictionary items based on the value difference.
We convert the sorted items back to a dictionary using the dict constructor and assign it to the res variable.
We print the sorted dictionary using the print function.
Python3
# Python3 code to demonstrate working of # Sort Dictionary by Value Difference# Using items() method and custom function for sorting# initializing dictionarytest_dict = {'gfg' : [34, 87], 'is' : [10, 13], 'best' : [19, 27], 'for' : [10, 50], 'neveropen' : [15, 45]}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# custom function for sortingdef sort_by_difference(item): key, value = item return abs(value[0] - value[1])# Sort Dictionary by Value Difference# Using items() method and custom function for sortingres = dict(sorted(test_dict.items(), key=sort_by_difference))# printing resultprint("The sorted dictionary : " + str(res)) |
The original dictionary is : {'gfg': [34, 87], 'is': [10, 13], 'best': [19, 27], 'for': [10, 50], 'neveropen': [15, 45]}
The sorted dictionary : {'is': [10, 13], 'best': [19, 27], 'neveropen': [15, 45], 'for': [10, 50], 'gfg': [34, 87]}
The time complexity of the sorting algorithm used in the sorted function is O(n log n) in the worst case, where n is the number of items in the dictionary.
The auxiliary space used in this approach is O(n), where n is the number of items in the dictionary.
