Given a dictionary, sort according to descended values, if similar values, then by keys lexicographically.
Input : test_dict = {“gfg” : 1, “is” : 1, “best” : 1, “for” : 1, “Lazyroar” : 1} Output : {“best” : 1, “is” : 1, “for” : 1, “Lazyroar” : 1, “gfg” : 1} Explanation : All values are equal, hence lexicographically sorted keys. Input : test_dict = {“gfg” : 5, “is” : 4, “best” : 3, “for” : 2, “Lazyroar” : 1} Output : {“gfg” : 5, “is” : 4, “best” : 3, “for” : 2, “Lazyroar” : 1} Explanation : All values are different, hence descending ordered sorted.
Method : Using sorted() + items() + dictionary comprehension + lambda
The combination of above functions can be used to solve this problem. In this, we perform task of sorting using sorted(), dictionary comprehension is used to remake dictionary, items() is used to extract items to sorted.
Python3
# Python3 code to demonstrate working of # Sort Dictionary by Values and Keys# Using sorted() + items() + dictionary comprehension + lambda# initializing dictionarytest_dict = {"Gfg" : 1, "is" : 3, "Best" : 2, "for" : 3, "Geeks" : 2}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# - sign for descended values, omit if low-high sorting requiredres = {val[0] : val[1] for val in sorted(test_dict.items(), key = lambda x: (-x[1], x[0]))}# printing result print("Sorted dictionary : " + str(res)) |
The original dictionary is : {'Gfg': 1, 'is': 3, 'Best': 2, 'for': 3, 'Geeks': 2}
Sorted dictionary : {'for': 3, 'is': 3, 'Best': 2, 'Geeks': 2, 'Gfg': 1}
Time Complexity: O(n*nlogn)
Auxiliary Space: O(n)
Method 2: Using operator.itemgetter() and sorted()
- Import the itemgetter() function from the operator module.
- Initialize the input dictionary test_dict.
- Use the sorted() function to sort the dictionary based on the keys and values, bypassing the itemgetter() function as a key.
- Assign the sorted dictionary to the res variable.
- Print the original and sorted dictionaries.
Python3
# importing itemgetter functionfrom operator import itemgetter# initializing dictionarytest_dict = {"Gfg": 1, "is": 3, "Best": 2, "for": 3, "Geeks": 2}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# sorting dictionary by keys and valuesres = dict(sorted(test_dict.items(), key=itemgetter(0, 1)))# printing resultprint("Sorted dictionary : " + str(res)) |
The original dictionary is : {'Gfg': 1, 'is': 3, 'Best': 2, 'for': 3, 'Geeks': 2}
Sorted dictionary : {'Best': 2, 'Geeks': 2, 'Gfg': 1, 'for': 3, 'is': 3}
Time Complexity: The time complexity of this method is O(n log n), where n is the number of items in the dictionary. This is because we are using the sorted() function, which has a time complexity of O(n log n).
Auxiliary Space: The auxiliary space used by this method is O(n), where n is the number of items in the dictionary. This is because we are creating a new dictionary res with the same number of items as the input dictionary test_dict.
