Given dictionary with tuple keys, sort dictionary items by tuple product of keys.
Input : test_dict = {(2, 3) : 3, (6, 3) : 9, (8, 4): 10, (10, 4): 12}
Output : {(2, 3) : 3, (6, 3) : 9, (8, 4): 10, (10, 4): 12}
Explanation : 6 < 18 < 32 < 40, key products hence retains order.Input : test_dict = {(20, 3) : 3, (6, 3) : 9, (8, 4): 10, (10, 4): 12}
Output : {(6, 3) : 9, (8, 4): 10, (10, 4): 12, (20, 3) : 3, }
Explanation : 18 < 32 < 40 < 60, key products hence adjusts order.
Method #1 : Using dictionary comprehension + lambda + sorted()
This is one of the ways in which this task can be performed. In this, we perform sort() using sorted() and lambda function is used to compute product over which sort can be performed.
Python3
# Python3 code to demonstrate working of # Sort dictionary by Tuple Key Product# Using dictionary comprehension + sorted() + lambda # initializing dictionarytest_dict = {(5, 6) : 3, (2, 3) : 9, (8, 4): 10, (6, 4): 12} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # sorted() over lambda computed product # dictionary comprehension reassigs dictionary by order res = {key: test_dict[key] for key in sorted(test_dict.keys(), key = lambda ele: ele[1] * ele[0])} # printing result print("The sorted dictionary : " + str(res)) |
The original dictionary is : {(5, 6): 3, (2, 3): 9, (8, 4): 10, (6, 4): 12}
The sorted dictionary : {(2, 3): 9, (6, 4): 12, (5, 6): 3, (8, 4): 10}
Method #2 : Using dict() + sorted() + lambda
The combination of above functions can be used to solve this problem. In this, similar method is used as above method. The only difference being items arrangement done using dict() rather than dictionary comprehension after computing keys ordering .
Python3
# Python3 code to demonstrate working of # Sort dictionary by Tuple Key Product# Using dict() + sorted() + lambda # initializing dictionarytest_dict = {(5, 6) : 3, (2, 3) : 9, (8, 4): 10, (6, 4): 12} # printing original dictionaryprint("The original dictionary is : " + str(test_dict)) # sorted() over lambda computed product # dict() used instead of dictionary comprehension for rearrangementres = dict(sorted(test_dict.items(), key = lambda ele: ele[0][1] * ele[0][0])) # printing result print("The sorted dictionary : " + str(res)) |
The original dictionary is : {(5, 6): 3, (2, 3): 9, (8, 4): 10, (6, 4): 12}
The sorted dictionary : {(2, 3): 9, (6, 4): 12, (5, 6): 3, (8, 4): 10}
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