Python – Sort dictionary by Tuple Key Product

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 dictionary
test_dict = {(5, 6) : 3, (2, 3) : 9, (8, 4): 10, (6, 4): 12}

# printing original dictionary
print("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))

Output

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 dictionary
test_dict = {(5, 6) : 3, (2, 3) : 9, (8, 4): 10, (6, 4): 12}

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

# sorted() over lambda computed product
# dict() used instead of dictionary comprehension for rearrangement
res = dict(sorted(test_dict.items(), key = lambda ele: ele[0][1] * ele[0][0]))

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

Output

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}