Sometimes while working with Python tuples, we can have a problem in which we need to extract all the pairs which are symmetric, i.e for any (x, y), we have (y, x) pair present. This kind of problem can have application in domains such as day-day programming and web development. Let’s discuss certain ways in which this task can be performed.
Input : test_list = [(6, 7), (2, 3), (7, 6)]
Output : {(6, 7)}Input : test_list = [(6, 7), (2, 3)]
Output : {}
Method #1: Using dictionary comprehension + set() The combination of above functionalities can be used to solve this problem. In this, we initially construct reverse pairs, and then compare with original list pairs, and extract one of equals. The set() is used to remove duplicates, to avoid unnecessary computations of elements.
Python3
# Python3 code to demonstrate working of # Extract Symmetric Tuples# Using dictionary comprehension + set()# initializing listtest_list = [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)]# printing original listprint("The original list is : " + str(test_list))# Extract Symmetric Tuples# Using dictionary comprehension + set()temp = set(test_list) & {(b, a) for a, b in test_list}res = {(a, b) for a, b in temp if a < b}# printing result print("The Symmetric tuples : " + str(res)) |
The original list is : [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)]
The Symmetric tuples : {(8, 9), (6, 7)}
Time complexity: O(n), where n is the length of the input list test_list.
Auxiliary space: O(n), where n is the length of the input list test_list. The temp set and res set both can contain up to n elements in the worst case.
Method #2 : Using Counter() + list comprehension This is yet another way in which this task can be performed. In this, we follow similar approach of constructing reverse pairs, but here, we count the equal elements, the element with count 2 is duplicate and matches the reversed tuples.
Python3
# Python3 code to demonstrate working of # Extract Symmetric Tuples# Using Counter() + list comprehensionfrom collections import Counter# initializing listtest_list = [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)]# printing original listprint("The original list is : " + str(test_list))# Extract Symmetric Tuples# Using Counter() + list comprehension<temp = [(sub[1], sub[0]) if sub[0] < sub[1] else sub for sub in test_list]cnts = Counter(temp)res = [key for key, val in cnts.items() if val == 2]# printing result print("The Symmetric tuples : " + str(res)) |
The original list is : [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)] The Symmetric tuples : [(7, 6), (9, 8)]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the input list.
Method 3: Using nested for loops and a temporary set
Steps:
Initialize the original list test_list.
Print the original list.
Initialize a temporary set temp_set to keep track of already seen tuples.
Initialize an empty list res to store the symmetric tuples.
Iterate through each tuple tpl in test_list.
If the tuple tpl or its reverse (tpl[1], tpl[0]) is already in temp_set, append it to res.
Otherwise, add the tuple tpl to temp_set.
Print the result res.
Python3
# Python3 code to demonstrate working of # Extract Symmetric Tuples# Using nested for loops and a temporary set# initializing listtest_list = [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)]# printing original listprint("The original list is : " + str(test_list))# Extract Symmetric Tuples# Using nested for loops and a temporary settemp_set = set()res = []for tpl in test_list: if tpl in temp_set or (tpl[1], tpl[0]) in temp_set: res.append(tpl) else: temp_set.add(tpl)# printing result print("The Symmetric tuples : " + str(res)) |
The original list is : [(6, 7), (2, 3), (7, 6), (9, 8), (10, 2), (8, 9)] The Symmetric tuples : [(7, 6), (8, 9)]
Time complexity: O(n^2), where n is the length of the original list.
Auxiliary space: O(n), where n is the length of the original list.
