The problem of removing the matching elements from two lists and constructing a new list having just the filtered elements not present in 2nd list has been discussed earlier, but sometimes, we have more than an elementary element, but a tuple as element of list. Handling such a case requires different type of handling. Lets discuss certain ways how this problem can be solved.
Method #1 : Using list comprehension
This particular task can be done using the list comprehension as shorthand for the for loop which we would have used. We just check for the existence of one tuple in another and make decision accordingly.
# Python3 code to demonstrate # filter repeated tuple # using list comprehension # initializing lists test_list1 = [('Geeks', 'for'), ('Geeks', 'is'), ('Computer', 'Science')] test_list2 = [('Geeks', 'for'), ('Geeks', 'is')] # printing original lists print("The original list 1 : " + str(test_list1)) print("The original list 1 : " + str(test_list2)) # using list comprehension # filter repeated tuple res = [sub for sub in test_list1 if sub not in test_list2] # print result print("The filtered list of tuples : " + str(res)) |
The original list 1 : [('Geeks', 'for'), ('Geeks', 'is'), ('Computer', 'Science')]
The original list 1 : [('Geeks', 'for'), ('Geeks', 'is')]
The filtered list of tuples : [('Computer', 'Science')]
Method #2 : Using set() + “-” operator
The task of getting the difference of two lists can also be done using the set that converts the list and then minus operator can be used to get the set difference.
# Python3 code to demonstrate # filter repeated tuple # using set() + "-" operator # initializing lists test_list1 = [('Geeks', 'for'), ('Geeks', 'is'), ('Computer', 'Science')] test_list2 = [('Geeks', 'for'), ('Geeks', 'is')] # printing original lists print("The original list 1 : " + str(test_list1)) print("The original list 1 : " + str(test_list2)) # using set() + "-" operator # filter repeated tuple res = list(set(test_list1) - set(test_list2)) # print result print("The filtered list of tuples : " + str(res)) |
The original list 1 : [('Geeks', 'for'), ('Geeks', 'is'), ('Computer', 'Science')]
The original list 1 : [('Geeks', 'for'), ('Geeks', 'is')]
The filtered list of tuples : [('Computer', 'Science')]
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