Removal of elements in list can be performed using many inbuilt functions. Removing all or just a single occurrence removal both functions are present in Python library. This article discusses to remove just the first K occurrences of elements matching particular condition.
Method #1: Naive Method We can append the elements that are matching condition after K occurrences of elements have been done and hence would perform the task similar to the removal.
Python3
# Python3 code to demonstrate # to remove first K elements matching condition# using Naive Method # initializing listtest_list = [3, 5, 1, 6, 7, 9, 8, 5]# printing original listprint ("The original list is : " + str(test_list))# using Naive Method # to remove first K elements matching condition # removes first 4 odd occurrencescounter = 1res = []for i in test_list: if counter > 4 or not (i % 2 != 0): res.append(i) else: counter += 1# printing resultprint ("The filtered list is : " + str(res)) |
The original list is : [3, 5, 1, 6, 7, 9, 8, 5] The filtered list is : [6, 9, 8, 5]
Time Complexity: O(n)
Space Compelxity: O(n)
Method #2: Using itertools.filterfalse() + itertools.count() This is different and elegant way to perform this particular task. It filters out all the numbers that become greater than K as counter reaches K and matches against the condition. This is one-liner and preferred method to achieve this task.
Python3
# Python3 code to demonstrate # to remove first K elements matching condition# using itertools.filterfalse() + itertools.count()from itertools import filterfalse, count# initializing listtest_list = [3, 5, 1, 6, 7, 9, 8, 5]# printing original listprint ("The original list is : " + str(test_list))# using itertools.filterfalse() + itertools.count()# to remove first K elements matching condition # removes first 4 odd occurrencesres = filterfalse(lambda i, counter = count(): i % 2 != 0 and next(counter) < 4, test_list)# printing resultprint ("The filtered list is : " + str(list(res))) |
The original list is : [3, 5, 1, 6, 7, 9, 8, 5] The filtered list is : [6, 9, 8, 5]
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #3: Use the filter() function and lambda function.
Step-by-step approach:
- Define a lambda function to check if an element is odd.
- Use the filter() function to filter out the first K elements that match the condition using the lambda function.
- Convert the filtered object returned by the filter() function to a list and return it.
Python3
# Python3 code to demonstrate # to remove first K elements matching condition# using filter() and lambda function# initializing listtest_list = [3, 5, 1, 6, 7, 9, 8, 5]# printing original listprint("The original list is : " + str(test_list))# using filter() and lambda function# to remove first K elements matching condition# removes first 4 odd occurrencesK = 4counter = Kdef filter_func(x): global counter if x % 2 != 0: counter -= 1 return counter < 0 else: return Trueres = list(filter(filter_func, test_list))# printing resultprint("The filtered list is : " + str(res)) |
The original list is : [3, 5, 1, 6, 7, 9, 8, 5] The filtered list is : [6, 9, 8, 5]
Time complexity: O(n) as we are iterating over the list once.
Auxiliary space: O(n) as we are creating a new list to store the filtered elements.
