Python | Remove first K elements matching some condition

Last Updated : 15 May, 2023

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 list
test_list = [3, 5, 1, 6, 7, 9, 8, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# using Naive Method 
# to remove first K elements matching condition 
# removes first 4 odd occurrences
counter = 1
res = []
for i in test_list:
    if counter > 4 or not (i % 2 != 0):
        res.append(i)
    else:
        counter += 1
 
# printing result
print ("The filtered list is : " + str(res))

Output:

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 list
test_list = [3, 5, 1, 6, 7, 9, 8, 5]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# using itertools.filterfalse() + itertools.count()
# to remove first K elements matching condition 
# removes first 4 odd occurrences
res = filterfalse(lambda i, counter = count(): i % 2 != 0 and
                                next(counter) < 4, test_list)
 
# printing result
print ("The filtered list is : " + str(list(res)))

Output:

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 list
test_list = [3, 5, 1, 6, 7, 9, 8, 5]
 
# printing original list
print("The original list is : " + str(test_list))
 
# using filter() and lambda function
# to remove first K elements matching condition
# removes first 4 odd occurrences
K = 4
counter = K
def filter_func(x):
    global counter
    if x % 2 != 0:
        counter -= 1
        return counter < 0
    else:
        return True
res = list(filter(filter_func, test_list))
 
# printing result
print("The filtered list is : " + str(res))

Output

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.

Suggest improvement

Python - Summation in Dual element Records List

Python | Initializing multiple lists

Share your thoughts in the comments