Working in-place
The list method remove removes the first occurrence of the argument from the list. So you can simply call it x amount of times on the list:
int_list = [1, 4, 3, 6, 4, 3, 2, 4, 8, 11] remove = 4 times = 2 for _ in range(times): int_list.remove(remove) print(int_list)
Will give:
[1, 3, 6, 3, 2, 4, 8, 11]
Handling errors
In case the element is not found, remove will raise an error. If you want to avoid that:
Check the input pre-hand by making sure there are enough elements to remove using the count method:
if int_list.count(remove) <= times: # rest of code
If you want to just remove all possible elements, add a check before the remove call:
for _ in range(times): if remove in int_list: int_list.remove(remove)
Returning a new list
You can of course simply create a copy of the input list (res = int_list.copy()) and then apply the in-place solution from above. Otherwise, to create a completely new list, use times as a counter for how many times to "skip" the wanted number:
int_list = [1, 4, 3, 6, 4, 3, 2, 4, 8, 11] remove = 4 times = 2 res = [] for num in int_list: if num == remove and times > 0: times -= 1 continue res.append(num) print(res)
In this case there is no error handling to make. We are simply ignoring the required element a given amount of times. If it doesn't exist at all, less, or more times than given - nothing will happen.
