How can I remove every the first number, last number and middle number in a list?

Question

So I'm trying to write a function chosen(lst:list[int]) -> list[int] that returns a list that contains all elements of lst but the front element, the middle element, and the last element. You may assume that the length of lst is greater than 3 and always odd.

For example, chosen([9,3,5,7,1]) returns [3,7] and chosen([0,2,7,0,0,5,0,0,0]) returns [2,7,0,5,0,0]. Here is my code so far....

def first(lst: list) -> list:
    if len(lst)%2 == 0:
        lst.remove(lst[0])
        lst.remove(lst[len(lst)//2])
        lst.remove(lst[-1])
    else:
        lst.remove(lst[0])
        lst.remove(lst[(len(lst) // 2)-1])
        lst.remove(lst[-1])
    return lst

What should I change to make sure this works??

Answer 1

One approach:

def chosen(lst):
    indices = [0, len(lst) // 2, len(lst) - 1]
    return [v for i, v in enumerate(lst) if i not in indices]


res = chosen([9,3,5,7,1])
print(res)

Output

[3, 7]

The idea is to first select the indices to remove and then simply filter out the elements at those indices.

Note that remove , actually removes the first item from the list whose value is equal to x. So your approach wont work with duplicated values.

Answer 2

Remove from the last otherwise indices won't be correct

l = [9,3,5,7,1]
indices_to_be_poped = [0, len(l)//2, len(l)-1]

for i in indices_to_be_poped[::-1]:
    l.pop(i)
    
    
print(l) #[3, 7]

Answer 3

You can use tuple unpacking to remove the first and the last item. Then pop the middle element.

def chosen(lst):
    _, *out, _ = lst
    out.pop(len(out)//2)
    return out

chosen([9,3,5,7,1])
# [3, 7]

Answer 4

You can use list.pop(index) function to remove element by providing index value.

NOte you need to provide index from the largest to lowest, as once you remove a element, list got reindexed, and element index changes

# your code goes here
def chosen(array: list):
    length = len(array)
    remove = [length-1, length//2, 0]
    
    for i in remove:
        array.pop(i)
    return array

assert chosen([0,2,7,0,0,5,0,0,0]) == [2,7,0,5,0,0]