Remove odd-indexed elements from list in Python

Question

I'm trying to remove the odd-indexed elements from my list (where zero is considered even) but removing them this way won't work because it throws off the index values.

lst = ['712490959', '2', '623726061', '2', '552157404', '2', '1285252944', '2', '1130181076', '2', '552157404', '3', '545600725', '0']


def remove_odd_elements(lst):
    i=0
    for element in lst:
        if i % 2 == 0:
            pass
        else:
            lst.remove(element)
        i = i + 1

How can I iterate over my list and cleanly remove those odd-indexed elements?

Answer 1

You can delete all odd items in one go using a slice:

del lst[1::2]

Demo:

>>> lst = ['712490959', '2', '623726061', '2', '552157404', '2', '1285252944', '2', '1130181076', '2', '552157404', '3', '545600725', '0']
>>> del lst[1::2]
>>> lst
['712490959', '623726061', '552157404', '1285252944', '1130181076', '552157404', '545600725']

You cannot delete elements from a list while you iterate over it, because the list iterator doesn't adjust as you delete items. See Loop "Forgets" to Remove Some Items what happens when you try.

An alternative would be to build a new list object to replace the old, using a list comprehension with enumerate() providing the indices:

lst = [v for i, v in enumerate(lst) if i % 2 == 0]

This keeps the even elements, rather than remove the odd elements.

Answer 2

Since you want to eliminate odd items and keep the even ones , you can use a filter as follows :

>>>filtered_lst=list(filter(lambda x : x % 2 ==0 , lst))

this approach has the overhead of creating a new list.