12

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?

Bill the Lizard
  • 386,424
  • 207
  • 554
  • 861
avereux
  • 530
  • 1
  • 5
  • 14
  • Fun fact: to remove all even (positioned) elements you can do: `for x in lst: lst.remove(x)`. To remove all odds do: `iter_lst = iter(lst); next(iter_lst); for x in iter_lst: lst.remove(x)`. So, basically, removing the `if` solves the problem... I hope I don't have to say to never actually use this stuff in real code. – Bakuriu Mar 05 '15 at 17:59

2 Answers2

37

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.

Community
  • 1
  • 1
Martijn Pieters
  • 963,270
  • 265
  • 3,804
  • 3,187
1

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.

jihed gasmi
  • 251
  • 3
  • 9