Python: remove odd number from a list

Question

I wrote a function to remove odd number from a list, like that:

def remove_odd(l):
    for i in l:
        if i % 2 != 0:
            l.remove(i)
    print l
    return l

remove_odd([4,5,4])
remove_odd([4,5,4,7,9,11])
remove_odd([4,5,4,7,9,11,12,13])

It returns:

[4, 4]
[4, 4, 9]
[4, 4, 9, 12]

-> wrong

but when I change to remove even number:

def remove_even(l):
    for i in l:
        if i % 2 == 0:
            l.remove(i)
    print l
    return l

remove_even([4,5,4])
remove_even([4,5,4,7,9,11])
remove_even([4,5,4,7,9,11,12,13])

The answer is OK:

[5]
[5, 7, 9, 11]
[5, 7, 9, 11, 13]

What is wrong with the remove_odd() func? I know people usually create the second list inside the func then append even number to that list, but can we solve this exercise with list.remove() ?

Thank you!

Answer 1

Your function is working in another way than you would expect. The for loop takes first element, than second etc., so when you remove one element, others change their positions and can be skipped by it (and that happens in your case) when they are preceded by another odd number.

If you insist on using .remove() method, you must operate on a copy instead, like this:

def remove_odd(1):
    for i in l[:]:
        if i % 2 != 0:
            l.remove(i)
    return l

(l[:] is a shallow copy of list l)

However, I think using list comprehension would be much clearer:

def remove_odd(l):
    return [x for x in l if x % 2 == 0]