Question

I am trying to iterate through a list checking every string in the list for a character.

test = [str(i) for i in range(100)]

for i in test:
    if '0' or '4' or '6' or '8' in str(i):
        test.remove(i)

I thought this would be fine but, the list is this after:

[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]

Where the ‘2’ is removed but, the ’41’ is not? I’ve noticed its only even numbers but, don’t know why.

Answer 1

You have two problems. First, this if statement is always taken.

if '0' or '4' or '6' or '8' in str(i):

Because the string 0 is a non-zero-length string, and is therefore True. 0 or 4 is 0 and therefore True. The rest of the statement doesn’t matter.

I expect you actually wanted to test whether each of those digits was in the string representation of the integer. As you have it written now, you are only testing whether 8 is in the string, and that’s not even getting tested because the expression evaluates to True before it even gets there. Something like this would work:

if any(x in i for x in '0468'):

By the way, str(i) is superfluous because your list is already a list of strings.

The other problem is that you are deleting items from the list you are iterating over. So, here’s what happens:

• The first item, 0, is removed, because your if statement is always taken.
• The second item, 1, becomes the first item, because you removed 0.
• The for loop goes on to the second item, which is now 2.

In other words, because you deleted 0, 1 is never tested. Thus, every other item (all the even numbers) are removed.

The easiest way to avoid this in your code is to iterate over a copy of the list:

for i in test[:]: