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Python Python Collections (2016, retired 2019) Lists Disemvowel

Posted by Max Green
Max Green
11,451 Points

Why does this only falter when there's two vowels in a row?

The generated combination of letters always seems to have two vowels paired next to each other. This is what's stumping me. For some reason my code doesn't catch the 2nd vowel in any pair of vowels. Help :S. I've been on this for a stupidly long amount of time.

disemvowel.py 
def disemvowel(word):
    new_list = list(word)
    for letter in new_list:
        if letter.upper() == "A":
            new_list.remove(letter)
            continue
        elif letter.upper() == "E":
            new_list.remove(letter)
            continue
        elif letter.upper() == "I":
            new_list.remove(letter)
            continue
        elif letter.upper() == "O":
            new_list.remove(letter)
            continue
        elif letter.upper() == "U":
            new_list.remove(letter)
            continue
    newest_list = "".join(new_list)
    return newest_list
Max Green
Max Green
11,451 Points

Ahhh yea I get it now. I never thought of building another word. Very clever :)

1 Answer

Louise St. Germain
Louise St. Germain
19,424 Points
Louise St. Germain
Louise St. Germain
19,424 Points

Hi Max,

Looks like as the for loop steps through your list, you're also modifying that very same list, so Python's indexing as it steps through the list gets out of sync when you start deleting elements on the fly. For example, say:

new_list = ['a', 'A', 'b', 'c']

The first "a" is at index 0, the second one "A" is at 1, and so on up to c at index = 3.

Python hits on the first "a" (index location 0) and it gets deleted. So now that the first letter is gone, the updated list is ['A', 'b','c'], with "A" now sitting at index 0, b at 1 (was 2 before), and c at 2 (was 3 before).

THEN Python says, hey, I've finished everything at index location 0 (which, as it turns out, has just been updated to a letter you haven't checked yet), and steps to index location 1 (which is now 'b'). It ends up skipping over that second A entirely - doesn't even evaluate it.

Your best bet would probably be to build your output in another variable as you go along, i.e., don't delete anything out of your original list, but just add a condition that says that if the letter is a consonant, add it to your new variable as you step through the original list (and if it's a vowel, don't add it). That way the indexing on the original list won't get messed up and it will evaluate every single letter.

Good luck! :-)

Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

Max, you could also use the slice notation to create a copy of new_list in the for loop statement. Deletions from new_list would no longer affect the for loop. Notice the [:] added to the end of new_list:

for letter in new_list[:]:

In, general, it is never a good practice to modify the container that you are looping over in a for loop.