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Daniel D'luyz
2,880 PointsThis editor does not check correctly the answer
I've tried 3 different solutions and they all work on the workspace, but this editor keeps telling me that the string I am returning contains the vowels.
This is the last solution I tried checking:
to_remove = ["a","A","e","E","i","I","o","O","u","U"]
def disemvowel(word):
while len(to_remove) > 0:
letter = to_remove[0]
word = word.replace(letter, '')
to_remove.remove(letter)
return word
And before that:
to_remove = ["a","A","e","E","i","I","o","O","u","U"]
def disemvowel(word):
word_list = list(word)
while len(to_remove) > 0:
letter = to_remove[0]
while True:
try:
word_list.remove(letter)
except ValueError:
to_remove.remove(letter)
break
return ''.join(word_list)
2 Answers
Jennifer Nordell
Treehouse TeacherHi there, Daniel D'luyz ! You're doing terrific and your logic is pretty solid. However, you haven't taken into account what happens when your disemvowel function is run more than once. Because you declared the list holding the vowels outside`of the function, once the function completes that list is no longer filled with vowels. It is, in fact, empty. So the second time that function is run, the vowels will not be removed.
To mitigate this, you can place the declaration of the to_remove list as the first line inside your function. That way, every time the function is run, the vowel list will be reset.
Hope this helps!
Daniel D'luyz
2,880 PointsYou are totally right! Thank you very much!