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Vetzi Code
Python Web Development Techdegree Student 3,014 PointsWhy does this not work?
it tried it on repl.it many times, worked just fine, but doesn't work when I check work?
new_word = []
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
def disemvowel(word):
for letter in word:
if letter not in vowels:
new_word.append(letter)
return ''.join(new_word)
1 Answer
Brendan Whiting
Front End Web Development Techdegree Graduate 84,735 PointsThe problem is that the new_word variable is in the global scope, so as the function is run more than once, it's altering the same variable again rather than starting from scratch. The challenge seems to be passing in multiple test cases.
Bring that variable inside the scope of the function and you should be fine.
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
def disemvowel(word):
new_word = [] # this variable is declared inside the function from scratch each time
for letter in word:
if letter not in vowels:
new_word.append(letter)
return ''.join(new_word)