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Python Python Collections (2016, retired 2019) Lists Disemvowel

Corey Buckley
Corey Buckley
1,257 Points
Posted by Corey Buckley
Corey Buckley
1,257 Points

Why does this not work?

I don't understand why this is adding every letter whether it's a vowel or not to the list new_word

disemvowel.py 
def disemvowel(word):
    new_word = []
    the_list = list(word)
    for letter in the_list:
        if letter != "A" or "a" or "E" or "e" or "I" or "i" or "O" or "o" or "U" or "u":
            new_word.append(letter)
    print(new_word)

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

Your test is probably not doing what you are expecting. Instead of comparing "letter" to each literal string, it is cchecking that it is not equal to "A", and then checking the "truthiness" of "a", and then the "truthiness" of "E" ... etc. And since any non-empty string is considered "truthy" by itself, the test always passes.

You could recode that test so each "or" joins a separate check of "letter" to each literal, but that would make it far more verbose. A much more compact way to do what you intended is to use the membership operator ("in") and invert it using "not":

        if letter not in "AaEeIiOoUu":

Also remember to "join" your list back into a string, and return it. You won't need to "print" anything.