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Python Python Collections (2016, retired 2019) Lists Disemvowel

Prisca Egbua
Prisca Egbua
1,167 Points
Posted by Prisca Egbua
Prisca Egbua
1,167 Points

This is my code so far but it has a bug which I have not been able to fix. I state this bug in the comment.

I am trying to create a function disemvowel that can remove the vowel from any iterable. Please, any suggestion on how to fix this bug? THANKS

disemvowel.py 
""" the bug is that when more than one vowel appear consequently in a word,
 it picks the first and skip the second vowel """
def disemvowel(word):
    word = word.lower()
    print(word)
    word=list(word)
    vowel = ['a','e','i','o','u']
    for letter in word:
        for sound in vowel:
            if letter == sound:
                word.remove(letter)
        print(vowel)
    word = "".join(word)
    print(word)

1 Answer

Kent Åsvang
Kent Åsvang
18,823 Points
Kent Åsvang
Kent Åsvang
18,823 Points

So this is what is happening - I'll show you my debugging process:

  1. First I added print-statements to check exactly how your algorithm worked: 

    def disemvowel(word):

    word = word.lower()
    word=list(word)

    vowels = ['a','e','i','o','u']

    for letter in word:
        print("Checking %s" % letter)

        for vowel in vowels:
            print("checking vowel %s against letter %s" % (vowel, letter))

            if letter == vowel:
                print("%s is in %s" % (letter, vowel))
                word.remove(letter)

    word = "".join(word)
    print("word that remains is %s" % word)

disemvowel("casaa")

And the output was: 

```terminal
    Checking c
    checking vowel a against letter c
    checking vowel e against letter c
    checking vowel i against letter c
    checking vowel o against letter c
    checking vowel u against letter c
    Checking a
    checking vowel a against letter a
    a is in a
    checking vowel e against letter a
    checking vowel i against letter a
    checking vowel o against letter a
    checking vowel u against letter a
    Checking a
    checking vowel a against letter a
    a is in a
    checking vowel e against letter a
    checking vowel i against letter a
    checking vowel o against letter a
    checking vowel u against letter a
    word that remains is csa

When you look through the output, you see that each time you remove a letter, your algorithm skips checking the next letter because the queue gets skewed - since you are changing the list that gets iterated through.

With that in mind, I created this algorithm: 

    def disemvowel(word):
        """ prints any given string without its vowels """
        vowels = ["a", "e", "i", "o", "u"]
        word = list(word.lower())
        no_vowels = ""

        for letter in word:
            if letter not in vowels:
                no_vowels += letter

        print(no_vowels)

Hope this helped

Prisca Egbua
Prisca Egbua
1,167 Points
Prisca Egbua
Prisca Egbua
1,167 Points

Thank you, it did help. After understanding your debugging process, I made a copy of the word and remove the vowel from the copy, so that the queue is not skewed.

def disemvowel(word): word = word.lower() word=list(word) word_copy = word[:] vowel = ['a','e','i','o','u'] for letter in word: for sound in vowel: if letter == sound: word_copy.remove(letter) word = "".join(word_copy) print(word)