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Python Python Collections (2016, retired 2019) Lists Disemvowel

David Gabriel
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David Gabriel
Python Web Development Techdegree Student 979 Points
Posted by David Gabriel
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David Gabriel
Python Web Development Techdegree Student 979 Points

Any help please ?

vowels = ("a", "e", "i", "o", "u")

def disemvowel(word): word = list(word) for letter in word: if letter.lower() in vowels: word.remove(letter) return word

disemvowel.py 
    vowels = ("a", "e", "i", "o", "u")

def disemvowel(word):
    word = list(word)
    for letter in word:
        if letter.lower() in vowels:
              word.remove(letter)
    return word

4 Answers

Gabbie Metheny
Gabbie Metheny
33,778 Points
Gabbie Metheny
Gabbie Metheny
33,778 Points

You're getting there! Let's work on your list. Right now you have:

letter = vowels[]

There's two problems here. First, let's rename it to new_word, since you're also using letter to represent each letter in word and variable confusion is going to break your code. Second, what you're going to need is an empty list, which looks like this:

new_word = []

This is the list you're going to add consonants to, and it will eventually become the word you'll return. One more thing, this variable needs to live inside your function, rather than in the global scope with the vowels tuple. If Kenneth tests more than one word against this function (which he will), letters will keep being added to the same list. For example:

disemvowel('apple)'    #returns 'ppl'
disemvowel('cheese')    #returns 'pplchs'

Now we can jump into your function.. Since you're appending consonants to a new list, rather than removing vowels from the old word, you need to use not in, rather than in. This way, if a letter is in your vowels tuple (so, it's a vowel), you'll disregard it. However, if it's not in vowels (so, it's a consonant), it will be added to new_word:

def disemvowel(word):
    new_word = []
    for letter in word:
        if letter.lower() not in vowels:
            new_word.append(letter)

Now you're ready to return new_word, except it's a list, not a string. You can join list items into a string with .join(), calling it on what you want your joiner to be (in this case, nothing, or an empty string) and passing it what you want joined (in this case, the letters you added to new_word):

return ''.join(new_word)

Let me know how it goes!

Gabbie Metheny
Gabbie Metheny
33,778 Points
Gabbie Metheny
Gabbie Metheny
33,778 Points

A couple things:

  1. First off, you have some unnecessary indentation on line 1 that is causing some problems. If you remove that, you should start to see more helpful error messages rather than the dreaded "Bummer! Try again!"
  2. Each time you remove an item from your list in your loop, the indexes change, and so some of your letters will never be checked. For example, if the word was "you," the y would start at index 0, o at index 1 and u at index 2. Your code would remove the o on the second time through the loop, but would bump the u down to index 1, the letter that was just checked, so the code wouldn't run a final time to check for that u. To get around this, rather than remove letters from the list, you could create an empty list and append the letters to the list if they are not in vowels.
  3. Currently, you are returning word, which is now a list, rather than a string. You'll need to join the letters together before returning them.

Let me know if you need further clarification!

David Gabriel
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.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points
David Gabriel
.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points

Hi Gabbie,

Thank you very much, but still no progress on the task vowels = ("a", "e", "i", "o","u") letter = vowels[] def disemvowel(word): for letter in word: if letter.lower() in word: letter.append(vowels) return false return true

David Gabriel
seal-mask
.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points
David Gabriel
.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points

vowels = ("a", "e", "i", "o", "u") new_word = list [] def disemvowel(word): for letter in word: if letter.lower() not in vowels: new_word.append(letter) return ''.join(new_word)

Gabbie Metheny
Gabbie Metheny
33,778 Points
Gabbie Metheny
Gabbie Metheny
33,778 Points

You're so close! Only one change (well, two, but on the same line):

new_word = list []

You don't need to say 'list' before your set of brackets. Simply assigning empty square brackets to the new_word variable will create an empty list. You also still need to move new_word inside your function, otherwise each word Kenneth tests will keep getting added to the same list and joined into a single string:

disemvowel('apple')    #returns 'ppl'
disemvowel('cheese')    #returns 'pplchs'
disemvowel('watermelon')    #returns 'pplchswtrmln'

And so on. So you just need to fix that line (removing the word list, and moving the variable from the global scope to the local scope), and your code should pass!

David Gabriel
seal-mask
.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points
David Gabriel
.a{fill-rule:evenodd;}techdegree
David Gabriel
Python Web Development Techdegree Student 979 Points

Hi Gabby, thanks for you patient and max support; please see

vowels = ("a", "e", "i", "o", "u") def disemvowel(word): new_word = [] for letter in word: if letter.lower() not in vowels: new_word.append(letter) return ' '.join(new_word)

Gabbie Metheny
Gabbie Metheny
33,778 Points
Gabbie Metheny
Gabbie Metheny
33,778 Points

Looks like the problem is on your return statement: the string that you're calling .join() on should be completely empty, but right now you have a space in there (' ' vs ''). You need the string to be totally empty, otherwise the letters in new_word will be joined with spaces in between each pair of letters ('ppl' vs 'p p l'). If that doesn't fix it, can you format your code using the Markdown Cheatsheet so I can see how your lines are indented?