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Python Python Collections (2016, retired 2019) Lists Removing Items From A List

My WORKING solution to the "Disemvowel" exercise is being rejected with the "Bummer..." notification...

This code gets the job done for the challenge posed, and yet the exercise prompter is rejecting my code/calling it "incorrect"...

HUWHY???

def disemvowel(word):
    #return word
    vowels = ["A","E","I","O","U"]
    word = word.upper()
    word = list(word)
    for letter in word:
        try:
            word.remove("A")
        except ValueError:
            pass
        try:
            word.remove("E")
        except ValueError:
            pass
        try:
            word.remove("I")
        except ValueError:
            pass
        try:
            word.remove("O")
        except ValueError:
            pass
        try:
            word.remove("U")
        except ValueError:
            pass
    word = ''.join(word).lower()
    print(word)

disemvowel("taouuoooiiieest")

7 Answers

You guys have a lot of code that can be taken out. Adam even goes through the effort of creating alist of vowels :p

def disemvowel(word):
    vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
    split_word = list(word)
    while vowels:
        try:
            split_word.remove(vowels[0])
        except ValueError:
            del vowels[0]
    word = "".join(split_word)


    return word

Remember to also pay attention to DRY (Don't Repeat Yourself)

If you find yourself typing the same thing over and over again, such as the try/except, take a loot at your code and see how you can make it a little more simple. This may not seem that important at this point in your careers/education, but as you move on to larger projects you're going to want things to be as concise as possible.

And Adam your last example there had return word up at the top and down at the bottom, you want to return after all the word is complete on the word.

Basically what my code does is loop through the list created by splitting the value of word. It loops through and each time it comes across the letter in vowels[0] (so a for starters), it then removes the letter from the vowels list once it hits the ValueError exception (so when there is no longer a letter there)

After it removes the letter it goes back through and tries to find the letter again. Remember he said it's only going to look for the first instance, so we want to make sure we continue to loop through until things are right. If it finds it it removes it just like before. If it doesn't it removes itself from the vowels list. Eventually vowels will run out of things to compare and we will have our end result.

The while loop evaluates if there are any values in vowels.

Andrew Patella
Andrew Patella
2,270 Points

Very slick, what a good idea Randy. would you mind looking at my code and telling my why I was getting rejected? I hate to admit I used your code to pass the challenge. but im still wondering why my attempt failed, any insight would be much appreciated. Thank You

def disemvowel(word):

    vowels = "aeiou"
    lis_word = list(word.lower())
    x = []

    for letter in lis_word:
        if letter not in vowels:
            x.append(letter)   

    word = ''.join(x)

    return word  

when I plug it into the teamTreeHouse workspace and print the function it seems to pop out all the right letters no matter what i use for the argument. I am really stumped on it

Steven Parker
Steven Parker
225,712 Points

The challenge expects letters that are not removed to remain unchanged, but this code converts them all to lower case.

Andrew Patella
Andrew Patella
2,270 Points

Ohhh ok, that makes sense, Thank you

This is brilliant, Randy! I wish I could have thought of it on my own.

tyler bucasas
tyler bucasas
2,453 Points

whats the function of this line in your code? word = "".join(split_word)

shubhamkt
shubhamkt
11,675 Points

In the function, you should do the following.

  1. Consider an empty string
  2. Make a list of Vowels( both lower and upper case)
  3. Don't change the 'word ' to list.
  4. Compare each letter with the array created (if x not in word) within a for loop.
  5. If the letter is not there, then append the empty string.
  6. return the word

I've tried making an adjustment, and I just keep getting the notification of "Bummer! Hmm, got back letters I wasn't expecting!"

STILL NEED HELP.

def disemvowel(word):
    return word
    vowels = ["a","e","i","o","u","A","E","I","O","U"]
    word = list(word)
    for letter in word:
        try:
            word.remove("a")
        except ValueError:
            pass
        try:
            word.remove("e")
        except ValueError:
            pass
        try:
            word.remove("i")
        except ValueError:
            pass
        try:
            word.remove("o")
        except ValueError:
            pass
        try:
            word.remove("u")
        except ValueError:
            pass
        try:
            word.remove("A")
        except ValueError:
            pass
        try:
            word.remove("E")
        except ValueError:
            pass
        try:
            word.remove("I")
        except ValueError:
            pass
        try:
            word.remove("O")
        except ValueError:
            pass
        try:
            word.remove("U")
        except ValueError:
            pass
        word = ''.join(word)
    return word
Minwook Jeong
Minwook Jeong
2,453 Points

I am also having trouble with that, I've added some advises from Steven on my work, Could you look at my work?? It works well. but not works in Challenge Question

          <p>
def disemvowel(word):
    word = word.lower()
    word = list(word)
    for letter in word:
        try:
            word.remove("a")
        except ValueError:
            pass
        try:
            word.remove("e")
        except ValueError:
            pass
        try:
            word.remove("i")
        except ValueError:
            pass
        try:
            word.remove("o")
        except ValueError:
            pass
        try:
            word.remove("u")
        except ValueError:
            pass
    word = "".join(word)
    return word
</p>
Steven Parker
Steven Parker
225,712 Points

The letters which are not removed should remain unchanged. But you lower-case the whole word, so any capital letters that are not vowels will not be correct in the output.

Also, I'm assuming those HTML paragraph tags are not really part of your code. The correct way to format code can be found in the Markdown Cheatsheet pop-up below the "Add an Answer" area. :arrow_heading_down: Or watch this video on code formatting.

Decided to go the route of appending letters which aren't in the list of Vowels. Working great.

def disemvowel(word):

    vowel = ['a','e','i','o','u','A','E','I','O','U']

    word_list = list(word)

    new_word = []

    for letter in word_list:
        if letter not in vowel:
            new_word.append(letter)

    return "".join(new_word)

This is my solution to pass this challenge, I was stuck and confused about string and list, and then I found String is immutable ( because my English is not very well so maybe I pass it in the video ). So I found this data type called slide to help my problem https://stackoverflow.com/questions/4324700/python-delete-a-character-from-a-string

I hope this will help someone!

def disemvowel(word):
    index =0
    for letter in word:
        if letter.lower() in ["a", "e", "i", "o","u"]: 
            word = word[:index] + word[index+1:]
        else:
            index +=1
    return word

P.s: omg slide is the next data type we will learn @@

Here is my solution to pass the challenge. ``ยด hope it helps

def disemvowel(word):
    vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
    list_word = list(word)
    for vowel in vowels:
        for char in word:
            if vowel == char:
                list_word.remove(vowel)
    word = "".join(list_word).strip()
    return word
Steven Parker
Steven Parker
225,712 Points

:warning: Treehouse strongly discourages posting of solutions without explanations, and sometimes the staff or moderators will redact them.