Disemvowel

Since 'remove' removes the first instance of a value from a list, how do i go about removing the second instance of 'a' from the word 'Insatiable'.

It can remove the 2nd occurrence of 'a' by calling remove.('a') 2 times, that's not the problem here.

Generally speaking, it's a very bad idea to write a for loop to iterate through each element of a list while simultaneously shrinking the length of list by removing its items in the loop body.

Here's what happens in the first 2 iterations of your for loop in disemvowel("Insatiable")

• in the 1st iteration, Check whether the first item in the list is a vowel.
• "I" is a vowel, it's removed from the list
• At the end of the 1st iteration, my_list becomes list("nsatiable").
• now in the 2nd iteration of the loop, check whether the second item in the list is a vowel
• "s" is not a vowel, therefore it's NOT removed.

You see the problem here? letter "n" was skipped and never get checked, that's because now with "I" removed, letter "n" is the 1st item in the list, and as far as the for loop is concerned, the first item on the list has been checked in the previous iteration, it doesn't care that the value of the 1st item has changed.

This is precisely why the second instance of "a" in the word doesn't get removed in your code. When there're 2 vowels back to back, the 2nd vowel will definitely be skipped checking after the removal of the 1st vowel. And the word you have here is Insatiable, the second occurance of "a" comes after letter "i", which happens to be a vowel also.

To avoid this problem, you can change your for loop into this

for char in word:

Now the code iterates through each letter in the String word, if a letter is found to be a vowel, it gets removed from my_list, problem solved.

Second, please add the try and except block, to say, 'No vowels were found in the word' if no vowels are present in the word.

I don't think that's necessary, the purpose of the disemvowel function is to strip away vowels, if any, from the argument; simply put, it's of no significance whether the word contains any vowels, we only need to make sure that the return value of this function is the same word w/ all possible occurrences of vowel removed.

PS:

I should also point out that, the grader is probably buggy, your code've passed the challenge, but it shouldn't have, as your implementation of disemvowel function has no return value.

From the challenge description

The function disemvowel takes a single word as a parameter and then returns that word at the end.

and the code template comes with 2 lines pre-written.

def disemvowel(word):
    return word

I do believe we should preserve these 2 lines and add more code in the middle to complete the implementation. Here's a slightly altered version of your disemvowel function definition.

def disemvowel(word):
    vowels = "aeiouAEIOU"
    my_list = list(word)
    for char in word:
        if char in vowels:
            my_list.remove(char)
    word = ''.join(my_list)
    return word

It'll pass the grader check, and more importantly, it complies with every requirement the challenge asks for.

Happy coding.