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Python Python Collections (Retired) Lists Redux Disemvoweled

Marius Wallin
Marius Wallin
11,464 Points
Posted by Marius Wallin
Marius Wallin
11,464 Points

Some help to understand the while in this one

vov = list("aeiouy") a_word = input("Remove vowels from my word: ") convert = list(a_word) output = []

for vow in vov: while True: try: convert.remove(vow) except: break output.append(''.join(convert).capitalize())

print(output)

The while loop inside the for loop, is it there to make sure that it will try to remove a vow until there is no vow left (it will only get an ValueError for each try) ant the try: returns false?

3 Answers

Daniel Petrov
Daniel Petrov
3,495 Points
Daniel Petrov
Daniel Petrov
3,495 Points

Yes, they give a different outcome indeed. I tried another approach which worked without using a try/catch block and actually uses fewer steps. In Keneth's example probably the reason using it just to teach us another way to use an exception.

word_list = ['installation', 'motivation', 'initegration', 'computation']
vowels_list = ['a', 'e', 'i', 'o', 'u']
new_list = []

for word in word_list:
    letter_list = list(word.lower())

    for letter in letter_list:
        while True:
            if letter in vowels_list and letter in letter_list:
                letter_list.remove(letter)
            else:
                break

    new_list.append(''.join(letter_list).capitalize())

print(new_list)
Arpit Singh
Arpit Singh
942 Points

will you please explain why this program is not removing all the variables

v = list('aeiou')

list1 =['adsgs','dgeiourhhhaaaaaa','howaejoihvnois'] e =[] for letter in list1: letter1 = list (letter) for l in letter1: while True: if l in v and l in letter1: letter1.remove(l) else: break e.append(''.join(letter1)) print(e)

Nathan Tallack
Nathan Tallack
22,159 Points
Nathan Tallack
Nathan Tallack
22,159 Points

Consider the two blocks of code below, the first one being the nested for while with the break out of the while, the second one being a for loop with a continue. Both have the same functional outcome.

vov = list("aeiouy") 
a_word = input("Remove vowels from my word: ") 
convert = list(a_word) 
output = []

for vow in vov: 
    while True: 
        try: 
            convert.remove(vow) 
        except: 
            break 
        output.append(''.join(convert).capitalize())

print(output)

vov = list("aeiouy") 
a_word = input("Remove vowels from my word: ") 
convert = list(a_word) 
output = []

for vow in vov: 
    try: 
        convert.remove(vow) 
    except: 
        continue
    output.append(''.join(convert).capitalize())

print(output)

I am not sure why they chose to use the nested loops in this video. Perhaps there is a reason to it that is not clear to us. But ultimately the lesson is about learning list mutation not control flow, so either way works for the purpose of the lesson. :)

Rui W
Rui W
6,269 Points
Rui W
Rui W
6,269 Points

Poked at these two different approaches. I think they are eventually giving different outcomes:

The first one, or the "nested for while", is going to keep removing each of the vowels (in the order of 'aeiouy') until it fails on doing so. Meanwhile the second logic, "for loop with a continue", will only perform deletion of each of the vowels for ONE TIME.

For instance, using the word 'people', and we want to remove all vowels (2 e and 1 o) in it:

First logic

The result will be ['Pople', 'Popl', 'Ppl'] since the output variable is trying to store converted word after each vowel deletion. Here we can see the last item in the result list, or say the final version of de-vowelled word is the thing we're expecting.

What the code is doing in for loop (ignored .capitalize())

  1. try 'people'.remove('a'), which will return an error since there's no 'a', thus break, jump to next vowel;
  2. try 'people'.remove('e'), since we have "while True" there, this indicates - don't stop until "convert.remove('e')" returns an error. Thus both 'e' will be removed and now we only have 'popl' instead of 'people';
  3. try 'popl'.remove('i'), which will return an error since there's no 'i', thus break, jump to next vowel;
  4. try 'popl'.remove('o'), easy to see, 'o' will be removed and we get 'ppl'; 5...6...There 's no 'u' or 'y' in the remaining of the string.

Second logic

The result will be ['Pople', 'Pple'], we can see the last item here is still having an 'e' in it. This is because the code will continue processing to the next vowel after a successful .remove('e').

What the code is doing in for loop (ignored .capitalize())

  1. try 'people'.remove('a'), which will return an error since there's no 'a', thus break, jump to next vowel;
  2. try 'people'.remove('e'), found an 'e' in the string, removed it; then the code realize we are still in the for loop, so, goes to the next vowel!
  3. try 'pople'.remove('i'), which will return an error since there's no 'i', thus break, jump to next vowel;
  4. try 'pople'.remove('o'), easy to see, 'o' will be removed and we get 'pple'; 5...6...There 's no 'u' or 'y' in the remaining of the string.
Marius Wallin
Marius Wallin
11,464 Points
Marius Wallin
Marius Wallin
11,464 Points

RUI - thanks. Thats a very good response and made it much more clearer! ;)