Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Collections (Retired) Lists Redux Disemvoweled

Posted by Felix Putra
.a{fill-rule:evenodd;}techdegree
Felix Putra
Python Web Development Techdegree Student 1,927 Points

Why doesn't my code remove the vowel letters?

Hi guys,

I've tried to make my own code, and here is the result. My code doesn't remove any of the vowel letters. In fact, it prints out the same thing as the variable letters_list. Any help will be very appreciated and thank you in advance!

https://w.trhou.se/aksnkdbgok

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points
Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

I found the following issues with your code.

  • A list() of a list returns the same list. That is:
>>> letters = ["abcde"]
>>> chunked = list(letters)
>>> letters
['abcde']
>>> chunked
['abcde']
>>> letters == chunked
True

If you wanted chunked to be ['a', 'b', 'c', ...] either start with letters equal to just the string "abcdefghijklmnopqrstuvwxyz" or set chunked = list(letters_list[0])

  • in the if/elif/else statement, the conditions should be using the called method .lower() (with parens). Otherwise, you are comparing the existence of a method called 'lower` with a string. This will always fail.

  • You do not have a way to increment your index i

  • At the end of the loop, the length is set to -1. This will fail the while condition on the first iteration. You probably wanted length -= 1 to decrement the length.

After correcting the above and rerunning:

  • printing chunked[i] at the end of the loop will print the current index even if it has been deleted. When the 'a' is deleted in the first iteration with i=0, the print(chunked[i]) will print 'b'

Good Luck!

Bob Buethe
Bob Buethe
3,276 Points
Bob Buethe
Bob Buethe
3,276 Points

As Chris said, list(letters_list) will just return the same string. You need to use list(letters_list[0])

Let's step through this:

Start:

chunked = ['a', 'b', 'c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']

length = 25

i = 0

FIRST PASS:

'a' is removed, so chunked = [ 'b', 'c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']

length is set to -1 (because the statement reads "length =- 1" instead of "length -= 1")

i is still 0

Since length is now less than 0, the procedure ends.

But what happens if "length =- 1" is changed to "length -= 1"? In that case:

FIRST PASS:

'a' is removed, so chunked = [ 'b', 'c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']

length is set to 24 i is still 0

SECOND PASS:

chunked[0] is 'b', so nothing is changed.

length is set to 23

i is still 0

THIRD PASS:

chunked[0] is still 'b', so nothing is changed.

length is set to 22

i is still 0

And so on, until length = -1. You're decrementing length, but not incrementing i, so the program keeps looking at chunked[0] at every pass.

But these loops aren't necessary, because remove will find the letter to remove wherever it is in the list. So all you need to do is:

from string import join
letters_list = ["abcdefghijklmnopqrstuvwxyz"]
chunked = list(letters_list[0])
chunked.remove('a').remove('e').remove('i').remove('o').remove('u')
print(join(chunked, '')