Python: Is there a way to remove a character without regard to capitalization?

Question

My code is supposed to remove vowels from the user inputted string. It works correctly, but I'm trying to see if there is a way to simplify my code. A screenshot of my code

I originally didn't have the line:

     elif i == "A" or i == "E" or i == "I" or i == "O" or i == "U":
        n.replace(i, "")
        output = output

but I added it because when my input contained capitalized vowels it would not remove the vowels because they were capitalized.

Is there a way to rewrite my code so I can remove the vowels that are both lowercase and capitalized without the extra lines?

Answer 1

(edited to fix)

Here is one way to do it:

word = input()

output = ""
for letter in word:
    if letter.casefold() not in "aeiou":
        output += letter

print(output)

to make it into a function:

def remove_vowels(word):
    output = ""
    for letter in word:
        if letter.casefold() not in "aeiou":
            output += letter
    return output

Answer 2

You can check in lowercase, this way you do not duplicate comparisons:

if i.lower() in 'aeiou':
    n.replace(i, "")

but you can write even simpler code with no loops using regular expression:

import re

def remove_vowels(n):
    return re.sub(r'[aueoi]', '', n, flags=re.IGNORECASE)

It basically removes any char from the list [a, u, e, o, i], adding flags=re.IGNORECASE matches both lower and upper case letters, and replaces each letter with '' basicaly removing it

Answer 3

You could use the regular expressions, in particular the built-in re.sub function:

import re
text = "ThE quick brown fOx jumps over the lazy dog"
re.sub("[aeiou]","", text, flags=re.IGNORECASE)

'Th qck brwn fx jmps vr th lzy dg'

The regular expression [aeiou] will match any vowel and then the flag re.IGNORECASE will allow the expression to match the upper case vowels as well. You could instead extend the pattern to [AEIOUaeiou].