There's a easier way to detect vowels in python?

Question

I was doing some exercises to train python and improve, etc. And then i got one that asked me to develop a program that read a letter and print if it is a vowel or a consonant.

if(letter== 'a' or letter == 'A' or letter == 'e' or letter == 'E' or letter == 'i' or letter == 'I' or letter == 'o' or letter == 'O' or letter == 'u' or letter == 'U'):
    print(f"{letter} it's a vow.")
elif((letter.isalpha()) == False):
    print('I said letter, not numbers')
else:
    print(f'{letter} it's a consonant.')

Anyway, this got really bigger than i wanted and it's really ugly also it's a pain in the ass type all those letters. I want to know if there's a way to detect if it's a vow or a consonant with python, importing or not a package to do that. Tried searching only but couldn't find anything.

You could do `letter.lower() in 'aeiou'` to check if it is a wovel. — Dani Mesejo, Sep 25 '19 at 13:46
Possible duplicate of [Count Vowels in String Python](https://stackoverflow.com/questions/19967001/count-vowels-in-string-python) — MyNameIsCaleb, Sep 25 '19 at 13:46
Check the link above, it's your exact question with many well written answers. — MyNameIsCaleb, Sep 25 '19 at 13:47
Don't forget about y! Edge cases always make things more interesting — erik258, Sep 25 '19 at 13:48

score 3 · Accepted Answer · answered Sep 25 '19 at 13:46

You can use the str.lower (or str.upper) method and the str in str operator:

# or, if letter.upper() in 'AEIOU':
if letter.lower() in 'aeiou':
    print(f"{letter} it's a vow.")
elif not letter.isalpha():
    print("I said letter, not numbers")
else:
    print(f"{letter} it's a consonant.")

There's a easier way to detect vowels in python?

1 Answers1