如果 String s 中没有字符是 Python 中的元音，则返回 True

Question

I've tried looking for answers but none seem to help. I've done:

def noVowel(s):
    'return True if string s contains no vowel, False otherwise'
    for char in s:
        if char.lower() not in 'aeiou':
            return True
        else:
            return False

No matter the string, it always returns True.

Answer 1

You've almost got it right, but the problem is, as soon as you see a character that is a non-vowel, you return True right then and there. You want to return True after you've made sure that all are non-vowel:

def noVowel(s):
    'return True if string s contains no vowel, False otherwise'
    for char in s:
        if char.lower() in 'aeiou':
            return False
    return True  # We didn't return False yet, so it must be all non-vowel.

It's important to remember that return stops the rest of the function from running, so only return if you're sure the function is done computing. In your case, we can safely return False once we see a vowel, even if we didn't check the whole string.

Answer 2

With any and short circuiting properties:

def noVowel(s):
    return not any(vowel in s.lower() for vowel in "aeiou")