Implement the function which should return `true` if given object is a vowel (meaning `a, e, i, o, u`), and `false` otherwise.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def is_vowel(s):
	

Solutions:
def is_vowel(s):
	return s.lower() in set('aeiou')


import re

def is_vowel(s):
	regex = '^[aeiou]{1}$'
	return bool(re.match(regex, re.escape(s), re.IGNORECASE))


import re

def is_vowel(stg):
	return bool(re.fullmatch('[aeiou]', stg, re.I))


def is_vowel(s):
	return True if s.lower() in ['a', 'e', 'i', 'u', 'o'] else False


is_vowel = set('aeiouAEIOU').__contains__


import re

def is_vowel(s):
	return True if re.match('^[AaEeIiOoUu](?!\\n)$', s) else False


def is_vowel(s):
	return bool(s) and s.lower() in 'uaeio'


import re
PATTERN = re.compile('[eaiou]', flags=re.I)

def is_vowel(s):
	return bool(PATTERN.fullmatch(s))


import re

def is_vowel(string):
	return bool(re.match('^[aeiou]\\Z', string, re.IGNORECASE))


import re

def is_vowel(s):
	if len(s) == 1:
		if s[0] in 'a e i o u A E I O U'.split():
			return True
	return False