Check presence of vowels in a string

by 
vowels = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"}
if any(char in vowels for char in word):
   ...

Note: This is better because it short circuits, as soon as it finds the vowel in the word. So, it doesn’t have to check all the characters unless there are no vowels in the string.

Edit: Ran a timeit test and found that, @falsetru’s answer is extremely fast, but with few optimizations, the re version beats everything else.

import re

vowels = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"}
pattern = re.compile("[AEIOUaeiou]")

def intersection():
    return bool(vowels.intersection("TWYNDYLLYNGS"))

def any_version():
    return any(char in vowels for char in "TWYNDYLLYNGS")

def re_version():
    return bool(pattern.search("TWYNDYLLYNGS"))

def disjoint():
    return vowels.isdisjoint("TWYNDYLLYNGS")

from timeit import timeit

print timeit("intersection()", "from __main__ import intersection, vowels")
print timeit("any_version()", "from __main__ import any_version, vowels")
print timeit("re_version()", "from __main__ import re_version, vowels")
print timeit("disjoint()", "from __main__ import disjoint, vowels")

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