vowels = {“a”, “e”, “i”, “o”, “u”, “A”, “E”, “I”, “O”, “U”} if any(char in vowels for char in word): …
这样比较好，因为它会在单词中找到元音后立即短路。因此，除非字符串中没有元音，否则不必检查所有字符。
了一个timeit测试，发现@falsetru的答案非常快，但是经过很少的优化，该re版本胜过其他所有功能。
使用set.isdisjoint（此方法找到匹配项后立即返回）：
>>> vowels = set('aeIoU') # set('aeIoUAEIoU') if you want case-insensitivty
>>> not vowels.isdisjoint('bcd')
False
>>> not vowels.isdisjoint('hello')
True
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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