目前写了判断元音的code如下
word = input()
count = 0
a = e = i = o = u = 0
for num in word.lower():
if num in 'a':
a = 1
elif num in 'e':
e = 1
elif num in 'i':
i = 1
elif num in 'o':
o = 1
elif num in 'u':
u = 1
count = a + e + i + o + u
print(count)
自己觉得这code看起来很笨......想问是不是有更好的写法，
另外也想问for num in word.lower()这行for的运作是怎么样呢？
知道word.lower()是变小写的意思，
但num in word.lower()的意思看了很久还是不太能理解> <

for num in word.lower(): if num in 'aeiou': count += 1(请自行排版) 另外你本来的写法是错的,aeiou的值没有累加

import recount=len(re.findall("[AEIOU]", word,re.I))

str.count(....)

[ch in word.lower() for ch in 'aeiou'].count(True)num in word.lower() #小写的word是否有num变量代表的字母

len( set("aeiou") & set(word) )

word='ptT is Back' for num in word作用如同for num in ['p','t','t',' ','i','s',' ',......]

感觉好leetcode