我发现了一些这样的例子，但我的问题略有不同。
我需要扫描一个句子
For letter in sentence
if letter == "a" or letter == "A": letterATotal = letterATotal + 1
elif letter == "e" or letter == "E": letterETotal = letterETotal + 1
等一直到 U。但是我需要将它们全部比较并打印包含最频繁的行，因此“最频繁的元音是 A 出现 5 次”。问题是我不知道如何显示实际的字母。我只能显示数字。有任何想法吗？
看这个：
>>> from collections import Counter
>>> mystr = "AaAaaEeEiOoouuu"
>>> a,b = Counter(c for c in mystr.lower() if c in "aeiou").most_common(1)[0]
>>> "the most frequent vowel is {} occurring {} times".format(a.upper(), b)
'the most frequent vowel is A occurring 5 times'
>>>
这是关于collections.Counter.
编辑：
这是正在发生的事情的分步演示：
>>> from collections import Counter
>>> mystr = "AaAaaEeEiOoouuu"
>>> Counter(c for c in mystr.lower() if c in "aeiou")
Counter({'a': 5, 'o': 3, 'e': 3, 'u': 3, 'i': 1})
>>> # Get the top three most common
>>> Counter(c for c in mystr.lower() if c in "aeiou").most_common(3)
[('a', 5), ('o', 3), ('e', 3)]
>>> # Get the most common
>>> Counter(c for c in mystr.lower() if c in "aeiou").most_common(1)
[('a', 5)]
>>> Counter(c for c in mystr.lower() if c in "aeiou").most_common(1)[0]
('a', 5)
>>> a,b = Counter(c for c in mystr.lower() if c in "aeiou").most_common(1)[0]
>>> a
'a'
>>> b
5
>>>
使用collections.Counter：
>>> from collections import Counter
>>> c = Counter()
>>> vowels = {'a', 'e', 'i', 'o', 'u'}
for x in 'aaaeeeeeefffddd':
if x.lower() in vowels:
c[x.lower()] += 1
...
>>> c
Counter({'e': 6, 'a': 3})
>>> letter, count = c.most_common()[0]
>>> letter, count
('e', 6)
the following code works if you are using Python 3 or above:
s = input ('Enter a word or sentence to find how many vowels:')
sub1 = 'a'
sub2 = 'e'
sub3 = 'i'
sub4 = 'o'
sub5 = 'u'
print ('The word you entered: ', s)
print ('The count for a: ', s.count(sub1))
print ('The count for e: ', s.count(sub2))
print ('The count for i: ', s.count(sub3))
print ('The count for o: ', s.count(sub4))
print ('The count for u: ', s.count(sub5))
通过句子字符串的子字符串，并在末尾增加索引为 0-4（对于 au）的整数数组，找到数组的最大值并让它打印正确的语句