我对手动输入的字符串计数感到有点困惑。我基本上是在尝试计算单词数和不带空格的字符数。另外，如果可以的话，谁能帮忙数一下元音？
这就是我目前所拥有的：
vowels = [ 'a','e','i','o','u','A','E','I','O','U']
constants= ['b','c','d','f','g','h','j','k','l','m','n','p','q','r','s','t','v','w','x','y','z']
s= input ('Enter a Sentence: ')
print ('Your sentence is:', s)
print ('Number of Characters', len(s));
print ('Number of Vowels', s);
【问题讨论】：
s = input("Enter a sentence: ")
word_count = len(s.split()) # count the words with split
char_count = len(s.replace(' ', '')) # count the chars having replaced spaces with ''
vowel_count = sum(1 for c in s if c.lower() in ['a','e','i','o','u']) # sum 1 for each vowel
更多信息：
在str.split：http://www.tutorialspoint.com/python/string_split.htm
在sum：
sum(sequence[, start]) -> value
Return the sum of a sequence of numbers (NOT strings) plus the value
of parameter 'start' (which defaults to 0). When the sequence is
empty, return start.
在str.replace:http://www.tutorialspoint.com/python/string_replace.htm
【讨论】：
vowels = [ 'a','e','i','o','u']
sentence = "Our World is a better place"
count_vow = 0
count_con = 0
for x in sentence:
if x.isalpha():
if x.lower() in vowels:
count_vow += 1
else:
count_con += 1
print count_vow,count_con
【讨论】：
count_vow += 1
元音：
执行此操作的基本方法是使用 for 循环并检查每个字符是否存在于字符串、列表或其他序列（本例中为元音）中。我认为这是您应该首先学习的方式，因为它对初学者来说最容易理解。
def how_many_vowels(text):
vowels = 'aeiou'
vowel_count = 0
for char in text:
if char.lower() in vowels:
vowel_count += 1
return vowel_count
一旦你了解更多并弄清楚list comprehensions，你就可以做到
def how_many_vowels(text):
vowels = 'aeiou'
vowels_in_text = [ch for ch in text if ch.lower() in vowels]
return len(vowels_in_text)
或如@totem 所写，使用 sum
def how_many_vowels(text):
vowels = 'aeiou'
vowel_count = sum(1 for ch in text if ch.lower() in vowels)
return vowel_count
【讨论】：