计算字符串python

Question

我对手动输入的字符串计数有点困惑。我基本上是在计算单词的数量和没有空格的字符的数量。如果可能的话，谁能帮助计算元音？

到目前为止，我只有这些：

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);

Answer 1

对于元音：

这样做的基本方法是使用for循环并检查每个字符是否存在于字符串、列表或其他序列中（本例中为元音）。我认为这是你应该首先学习的方法，因为这对初学者来说是最容易理解的。

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 

一旦你了解了更多，并理解了列表的含义，你就可以做到了

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)

或者像图腾写的那样，用和

def how_many_vowels(text):
  vowels = 'aeiou'
  vowel_count = sum(1 for ch in text if ch.lower() in vowels)
  return vowel_count

Answer 2

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

Answer 3

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上：http://www.tutorialspoint.com/python/string_split.htm

关于和：

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上：http://www.tutorialspoint.com/python/string_replace.htm