Question

s = input("Enter a sentence: ")
emp = ""
vow = 0
for i in s: 
    if i.isspace() == False:
        emp = emp+i
    else:
        for j in emp:
            if j in 'aeiouAEIOU':
                vow = vow+1
        print("The number of vowels in", emp, "are", vow)
        emp = ""

它只给出第一个单词的元音。

Answer 1

您的代码无效，因为变量vow永远不会重置为0.您应该为每个遇到的新单词重置它，否则它会打印字符串中元音的总数。

更好的解决方案是使用split()代替isspace()，因此词语会自动生成＆＃34;分离。

然后，你可以迭代每个单词，并总结出作为元音的字母。

words = s.split()
for word in words:
    vow = sum(letter in 'aeiou' for letter in word.lower())
    print("The number of vowels in", word, "are", vow)

Answer 2

将re模块用于正则表达式。

import re

sentence = r'Hello World, how many vowels are in this sentence by word?'

for word in sentence.split():
    s = re.findall(r'[aeiou]', word, flags=re.IGNORECASE)
    print(word,'has',len(s),'vowels.')

Answer 3

您可以使用sum或len个功能。

>>> s="asdf 14oji iii34"
>>> sum(1 for i in s if i in 'aeiouAEIOU')
6
>>>

或

>>> len([i for i in s if i in 'aeiouAEIOU'])
6
>>>

Answer 4

您也可以这样做：

sentence = input("Enter a sentence: ")
vowels = "aeiouAEIOU"

for word in sentence.split():
    vowel_count = 0
    for letter in word:
        if letter in vowels:
            vowel_count += 1

    print("The number of vowels in %s is: %d" % (word, vowel_count))

如何在字符串的每个单词中找到元音的数量？

4 个答案: