我的用于计算用户输入语句中的元音数量的程序不会对同一元音进行两次计数。 我该如何解决？

Question

print ("Sentence analysis")

Sentence = (input("Please enter a sentence"))


def WordCount(Sentence):
    words = (Sentence.count(' ')+1)
    print ("There are", words ,"words in this sentence")
WordCount(Sentence)

The code above is fine and used to count how many words in the input sentence.

vowels = ['a','e','i','o','u']
count=0
for v in vowels:
    if v in Sentence:
        count+=1


print (count)

When running, say if I input aaeiou would only count 5 vowels whereas there are 6. How do I fix this?

Answer 1

Use .count() :

count = 0
for v in vowels:
    count += Sentence.count(v)

Or better:

count = sum(Sentence.count(v) for v in vowels)

Answer 2

That is because you are doing your check in reverse. You want to go over your sentence and check each letter against vowels:

vowels = ['a','e','i','o','u']
count=0
for s in Sentence:
    if s in vowels:
        count+=1


print (count)

For a nicer approach, however, check out @zondo's answer.

Answer 3

The problem you are having is that when you loop through v in vowels and check if v is in Sentence it only checks if it v is present in Sentance not how many times. If you flip it around so it checks through Sentence first and check each letter to see if it is in vowels it will check all of the letters in Sentence .

print ("Sentence analysis")

Sentence = (input("Please enter a sentence"))

vowels = ['a','e','i','o','u']
count=0
for letter in Sentence:
    if letter in vowels:
        count+=1

print (count)

Answer 4

For the first code can be simplified the WordCount() function as:

print(len(Sentence.split()))

or:

import re
print(len(re.findall(r'\w+', Sentence)))