python - multi_vowel_words 函数返回具有 3 个或更多连续元音的所有单词

Question

import re
    def multi_vowel_words(text):
        pattern =r"\b\w[aeiou]{3,},?\s?.*\w[aeiou]{3,}.*\b"
        result = re.findall(pattern, text)
        return result

我在哪里做错了

运行我的代码后，我得到了这个：

[]
['queen is courageous and gracious']
['quietly and await their delicious dinner']
[]
[]

#below this are desired outputs
print(multi_vowel_words("Life is beautiful")) 
# ['beautiful']

print(multi_vowel_words("Obviously, the queen is courageous and 
gracious.")) 

# ['Obviously', 'queen', 'courageous', 'gracious']

print(multi_vowel_words("The rambunctious children had to sit quietly and 
await their delicious 
dinner.")) 
# ['rambunctious', 'quietly', 'delicious']

print(multi_vowel_words("The order of a data queue is First In First Out 
(FIFO)")) 
   # ['queue']

print(multi_vowel_words("Hello world!")) 
   # []
print(multi_vowel_words("The order of a data queue is First In First Out 

(FIFO)")) # ['queue']

print(multi_vowel_words("Hello world!")) # []

Answer 1

试图尝试回答这个问题。一个简单的模式来检查三个连续的元音怎么样

def multi_vowel_words(text):
    pattern =r"\b\w*[aeiou]{3,}\w*\b"
    result = re.findall(pattern, text)
    return result

Answer 2

每个人都在使用\w匹配元音的前后字符。这是错误的，因为\w匹配任何单个字母、数字或下划线（与 相同[a-zA-Z0-9_]）。正确的模式应该是：

pattern = r"\b[A-Za-z]*[aeiou]{3,}[A-Za-z]*\b"

Answer 3

pattern = r"\w*[aeiou]{3,}\w*"

这将起作用。

Answer 4

假设我猜对了你的意图，这就是你需要的正则表达式：

r"\b\w*[aeiou]{3,}\w*\b"