提取连续使用元音符号的单词

你可以这样做,

texts = ['ane', 'mood', 'xao', 'pqr', 'aa']
signs = ['a', 'e', 'i', 'o', 'u']

for i in texts:
    for x in zip(i, i[1:]):
        if all(check in signs for check in x):
            print(f"double vowel sign exists in {i}")

输出

double vowel sign exists in mood
double vowel sign exists in xao
double vowel sign exists in aa

不是最简洁的,但可读性很强。 一旦找到双元音,就停止迭代一个单词。

texts = ["ane", "mood", "xao", "pqr", "aa"]
signs = ["a", "e", "i", "o", "u"]


def has_two_consecutive_vowels(word: str) -> bool:
    counter = 0
    for letter in word:
        counter = counter + 1 if letter in signs else 0
        if counter > 1:
            return True
    return False


for word in texts:
    if has_two_consecutive_vowels(word):
        print(f"double vowel sign exists in {word}")

请尝试以下解决方案。我们在中的每个字符串上循环 texts 并将每个字符串拆分为一个字符列表,如 ['a', 'n', 'e'] 对于 "ane" 。然后你可以循环浏览这个列表,检查每个索引和下一个索引处的字符是否也是元音。我们将这些字符串附加到一个空列表中,然后可以打印它们。

texts = ['ane', 'mood', 'xao', 'pqr', 'aa']
signs = ['a', 'e', 'i', 'o', 'u']
consecutive_vowels = []

for i in texts:
    list_i = list(i)
    for j in range(len(list_i)-1):
        if (list_i[j] in signs) and (list_i[j + 1] in signs):
            consecutive_vowels.append(i)
            
for word in consecutive_vowels:
    print("double vowel sign exists in {}".format(word))

输出

double vowel sign exists in mood
double vowel sign exists in xao
double vowel sign exists in aa

这是一个正则表达式版本。它只是检查双元音,如果找到了,就会保留单词。

dbl = __import__('re').compile(r'[aeiou]{2}')
txt = ('ane', 'mood', 'xao', 'pqr', 'aa')
mtc = [w for w in txt if dbl.search(w)]
print('double vowels in:', *mtc, sep='\n\t') 

您可以使用正则表达式来定义所需的模式。

https://docs.python.org/3/library/re.html

import re             
pattern = ".*[" + ''.join(signs) + "]{2,}"
for _text in texts:
    if re.search(pattern, _text):
        print(f"double vowel sign exists in {_text}")
        
double vowel sign exists in mood
double vowel sign exists in xao
double vowel sign exists in aa