python - python中的列表切片问题

Question

这段代码看起来应该可以工作。它对“条纹”(字母-辅音-字母等)的单词数进行求和，然后返回总和。但是，当我用 print (striped("My name is ...") ) 测试它时，它只计算 my 和 is 并给我总和是 2 而不是 3...为什么缺少 name？

VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"


def striped(text):

    my_sum = 0
    text = text.replace(".", " ").replace(",", " ").split()
    vowels = VOWELS.lower()
    consonants = CONSONANTS.lower()

    for word in text:
        word = word.lower()
        if ((word[::2] in vowels and word[1::2] in consonants)\
        or (word[::2] in consonants and word[1::2] in vowels))\
        and len(word) > 1:
            print (word)
            my_sum += 1

    return my_sum        

Answer 1

这是一个带有列表的解决方案。您的代码的问题在于，当您使用 [::2] 时，长度超过两个字符的单词会返回子字符串，而不是测试它们是否包含在 元音 中的单个字符/常量。通过先将其转换为列表，您可以检查列表中的每个项目是否包含在相应的字符集中。

VOWELS = "AEIOUY"
CONSONANTS = "BCDFGHJKLMNPQRSTVWXZ"


def striped(text):

    my_sum = 0
    text = text.replace(".", " ").replace(",", " ").split()
    vowels = VOWELS.lower()
    consonants = CONSONANTS.lower()

    for word in text:
        word = word.lower()

        if ((all(c in vowels for c in list(word[::2]))\
            and all(c in consonants for c in list(word[1::2])))\
        or (all(c in consonants for c in list(word[::2]))\
            and all(c in vowels for c in list(word[1::2]))))\
        and len(word) > 1:
            print (word)
            my_sum += 1

    return my_sum

print striped("My name is")