擅长:python、mysql、java
<p>首先,您需要创建一个没有您想要忽略的字符的新字符串(看一下<a href="https://docs.python.org/3.1/library/string.html" rel="nofollow noreferrer">string library</a>,尤其是{<cd1>}),然后<code>split()</code>将得到的字符串(句子)分成子字符串(单词)。除此之外,我建议使用<a href="https://docs.python.org/3/library/typing.html" rel="nofollow noreferrer">type annotation</a>,而不是像这样的注释。在</p>
<pre><code>def words_of_length(n: int, s: str) -> list:
return [x for x in ''.join(char for char in s if char not in __import__('string').punctuation).split() if len(x) == n]
>>> words_of_length(3, 'Guido? van, rossum. is the best!'))
['van', 'the']
</code></pre>
<p>或者,您可以用您想忽略的字符定义一个变量,而不是<code>string.punctuation</code>。在</p>