<p>可能这不是一个好看的解决方案,但它将剥离所有不可打印和空白字符:</p>
<pre><code>from itertools import takewhile
check = lambda c: not c.isprintable() or c.isspace()
result = s[len(tuple(takewhile(check, s))): -len(tuple(takewhile(check, reversed(s))))]
# OR to not waste memory for tuple
result = s[sum(1 for _ in takewhile(check, s)): -sum(1 for _ in takewhile(check, reversed(s)))]
</code></pre>
<p>在这里,我使用<a href="https://docs.python.org/3/library/itertools.html#itertools.takewhile" rel="nofollow noreferrer">^{<cd1>}</a>从字符串开始获取所有不可打印(<a href="https://docs.python.org/3/library/stdtypes.html#str.isprintable" rel="nofollow noreferrer">^{<cd2>}</a>)和空格(<a href="https://docs.python.org/3/library/stdtypes.html#str.isspace" rel="nofollow noreferrer">^{<cd3>}</a>)字符。然后使用<a href="https://docs.python.org/3/library/functions.html#len" rel="nofollow noreferrer">^{<cd4>}</a>(或带有列表理解的选项中的<a href="https://docs.python.org/3/library/functions.html#sum" rel="nofollow noreferrer">^{<cd5>}</a>),我得到了诸如字符的数量,并将该数量用作字符串切片中的开始索引。为了获得结束索引,我使用了相同的方法,但使用了反向(<a href="https://docs.python.org/3/library/functions.html#reversed" rel="nofollow noreferrer">^{<cd6>}</a>)字符串</p>