擅长:python、mysql、java
<p>或者,使用<a href="https://docs.python.org/2/library/re.html#re.sub" rel="nofollow noreferrer">^{<cd1>}</a>和一个<em>替换函数</em>:</p>
<pre><code>>>> import re
s = 'hello <wolfrevokcats>, how <t uoy era>oday?'
>>> re.sub(r"<(.*?)>", lambda match: match.group(1)[::-1], s)
'hello stackoverflow, how are you today?'
</code></pre>
<p>其中<code>.*?</code>将以<a href="https://stackoverflow.com/questions/2301285/what-do-lazy-and-greedy-mean-in-the-context-of-regular-expressions">non-greedy</a>方式匹配任意数量的任何字符。它周围的括号可以帮助我们将它捕获到<a href="https://docs.python.org/2/howto/regex.html#grouping" rel="nofollow noreferrer">group</a>中,然后在替换函数-<code>match.group(1)</code>中引用它。<code>[::-1]</code>切片表示法<a href="https://stackoverflow.com/a/931095/771848">reverses a string</a>。在</p>