<p>将<code>re.split</code>与</p>
<pre><code>(?<=[a-z])(?=[A-Z])
</code></pre>
<p>见<a href="https://regex101.com/r/Ojh7G0/1" rel="nofollow noreferrer">proof</a></p>
<p><strong>解释</strong></p>
<pre><code>
(?<= look behind to see if there is:
[a-z] any character of: 'a' to 'z'
) end of look-behind
(?= look ahead to see if there is:
[A-Z] any character of: 'A' to 'Z'
) end of look-ahead
</code></pre>
<p><a href="https://tio.run/##K6gsycjPM9ZNBtP//2fmFuQXlSgUpXIVJJaUpBblKdgqFClp2NvYRifqVsVqatjbRjvqRsVqKnGVpBaXAGWVQkKAjOCSosy8dEdHJa4CIKNEoyhVr7ggJ7NEA2qMjgJIuabm//8A" rel="nofollow noreferrer">Python code</a>:</p>
<pre><code>import re
pattern = r"(?<=[a-z])(?=[A-Z])"
test = "TTestStringAA"
print(re.split(pattern, test))
</code></pre>
<p>结果:</p>
<pre><code>['TTest', 'String', 'AA']
</code></pre>