擅长:python、mysql、java
<p>您可以使用<a href="https://docs.python.org/3/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">^{<cd1>}</a>:</p>
<pre><code>>>> from itertools import groupby
>>> example_list=['This', 'is', 'QQQQQ', 'an', 'QQQQQ', 'example', 'list', 'QQQQQ', '.']
>>> [' '.join(g) for k, g in groupby(example_list, lambda x: x == 'QQQQQ') if not k]
['This is', 'an', 'example list', '.']
</code></pre>
<p>或者甚至用<a href="https://docs.python.org/3/reference/datamodel.html#object.__eq__" rel="nofollow noreferrer">^{<cd2>}</a>比较,正如<a href="https://stackoverflow.com/questions/51634279/how-to-reorganize-a-list-in-specific-way-in-python?noredirect=1#comment90233044_51634393">@tobias_k</a>在评论中所建议的:</p>
<pre><code>>>> [' '.join(g) for k, g in groupby(example_list, key='QQQQQ'.__eq__) if not k]
['This is', 'an', 'example list', '.']
</code></pre>