擅长:python、mysql、java
<p>您可以使用<a href="https://docs.python.org/2/library/itertools.html#itertools.groupby" rel="nofollow">^{<cd1>}</a>来实现:</p>
<pre><code>>>> l = [['boy','121','is a male child'],['boy','121','is male'],['boy','121','is a child'],['girl','122','is a female child'],['girl','122','is a child']]
>>> import itertools
>>> [k+[m[2] for m in v] for k,v in itertools.groupby(l,key = lambda x:x[:2])]
[['boy', '121', 'is a male child', 'is male', 'is a child'], ['girl', '122', 'is a female child', 'is a child']]
</code></pre>
<p>从文档中</p>
<blockquote>
<pre><code>itertools.groupby(iterable[, key])
</code></pre>
<p>Make an iterator that returns consecutive keys and groups from the iterable. The key is a function computing a key value for each
element.</p>
</blockquote>