擅长:python、mysql、java
<pre><code>In [4]: ["".join(pair) for pair in zip(* 2 * [iter(s)])]
Out[4]: ['aa', 'bb', 'cc']
</code></pre>
<p>请参见:<a href="https://stackoverflow.com/questions/2233204/how-does-zipitersn-work-in-python">How does zip(*[iter(s)]*n) work in Python?</a>以了解对相同<code>str</code>语法的奇怪“2-<code>iter</code>”的解释。在</p>
<hr/>
<p>你在评论中说你想“拥有最快的执行速度”,我不能向你保证这个实现,但是你可以使用<a href="http://docs.python.org/library/timeit.html" rel="nofollow noreferrer">^{<cd3>}</a>来测量</em>的执行。当然记得<a href="http://en.wikiquote.org/wiki/Donald_Knuth#Computer_Programming_as_an_Art_.281974.29." rel="nofollow noreferrer">what Donald Knuth said about premature optimisation</a>。对于手头的问题(现在你已经揭示了它),我想你会发现{<cd4>}很难克服。在</p>
^{pr2}$
<p>比较</p>
<pre><code>python3.2 -m timeit -c '
s = "aabbcc"
r, g, b = s[0:2], s[2:4], s[4:6]
'
1000000 loops, best of 3: 1.2 usec per loop
</code></pre>