擅长:python、mysql、java
<p>我建议不要使用join to string,然后再拆分答案。<code>groupby</code>(或等效for loop)更好</p>
<pre><code>>>> from itertools import groupby
>>> my_list = ['A > 1', 'B > 2', '\n', 'C > 3', 'D > 4', '\n', 'E > 5', 'F > 6', '\n']
>>> [list(g) for k, g in groupby(my_list, "\n".__ne__) if k]
[['A > 1', 'B > 2'], ['C > 3', 'D > 4'], ['E > 5', 'F > 6']]
</code></pre>
<p>你可以像这样直接进入最后一步</p>
^{pr2}$