擅长:python、mysql、java
<p>这里有一个用<a href="https://docs.python.org/2/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">^{<cd1>}</a>实现的方法</p>
<pre><code>from itertools import groupby
ls = []
for key, group in groupby(lst, lambda x: x != "TAG"):
if key:
ls.append(list(group))
print ls
# [[2, 3, 5], [6, 7, 3, 2, 6], [9, 9, 8, 3]]
</code></pre>
<p>这也可以理解为:</p>
<pre><code>ls = [list(group) for key, group in groupby(lst, lambda x: x != "TAG") if key]
</code></pre>