擅长:python、mysql、java
<p>这里有一个<code>itertools.groupby</code>解决方案:</p>
<pre><code>from itertools import groupby
from operator import itemgetter
li = [('x', 'y'), ('s', 'e'), ('s, 'a'), ('x', 'z')]
[p for k, g in groupby(sorted(li, reverse=True), itemgetter(0)) for p in reversed(list(g))]
# [('x', 'y'), ('x', 'z'), ('s', 'a'), ('s', 'e')]
</code></pre>