擅长:python、mysql、java
<p>这里有一个易于扩展的简单解决方案:</p>
<pre><code>def filterls(ls, opts):
"""
ls - list
opts - dict - {id: match_info}
"""
results = []
for l in ls:
for (i, t) in opts.items():
if l[i] != t:
break
else:
results.append(l)
return results
</code></pre>
<p>例如:</p>
^{pr2}$