擅长:python、mysql、java
<p>这个怎么样:</p>
<pre><code>from random import shuffle
items = [
["e",None,None],
["pork","pork.png","meat"],
["beef","b.png","meat"],
["cheese","c.png","not"],
]
items_list = ["e","beef","pork","beef"]
shuffle(items_list)
for item in items_list:
for orig_item in items:
try:
orig_item.index(item)
print(item, orig_item)
except ValueError:
pass
</code></pre>
<p>输出:</p>
<pre><code>beef ['beef', 'b.png', 'meat']
pork ['pork', 'pork.png', 'meat']
e ['e', None, None]
beef ['beef', 'b.png', 'meat']
</code></pre>
<p>这可能比@pm2ring的答案慢(我没有测量),因为有两个循环和异常处理。你知道吗</p>
<p>使用<code>orig_item[1]</code>访问子列表的第二个元素</p>