擅长:python、mysql、java
<p>您可以使用列表:</p>
<pre><code>>>> seq = [['String1', 4], ['String2', 8], ['String3', 15], ['String4', 8]]
>>> result = [s for s, i in seq if i == 8]
>>> print result
['String2', 'String4']
</code></pre>
<hr/>
<p>或者,如果您可以控制数据结构的格式,我建议您使用字典。那么获取字符串很容易:</p>
<pre><code>>>> d = {8: ['String2', 'String4'], 4: ['String1'], 15: ['String3']}
>>> print d[8]
['String2', 'String4']
</code></pre>
<p>首先将数据转换成这种格式是相当简单的,尽管我不认为一行就可以干净利落地完成。你知道吗</p>
<pre><code>>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for s, i in seq:
... d[i].append(s)
...
>>> print d
defaultdict(<type 'list'>, {8: ['String2', 'String4'], 4: ['String1'], 15: ['String3']})
</code></pre>
<p>或者</p>
<pre><code>>>> d = {}
>>> for s,i in seq:
... d.setdefault(i, []).append(s)
...
>>> print d
{8: ['String2', 'String4'], 4: ['String1'], 15: ['String3']}
</code></pre>