擅长:python、mysql、java
<p>为了便于阅读,我会创建一个<code>l1</code>的临时字典,以便更容易地查找键</p>
<p>解压<code>l2</code>中的元组,并将键与<code>l1</code>-字典中的键进行比较。如果匹配并将所有内容包装到列表中,则创建一个新元组</p>
<pre><code>l1_map = {k: v for k, v in l1}
l2_updated = [(key, num, [l1_map[key]]) for key, num in l2 if key in l1_map]
</code></pre>
<p>结果:</p>
<pre><code>[('Total Amount', ['593.52'], [9]),
('Total (words are in between)', ['593.52'], [6])]
</code></pre>