擅长:python、mysql、java
<p>您只需像这样比较两个列表:循环遍历<code>findKBs</code>的值,如果它们不在<code>df['A'][0]</code>中,则将它们分配给新列表</p>
<pre><code>df['C'] = [[x for x in findKBs if x not in df['A'][0]]]
</code></pre>
<p>结果:</p>
<pre><code> A B C
0 [KB4525236, KB4485447, KB4520724, KB3192137, K... [a, b] [KB4525202]
</code></pre>