擅长:python、mysql、java
<p>我根据Vroomfondel的回答得出了一个答案:</p>
<pre><code>dict1 = {'a':[1,2,3], 'b':[1,2,3,4], 'c':[1,2]}
dict2 = {item: [key for key in dict1 if item in dict1[key]] for value in dict1.values() for item in value}
</code></pre>
<p>这不是最快的,但它是一个单一的班轮,它不是最慢的选择提出!在</p>
^{pr2}$
<p>打印输出:</p>
<pre><code> Vroomfondel1 0.202519893646
Vroomfondel2 0.164724111557
***Vroomfondel2 mod 0.114083051682
mhlester1 0.0599339008331
mhlester1 mod 0.091933965683
mhlester2 0.0900268554688
initial 0.0953099727631
</code></pre>