擅长:python、mysql、java
<p>你的问题不太清楚,但这是你想要的吗?在</p>
<pre><code>>>> D = df.groupby(['User','ASIN'])['Rating'].apply(list).to_dict()
>>> {key[0]:{key[1]:val} for key, val in D.items()}
{('A23VKINWRY6J92', '1476783284'): [5], ('A3HC4SRK7B2AXR', '1496177029'): [5], ('AE12HJWB5ODOD', 'B00K2GAUC0'): [4], ('AL4RYO265J1G', '061579615X'): [3]}
</code></pre>
<p>因此,如果将其分配给<code>my_dict</code>,则</p>
^{pr2}$
<p>等等</p>