回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>鉴于此<code>DataFrame</code>:</p>
<pre><code>import pandas as pd
first=[0,1,2,3,4]
second=[10.2,5.7,7.4,17.1,86.11]
third=['a','b','c','d','e']
fourth=['z','zz','zzz','zzzz','zzzzz']
df=pd.DataFrame({'first':first,'second':second,'third':third,'fourth':fourth})
df=df[['first','second','third','fourth']]
first second third fourth
0 0 10.20 a z
1 1 5.70 b zz
2 2 7.40 c zzz
3 3 17.10 d zzzz
4 4 86.11 e zzzzz
</code></pre>
<p>我可以使用<code>df</code>创建字典</p>
<pre><code>a=df.set_index('first')['second'].to_dict()
</code></pre>
<p>所以我可以决定什么是<code>keys</code>,什么是<code>values</code>。但是如果你想让<code>values</code>成为列的列表,比如<code>second</code>和<code>third</code>,该怎么办?</p>
<p>如果我试试这个</p>
<pre><code>b=df.set_index('first')[['second','third']].to_dict()
</code></pre>
<p>我有一本奇怪的字典</p>
<pre><code>{'second': {0: 10.199999999999999,
1: 5.7000000000000002,
2: 7.4000000000000004,
3: 17.100000000000001,
4: 86.109999999999999},
'third': {0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e'}}
</code></pre>
<p>相反,我想要一本列表字典</p>
<pre><code>{0: [10.199999999999999,a],
1: [5.7000000000000002,b],
2: [7.4000000000000004,c],
3: [17.100000000000001,d],
4: [86.109999999999999,e]}
</code></pre>
<p><strong>如何处理?</strong></p>