擅长:python、mysql、java
<pre><code>print( df.groupby(['code1', 'code2'], as_index=False).agg('sum') )
</code></pre>
<p>印刷品:</p>
<pre><code> code1 code2 Names
0 A k [EUGENIO NETO, JUAN MATIAS SERAGOPIAN, SIMONE ...
1 B l [EUGENIO LUPORINI NETO]
</code></pre>
<hr/>
<p>编辑:具有<code>itertools.chain</code>的解决方案:</p>
<pre><code>from itertools import chain
df=pd.DataFrame({'code1':["A","B","A"],"code2":["k","l","k"],'Names':[['EUGENIO NETO','JUAN MATIAS SERAGOPIAN'],['EUGENIO LUPORINI NETO'],['SIMONE FANKHAUSER','ALEX SOUZA']]})
print( df.groupby(['code1', 'code2'], as_index=False).agg(lambda x: list(chain.from_iterable(x))) )
</code></pre>