如果你想这样做Pandas，你可以这样做：

df[df.assign(NC = df.isnull().sum(1)).groupby('id')['NC'].transform(lambda x: x == x.min())]

输出：

row_number  id firstname middlename lastname
1           1   1      John      Jacob      Doe
2           2   2    Alison      Marie    Smith

决胜局：
添加一行：

df.loc[4,['row_number','id','firstname','middlename','lastname']] = ['4',2,'Mary','Maxine','Maxwell']

然后使用groupby、transform和idxmin：

df[df.index == df.assign(NC = df.isnull().sum(1)).groupby('id')['NC'].transform('idxmin')]

输出：

row_number id firstname middlename lastname
1          1  1      John      Jacob      Doe
2          2  2    Alison      Marie    Smith

哦，你想要null值最少的行。我建议：

select t.*
from (select t.*,
             dense_rank() over (order by (case when firstname is null then 1 else 0 end) + 
                                         (case when middlename is null then 1 else 0 end) + 
                                         (case when lastname is null then 1 else 0 end)
                               ) as seqnum

      from t
     ) t
where seqnum = 1;

这是ANSI标准的SQL。