如何找到列值之差最大的那对行按某个列值分组

Question

我有一个table类似于下面的，

我想找到每个会话的性能差异最大的一对中心。所以，我想要一个像下面这样的 table，

我正在努力为 h2 数据库形成正确的查询。

Answer 1

您可以自行加入 table 并使用相关子查询进行过滤：

select 
    t1.session,
    t1.center center1,
    t2.center center2,
    t1.performance - t2.performance performance
from mytable t1
inner join mytable t2 on t1.session = t2.session
where t1.performance - t2.performance = (
    select max(t11.performance - t22.performance)
    from mytable t11
    inner join mytable t22 on t11.session = t22.session
    where t11.session = t1.session
)

或者您可以使用 window 函数：

select *
from (
    select 
        t1.session,
        t1.center center1,
        t2.center center2,
        t1.performance - t2.performance performance,
        rank() over(partition by t1.session order by t1.performance - t2.performance desc)rn
    from mytable t1
    inner join mytable t2 on t1.session = t2.session
) t
where rn = 1

Answer 2

H2 可能有点棘手。我认为这符合您的要求：

select t.session,
       max(case when performance = min_performance then center end) as center_1,
       max(case when performance = max_performance then center end) as center_2,
       (max_performance - min_performance) as performance_difference
from (select t.*
             min(performance) over (partition by session) as min_performance,
             max(performance) over (partition by session) as max_performance
      from t
     ) t
group by t.session, min_performance, max_performance;

如何找到列值之差最大的那对行按某个列值分组

How to find the pair of rows where the difference of columns value is maximum group by some column value

sql

h2