我的查询需要很长时间才能完成查找列值差异最大的行对由另一列分组

Question

说，我有一个 table 这样的：

我想找到每个会话中性能差异最高的一对中心，如下所示：

我有以下查询，

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
)

它可以工作，但需要很长时间，对于 20 列和 200 行的 table 几分钟。 如何修改查询以更快地实现相同的输出？

Answer 1

使用row_number():

select session, center1, center2, performance
from (select t1.center as center1, t2.center as center2,
             (t1.performance - t2.performance) as performance,
             row_number() over (partition by t1.session order by (t1.performance - t2.performance) desc) as seqnum
      from mytable t1 join
           mytable t2
           on t1.session = t2.session
where seqnum = 1;

或者为了更好的表现。最大差值是最大值减去最小值。你想要中心，这是一个没有子查询的方法：

select session,
       max(case when seqnum_desc = 1 then center end) as center1,
       max(case when seqnum_asc = 1 then center end) as center2,
       max(performance) - min(performance)
from (select t.*,
             row_number() over (partition by session order by performance) as seqnum_asc,
             row_number() over (partition by session order by performance desc) as seqnum_desc
      from mytable t
where 1 in (seqnum_asc, seqnum_desc)
group by session

Answer 2

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 = (SELECT MAX(performance) 
                            FROM mytable t3 WHERE t3.session = t1.session)
      AND t2.performance = (SELECT MIN(performance) 
                            FROM mytable t3 WHERE t3.session = t2.session)

     // Im thinking this will solve the border case when performance is a tie 
     // and difference 0 will return 2 rows

     AND (CASE WHEN t1.performance = t2.performance 
               THEN CASE WHEN t1.center < t2.center
                         THEN 1
                         ELSE 0
                    END
               ELSE 1
          END) = 1

只要您在 performance 和 session 上有索引就可以了。

Answer 3

按会话分组并采用组的最小和最大性能似乎合乎逻辑。不幸的是，实际的中心在这里需要 subquery/join。

select g.session as Session,
    (select min(center) from mytable
     where session = g.session and performance = g.maxim) as Center1,
    (select min(center) from mytable
     where session = g.session and performance = g.minim) as Center2,
    g.maxim - g.minim as Performance
from (select 
        t1.session,
        min(t1.performance) as minim,
        max(t1.performance) as maxim
    from mytable t1
    group by t1.session)
    as g

确保会话和性能指标。

Answer 4

select distinct(session) * from (
select  t1.session, t1.center, t2.center, 
(case when t1.performance > t2.performance then (t1.performance-t2.performance) else (t2.performance-t1.performance))as performance_diff 
from mytable t1, mytable t2 
where t1.session=t2.session and t1.center!=t2.center) as T1 order by session,performance_diff desc limit 1;

我的查询需要很长时间才能完成查找列值差异最大的行对由另一列分组

My query is taking too long to finish for finding the pair of rows where the difference of columns value is maximum group by another column

java

sql

h2