前2名的平均值

Question

我想获取每个 policyid 的前2个 limit1 的平均值。我需要我的结果表也要有 objectid 。

Limit1 和 objectid 来自表p_coverage。 策略来自表p_risk。 表p_item是p_risk和p_coverage之间的链接表。

我认为应该建立查询的方式是：在每个 policyid 中创建 limit1 的排名。然后取平均数top2。

但是排名不起作用，并且给出错误的结果。如果我从ONE表中获取列，我的查询将起作用，但是一旦在它们之间添加联接，它将给出错误的排名。

SELECT policyid, limit1,  /*pcob,*/ RANK() OVER(PARTITION BY policyid ORDER BY limit1 DESC) AS rn
FROM   (SELECT policyid, limit1/*, pc.objectid ASpcob*/ 
        FROM p_risk pr 
        LEFT JOIN p_item 
          ON pr.objectid=p_item.riskobjectid 
        LEFT JOIN p_coverage pc 
          ON p_item.objectid=pc.insuranceitemid) AS s
        ) AS SubQueryAlias
GROUP BY
        policyid, limit1/*, pcob*/, rn
        ORDER BY rn,policyid,limit1 DESC

图片结尾的桌子是我想要的。第一个表是对Golden Linoff的查询结果

Answer 1

感谢先前的评论！我成功地做了我想要的。有查询

 select b.policyid, avg(b.limit1) as avg_top2_limit1 from(
 SELECT distinct(policyid) policyid, limit1 
 FROM (SELECT policyid, limit1,
         Dense_rank() OVER (PARTITION BY policyid ORDER BY limit1 DESC) as 
seqnum
  FROM p_risk pr LEFT JOIN
       p_item i
       ON pr.objectid = i.riskobjectid LEFT JOIN
       p_coverage pc 
       ON i.objectid = pc.insuranceitemid) AS s

 WHERE seqnum <= 2 ) as b
 GROUP BY policyid`

Answer 2

如果我理解正确，您希望子查询中的ROW_NUMBER()和 then 在外部查询中进行汇总和过滤：

SELECT policyid, AVG(limit1) as avg_top2_limit1
FROM (SELECT policyid, limit1,
             DENSE_RANK() OVER (PARTITION BY policyid ORDER BY limit1 DESC) as seqnum
      FROM p_risk pr LEFT JOIN
           p_item i
           ON pr.objectid = i.riskobjectid LEFT JOIN
           p_coverage pc 
           ON i.objectid = pc.insuranceitemid) AS s
        ) p
WHERE seqnum <= 2
GROUP BY policyid