Question

我有一个按平均值排名的结果集。

LOCATION      JOB TILE    AVG    RANK
SEATTLE      PRESIDENT    10.2    1
SEATTLE         VP        8.5     2
SEATTLE       SN. VP      5.4     3
CHICAGO      SALESMAN     7.6     1
CHICAGO      ENGINEER     3.2     2
CHICAGO      PRESIDENT    1.5     3

我需要的下一部分是每个LOCATION组的第一个和最后一个排名平均值加上JOB TITLE。例如：

LOCATION  JOBTITLE   AVG
SEATTLE   PRESIDENT  10.2
SEATTLE   SN. VP     5.4  
CHICAGO   SALESMAN   7.6
CHICAGO   PRESIDENT  1.5

使用Redshift SQL有一种简单的方法吗？仅仅是一个FYI，AVG字段是在查询中完成的聚合。

Answer 1

您可以计算每个位置的最小和最大平均值，并将其与平均值连接到原始查询，以获得所需的行。

with averages as (<your query>)
,minandmax as (select location, min(avgcol) minavg, max(avgcol) maxavg
               from averages
               group by location)
select a.location,a.job_title, a.avgcol
from averages a
join minandmax m on m.location = a.location and (m.maxavg=a.avgcol or m.minavg=a.avgcol)

Answer 2

您可以使用窗口功能执行此操作：

with t as (
      <your query here>
     )
select t.*
from (select t.*,
             max(rank) over (partition by location) as maxr
      from t
     ) t
where rank in (1, maxr);

选取第一个和最后一个平均排名

2 个答案: