想要从 ORACLE SQL 中的 2 个单独的列中为一个团队提供平均点

Posted 2023-04-14

【中文标题】想要从 ORACLE SQL 中的 2 个单独的列中为一个团队提供平均点

【英文标题】：want avg point for one team from 2 separate columns in ORACLE SQL

【发布时间】：2020-05-10 12:43:33

【问题描述】：

游戏桌：

GAME_ID - NUMBER    
GAME_DATE - DATE
ARENA - VARCHAR2(20)
HOME - VARCHAR2(30)
AWAY - VARCHAR2(30)
"HOME POINT" - NUMBER
"AWAY POINT" - NUMBER

HOME 和 AWAY 标识球队名称。 “HOME POINT”和“AWAY POINT”对应于这些列。例如（仅表示为 4 列），我想获得一支球队的平均分。考虑下面的示例，我想获得 AAA、BBB、CCC 和 DDD 团队的 AVG 积分。

AAA
BBB
90 <- got by AAA
100 <- got by BBB
------------------------
AAA
CCC
99 <- got by AAA
82 <- got by CCC
------------------------
AAA 
DDD
100 <- got by AAA
78 <- got by CCC

我尝试了下面的代码，但结果并不理想：

SELECT HOME, SUM("HOME POINT") + SUM("AWAY POINT") / COUNT(*)
FROM GAME 
GROUP BY HOME

【参考方案1】：

取消透视数据并聚合一种方法使用union all：

select team, avg(points)
from ((select home as team, home_points as points from games) union all
      (select away as team, away_points as points from games)
     ) t
group by team;

在 Oracle 12C+ 中，我建议使用横向连接而不是 union all。

【参考方案2】：

下面是横向连接的方法；从版本 12c 开始，此语法在 Oracle 中可用，在这种情况下，它比 union all 更有效，因为它不需要扫描表两次：

select t.team, avg(t.points) avg_points
from games g
cross apply (
    select home team, home_points points from dual
    union all select away, away_points points from dual
) t
group by t.team

旁注：

我建议将点存储为int 而不是数字（大概，点没有小数部分）

使用不需要引用的列名（home_points 优于 "home point"）

这个 demo on DB Fiddle 与您的示例数据产生：

团队 | AVG_POINTS :--- | ---------: AAA | 96.33 血脑屏障 | 100 CCC | 82 DDD | 78

【参考方案3】：

with points as (
select home as team, sum("HOME POINT") as point_sum,
count("HOME POINT") as point_qty
from game 
group by home
having count("HOME POINT")!=0

union all

select away as team, sum("AWAY POINT") as point_sum,
count("AWAY POINT") as point_qty
from game 
group by away
having count("AWAY POINT")!=0
)

select team, sum(point_sum)/sum(point_qty) as avg_point
from points
group by team;