Question

我有一个团队统计数据库，我想规范化某个统计数据的总和。例如，我有一个冗余记录表，如下所示：

team_name | team_code | player | steal | block
Team A    | TA        | PA     | 100   | 100
Team A    | TA        | PA     | 100   | 100
Team A    | TA        | PA     | 100   | 100
Team B    | TB        | PB     | 200   | 200
Team C    | TC        | PC     | 300   | 300
Team C    | TC        | PC     | 300   | 300

我想让所有球员的球队得分（抢断+盖帽）除以所有球队的平均值（抢断+盖帽）。

例如，我需要一个输出看起来像这样（注意：我在norm_def_total所需的表达式中添加了）：

team_name | team_code | steal_sum | block_sum | def_total | norm_def_total
Team A    | TA        | 300       | 300       | 600       | 600/(avg(def_total))
Team B    | TB        | 200       | 200       | 400       | 400/(avg(def_total))
Team C    | TC        | 600       | 600       | 1200      | 1200/(avg(def_total))

这是我的尝试：

SELECT b.team_name, b.team_code, b.def_total/AVG(b.def_total) normalized_def FROM
(
    SELECT  team_name,
        team_code,
        SUM(steal) steal_sum,
        SUM(block) block_sum,
        CAST(SUM(steal) + SUM(block) AS double precision) def_total
    FROM gamelog
    WHERE team_name IN (SELECT DISTINCT (team_name) FROM gamelog)
    GROUP BY team_name,
             team_code
    ORDER BY def_total DESC
) b
GROUP BY
    b.team_name,
    b.team_code,
    b.def_total

从上面来看，我的问题是b.def_total / AVG（b.def_total）总是返回1.我认为它只是为特定团队拍摄AVG，但我希望所有团队的AVG都来自b。如何从计算中使用b获得实际平均值？

Answer 1

试试这个。

SELECT b.team_name,
       b.team_code,
       Sum(def_total) / Avg(b.def_total)OVER(partition BY NULL)
FROM   (SELECT team_name,
               team_code,
               Sum(steal)                                        steal_sum,
               Sum(block)                                        block_sum,
               Cast(Sum(steal) + Sum(block) AS DOUBLE PRECISION) def_total
        FROM   gamelog
        WHERE  team_name IN (SELECT DISTINCT ( team_name )
                             FROM   gamelog)
        GROUP  BY team_name,
                  team_code
        ORDER  BY def_total DESC) b
GROUP  BY b.team_name,
          b.team_code

Answer 2

尝试类似：

SELECT team_name, 
       team_code, 
       sum(steal) as steal_sum, 
       sum(block) AS block_sum, 
       sum(steal) + sum(block) AS def_total, 
       ((sum(steal) + sum(block)) / sum(steal) + sum(block)) / count(*) AS norm_def_total
FROM   gamelog
GROUP BY team_name, team_code

计算和使用子查询中的列平均值

2 个答案: