计算总和，平均和DB 90％/ 2

如果我有以下的表称为stats：计算总和，平均和DB 90％/ 2

type duration 
------ --------- 
get  10 
get  15 
get  20 
put  12 
put  14 
put  16

我知道我能得到sum()和average()用类似的查询：

select 
    type, 
    sum(duration) as "total duration", 
    avg(duration) as "average duration" 
from 
    stats 
group by 
    type

而且我还可以使用窗口函数功能获得90％的持续时间和max()：

select 
    type, 
    avg("dur") as "average duration of the fastest 90%", 
    max("dur") as "max duration of the fastest 90%" 
from 
(
    select 
     type, 
     duration as "dur", 
     row_number() over (partition by type order by duration asc) as "seqnum" 
     count(*) over (partition by type) as "totnum" 
    from 
     stats 
) 
where 
    "seqnum" <= (.9 * "totnum") 
group by 
    type

但我很努力地了解如何将两者合并，以便我可以有一个查询返回：type，total duration，average duration,average duration of the fastest 90%,max duration of the fastest 90%？

来源

2016-12-14 Matt Felzani

使用条件汇总：

select type, 
     sum(duration) as "total duration", 
     avg(duration) as "average duration", 
     avg(case when "seqnum" <= (0.9 * "totnum") then "dur" end) as "average duration of the fastest 90%", 
     max(case when "seqnum" <= (0.9 * "totnum") then "dur" end) as "max duration of the fastest 90%" 
from (select type, duration as "dur", 
      row_number() over (partition by type order by duration asc) as "seqnum", 
      count(*) over (partition by type) as "totnum" 
     from stats 
    ) s 
group by type;

来源

2016-12-14 21:18:39

因此，'seqnum“<（0.9 *”totnum“）时的情况”then“dur”end“意思是”如果符合标准，那么在聚合函数中使用它，否则不要使用它？ –

@MattFelzani。。。基本上。它实际上意味着评估表达式; 'else'子句是'NULL'。大多数聚合函数忽略NULL值。 –

你可以试试这个。

select distinct 
    type, 
    avg("dur") over(partition by type) as "average duration of the fastest 90%", 
    max("dur") over(partition by type) as "max duration of the fastest 90%", 
    "total duration", 
    "average duration" 
from 
(
    select 
     type, 
     duration as "dur", 
     row_number() over (partition by type order by duration asc) as "seqnum", 
     count(*) over (partition by type) as "totnum", 
     sum(duration) over(partition by type) as "total duration", 
     avg(duration) over(partition by type) as "average duration" 
    from 
     stats 
) x 
where 
    "seqnum" <= (.9 * "totnum")

来源

2016-12-14 21:10:43

，但不会是给我的四个统计的仅仅是90％？我想要100％的总体和平均水平，其他两个只有90％？ –

它将为您提供100％的总计和平均值，因为在内部查询中给定类型的所有行的总和（持续时间）（按类型划分）和平均值（持续时间）（按类型划分）将相同。你可以运行内部查询并检查。 –

啊，我刚刚看到那部分。这很酷。基本上*所有行都带有这两个100％的字段。一旦我证明出来，我会试一试并奖励梦寐以求的复选框。谢谢！ –

计算总和，平均和DB 90％/ 2

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