如何汇总组中最早和最多的记录？答案

【问题标题】：How can I sum the earliest and the most record in a group?如何汇总组中最早和最多的记录？
【发布时间】：2016-12-21 11:04:15
【问题描述】：

我有一张桌子，上面有 uid、date 和 value。

我想按 uid 分组，并得到最早找到的value 和最近的value 之间的差异。如何在 ActiveRecord 中或仅在 SQL 中执行此操作？

示例数据：

UID   DATE        VALUE
------------------------------
a    | 2016-10-01 | 5
a    | 2016-8-01 | 10
a    | 2016-12-01 | 15
b    | 2016-10-01 | 5
b    | 2016-8-01 | 10
c    | 2016-12-01 | 15

想要的结果

UID   DELTA (absolute value)     
------------------------------
a    | 5
b    | 5
c    | 15

【问题讨论】：

亲爱的，添加数据集和所需的结果集。你已经写了 350 个问题，请写得更好；）
我认为 a = 10 的期望结果？
@Mr.Bhosale 最早也不是最不重要

标签： sql postgresql select activerecord ruby-on-rails-5

【解决方案1】：

select uid, abs(max(latest) - max(first)) diff
from
(select uid,
  case when (row_number() over 
           (partition by uid order by date asc)) = 1 
           then value end first,
  case when (row_number() over 
           (partition by uid order by date desc)) = 1 
           then value end latest
from table
) t
group by uid;

【讨论】：

【解决方案2】：

我对 activerecord 没有经验，但是在普通 SQL 中，您可以使用 row_number 窗口函数来查找第一条和最后一条记录。例如：

WITH first_and_last AS (
    SELECT uid, 
           value,
           ROW_NUMBER() OVER (PARTITION BY uid ORDER BY date ASC) as r_asc
           ROW_NUMBER() OVER (PARTITION BY uid ORDER BY date DESC) as r_desc
    FROM   mytable)
SELECT a.uid, a.value - b.value
FROM   first_and_last a
JOIN   first_and_last b ON a.uid = b.uid AND a.r_desc = 1 AND b.r_asc = 1

【讨论】：

【解决方案3】：

如果您要在聚合之前使用窗口函数，请使用first_value() 和/或last_value()：

select uid, abs(max(value_first) - max(value_last)) as diff
from (select uid,
             first_value(value) over (partition by uid order by date asc) as value_first,
             first_value(value) over (partition by uid order by date desc) as value_last
      from table
     ) t
group by uid;

如果您想使用数组聚合函数，您也可以完全不使用子查询来执行此操作。

【讨论】：

【解决方案4】：

with
  t(i,d,v) as (
    values
      ('a'::text, '2016-10-01'::date, 5::int),
      ('a', '2016-08-01', 10),
      ('a', '2016-12-01', 15),
      ('b', '2016-10-01', 5),
      ('b', '2016-08-01', 10),
      ('c', '2016-12-01', 15)),
  e as (
    select distinct on (i) * from t order by i, d),
  l as (
    select distinct on (i) * from t order by i, d desc)
select
  e.i,
  abs(e.v - case when e.d=l.d then 0 else l.v end) as diff
from
  e join l using(i);

最终查询中的case 需要处理像c uid 这样的情况，其中行集中只有一行（当然它不处理具有相同日期的多行的情况，所以你可以如果存在，请使用“id”）。

【讨论】：

【解决方案5】：

检查一下。

        select distinct a.UID,
        -- ,a."VALUE",b."VALUE",
        abs(coalesce (a."VALUE",0)-coalesce(b."VALUE",0)) 
        AS "DELTA (absolute value)"
        from
        (select * from 
        (select UID,"VALUE",ROW_NUMBER() over(PARTITION by uid order by rnk desc ) as rnk from 
        (
        select UID,"VALUE", date,ROW_NUMBER() over(PARTITION by uid order by  (select 1 ) )as rnk
        from  edata
        )a  )a where rnk='1' )a
        left join 
        ( select * FROM 
        (select UID,"VALUE",ROW_NUMBER() over(PARTITION by uid order by rnk desc ) as rnk from 
        (
        select UID,"VALUE", date,ROW_NUMBER() over(PARTITION by uid order by  (select 1 ) )as rnk
        from  edata
        )
        b )b  where    b.rnk=2 )B
        on a.uid=b.uid ORDER BY UID

也可以在这里尝试 postgresql Demo。

以上代码在 SQL server 和 postgresql 上运行良好。

输出：

【讨论】：