SUM distinct 内窗函数答案

【问题标题】：SUM distinct inside window functionSUM distinct 内窗函数
【发布时间】：2021-11-13 18:49:26
【问题描述】：

我有一个类似这样的数据结构：

CREATE TABLE some_table (
    dude_id INTEGER, 
    main_date TIMESTAMP, 
    how_many INTEGER, 
    how_much NUMERIC(5,2), 
    their_ids INTEGER[]
)

这是我目前得到的查询

SELECT
    dude_id,
    main_date,
    how_many,
    how_much,
    their_ids,
    SUM(how_many) OVER (PARTITION BY dude_id ORDER BY main_date) AS count_stuff_WRONG,
    SUM(how_much) OVER (PARTITION BY dude_id ORDER BY main_date) AS cumulative_sum_WRONG
FROM some_table

这是我想要达到的结果：

dude_id	main_date	how_many	how_much	their_ids	count_stuff_EXPECTED	cumulative_sum_EXPECTED	count_stuff_WRONG	cumulative_sum_WRONG
38	2019-06-14	1	6	373	1	6	1	6
38	2019-07-15	1	7	374	2	13 (6+7)	2	13 (6+7)
38	2019-07-16	1	8	375	3	21 (6+7+8)	3	21 (6+7+8)
38	2020-06-14	1	16	373	3	31 (7+8+16)	4	37 (6+7+8+16)
38	2020-07-15	1	17	374	3	41 (8+16+17)	5	54 (6+7+8+16+17)
38	2020-07-16	1	18	375	3	51 (16+17+18)	6	72 (6+7+8+16+17+18)

count_stuff_EXPECTED 和 cumulative_sum_EXPECTED 列是我想要得到的，count_stuff_WRONG 和 cumulative_sum_WRONG 列是我当前查询返回的。

换句话说，我想获得每个 main_date 的累积值，但不计算/求和多次相同的their_ids。所以以第 4 行为例，窗口分区有不止一次的their_ids {373}，所以它应该只考虑最近的一个（第 4 行）而不考虑第一次出现（第 1 行）

注意：无需在查询中显示总和是如何计算的，为了清楚起见，我只是将其放在括号中。

我尝试过使用

SUM(DISTINCT how_many) over (PARTITION BY dude_id ORDER BY main_date) as count_stuff

但是得到了

错误：没有为窗口函数实现 DISTINCT

SQL 小提琴：http://sqlfiddle.com/#!17/44850/2

【问题讨论】：

标签： postgresql aggregate-functions distinct window-functions

【解决方案1】：

也许这不是最好的答案，但我认为它可以达到你的目标。

首先，我自己创建一个表数据的INNER JOIN，以获得每行的总和（每个 their_ids 的更新记录）。然后我GROUP BY 丢弃“重复”的行，我终于得到了总和。

WITH sq1 AS (SELECT d1.dude_id, d1.main_date, d1.how_many, d1.how_much, d1.their_ids,
                    d2.dude_id AS d2_dude_id, d2.main_date AS d2_main_date, d2.their_ids AS d2_their_ids,
                   FIRST_VALUE(d2.how_much) OVER (PARTITION BY d1.dude_id, d1.main_date, d1.their_ids, d2.their_ids ORDER BY d2.main_date DESC) AS how_much_to_sum
             FROM data d1
             INNER JOIN data d2 ON d1.dude_id = d2.dude_id AND d1.main_date >= d2.main_date),
            
     sq2 AS (SELECT dude_id, main_date, how_many, how_much, their_ids, d2_dude_id, d2_their_ids, MIN(how_much_to_sum) AS how_much_to_sum
             FROM sq1
             GROUP BY dude_id, main_date, how_many, how_much, their_ids, d2_dude_id, d2_their_ids)
             
select
      dude_id,
      main_date,
      how_many,
      how_much,
      their_ids,
      SUM(how_many) AS count_stuff,
      SUM(how_much_to_sum) AS cumulative_sum
      
from sq2
GROUP BY dude_id, main_date, how_many, how_much, their_ids
ORDER BY main_date;

【讨论】：