如何对具有分组列的整数进行聚合，我只希望包含一些列？

Question

我有一张表prices 保存一些产品的所有价格：

CREATE TABLE prices (
  id INT,
  product_id INT, /*Foreign key*/
  created_at TIMESTAMP,
  price INT
);

product_id 的第一个实体是它的初始销售价格。如果产品随后减少，则将添加一个新实体。

我想找出所有产品每天的平均价格和总价格变化。

这是一些示例数据：

INSERT INTO prices (id, product_id, created_at, price) VALUES (1, 1, '2020-01-01', 11000);
INSERT INTO prices (id, product_id, created_at, price) VALUES (2, 2, '2020-01-01', 3999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (3, 3, '2020-01-01', 9999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (4, 4, '2020-01-01', 2000);
INSERT INTO prices (id, product_id, created_at, price) VALUES (5, 1, '2020-01-02', 9999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (6, 2, '2020-01-02', 2999);    
INSERT INTO prices (id, product_id, created_at, price) VALUES (7, 5, '2020-01-02', 2999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (8, 1, '2020-01-03', 8999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (9, 1, '2020-01-03 10:00:00', 7000);
INSERT INTO prices (id, product_id, created_at, price) VALUES (10, 5, '2020-01-03', 4000);
INSERT INTO prices (id, product_id, created_at, price) VALUES (11, 6, '2020-01-03', 3999);
INSERT INTO prices (id, product_id, created_at, price) VALUES (12, 3, '2020-01-03', 6999);

预期的结果应该是：

date       mean_price_change    total_price_change
2020-01-01 0                    0
2020-01-02 1000.5               2001
2020-01-03 1666                 4998

说明：

• “2020-01-01”的平均降价和总计为 0，因为所有产品在该日期都是新的。
• 在“2020-01-02”上，平均价格变化为：(11000-9999 + 3999-2999)/2 = 1000.5，因为 product_id 1 和 2 都已减少到 9999 和 2999天，它们之前的价格是 11000 和 3999，总共减少了：(11000-9999 + 3999-2999) = 2001。
• 在“2020-01-03”上，仅更改了 product_id 1、3 和 5。 1 在一天中的两个不同时间：9999 => 8999 => 7000（最后一次执政）和3：从 9999 => 6999 a 然后 5：从 2999 => 4000 上升。这给出了总计：(9999-7000 + 9999-6999 + 2999-4000) = 4998 当天平均降价：1666

我这里也添加了数据：https://www.db-fiddle.com/f/tJgoKFMJxcyg5gLDZMEP77/1

我说要玩一些DISTINCT ON，但似乎没有这样做......

Answer 1

你似乎想要lag() 和聚合：

select created_at, avg(prev_price - price), sum(prev_price - price)
from (select p.*, lag(price) over (partition by product_id order by created_at) as prev_price
      from prices p
     ) p
group by created_at
order by created_at;

您在 2020 年 1 月 3 日有两个产品 1 的价格。一旦我解决了这个问题，我会得到与你的问题相同的结果。 Here 是 dbfiddle。

编辑：

每天处理多个价格：

select created_at, avg(prev_price - price), sum(prev_price - price)
from (select p.*, lag(price) over (partition by product_id order by created_at) as prev_price
      from (select distinct on (product_id, created_at::date) p.*
            from prices p
            order by product_id, created_at::date
           ) p
     ) p
group by created_at
order by created_at;

Answer 2

试试这个

select 
created_at, 
avg(change),
sum(change)
from
(
    with cte as 
    (
    select 
    id, 
    product_id,
    created_at,
    lag(created_at) over(order by product_id, created_at) as last_date,
    price
    from prices
    )
    select
    c.id,
    c.product_id,
    c.created_at,
    c.last_date,
    p.price as last_price,
    c.price,
    COALESCE(p.price - c.price,0) as change
    from cte c
    left join prices p on c.product_id =p.product_id  and c.last_date =p.created_at
  where p.price != c.price or p.price is null
) tmp
group by created_at
order by created_at

Answer 3

下面的查询跟踪所有价格变化，请注意我们加入当前和早期基于

• 他们的产品是一样的
• earlier 确实比 current 早
• earlier 是比当前日期早的最新项目
• current 是它自己日期的最新项目

select today.product_id, (today.price - coalesce(earlier.price)), today.created_at as difference from prices current join prices earlier on today.product_id = earlier.product_id and earlier.created_at < current.created_at where not exists ( select 1 from prices later where later.product_id = today.product_id and ( ((today.created_at = later.created_at) and (today.id < later.id)) or ((earlier.created_at <= later.created_at) and (earlier.id < later.id)) ) );

现在，让我们做一些聚合：

select created_at, avg(today.price - coalesce(earlier.price)) as mean, sum(today.price - coalesce(earlier.price)) as total
from prices current
left join prices earlier
on today.product_id = earlier.product_id and earlier.created_at < current.created_at
where not exists (
    select 1
    from prices later
    where later.product_id = today.product_id and
    (
     ((today.created_at = later.created_at) and (today.id < later.id)) or
     ((earlier.created_at <= later.created_at) and (earlier.id < later.id))
    )
)
group by created_at
order by created_at;