当前行为null时，将最后一个值用于PostgreSQL时间序列表

如何解决当前行为null时，将最后一个值用于PostgreSQL时间序列表

我遇到一个找不到最佳解决方案的问题。因此，想法是从商店列表中获取商品列表在每个给定时间的价格，但是由于价格是在不同的时间记录的，因此在按时间分组时我会得到一些空值，并且还会得到一个数组o值。因此，它需要采取几个步骤才能获得我所需要的。我想知道是否有人知道更好，更快的方法来实现这一目标。贝娄是我最初的PostgreSQL表，当然，这只是它的摘录：

初始表

所需结果（中间表和最后一个）

下面是提供我想要的结果的PostgreSQL sql代码，但这看起来非常昂贵：

SELECT times,first_value(price_yami_egg)  OVER (PARTITION BY partition_price_yami_egg order by time) as price_yami_egg
    first_value(price_yami_salt)  OVER (PARTITION BY partition_price_yami_salt order by time) as price_yami_salt
    first_value(price_dobl_egg)  OVER (PARTITION BY partition_price_dobl_egg order by time) as price_dobl_egg
    first_value(price_dobl_salt)  OVER (PARTITION BY partition_price_dobl_salt order by time) as price_dobl_salt

    FROM(
    SELECT time,min(price_yami_egg) as price_yami_egg,sum(case when min(price_yami_egg) is not null then 1 end) over (order by times) as partition_price_yami_egg
        min(price_yami_salt) as price_yami_salt,sum(case when min(price_yami_salt) is not null then 1 end) over (order by times) as partition_price_yami_salt
        min(price_dobl_egg) as price_dobl_egg,sum(case when min(price_dobl_egg) is not null then 1 end) over (order by times) as partition_price_dobl_egg
        min(price_dobl_salt) as price_dobl_salt,sum(case when min(price_dobl_salt) is not null then 1 end) over (order by times) as partition_price_dobl_salt
        FROM ( 
            SELECT "time" AS times,CASE WHEN  shop_name::text = 'yami'::text AND product_name::text = 'egg'::text THEN price END AS price_yami_egg
                CASE WHEN  shop_name::text = 'yami'::text AND product_name::text = 'salt'::text THEN price END AS price_yami_salt
                CASE WHEN  shop_name::text = 'dobl'::text AND product_name::text = 'egg'::text THEN price END AS price_dobl_egg
                CASE WHEN  shop_name::text = 'dobl'::text AND product_name::text = 'salt'::text THEN price END AS price_dobl_salt

                FROM shop sh
                 ) S

          GROUP BY time
          ORDER BY time) SS

解决方法

您只想聚合吗？

select time,min(price) filter (where shop_name = 'Yami' and product_name = 'EGG'),min(price) filter (where shop_name = 'Yami' and product_name = 'SALT'),min(price) filter (where shop_name = 'Dobl' and product_name = 'EGG'),min(price) filter (where shop_name = 'Dobl' and product_name = 'SALT')
from shop s
group by time;

如果。您关心的是结果中的NULL值，然后可以将其填充。这有点棘手，但想法是：

with t as (
      select time,min(price) filter (where shop_name = 'Yami' and product_name = 'EGG') as yami_egg,min(price) filter (where shop_name = 'Yami' and product_name = 'SALT') as yami_salt,min(price) filter (where shop_name = 'Dobl' and product_name = 'EGG') as dobl_egg,min(price) filter (where shop_name = 'Dobl' and product_name = 'SALT') as dobl_salt
      from shop s
      group by time
     )
select s.*,max(yaml_egg) over (yaml_egg_grp) as imputed_yaml_egg,max(yaml_salt) over (yaml_egg_grp) as imputed_yaml_salt,max(dobl_egg) over (yaml_egg_grp) as imputed_dobl_egg,max(dobl_salt) over (yaml_egg_grp) as imputed_dobl_salt
from (select s.*,count(yaml_egg) over (order by time) as yaml_egg_grp,count(yaml_salt) over (order by time) as yaml_egg_grp,count(dobl_egg) over (order by time) as dobl_egg_grp,count(dobl_salt) over (order by time) as dobl_salt_grp
      from s
     ) s