我在PostgreSQL数据库中有一个表，如下所示：
stid | e5 | e10 | diesel | date
-----+------+------+--------+------------------------
e850 | 1300 | 1400 | 1500 | 2016-05-02 05:30:01+02
e850 | 1400 | 1500 | 1700 | 2016-05-02 08:30:01+02
e850 | 1300 | 1400 | 1500 | 2016-05-02 21:00:01+02
e850 | 1200 | 1300 | 1350 | 2016-05-03 10:30:01+02
e850 | 1300 | 1400 | 1500 | 2016-05-03 21:00:01+02
954d | 1200 | 1100 | 1300 | 2016-05-02 03:30:01+02
954d | 1300 | 1100 | 1300 | 2016-05-02 15:00:01+02
954d | 1400 | 1800 | 1400 | 2016-05-02 22:30:01+02
954d | 1700 | 1900 | 1400 | 2016-05-03 09:30:01+02
954d | 1500 | 1900 | 1200 | 2016-05-03 23:30:01+02
所以我有独特的ID（stid），价格（e5，e10，柴油）和时间戳（日期），表明价格何时推出。现在我想计算每天的平均价格和stid，加权价格的持续时间。我只想考虑上午8点到晚上8点之间的时间。
要计算sti8 e850的e5加权平均价格以及2016-05-02日上午8点到晚上8点之间的加权平均价格，我将执行以下操作：
(1300 * 1801 + 1400 * 41399) / 43200 = 1395.83102
1300 is the price that was set at 5:30:01 am and 1801 is the duration in
seconds between 8 am and 8:30:01 am.
1400 is the price that was set at 8:30:01 am and 41399 is the duration in
seconds between 8:30:01 am and 8 pm.
最后，我希望有一个看起来像这样的表：
stid | date | average_e5 | average_e10 | average_diesel
-----+------------+------------+-------------+---------------
e850 | 2016-05-02 | 1395.83102 | 1495.83102 | 1691.66204
e850 | 2016-05-03 | 1220.83565 | 1320.83565 | 1381.25347
954d | 2016-05-02 | 1241.66435 | 1100 | 1300
954d | 2016-05-03 | 1662.49306 | 1887.49769 | 1400
以下代码来自Vao Tsun的答案几乎可以解决我所寻找的一切问题。但是，如果在上午8点之前或晚上8点之后没有价格且一天没有价格，我就无法获得我正在寻找的加权平均值。但是，如果在上午8点之前或晚上8点之后没有价格的情况下创建虚拟条目，我就能够解决这个问题。
我使用以下代码创建了一个名为mytable2的新表，它包含虚拟条目。
DROP TABLE IF EXISTS mytable2;
CREATE TABLE mytable2 AS SELECT * FROM mytable;
WITH c AS (
SELECT
*,
LAG(date) OVER(PARTITION BY stid ORDER BY date) AS lag_date,
LAG(e5) OVER(PARTITION BY stid ORDER BY date) AS lag_e5,
LAG(e10) OVER(PARTITION BY stid ORDER BY date) AS lag_e10,
LAG(diesel) OVER(PARTITION BY stid ORDER BY date) AS lag_diesel
FROM mytable
)
INSERT INTO mytable2
SELECT
stid,
lag_e5 AS e5,
lag_e10 AS e10,
lag_diesel AS diesel,
date_trunc('day', date) + '0 hours'::interval AS date
FROM c WHERE lag_date < date_trunc('day', date) + '0 hours'::interval
AND date > date_trunc('day', date) + '8 hours'::interval;
WITH d AS (
SELECT
*,
LEAD(date) OVER(PARTITION BY stid ORDER BY date) AS lead_date
FROM mytable
)
INSERT INTO mytable2
SELECT
stid,
e5,
e10,
diesel,
date_trunc('day', date) + '23 hours'::interval AS date
FROM d WHERE lead_date >= date_trunc('day', date) + '24 hours'::interval
AND date < date_trunc('day', date) + '20 hours'::interval;
然后我可以运行Vao Tsun的答案中的代码来获得所需的加权平均值。我只将mytable更改为mytable2，以使用带有添加的虚拟条目的表。
with a as (
select *
, case
when date < date_trunc('day', date) + '8 hours'::interval then date_trunc('day', date) + '8 hours'::interval
when date > date_trunc('day', date) + '20 hours'::interval then date_trunc('day', date) + '20 hours'::interval
else date
end d
, date_trunc('day', date) dt
from mytable2
)
, b as (
select stid, e5, e10, diesel,date,d, dt
, extract(epoch from lead(d) over (partition by stid,dt order by stid,d) - d) diff
from a
)
select DISTINCT
stid, dt,sum(e5*diff*1.0) over (partition by stid,dt)/sum(diff) over (partition by stid,dt) e5_weight_avg
from b
order by stid desc, dt;
stid | dt | e5_weight_avg
-----+---------------------+-----------------
e850 | 2016-05-02 00:00:00 | 1395.83101851852
e850 | 2016-05-03 00:00:00 | 1220.83564814815
954d | 2016-05-02 00:00:00 | 1241.66435185185
954d | 2016-05-03 00:00:00 | 1662.49305555556
此处的代码也可以在rextester
找到答案 0 :(得分：2)
我做了一些不需要CTE，以使其更具可读性：
t=# with a as (
select *
, case
when date < date_trunc('day', date) + '8 hours'::interval then date_trunc('day', date) + '8 hours'::interval
when date > date_trunc('day', date) + '20 hours'::interval then date_trunc('day', date) + '20 hours'::interval
else date
end d
, date_trunc('day', date) dt
from mytable
)
, b as (
select stid, e5, e10, diesel,date,d, dt
, extract(epoch from lead(d) over (partition by stid,dt order by stid,d) - d) diff
from a
)
select
stid, e5,date,d, diff,sum(e5*diff*1.0) over (partition by stid,dt)/sum(diff) over (partition by stid,dt) e5_weight_avg
from b
order by stid desc, date;
stid | e5 | date | d | diff | e5_weight_avg
------+---------+---------------------+---------------------+-------+------------------
e850 | 1300.00 | 2016-05-02 05:30:01 | 2016-05-02 08:00:00 | 1801 | 1395.83101851852
e850 | 1400.00 | 2016-05-02 08:30:01 | 2016-05-02 08:30:01 | 41399 | 1395.83101851852
e850 | 1300.00 | 2016-05-02 21:00:01 | 2016-05-02 20:00:00 | | 1395.83101851852
e850 | 1200.00 | 2016-05-03 10:30:01 | 2016-05-03 10:30:01 | 34199 | 1200
e850 | 1300.00 | 2016-05-03 21:00:01 | 2016-05-03 20:00:00 | | 1200
954d | 1200.00 | 2016-05-02 03:30:01 | 2016-05-02 08:00:00 | 25201 | 1241.66435185185
954d | 1300.00 | 2016-05-02 15:00:01 | 2016-05-02 15:00:01 | 17999 | 1241.66435185185
954d | 1400.00 | 2016-05-02 22:30:01 | 2016-05-02 20:00:00 | | 1241.66435185185
954d | 1700.00 | 2016-05-03 09:30:01 | 2016-05-03 09:30:01 | 37799 | 1700
954d | 1500.00 | 2016-05-03 23:30:01 | 2016-05-03 20:00:00 | | 1700
(10 rows)
因此，跳过中间步骤：
t=# with a as (
select *
, case
when date < date_trunc('day', date) + '8 hours'::interval then date_trunc('day', date) + '8 hours'::interval
when date > date_trunc('day', date) + '20 hours'::interval then date_trunc('day', date) + '20 hours'::interval
else date
end d
, date_trunc('day', date) dt
from mytable
)
, b as (
select stid, e5, e10, diesel,date,d, dt
, extract(epoch from lead(d) over (partition by stid,dt order by stid,d) - d) diff
from a
)
select DISTINCT
stid, dt,sum(e5*diff*1.0) over (partition by stid,dt)/sum(diff) over (partition by stid,dt) e5_weight_avg
from b
order by stid desc, dt;
stid | dt | e5_weight_avg
------+---------------------+------------------
e850 | 2016-05-02 00:00:00 | 1395.83101851852
e850 | 2016-05-03 00:00:00 | 1200
954d | 2016-05-02 00:00:00 | 1241.66435185185
954d | 2016-05-03 00:00:00 | 1700
(4 rows)