Question

我的表格结构如下：

item_id    first_observed    last_observed    price
1          2016-10-21        2016-10-27       121
1          2016-10-28        2016-10-31       145
2          2016-10-22        2016-10-28       135
2          2016-10-29        2016-10-30       169

我想要的是获取每天的平均价格。我显然不能仅仅按first_observed或last_observed分组。 Postgres是否提供执行此操作的明智方法？

预期输出如下：

date        avg(price)
2016-10-21  121
2016-10-22  128
2016-10-23  128
2016-10-24  128
2016-10-25  128
2016-10-26  128
2016-10-27  128
2016-10-28  140
2016-10-29  157
2016-10-30  157
2016-10-31  157

我也可以这样输出（都很好）：

start       end         avg(price)
2016-10-21  2016-10-21  121
2016-10-22  2016-10-27  128
2016-10-28  2016-10-28  140
2016-10-29  2016-10-31  157

Answer 1

WITH ObserveDates (ObserveDate) AS (
    SELECT * FROM generate_series((SELECT MIN(first_observed) FROM T), (SELECT MAX(last_observed) FROM T), '1 days')
)
SELECT ObserveDate, AVG(Price)
FROM ObserveDates
JOIN T ON ObserveDate BETWEEN first_observed AND last_observed
GROUP BY ObserveDate
ORDER BY ObserveDate

Answer 2

demo:db<>fiddle

<driverClassName>net.sf.log4jdbc.DriverSpy</driverClassName>允许您扩展日期范围：

第一步：

generate_series

扩大日期范围
分组AVG汇总的日期

第二步

SELECT 
    generate_series(first_observed, last_observed, interval '1 day')::date as observed, 
    AVG(price)::int as avg_price
FROM items
GROUP BY observed
ORDER BY observed

按avg_price分组以获取日期SELECT MIN(observed) as start, MAX(observed) as end, avg_price FROM ( -- <first step as subquery> )s GROUP BY avg_price ORDER BY start / MIN

PostgreSQL：每天平均间隔

2 个答案: