我在使用BigQuery / SQL进行移动平均时遇到麻烦,我使用了表'SCORES',在使用用户对数据进行分组时,我需要进行30天移动平均,问题是我的日期不是连续的,例如有差距。
下面是我当前的代码:
SELECT user, date,
AVG(score) OVER (PARTITION BY user ORDER BY date)
FROM SCORES;
我不知道如何在该行中添加日期限制,或者甚至不可能。
我当前的表格如下所示,但当然会有更多的用户:
user date score
AA 13/02/2018 2.00
AA 15/02/2018 3.00
AA 17/02/2018 4.00
AA 01/03/2018 5.00
AA 28/03/2018 6.00
然后我需要它成为这个,
user date score 30D Avg
AA 13/02/2018 2.00 2.00
AA 15/02/2018 3.00 2.50
AA 17/02/2018 4.00 3.00
AA 01/03/2018 5.00 3.50
AA 28/03/2018 6.00 5.50
在最后一行中,由于日期的原因,它仅向后测量(向后最多30D)是否有任何方法可以在SQL中实现此功能,或者我要求太多?
答案 0 :(得分:2)
您要使用range between
。为此,您需要一个整数,所以:
select s.*,
avg(score) over (partition by user
order by days
range between 29 preceding and current row
) as avg_30day
from (select s.*, date_diff(s.date, date('2000-01-01'), day) as days
from scores s
) s;
date_diff()
的替代方法是unix_date()
:
select s.*,
avg(score) over (partition by user
order by unix_days
range between 29 preceding and current row
) as avg_30day
from (select s.*, unix_date(s.date) as unix_days
from scores s
) s;
答案 1 :(得分:0)
以下是用于BigQuery标准SQL
#standardSQL
SELECT *,
AVG(score) OVER (
PARTITION BY user
ORDER BY UNIX_DATE(PARSE_DATE('%d/%m/%Y', date))
RANGE BETWEEN 29 PRECEDING AND CURRENT ROW
) AS avg_30day
FROM `project.dataset.scores`
您可以使用问题中的虚拟数据来测试/玩游戏
#standardSQL
WITH `project.dataset.scores` AS (
SELECT 'AA' user, '13/02/2018' date, 2.00 score UNION ALL
SELECT 'AA', '15/02/2018', 3.00 UNION ALL
SELECT 'AA', '17/02/2018', 4.00 UNION ALL
SELECT 'AA', '01/03/2018', 5.00 UNION ALL
SELECT 'AA', '28/03/2018', 6.00
)
SELECT *,
AVG(score) OVER (
PARTITION BY user
ORDER BY UNIX_DATE(PARSE_DATE('%d/%m/%Y', date))
RANGE BETWEEN 29 PRECEDING AND CURRENT ROW
) AS avg_30day
FROM `project.dataset.scores`
结果
Row user date score avg_30day
1 AA 13/02/2018 2.0 2.0
2 AA 15/02/2018 3.0 2.5
3 AA 17/02/2018 4.0 3.0
4 AA 01/03/2018 5.0 3.5
5 AA 28/03/2018 6.0 5.5