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SQL Server 滚动排名

转载 作者:行者123 更新时间:2023-12-02 01:02:11 25 4
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我有一张不同日期的股票价格表。

DECLARE @table TABLE (MyDate DATE, Ticker VarChar(6), Price Decimal (6,2))

INSERT INTO @Table
VALUES ('1/1/13' , 'ABC' , '100.00')
,('1/2/13' , 'ABC' , '101.50')
,('1/3/13' , 'ABC' , '99.80')
,('1/4/13' , 'ABC' , '95.50')
,('1/5/13' , 'ABC' , '78.00')
,('1/1/13' , 'JKL' , '34.57')
,('1/2/13' , 'JKL' , '33.99')
,('1/3/13' , 'JKL' , '31.85')
,('1/4/13' , 'JKL' , '30.11')
,('1/5/13' , 'JKL' , '45.00')
,('1/1/13' , 'XYZ' , '11.50')
,('1/2/13' , 'XYZ' , '12.10')
,('1/3/13' , 'XYZ' , '17.15')
,('1/4/13' , 'XYZ' , '14.10')
,('1/5/13' , 'XYZ' , '15.55')

我想计算每个 Ticker 和每个日期的连续 3 天排名(基于今天和前两天)。

我可以创建静态排名,但无法生成动态滚动排名。

SELECT 
*,
RANK() OVER (PARTITION BY Ticker ORDER BY Price DESC) AS Rank_3d
FROM
@table
ORDER BY
Ticker ASC, MyDate DESC

输出:

MyDate  Ticker  PRICE   Rank_3d
-------------------------------
2013-01-05 ABC 78.00 5
2013-01-04 ABC 95.50 4
2013-01-03 ABC 99.80 3
2013-01-02 ABC 101.50 1
2013-01-01 ABC 100.00 2
2013-01-05 JKL 45.00 1
2013-01-04 JKL 30.11 5
2013-01-03 JKL 31.85 4
2013-01-02 JKL 33.99 3
2013-01-01 JKL 34.57 2
2013-01-05 XYZ 15.55 2
2013-01-04 XYZ 14.10 3
2013-01-03 XYZ 17.15 1
2013-01-02 XYZ 12.10 4
2013-01-01 XYZ 11.50 5

我的输出应该是这样的:

MyDate     Ticker    Price    Rank_3d
-------------------------------------
1/3/13 ABC 99.80 3
1/4/13 ABC 95.50 3
1/5/13 ABC 78.00 3
1/3/13 JKL 31.85 3
1/4/13 JKL 30.11 3
1/5/13 JKL 45.00 1
1/3/13 XYZ 17.15 1
1/4/13 XYZ 14.10 2
1/5/13 XYZ 15.55 2

我错过了什么?

最佳答案

使用排名函数没有“简单”的方法来做到这一点。

对于您的特定问题,使用 lag() 并不难。前两天(缺失值)和有关系时,逻辑有点棘手。但这是一种方法:

select t.*,
(case when price = max(price) over (partition by ticker order by date rows between 2 preceding and current row)
then 1
when price = min(price) over (partition by ticker order by date rows between 2 preceding and current row)
then 3
else 2
end) as rank_3day
from @table t;

即使使用 apply 逻辑也很棘手:

select t.*, t2.rank
from @table t cross apply
(select count(*) as rank
from (select top (3) t2.*
from @table t2
where t2.date <= t.date
order by t2
) t2
where t2.price >= t.price
) t2

关于SQL Server 滚动排名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49808249/

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