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sql-server - 递归检索前一条记录的 LAG() 值

转载 作者:行者123 更新时间:2023-12-02 19:20:44 24 4
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我进行了以下涉及 LAG() 的计算:

(lag(fValue,1,fValue) OVER (PARTITION BY Cluster ORDER BY iSequence) + fValue)/2 as fValueAjusted

它采用前一个(基于 iSequence)记录的 fValue,与当前记录的 fValue 相加,然后除以 2。

但是,我必须使用上一条记录的 fValueAjusted 来代替 fValue。

这意味着第一条记录的 fValueAjusted 将是它自己的 fValue。第二条记录的 fValueAjusted 将基于第一条记录的 fValue。并且,从第三条记录开始,将根据前一条记录的fValueAjusted进行计算。

我需要在前一个记录的 fValueAjusted 上递归计算 fValueAjusted。我不知道该怎么做。

更新 1:这是源数据的示例。真实表有数百条记录和80个簇。

CREATE TABLE dbo.example (
iUnity int NOT NULL,
Cluster char(2) NOT NULL,
fValue float NOT NULL
)

15 A1 150
17 A1 170
21 B2 210
23 B2 230
71 C3 710

这是计算序列的 CTE:

WITH cteSequencing AS (
SELECT
iUnity,Cluster
,fValue as fValueOriginal
,row_number() OVER (PARTITION BY Cluster ORDER BY fValueOriginal) as iSequence
FROM dbo.example
)

如果 fValueAjusted 根据 fValueOriginal 计算,则查询将如下所示:

SELECT
iUnity,Cluster,fValueOriginal
,(
lag(fValue,1,fValue) OVER (PARTITION BY Cluster ORDER BY iSequence)
+ fValueOriginal
) / 2 as fValueAjusted
FROM cteSequencing

但是一条记录的fValueAjusted必须根据前一条记录的fValueAjusted计算。它会是这样的:

SELECT
iUnity,Cluster,fValueOriginal
,(
lag(fValueAjusted,1,fValueOriginal) OVER (PARTITION BY Cluster ORDER BY iSequence)
+ fValueOriginal
) / 2 as fValueAjusted
FROM cteSequencing

当然,fValueAjusted 在执行时不可用。 LAG() 必须递归地计算一条记录的列,然后提供该列供下一条记录使用。

最佳答案

更新:原始答案不正确

正确的是:

代码使用递归 CTE

CREATE TABLE #example (
iUnity int NOT NULL,
Cluster char(2) NOT NULL,
fValue float NOT NULL
)
INSERT INTO #example
VALUES
( 15, 'A1', 150 ),
( 16, 'A1', 170 ),
( 17, 'A1', 190 ),
( 18, 'A1', 210 ),
( 21, 'B2', 210 ),
( 23, 'B2', 230 ),
( 71, 'C3', 710 )

WITH cteSequencing AS (
-- Get Values Order
SELECT iUnity, Cluster, fValue, fValue AS fValueAjusted,
ROW_NUMBER() OVER (PARTITION BY Cluster ORDER BY fValue) AS iSequence
FROM #example
),
Recursion AS(
-- Anchor - the first value in clusters
SELECT iUnity, Cluster, fValue, fValueAjusted, iSequence
FROM cteSequencing
WHERE iSequence = 1
UNION ALL
-- Calculate next value based on the previous
SELECT b.iUnity As iUnity, b.Cluster, b.fValue,
( a.fValueAjusted + b.fValue ) / 2 AS fValueAjusted,
b.iSequence
FROM Recursion AS a
INNER JOIN cteSequencing AS b ON a.iSequence + 1 = b.iSequence AND a.Cluster = b.Cluster
)
SELECT * FROM Recursion ORDER BY Cluster, fValue

-- Manually check results
SELECT ( 150 + 170 ) / 2
SELECT ( 190 + 160 ) / 2
SELECT ( 190 + 170 ) / 2

输出:

iUnity      Cluster fValue                 fValueAjusted          iSequence
----------- ------- ---------------------- ---------------------- --------------------
15 A1 150 150 1
16 A1 170 160 2
17 A1 190 175 3
18 A1 210 192.5 4
21 B2 210 210 1
23 B2 230 220 2
71 C3 710 710 1

更新

如果你遇到

The maximum recursion 100 has been exhausted before statement completion

错误,则使用OPTION (MAXRECURSION xxx)设置更大的递归限制,最多 32,767。

关于sql-server - 递归检索前一条记录的 LAG() 值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45016753/

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