如何计算运行乘法

Question

我有两张桌子

WAC表

ID  wac_inc             item
--  -----------------   ----
1   2.310000000000000   A
2   1.100000000000000   A
3   2.130000000000000   A
4   1.340000000000000   A

基线表

item    baseline
----    ------------------
A       10.000000000000000

预期结果

ID  wac_inc             item    Running_Mul   
--  -----------------   ----    -----------
1   2.310000000000000   A       10.231     --  10 * (1+(2.310000000000000/100))
2   1.100000000000000   A       10.343541  --  10.231 * (1+(1.100000000000000/100))
3   2.130000000000000   A       10.563858  --  10.343541 * (1+(2.130000000000000/100))
4   1.340000000000000   A       10.705413  --  10.563858 * (1+(1.340000000000000/100))

找到的公式running_mul是

基线*(1 +(wac_inc/100))

SQLFIDDLE

这里对于每一行,前一行的Running_Mul值是baseline和第一行的baseline来源baseline table.

希望我说清楚.AFAIK我们可以使用,CURSOR但我想尽可能避免 RBAR.任何人都可以建议我做更好的方法.

Answer 1

尝试:

DECLARE @t TABLE
    (
      ID INT ,
      wac DECIMAL(30, 10) ,
      item CHAR(1)
    )
DECLARE @b TABLE
    (
      item CHAR(1) ,
      baseline DECIMAL(30, 10)
    )

INSERT  INTO @t
VALUES  ( 1, 2.31, 'A' ),
        ( 2, 1.10, 'A' ),
        ( 3, 2.13, 'A' ),
        ( 4, 1.34, 'A' )


INSERT  INTO @b
VALUES  ( 'A', 10 );


WITH    ordercte
          AS ( SELECT   * ,
                        ROW_NUMBER() OVER ( PARTITION BY item ORDER BY ID ) AS rn
               FROM     @t
             ),
        rec
          AS ( SELECT   t.item ,
                        t.ID ,
                        t.wac ,
                        t.rn ,
                        b.baseline * ( 1 + ( t.wac / 100 ) ) AS m
               FROM     ordercte t
                        JOIN @b b ON b.item = t.item
               WHERE    t.rn = 1
               UNION ALL
               SELECT   t.item ,
                        t.ID ,
                        t.wac ,
                        t.rn ,
                        c.m * ( 1 + ( t.wac / 100 ) )
               FROM     ordercte t
                        JOIN rec c ON t.item = c.item
                                      AND t.rn = c.rn + 1
             )
    SELECT  id ,
            wac ,
            item ,
            m
    FROM    rec

输出:

id  wac             item    m
1   2.3100000000    A       10.231000
2   1.1000000000    A       10.343541
3   2.1300000000    A       10.563858
4   1.3400000000    A       10.705414

EDIT1

我试图实现LOG EXP技巧,但除非@usr引导我解决方案,否则无法管理.所以用户@usr的所有积分:

WITH    ordercte
          AS ( SELECT   t.ID ,
                        t.wac ,
                        t.item ,
                        b.baseline ,
                        ROW_NUMBER() OVER ( PARTITION BY t.item ORDER BY ID ) AS rn
               FROM     @t t
                        JOIN @b b ON b.item = t.item
             )
    SELECT  baseline
            * EXP(SUM(LOG(( 1 + ( wac / 100 ) ))) OVER ( PARTITION BY item ORDER BY rn )) AS m
    FROM    ordercte

要不就:

SELECT  t.ID, t.wac, t.item, baseline
        * EXP(SUM(LOG(( 1 + ( wac / 100 ) ))) OVER ( PARTITION BY t.item ORDER BY t.ID )) AS m
FROM    @t t
        JOIN @b b ON b.item = t.item  

如果ID是您订购的字段.

输出:

ID  wac             item    m
1   2.3100000000    A       10.231
2   1.1000000000    A       10.343541
3   2.1300000000    A       10.5638584233
4   1.3400000000    A       10.7054141261722

EDIT2

对于SQL 2008,请使用:

WITH    cte
          AS ( SELECT   t.ID ,
                        t.wac ,
                        t.item ,
                        baseline ,
                        ( SELECT    SUM(LOG(( 1 + ( wac / 100 ) )))
                          FROM      @t it
                          WHERE     it.item = t.item AND it.ID <= t.ID
                        ) AS e
               FROM     @t t
                        JOIN @b b ON b.item = t.item
             )
    SELECT  ID, wac, item, baseline * EXP(e) AS m
    FROM    cte

EDIT3

这是SQL Server 2008的完整解决方案,使用NULL和负值拨号:

WITH    cte
          AS ( SELECT   t.ID ,
                        t.wac ,
                        t.item ,
                        b.baseline , 
                        ca.e,
                        ca.n,
                        ca.m
               FROM     @t t
               JOIN @b b ON b.item = t.item
               CROSS APPLY(SELECT   SUM(LOG(ABS(NULLIF( 1 +  wac / 100 , 0)))) as e,
                                    SUM(SIGN(CASE WHEN 1 +  wac / 100 < 0 THEN 1 ELSE 0 END)) AS n,
                                    MIN(ABS(1 +  wac / 100)) AS m
                          FROM      @t it
                          WHERE     it.item = t.item AND it.ID <= t.ID
                          ) ca
             )
    SELECT  ID, wac, item, baseline *
                        CASE
                            WHEN m = 0 THEN 0
                            WHEN n % 2 = 1 THEN -1 * EXP(e)
                            ELSE EXP(e) 
                        END as Result
    FROM    cte

Answer 2

您可以使用以下数学技巧将一系列乘法转换为一系列加法:

exp(log(a) + log(b)) = a * b

所以MUL(a)是EXP(SUM(LOG(a))).

SELECT SUM(val) AS [Sum], EXP(SUM(LOG(val))) AS Product
FROM (VALUES 
    (1), (2), (3), (4)
) x(val)

这会发出sum = 10, product = 24.

潜在的问题是四舍五入的错误和零因素.

您现在可以使用常用方法之一来实现运行聚合,例如窗口函数.这是一个已解决的问题.

Answer 3

为了完整起见，这里是 SQL Server 2012 的完整解决方案，它使用EXP(SUM(LOG(val)))@usr 在另一个答案中建议的技巧。

WITH
CTE
AS
(
    SELECT
        0 AS ID
        ,item
        ,baseline AS wac_inc
        ,baseline AS m
    FROM baseline

    UNION ALL

    SELECT
        ID
        ,item
        ,wac_inc
        ,1 + wac_inc/100 AS m
    FROM wac
)
SELECT
    ID
    ,item
    ,wac_inc
    ,m
    ,EXP(SUM(LOG(m)) OVER (PARTITION BY item ORDER BY ID ROWS UNBOUNDED PRECEDING)) AS MulRows
FROM CTE;

结果集

ID  item wac_inc            m                   MulRows
0   A   10.000000000000000  10.000000000000000  10
1   A   2.310000000000000   1.023100000000000   10.231
2   A   1.100000000000000   1.011000000000000   10.343541
3   A   2.130000000000000   1.021300000000000   10.5638584233
4   A   1.340000000000000   1.013400000000000   10.7054141261722

如果 SQL Server 2012 可用，则此窗口SUM非常有效。对于以前的版本，任何基于集合的解决方案都会导致O(n*n)复杂性，这意味着游标将是更好的方法。这是 Aaron Bertrand 写的一篇非常好的文章，比较了计算运行总计的不同方法：http://sqlperformance.com/2012/07/t-sql-queries/running-totals 或者 SO 问题：计算运行总计/运行余额

当然，如果您的表很小，那么由于游标开销，基于集合的O(n*n)复杂解决方案可能比使用游标的解决方案运行得更快，因此您需要使用真实数据检查性能。O(n)