Question

我已经创建了一个指数价格水平表（例如，S＆amp; P 500），我想计算每日回报率。表结构如下所示：

Date        Value
2009-07-02  880.167341
2009-07-03  882.235134
2009-07-06  881.338052
2009-07-07  863.731494
2009-07-08  862.458985

我想计算索引的每日算术回报（即百分比回报），定义为：

Daily Return = P(2)/P(1) - 1

其中P代表这种情况下的指数值。给定上面的输入表，所需的输出将如下所示：

    Date        Return
    2009-07-03  0.002349318
    2009-07-06  -0.001016829
    2009-07-07  -0.019977077
    2009-07-08  -0.001473269

我觉得自我加入会起作用，但我不确定将第二张桌子上的日期增加到周末的最佳方法。

关于最佳方式的任何想法？

Answer 1

WITH cteRank AS (
    SELECT [Date], Value, 
           ROW_NUMBER() OVER(ORDER BY [Date]) AS RowNum
        FROM YourTable
)
SELECT c1.[Date], c1.Value/c2.Value - 1 as [Return]
    from cteRank c1
        inner join cteRank c2
            on c1.RowNum - 1 = c2.RowNum
    where c1.RowNum > 1

Answer 2

一个简单的交叉申请

SELECT
  Tlater.Date, (Tlater.Value / TPrev2.Value) - 1
FROM
   MyTable Tlater
   CROSS APPLY
   (
    SELECT TOP 1 TPrev.Value 
    FROM MyTable TPrev
    WHERE TPrev.Date < Tlater.Date
    ORDER BY TPrev.Date
   ) TPrev2

注意：这在使用LAG的Denali（SQL Server 2012）中变得微不足道（未经测试，可能需要CTE）

SELECT 
    OrderDate, 
    (Value / (LAG(Value) OVER (ORDER BY Date))) -1
FROM
    MyTable

或者

;WITH cPairs AS
(
   SELECT 
      Date, 
      Value AS Curr,
      LAG(Value) OVER (ORDER BY Date) AS Prev
   FROM
    MyTable
)
SELECT 
    Date, 
    (Curr / Prev) -1
FROM
    cPairs

Answer 3

如果您使用的是2005+，则可以将ROW_NUMBER功能与CTE结合使用：

;with RowNums as ( select *, row_number() over (order by date) as RN from table )
select *, r1.Value / r.Value - 1 as Return
from RowNums r
inner join RowNums r1 
  on r.RN = r1.RN - 1

从值表计算算术返回值

3 个答案: