假设您有一个包含列,Date,GroupID,X和Y的表.
CREATE TABLE #sample
(
[Date] DATETIME,GroupID INT,X FLOAT,Y FLOAT
)
DECLARE @date DATETIME = getdate()
INSERT INTO #sample VALUES(@date,1,3)
INSERT INTO #sample VALUES(DATEADD(d,@date),1)
INSERT INTO #sample VALUES(DATEADD(d,2,4,2)
INSERT INTO #sample VALUES(DATEADD(d,3,6,4)
INSERT INTO #sample VALUES(DATEADD(d,5,7,5)
INSERT INTO #sample VALUES(DATEADD(d,6)
并且您想要计算每个组的X和Y的相关性.目前我使用的CTE有点乱：
;WITH DataAvgStd
AS (SELECT GroupID,AVG(X) AS XAvg,AVG(Y) AS YAvg,STDEV(X) AS XStdev,STDEV(Y) AS YSTDev,COUNT(*) AS SampleSize
FROM #sample
GROUP BY GroupID),ExpectedVal
AS (SELECT s.GroupID,SUM(( X - XAvg ) * ( Y - YAvg )) AS ExpectedValue
FROM #sample s
JOIN DataAvgStd das
ON s.GroupID = das.GroupID
GROUP BY s.GroupID)
SELECT das.GroupID,ev.ExpectedValue / ( das.SampleSize - 1 ) / ( das.XStdev * das.YSTDev )
AS
Correlation
FROM DataAvgStd das
JOIN ExpectedVal ev
ON das.GroupID = ev.GroupID
DROP TABLE #sample
似乎应该有一种方法可以使用OVER和PARTITION一次性执行此操作而不需要任何子查询.理想情况下,Tsql会有一个函数,所以你可以写：
SELECT GroupID,CORR(X,Y) OVER(PARTITION BY GroupID) FROM #sample GROUP BY GroupID
解决方法
使用这个corellation公式,即使使用over(),也无法避免所有嵌套查询.问题是你不能在同一个查询中反复使用这两个组,也不能有嵌套的聚合函数,例如sum(x – avg(x)).因此,在最佳情况下,根据您的数据,您至少需要保留.
你的代码看起来就像那样
;WITH DataAvgStd
AS (SELECT GroupID,STDEV(X) over(partition by GroupID) AS XStdev,STDEV(Y) over(partition by GroupID) AS YSTDev,COUNT(*) over(partition by GroupID) AS SampleSize,( X - AVG(X) over(partition by GroupID)) * ( Y - AVG(Y) over(partition by GroupID)) AS ExpectedValue
FROM #sample s)
SELECT distinct GroupID,SUM(ExpectedValue) over(partition by GroupID) / (SampleSize - 1 ) / ( XStdev * YSTDev ) AS Correlation
FROM DataAvgStd
另一种方法是使用等效公式进行相关,如Wikipedia所述.
这可以写成
SELECT GroupID,Correlation=(COUNT(*) * SUM(X * Y) - SUM(X) * SUM(Y)) /
(SQRT(COUNT(*) * SUM(X * X) - SUM(X) * SUM(x))
* SQRT(COUNT(*) * SUM(Y* Y) - SUM(Y) * SUM(Y)))
FROM #sample s
GROUP BY GroupID;
