从表中查找中间值，按日期分组SQLServer

Question

我有一个复杂的问题，我试图解决。请耐心等待，随时提出任何问题。我是相当新的SQL和有这个困难...

我需要计算一组值的中位数。现在，这些值不会在表格中给出。这些值是根据按日期分组的每小时发生的表派生的。

下面是从中汇总数据的示例表。

CREATE TABLE Table22(
    Request_Number BIGINT NOT NULL 
    ,Request_Received_Date DATETIME NOT NULL 
); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016311446,'8/9/16 9:56'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20163612157,'9/6/16 9:17'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016384250,'9/12/16 14:52'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20162920101,'4/19/16 8:11'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016418170,'10/6/16 12:28'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016392953,'9/6/16 12:39'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20164123416,'10/6/16 15:05'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016335972,'8/9/16 7:49'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20162622951,'9/6/16 9:57'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20163913504,'9/6/16 9:47'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20163211326,'9/6/16 12:38'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20163610132,'8/30/16 16:34'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20164119560,'10/6/16 15:53'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016334416,'8/10/16 11:06'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20164320028,'10/6/16 15:27'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (20163515193,'8/24/16 19:50'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016159834,'4/19/16 13:21'); 
INSERT INTO Table22(Request_Number,Request_Received_Date) VALUES (2016178443,'4/19/16 13:05'); 

该表有2列：Request_Number和Request_Received_Date。 Request_Number不是唯一的，而且是无关紧要的。我正在查找在该日期（24小时）内特定日期和小时内收到多少个请求。每次有一个日期条目时，这被计为一个事件（TicketCount）。我可以使用COUNT语句从Request_received_date和按日期和小时计数*。

我就是这样做的，我的脚本中创建的临时表：

CREATE TABLE #z (ForDate date, OnHour int, TicketCount int) 
INSERT INTO #z (ForDate, OnHour, TicketCount)   
SELECT CAST(Request_received_date as DATE) AS 'ForDate', 
       DATEPART(hh, request_received_date) AS 'OnHour', 
       COUNT(*) AS TicketCount /*Hourly Ticket Count Column*/ 
       FROM Table22 
       GROUP BY CAST(request_received_date as DATE), DATEPART(hh, request_received_date) 
       ORDER BY ForDate Desc, OnHour ASC 

SELECT * FROM #z order by ForDate Desc, OnHour ASC 

现在我有困难的时候发现，每天数的中值。我已经尝试了许多不同的中位数计算公式，并且能够使他们的大部分工作。许多不同的中值计算示例可以在这里找到 https://sqlperformance.com/2012/08/t-sql-queries/median

我喜欢这段脚本来找到中位数。寻找中位数的脚本很简单。但它找到了Request_Received_Date的所有值的中位数。我无法找到在这里使用group by date子句的方法。

DECLARE @Median DECIMAL (12,2); 

SELECT @Median = (
    (SELECT MAX(TicketCount) FROM 
    (SELECT TOP 50 PERCENT TicketCount FROM #z ORDER BY TicketCount) AS BottomHalf) 
    + 
    (SELECT MIN(TicketCount) FROM 
    (SELECT TOP 50 PERCENT TicketCount FROM #z ORDER BY TicketCount DESC) AS TopHalf))/2; 

SELECT @Median 

任何帮助将非常感激。

预期的结果是这样的：

ForDate Median 
10/6/2016 2 
9/12/2016 1 
9/6/2016 2.5 
8/30/2016 1 
8/24/2016 1 
8/10/2016 1 
8/9/2016 1 
4/19/2016 1.5 

Answer 1

怎么这样呢？ （只有当你使用SQL Server 2012或以上的人士）

SELECT DISTINCT ForDate, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY TicketCount) OVER (PARTITION BY ForDate) AS Median 
FROM #z; 

---

总之，SQL-Server有两种方法来计算值，你可以在这里读到它：https://msdn.microsoft.com/en-us/library/hh231327.aspx

您可以比较他们都在这里与代码在这里：

SELECT DISTINCT 
    ForDate 
    , PERCENTILE_DISC(0.5) WITHIN GROUP (ORDER BY TicketCount) OVER (PARTITION BY ForDate) AS MedianDisc 
    , PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY TicketCount) OVER (PARTITION BY ForDate) AS MedianCont 
FROM 
    #z;