从 table 中查找中值，按日期分组 SQL Server

Question

我有一个复杂的问题要解决。请耐心等待，随时提出任何问题。我对 SQL 很陌生，对此有困难...

我需要计算一组值的中位数。现在这些值没有在 table 中给出。这些值源自 table，基于按日期分组的每小时出现次数。

这是汇集数据的示例 table。

   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');

Table 有 2 列：Request_Number 和 Request_Received_Date。 Request_Number 不是唯一的，有点无关紧要。我正在寻找特定日期（24 小时）内每小时收到多少请求。每次有日期条目时，都计为一次 (TicketCount)。我可以使用 COUNT 语句从 Request_received_date 开始计算 * 并按日期和小时分组。

我就是这么做的，并在我的脚本中创建了一个临时 table：

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

现在我很难找到每天计数的中值。我已经尝试了许多不同的中位数计算公式，并且能够使大多数公式起作用。可以在此处找到许多不同的中位数计算示例 https://sqlperformance.com/2012/08/t-sql-queries/median

我喜欢这段找中位数的脚本。查找中位数的脚本很简单。但它会找到 Request_Received_Date 的所有值的中位数。我无法在这里找到使用 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

任何帮助将不胜感激。

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

这样的事情怎么样？（仅适用于 SQL Server 2012 或更高版本）

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

简而言之，SQL-服务器有两种计算中位数的方法，你可以在这里阅读：https://msdn.microsoft.com/en-us/library/hh231327.aspx

在这种情况下，您可以将它们与此处的代码进行比较：

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;

从 table 中查找中值，按日期分组 SQL Server

Finding the Median value from a table, Group By Date SQLServer

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