计算下四分位数和上四分位数的精确值

Question

我在网上到处找，似乎没有符合我情况的答案。

我正在努力计算 SQL 服务器中的确切下四分位数和上四分位数。我知道 SQL 服务器有一个有助于计算四分位数的内置函数，即 NTILE 函数。但这对我来说还不够。

给定以下 table 值（请注意 table 包含的产品和价格比下面 table 中的更多）：

AveragePrice	ProductNumber	Year
45.7820	2	2015
46.0142	2	2016
59.0133	2	2017
60.1707	2	2018
62.6600	2	2019

我是运行以下查询：

SELECT 
    AveragePrice
    ,NTILE(4) OVER (
        PARTITION BY ProductNumber ORDER BY AveragePrice
        ) AS Quartile
FROM products

结果如下：

AveragePrice	Quartile
45.7820	1
46.0142	1
59.0133	2
60.1707	3
62.6600	4

对于完整的上下文，整个查询如下所示：

    SELECT ProductNumber
    ,MIN(AveragePrice) Minimum
    ,MAX(CASE 
            WHEN Quartile = 1
                THEN AveragePrice
            END) AS Quartile_1
    ,
    MAX(CASE 
            WHEN Quartile = 3
                THEN AveragePrice
            END) AS Quartile_3
    ,MAX(AveragePrice) Maximum
    ,COUNT(Quartile) AS 'Number of items'
FROM (
    SELECT ProductNumber
        ,AveragePrice   
        ,NTILE(4) OVER (
            PARTITION BY ProductNumber ORDER BY ProductNumber
            ) AS Quartile
    FROM #temp_products
    
    ) Vals
GROUP BY ProductNumber
ORDER BY ProductNumber

但是当我手动计算四分位数时，第一个四分位数应该是：45.8981（在这种特殊情况下第一行和第二行的平均值）而不是 46.0142。

第三个四分位数应该是 61.41535（在这种特殊情况下第三个和第二个四分位数的平均值）而不是 60.1707。

所以说清楚。这是存储过程的一部分，其中计算多个价格组并将其聚合到包含平均价格的组中。我需要根据这些按产品编号分组的平均价格计算上四分位数和下四分位数。结果集应包含产品编号、下四分位数和上四分位数。有人可以帮助我或指导我正确的方向吗？

Answer 1

在某些情况下，

NTILE() 以一些奇怪的方式四舍五入。我宁愿使用带等级的整数除法进行分组。此解决方案适用于任意数量的值，并在需要时使用经过深思熟虑的平均值。

LEAD 是捕捉下一行值的魔法窗口函数

select *
    ,[Q] = case when [rank] in ((N+3)/4 ,(N+1)/2, (3*N+1)/4) then
                case [decimal] 
                when 0    then AveragePrice
                when 0.25 then /*pondered avg*/(3*AveragePrice +  LEAD(AveragePrice,1,null)over(PARTITION BY ProductNumber ORDER BY AveragePrice)) / 4
                when 0.5  then /*simple avg*/(    AveragePrice +  LEAD(AveragePrice,1,null)over(PARTITION BY ProductNumber ORDER BY AveragePrice)) / 2
                when 0.75 then /*pondered avg*/(  AveragePrice +3*LEAD(AveragePrice,1,null)over(PARTITION BY ProductNumber ORDER BY AveragePrice)) / 4 
                end
           end
from(
    select  *
        ,[rank]     = ROW_NUMBER()over(PARTITION BY ProductNumber ORDER BY AveragePrice)
        ,[N]        = SUM(1)over()
        ,[group4]   = ((ROW_NUMBER()over(PARTITION BY ProductNumber ORDER BY AveragePrice)-1 )*4 / SUM(1)over())
        ,[decimal]  = case /*rank*/ROW_NUMBER()over(PARTITION BY ProductNumber ORDER BY AveragePrice) 
                        when /*Q1*/  (SUM(1)over()+3)/4   then   (SUM(1)over()+3)/4.0 - FLOOR((SUM(1)over()+3)/4.0) 
                        when /*Q2*/  (SUM(1)over()+1)/2   then   (SUM(1)over()+1)/2.0 - FLOOR((SUM(1)over()+1)/2.0)
                        when /*Q3*/(3*SUM(1)over()+1)/4   then (3*SUM(1)over()+1)/4.0 - FLOOR((3*SUM(1)over()+1)/4.0)
                      end
    from
    (values(45.7820,2,2015),(46.0142,2,2016),(59.0133,2,2017),(60.1707,2,2018),(62.6600,2,2019))a(AveragePrice,ProductNumber,Year)
  )a

AveragePrice	ProductNumber	Year	rank	N	group4	decimal	Q
45.7820	2	2015	1	5	0	NULL	NULL
46.0142	2	2016	2	5	0	0.000000	46.014200
59.0133	2	2017	3	5	1	0.000000	59.013300
60.1707	2	2018	4	5	2	0.000000	60.170700
62.6600	2	2019	5	5	3	NULL	NULL

Answer 2

好的，受此启发 post 我设法构建了一个实际计算精确四分位数的查询：

  -- ; since it is being used in a sp
    ;WITH quartile_data AS (
    
        SELECT Price,
             ProductNumber
        FROM (values(29.4785,2,2015),(30.0000,2,2016),(33.4762,2,2017),(35.2917,2,2018),(35.8731,2,2018),(36.2475,2,2018),(37.9790,2,2018),(39.5846,2,2018),(67.4443,2,2018))sales(Price,ProductNumber)
     
    )
   --Aggregate into a single record for each group, using MAX to select the non-null 
   --detail value for each column
   -- ISNULL check to include groups with three values as well
    SELECT ProductNumber, 
        (Max(Q1NextVal) - MAX(Q1Val)) * Max(Q1Frac) + Max(Q1Val) as [Q1],
        (Max(MidVal1) + Max(MidVal2)) / 2 [Median],
        (ISNULL(Max(Q3NextVal),0) - MAX(Q3Val)) * Max(Q3Frac) + Max(Q3Val) as [Q3]
    
    -- save the result into a temp table
    INTO #my_temp_table
    FROM (
--Expose the detail values for only the records at the index values 
--generated by the summary subquery. All other values are left as NULL. som NULL.
        SELECT detail.ProductNumber, 
            CASE WHEN RowNum = Q1Idx THEN Price ELSE NULL END Q1Val,
            CASE WHEN RowNum = Q1Idx + 1 THEN Price ELSE NULL END Q1NextVal,
            CASE WHEN RowNum = Q3Idx THEN Price ELSE NULL END Q3Val,
            CASE WHEN RowNum = Q3Idx + 1 THEN Price ELSE NULL END Q3NextVal,
            Q1Frac,
            Q3Frac,
            CASE WHEN RowNum = MidPt1 THEN Price ELSE NULL END MidVal1,
            CASE WHEN RowNum = MidPt2 THEN Price ELSE NULL END MidVal2
        FROM
               --Calculate a row number sorted by measure for each group.
            (SELECT *,  ROW_NUMBER() OVER (PARTITION BY ProductNumber ORDER BY Price) RowNum
            FROM  quartile_data) AS detail
    
        INNER JOIN (
    --Summarize to find index numbers and fractions we need to use to locate 
    --the values at the quartile points.
    -- The modulus operator is used to sum the correct number if the number of rows in the group is even or uneven
            SELECT ProductNumber, 
                FLOOR((COUNT(*) + IIF((COUNT(*) % 2 = 0), 2,1)) / 4.0) Q1Idx,
                ((COUNT(*) + IIF((COUNT(*) % 2 = 0), 2,1)) / 4.0) - FLOOR((COUNT(*) + IIF((COUNT(*) % 2 = 0), 2,1)) / 4.0) Q1Frac,
                (COUNT(*) + 1) / 2 AS MidPt1,
                (COUNT(*) + 2) / 2 AS Midpt2,
                  FLOOR((COUNT(*) * 3 + IIF((COUNT(*) % 2 = 0), 2,3)) / 4.0) Q3Idx,
                ((COUNT(*) * 3 + IIF((COUNT(*) % 2 = 0), 2,3)) / 4.0) - FLOOR((COUNT(*) * 3 + IIF((COUNT(*) % 2 = 0), 2,3)) / 4.0) Q3Frac
            FROM  quartile_data
            GROUP BY ProductNumber
            HAVING COUNT(*) > 1
        
        ) AS summary ON detail.ProductNumber  = summary.ProductNumber
    
    ) AS step_two
    GROUP BY ProductNumber
   -- Include only groups with more than 2 rows
    HAVING count(*) > 2

以下价格： 29.4785 30.0000 33.4762 35.2917 35.8731 36.2475 37.9790 39.5846 67.4443

给出正确的值：Q1 = 31.7381000000 和 Q3 = 38.7818000000

使用这个online tool

验证

计算下四分位数和上四分位数的精确值

Calculate Exact Values Of Lower and Upper Quartiles

sql

tsql

sql-server

algorithm

quartile