T-SQL 中的移动中位数、众数
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【中文标题】T-SQL 中的移动中位数、众数【英文标题】：Moving Median, Mode in T-SQL 【发布时间】：2015-04-09 15:31:49 【问题描述】：我使用的是 SQL Server 2012，我知道计算移动平均线非常简单。 但我需要的是像这样获取定义的窗口框架的模式和中位数（当前行前面的窗口为 2；月份唯一）：
MONTH | CODE | MEDIAN | MODE
1 0 0 0
2 3 1.5 0
3 2 2 0
4 2 2 2
5 2 2 2
6 5 2 2
7 3 3 2
如果有几个值符合模式，则选择第一个。
【问题讨论】：
这是什么意思？ “当前行之前的窗口为 2”？如果您发布的代码是示例数据，您能否发布该数据所需的结果？ 中位数和众数列包含所需的结果数据。 那么，在这种情况下，执行 SELECT median, mode FROM MyTable，您将得到您想要的结果。 【参考方案1】：我彻底评论了我的代码。在我的模式计算中阅读我的 cmets，让我知道它需要调整。总的来说，这是一个相对简单的查询。它只是有很多丑陋的子查询，并且有很多 cmets。看看吧：
DECLARE @Table TABLE ([Month] INT,[Code] INT);
INSERT INTO @Table
VALUES (1,0),
(2,3),
(3,2),
(4,2), --Try commenting this out to test my special mode thingymajig
(5,2),
(6,5),
(7,3);
WITH CTE
AS
(
SELECT ROW_NUMBER() OVER (ORDER BY [Month]) row_num,
[Month],
CAST(Code AS FLOAT) Code
FROM @Table
)
SELECT [Month],
Code,
ISNULL((
SELECT CASE
--When there is only one previous value at row_num = 2, find Mean of first two codes
WHEN A.row_num = 2 THEN (LAG(B.code,1) OVER (ORDER BY [Code]) + B.Code)/2.0
--Else find middle code value of current and previous two rows
ELSE B.Code
END
FROM CTE B
--How subquery relates to outer query
WHERE B.row_num BETWEEN A.row_num - 2 AND A.row_num
ORDER BY B.[Code]
--Order by code and offset by 1 so don't select the lowest value, but fetch the one above the lowest value
OFFSET 1 ROW FETCH NEXT 1 ROW ONLY),
0) AS Median,
--I did mode a little different
--Instead of Avg(D.Code) you could list the values because with mode,
--If there's a tie with more than one of each number, you have multiple modes
--Instead of doing that, I simply return the mean of the tied modes
--When there's one, it doesn't change anything.
--If you were to delete the month 4, then your number of Codes 2 and number of Codes 3 would be the same in the last row.
--Proper mode would be 2,3. I instead average them out to be 2.5.
ISNULL((
SELECT AVG(D.Code)
FROM (
SELECT C.Code,
COUNT(*) cnt,
DENSE_RANK() OVER (ORDER BY COUNT(*) DESC) dnse_rank
FROM CTE C
WHERE C.row_num <= A.row_num
GROUP BY C.Code
HAVING COUNT(*) > 1) D
WHERE D.dnse_rank = 1),
0) AS Mode
FROM CTE A
结果：
Month Code Median Mode
----------- ---------------------- ---------------------- ----------------------
1 0 0 0
2 3 1.5 0
3 2 2 0
4 2 2 2
5 2 2 2
6 5 2 2
7 3 3 2
【讨论】：
【参考方案2】：如果我正确理解您的要求，您的源表包含 MONTH 和 CODE 列，并且您想要计算 MEDIAN 和 MODE。
下面的查询计算 MEDIAN 和 MODE，移动窗口
-----------------------------------------------------
--Demo data
-----------------------------------------------------
CREATE TABLE #Data(
[Month] INT NOT NULL,
[Code] INT NOT NULL,
CONSTRAINT [PK_Data] PRIMARY KEY CLUSTERED
(
[Month] ASC
));
INSERT #Data
([Month],[Code])
VALUES
(1,0),
(2,3),
(3,2),
(4,2),
(5,2),
(6,5),
(7,3);
-----------------------------------------------------
--Query
-----------------------------------------------------
DECLARE @PrecedingRowsLimit INT = 2;
WITH [MPos] AS
(
SELECT [R].[Month]
, [RB].[Month] AS [SubId]
, [RB].[Code]
, ROW_NUMBER() OVER(PARTITION BY [R].[Month] ORDER BY [RB].[Code]) AS [RowNumberInPartition]
, CASE
WHEN [R].[Count] % 2 = 1 THEN ([R].[Count] + 1) / 2
ELSE NULL
END AS [MedianPosition]
, CASE
WHEN [R].[Count] % 2 = 0 THEN [R].[Count] / 2
ELSE NULL
END AS [MedianPosition1]
, CASE
WHEN [R].[Count] % 2 = 0 THEN [R].[Count] / 2 + 1
ELSE NULL
END AS [MedianPosition2]
FROM
(
SELECT [RC].[Month]
, [RC].[RowNumber]
, CASE WHEN [RC].[Count] > @PrecedingRowsLimit + 1 THEN @PrecedingRowsLimit + 1 ELSE [RC].[Count] END AS [Count]
FROM
(
SELECT [Month]
, ROW_NUMBER() OVER(ORDER BY [Month]) AS [RowNumber]
, ROW_NUMBER() OVER(ORDER BY [Month]) AS [Count]
FROM #Data
) [RC]
) [R]
INNER JOIN #Data [RB]
ON [R].[Month] >= [RB].[Month]
AND [RB].[Month] >= [R].[RowNumber] - @PrecedingRowsLimit
)
SELECT DISTINCT [M].[Month]
, [ORIG].[Code]
, COALESCE([ME].[Code],([M1].[Code] + [M2].[Code]) / 2.0) AS [Median]
, [MOD].[Mode]
FROM [MPos] [M]
LEFT JOIN [MPOS] [ME]
ON [M].[Month] = [ME].[Month]
AND [M].[MedianPosition] = [ME].[RowNumberInPartition]
LEFT JOIN [MPOS] [M1]
ON [M].[Month] = [M1].[Month]
AND [M].[MedianPosition1] = [M1].[RowNumberInPartition]
LEFT JOIN [MPOS] [M2]
ON [M].[Month] = [M2].[Month]
AND [M].[MedianPosition2] = [M2].[RowNumberInPartition]
INNER JOIN
(
SELECT [MG].[Month]
, FIRST_VALUE([MG].[Code]) OVER (PARTITION BY [MG].[Month] ORDER BY [MG].[Count] DESC , [MG].[SubId] ASC) AS [Mode]
FROM
(
SELECT [Month] , MIN([SubId]) AS [SubId], [Code] , COUNT(1) AS [Count]
FROM [MPOS]
GROUP BY [Month] , [Code]
) [MG]
) [MOD]
ON [M].[Month] = [MOD].[Month]
INNER JOIN #Data [ORIG]
ON [ORIG].[Month] = [M].[Month]
ORDER BY [M].[Month];
【讨论】：
这是一些严肃的代码，但可能有点太复杂了。查看我的代码。我在两个相关子查询中进行了所有计算。 是的，@Stephan，你的代码更干净。尤其是FETCH NEXT 1 ROW ONLY在中位数计算中的使用非常简洁。
它是在 SQL 2012 中引入的，并不是最有用的。我很高兴我终于有机会使用它！ :)以上是关于T-SQL 中的移动中位数、众数的主要内容，如果未能解决你的问题，请参考以下文章