Sql server T-SQL中的移动中值模式
我使用的是SQL Server 2012，我知道计算移动平均值非常简单。 但我需要的是得到一个定义的窗口框架的模式和中位数，就像这样（在当前行之前有一个2的窗口；月份唯一）：Sql server T-SQL中的移动中值模式,sql-server,tsql,window-functions,Sql Server,Tsql,Window Functions,我使用的是SQL Server 2012，我知道计算移动平均值非常简单。 但我需要的是得到一个定义的窗口框架的模式和中位数，就像这样（在当前行之前有一个2的窗口；月份唯一）： 如果有几个值符合模式，则选择第一个。如果我正确理解了您的要求，则源表包含月份和代码列，您需要计算中值和模式 下面的查询使用移动窗口计算中值和模式我对代码进行了彻底的注释。阅读我对模式计算的评论，让我知道它需要调整。总的来说，这是一个相对简单的查询。它只是有很多难看的子查询和很多注释。请查看： DECLARE @Table
如果有几个值符合模式，则选择第一个。如果我正确理解了您的要求，则源表包含月份和代码列，您需要计算中值和模式
下面的查询使用移动窗口计算中值和模式我对代码进行了彻底的注释。阅读我对模式计算的评论，让我知道它需要调整。总的来说，这是一个相对简单的查询。它只是有很多难看的子查询和很多注释。请查看：
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
这是什么意思？“当前行前面有2个窗口”？如果您发布的代码是示例数据，您可以发布该数据的所需结果吗？中间值和模式列包含所需结果数据。在这种情况下，请从MyTable中选择中间值、模式，您将获得所需结果。这是一些严肃的代码，但可能有点太复杂。查看我的代码。我在两个相关的子查询中完成了所有的计算。是的，@Stephan，你的代码更干净了。尤其是
仅获取下一行-----------------------------------------------------
--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];
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
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