有没有办法将这些温度测量分组到连续组？ 我想得到小组，时差和数在0-7到8-12之间，超过12
Date Heat
01/01/2012 12:00 8
01/01/2012 12:03 9
01/01/2012 12:06 5
01/01/2012 12:09 3
01/01/2012 12:12 6
01/01/2012 12:15 7
01/01/2012 12:18 1
01/01/2012 12:21 12
01/01/2012 12:24 28
01/01/2012 12:27 25
01/01/2012 12:30 20
01/01/2012 12:33 20
01/01/2012 12:36 20
01/01/2012 12:39 12
01/01/2012 12:42 6
01/01/2012 12:45 3
01/01/2012 12:48 5
01/01/2012 12:51 7
01/01/2012 12:54 11
01/01/2012 12:57 12
01/01/2012 13:00 6
结果应为：
0-7 (01/01/2012 12:06-01/01/2012 12:18) 5
/* Rows of dataset:
01/01/2012 12:06 5
01/01/2012 12:09 3
01/01/2012 12:12 6
01/01/2012 12:15 7
01/01/2012 12:18 1
*/
0-7 (01/01/2012 12:42-01/01/2012 12:51) 5
/* Rows of dataset:
01/01/2012 12:42 6
01/01/2012 12:45 3
01/01/2012 12:48 5
01/01/2012 12:51 7
*/
8-12 (01/01/2012 12:00-01/01/2012 12:03) 2
/* Rows of dataset:
01/01/2012 12:00 8
01/01/2012 12:03 9
*/
more then 12 (01/01/2012 12:24-01/01/2012 12:36) 5
/* Rows of dataset:
01/01/2012 12:24 28
01/01/2012 12:27 25
01/01/2012 12:30 20
01/01/2012 12:33 20
01/01/2012 12:36 20
*/
8-12 (01/01/2012 12:21) 1
/* Rows of dataset:
01/01/2012 12:21 12 */
答案 0 :(得分：4)
注意：由于RANK/DENSE_RANK的处理顺序为PARTITION BY，然后是ORDER BY，因此在这种情况下这些功能无效。也许，在某个时间点，MS将引入补充语法：
[DENSE_]RANK() OVER(ORDER BY fields PARTITION BY fields) {}首先处理ORDER BY，然后PARTITION BY。
1）第一个解决方案（SQL2005 +）
DECLARE @TestData TABLE
(
Dt SMALLDATETIME PRIMARY KEY,
Heat TINYINT NOT NULL
);
INSERT @TestData(Dt, Heat)
VALUES
SELECT '2012-01-01T12:00:00', 8 UNION ALL SELECT '2012-01-01T12:03:00', 9 UNION ALL SELECT '2012-01-01T12:06:00', 5
UNION ALL SELECT '2012-01-01T12:09:00', 3 UNION ALL SELECT '2012-01-01T12:12:00', 6 UNION ALL SELECT '2012-01-01T12:15:00', 7
UNION ALL SELECT '2012-01-01T12:18:00', 1 UNION ALL SELECT '2012-01-01T12:21:00', 12 UNION ALL SELECT '2012-01-01T12:24:00', 28
UNION ALL SELECT '2012-01-01T12:27:00', 25 UNION ALL SELECT '2012-01-01T12:30:00', 20 UNION ALL SELECT '2012-01-01T12:33:00', 20
UNION ALL SELECT '2012-01-01T12:36:00', 20 UNION ALL SELECT '2012-01-01T12:39:00', 12 UNION ALL SELECT '2012-01-01T12:42:00', 6
UNION ALL SELECT '2012-01-01T12:45:00', 3 UNION ALL SELECT '2012-01-01T12:48:00', 5 UNION ALL SELECT '2012-01-01T12:51:00', 7
UNION ALL SELECT '2012-01-01T12:54:00', 11 UNION ALL SELECT '2012-01-01T12:57:00', 12 UNION ALL SELECT '2012-01-01 13:00:00', 6;
SET STATISTICS IO ON;
WITH CteSource
AS
(
SELECT a.*,
CASE
WHEN a.Heat >= 0 AND a.Heat <= 7 THEN 1
WHEN a.Heat >= 8 AND a.Heat <= 12 THEN 2
WHEN a.Heat > 12 THEN 3
END AS Grp,
ROW_NUMBER() OVER(ORDER BY a.Dt) AS RowNum
FROM @TestData a
), CteRecursive
AS
(
SELECT s.RowNum,
s.Dt,
s.Heat,
s.Grp,
1 AS DENSE_RANK_OVER_ORDERBY_PARTITIONBY
FROM CteSource s
WHERE s.RowNum = 1
UNION ALL
SELECT crt.RowNum,
crt.Dt,
crt.Heat,
crt.Grp,
CASE
WHEN crt.Grp = prev.Grp THEN prev.DENSE_RANK_OVER_ORDERBY_PARTITIONBY
ELSE prev.DENSE_RANK_OVER_ORDERBY_PARTITIONBY + 1
END
FROM CteSource crt
INNER JOIN CteRecursive prev ON crt.RowNum = prev.RowNum + 1
)
SELECT r.DENSE_RANK_OVER_ORDERBY_PARTITIONBY,
MAX(r.Grp) AS Grp,
COUNT(*) AS Cnt,
MIN(r.Dt) AS MinDt,
MAX(r.Dt) AS MaxDt
FROM CteRecursive r
GROUP BY r.DENSE_RANK_OVER_ORDERBY_PARTITIONBY;
结果：
DENSE_RANK_OVER_ORDERBY_PARTITIONBY Grp Cnt MinDt MaxDt
----------------------------------- ----------- ----------- ----------------------- -----------------------
1 2 2 2012-01-01 12:00:00 2012-01-01 12:03:00
2 1 5 2012-01-01 12:06:00 2012-01-01 12:18:00
3 2 1 2012-01-01 12:21:00 2012-01-01 12:21:00
4 3 5 2012-01-01 12:24:00 2012-01-01 12:36:00
5 2 1 2012-01-01 12:39:00 2012-01-01 12:39:00
6 1 4 2012-01-01 12:42:00 2012-01-01 12:51:00
7 2 2 2012-01-01 12:54:00 2012-01-01 12:57:00
8 1 1 2012-01-01 13:00:00 2012-01-01 13:00:00
2）第二个解决方案（SQL2012;更好的性能）
SELECT d.DENSE_RANK_OVER_ORDERBY_PARTITIONBY,
MAX(d.Grp) AS Grp,
MIN(d.Dt) AS MinDt,
MAX(d.Dt) AS MaxDt
FROM
(
SELECT c.*,
1+SUM(c.IsNewGroup) OVER(ORDER BY c.Dt) AS DENSE_RANK_OVER_ORDERBY_PARTITIONBY
FROM
(
SELECT b.*,
CASE
WHEN LAG(b.Grp) OVER(ORDER BY b.Dt) <> b.Grp THEN 1
ELSE 0
END
AS IsNewGroup
FROM
(
SELECT a.*,
CASE
WHEN a.Heat >= 0 AND a.Heat <= 7 THEN 1
WHEN a.Heat >= 8 AND a.Heat <= 12 THEN 2
WHEN a.Heat > 12 THEN 3
END AS Grp
FROM @TestData a
) b
) c
) d
GROUP BY d.DENSE_RANK_OVER_ORDERBY_PARTITIONBY;
答案 1 :(得分：1)
以下是SQL Server 2005或更新版本的替代解决方案：
WITH auxiliary (HeatID, MinHeat, MaxHeat, HeatDescr) AS (
SELECT 1, 0 , 7 , '0-7' UNION ALL
SELECT 2, 8 , 12 , '8-12' UNION ALL
SELECT 3, 13, NULL, 'more than 12'
),
datagrouped AS (
SELECT
d.*,
a.HeatDescr,
grp = ROW_NUMBER() OVER ( ORDER BY d.Date)
- ROW_NUMBER() OVER (PARTITION BY a.HeatID ORDER BY d.Date)
FROM data d
INNER JOIN auxiliary a
ON d.Heat BETWEEN a.MinHeat AND ISNULL(a.MaxHeat, 0x7fffffff)
)
SELECT
HeatDescr,
DateFrom = MIN(Date),
DateTo = MAX(Date),
ItemCount = COUNT(*)
FROM datagrouped
GROUP BY
HeatDescr, grp
ORDER BY
MIN(Date)
data的定义如下：
CREATE TABLE data (Date datetime, Heat int);
INSERT INTO data (Date, Heat)
SELECT '01/01/2012 12:00', 8 UNION ALL
SELECT '01/01/2012 12:03', 9 UNION ALL
SELECT '01/01/2012 12:06', 5 UNION ALL
SELECT '01/01/2012 12:09', 3 UNION ALL
SELECT '01/01/2012 12:12', 6 UNION ALL
SELECT '01/01/2012 12:15', 7 UNION ALL
SELECT '01/01/2012 12:18', 1 UNION ALL
SELECT '01/01/2012 12:21', 12 UNION ALL
SELECT '01/01/2012 12:24', 28 UNION ALL
SELECT '01/01/2012 12:27', 25 UNION ALL
SELECT '01/01/2012 12:30', 20 UNION ALL
SELECT '01/01/2012 12:33', 20 UNION ALL
SELECT '01/01/2012 12:36', 20 UNION ALL
SELECT '01/01/2012 12:39', 12 UNION ALL
SELECT '01/01/2012 12:42', 6 UNION ALL
SELECT '01/01/2012 12:45', 3 UNION ALL
SELECT '01/01/2012 12:48', 5 UNION ALL
SELECT '01/01/2012 12:51', 7 UNION ALL
SELECT '01/01/2012 12:54', 11 UNION ALL
SELECT '01/01/2012 12:57', 12 UNION ALL
SELECT '01/01/2012 13:00', 6;
对于上面的示例，查询提供以下输出：
HeatDescr DateFrom DateTo ItemCount
------------ ------------------- ------------------- ---------
8-12 2012-01-01 12:00:00 2012-01-01 12:03:00 2
0-7 2012-01-01 12:06:00 2012-01-01 12:18:00 5
8-12 2012-01-01 12:21:00 2012-01-01 12:21:00 1
more than 12 2012-01-01 12:24:00 2012-01-01 12:36:00 5
8-12 2012-01-01 12:39:00 2012-01-01 12:39:00 1
0-7 2012-01-01 12:42:00 2012-01-01 12:51:00 4
8-12 2012-01-01 12:54:00 2012-01-01 12:57:00 2
0-7 2012-01-01 13:00:00 2012-01-01 13:00:00 1
答案 2 :(得分：0)
你应该使用RANK（）
达到目标http://msdn.microsoft.com/en-us/library/ms176102.aspx
像
这样的东西SELECT date, heat, RANK() OVER (PARTITION BY heat ORDER BY date DESC) AS Rank
FROM tbl
然后你可以在之后对它进行分组，或者根据你的结果进行更多的子选择和联合。