Question

我有一张桌子，如下所示：

我需要按Cat和Timestamp对数据进行分组，并按组进行计数。小组被定义为5分钟的动态时间窗，这意味着它可以跨越不同的时间。

查询结果应提供以下内容：

以黄色查看第一个表组。这些组应被检测并计为一个，而未突出显示的组也应计为一个

现在我在Stackoverflow上阅读了许多解决方案，以下是我尝试过的相关解决方案：

创建5分钟的时间间隔组-这不起作用，因为跨越不同小时的时间戳记与同一组不匹配
使用ROWNUMBER（）OVER（按类别按时间戳排序）并加入t1.Cat = t2.Cat和t1.rn + 1 = t2.rn。按DATEDIFF过滤。这是行不通的，因为只能检测到两个对。如果5分钟内连续5个时间戳怎么办？

我将非常感谢对此的任何帮助

有关ascii表中的原始数据，请参见下文

原始数据

+---------------------+----------+
|      Timestamp      | Category |
+---------------------+----------+
| 2018-10-01 04:06:12 | Cat1     |
| 2018-10-01 05:07:18 | Cat1     |
| 2018-10-01 05:07:19 | Cat1     |
| 2018-10-01 05:07:20 | Cat1     |
| 2018-10-01 06:09:29 | Cat1     |
| 2018-10-01 07:24:12 | Cat2     |
| 2018-10-01 07:30:43 | Cat2     |
| 2018-10-01 07:59:13 | Cat2     |
| 2018-10-01 08:02:15 | Cat2     |
| 2018-10-01 10:09:25 | Cat2     |
| 2018-10-01 11:13:42 | Cat2     |
+---------------------+----------+

Answer 1

这是一种实现方式

第一步，根据先前的时间戳记值是否在5分钟之内将记录分类。如果是，则为它分配一个row_number。

这将是如下获得您的值

+---------------------+----------+-----------+
|     timestamp1      | category | grps_of_5 |
+---------------------+----------+-----------+
| 01/10/2018 05:06:12 | Cat1     |         1 |
| 01/10/2018 05:07:18 | Cat1     |           |
| 01/10/2018 05:07:19 | Cat1     |           |
| 01/10/2018 05:07:20 | Cat1     |           |
| 01/10/2018 06:09:29 | Cat1     |         5 |
| 01/10/2018 07:24:12 | Cat2     |         1 |
| 01/10/2018 07:30:43 | Cat2     |         2 |
| 01/10/2018 07:59:13 | Cat2     |         3 |
| 01/10/2018 08:02:15 | Cat2     |           |
| 01/10/2018 10:09:25 | Cat2     |         5 |
| 01/10/2018 11:13:42 | Cat2     |         6 |
+---------------------+----------+-----------+


After that i "copy" the values to fill up the nulls in groups using
max(grps_of_5) over(partition by category order by timestamp1)


This is done in the curated_data block and will look like this

+---------------------+----------+-----------+---------+
|     timestamp1      | category | grps_of_5 | max_val |
+---------------------+----------+-----------+---------+
| 01/10/2018 04:06:12 | Cat1     |         1 |       1 |
| 01/10/2018 05:07:18 | Cat1     |         2 |       2 |
| 01/10/2018 05:07:19 | Cat1     |           |       2 |
| 01/10/2018 05:07:20 | Cat1     |           |       2 |
| 01/10/2018 06:09:29 | Cat1     |         5 |       5 |
| 01/10/2018 07:24:12 | Cat2     |         1 |       1 |
| 01/10/2018 07:30:43 | Cat2     |         2 |       2 |
| 01/10/2018 07:59:13 | Cat2     |         3 |       3 |
| 01/10/2018 08:02:15 | Cat2     |           |       3 |
| 01/10/2018 10:09:25 | Cat2     |         5 |       5 |
| 01/10/2018 11:13:42 | Cat2     |         6 |       6 |
+---------------------+----------+-----------+---------+


After that i am counting the distinct max_val which will tell count all 5 minute intervals as a single group and others seperately.

with raw_data
  as(select timestamp1
            ,category
            ,case when datediff(mi,lag(timestamp1) over(partition by category order by timestamp1),timestamp1) >5 
                    or lag(timestamp1) over(partition by category order by timestamp1) is null
                  then row_number() over(partition by category order by timestamp1)                  
              end as grps_of_5
       from t  
     )
   ,curated_data
      as (select max(grps_of_5) over(partition by category order by timestamp1) as max_val
                 ,x.*
            from raw_data x
           )
 select category,count(distinct max_val) as cnt
   from curated_data
group by category            

+----------+------+
| category | cnt2 |
+----------+------+
| Cat1     |    3 |
| Cat2     |    5 |
+----------+------+

编辑版本

演示链接

https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=556e0ec16bb040b96b637e3da3e8178b

Answer 2

这很容易通过LAG完成：

DECLARE @t TABLE (timestamp DATETIME, category VARCHAR(100));
INSERT INTO @t VALUES
('2018-10-01 04:06:12', 'CAT1'),
('2018-10-01 05:07:18', 'CAT1'),
('2018-10-01 05:07:19', 'CAT1'),
('2018-10-01 05:07:20', 'CAT1'),
('2018-10-01 06:09:29', 'CAT1'),
('2018-10-01 07:24:12', 'CAT2'),
('2018-10-01 07:30:43', 'CAT2'),
('2018-10-01 07:59:13', 'CAT2'),
('2018-10-01 08:02:15', 'CAT2'),
('2018-10-01 10:09:25', 'CAT2'),
('2018-10-01 11:13:42', 'CAT2');

WITH cte1 AS (
    SELECT timestamp, category, CASE WHEN LAG(timestamp) OVER (PARTITION BY category ORDER BY timestamp) > DATEADD(MINUTE, -5, timestamp) THEN 0 ELSE 1 END AS chg
    FROM @t
)
SELECT category, COUNT(CASE WHEN chg = 1 THEN 1 END)
FROM cte1
GROUP BY category

要了解其工作原理，请重点关注chg列的计算方式，并查看cte的结果：

timestamp                  category    chg
2018-10-01 04:06:12.000    CAT1        1
2018-10-01 05:07:18.000    CAT1        1
2018-10-01 05:07:19.000    CAT1        0
2018-10-01 05:07:20.000    CAT1        0
2018-10-01 06:09:29.000    CAT1        1
2018-10-01 07:24:12.000    CAT2        1
2018-10-01 07:30:43.000    CAT2        1
2018-10-01 07:59:13.000    CAT2        1
2018-10-01 08:02:15.000    CAT2        0
2018-10-01 10:09:25.000    CAT2        1
2018-10-01 11:13:42.000    CAT2        1

Answer 3

请尝试以下代码：

SELECT * INTO #temp
FROM(
    SELECT '2018-10-01 05:06:12' AS Timestamp , 'Cat1' AS Category   
    UNION ALL
    SELECT '2018-10-01 05:07:18' AS Timestamp , 'Cat1' AS Category  
    UNION ALL
    SELECT '2018-10-01 05:07:19' AS Timestamp , 'Cat1' AS Category  
    UNION ALL
    SELECT '2018-10-01 05:07:20' AS Timestamp , 'Cat1' AS Category 
    UNION ALL
    SELECT '2018-10-01 06:09:29' AS Timestamp , 'Cat1' AS Category 
    UNION ALL
    SELECT '2018-10-01 07:24:12' AS Timestamp , 'Cat2' AS Category   
    UNION ALL
    SELECT '2018-10-01 07:30:43' AS Timestamp , 'Cat2' AS Category  
    UNION ALL
    SELECT '2018-10-01 07:59:13' AS Timestamp , 'Cat2' AS Category  
    UNION ALL
    SELECT '2018-10-01 08:02:15' AS Timestamp , 'Cat2' AS Category 
    UNION ALL
    SELECT '2018-10-01 10:09:25' AS Timestamp , 'Cat2' AS Category 
   UNION ALL
    SELECT '2018-10-01 11:13:42' AS Timestamp , 'Cat2' AS Category 
) AS T

SELECT  Category AS [Group], COUNT(CONVERT(DATE,Timestamp)) AS [Count]  FROM #temp GROUP By Category

每组间隔时间少于5分钟的计数

3 个答案: