按值对行进行分组和计数,直到更改为止

Question

我有一个表,消息在发生时存储.通常有一条消息'A',有时A被单个消息'B'分隔.现在我想对值进行分组,以便我能够分析它们,例如找到最长的'A'条纹或'A'条纹的分布.

我已经尝试过COUNT-OVER查询,但是仍在为每条消息计算.

SELECT message, COUNT(*) OVER (ORDER BY Timestamp RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)

这是我的示例数据:

Timestamp        Message
20150329 00:00   A
20150329 00:01   A
20150329 00:02   B
20150329 00:03   A
20150329 00:04   A
20150329 00:05   A
20150329 00:06   B

我想要跟随输出

Message    COUNT
A          2
B          1
A          3
B          1

Answer 1

那很有趣:)

;WITH cte as (
SELECT Messages.Message, Timestamp, 
ROW_NUMBER() OVER(PARTITION BY Message ORDER BY Timestamp) AS gn,
ROW_NUMBER() OVER (ORDER BY Timestamp) AS rn
FROM Messages
), cte2 AS (
SELECT Message, Timestamp, gn, rn, gn - rn  as gb
FROM cte 
), cte3 AS (
SELECT Message, MIN(Timestamp) As Ts, COUNT(1) as Cnt
FROM cte2
GROUP BY Message, gb)
SELECT Message, Cnt FROM cte3
ORDER BY Ts

这是结果集:

  Message   Cnt
    A   2
    B   1
    A   3
    B   1

查询可能会更短,但我会以这种方式发布,以便您可以看到正在发生的事情.结果完全按照要求.这是最重要的部分,gn - rn这个想法是对每个分区中的行进行编号,同时对整个集合中的行进行编号,然后如果从另一个中减去一个,您将获得每个组的"等级".

;WITH cte as (
SELECT Messages.Message, Timestamp, 
ROW_NUMBER() OVER(PARTITION BY Message ORDER BY Timestamp) AS gn,
ROW_NUMBER() OVER (ORDER BY Timestamp) AS rn
FROM Messages
), cte2 AS (
SELECT Message, Timestamp, gn, rn, gn - rn  as gb
FROM cte 
)
SELECT * FROM cte2

Message Timestamp           gn  rn  gb
A   2015-03-29 00:00:00.000 1   1   0
A   2015-03-29 00:01:00.000 2   2   0
B   2015-03-29 00:02:00.000 1   3   -2
A   2015-03-29 00:03:00.000 3   4   -1
A   2015-03-29 00:04:00.000 4   5   -1
A   2015-03-29 00:05:00.000 5   6   -1
B   2015-03-29 00:06:00.000 2   7   -5

Answer 2

这是一个更小的解决方案：

DECLARE @t TABLE ( d DATE, m CHAR(1) )

INSERT  INTO @t
VALUES  ( '20150301', 'A' ),
        ( '20150302', 'A' ),
        ( '20150303', 'B' ),
        ( '20150304', 'A' ),
        ( '20150305', 'A' ),
        ( '20150306', 'A' ),
        ( '20150307', 'B' );

WITH 
c1 AS(SELECT d, m, IIF(LAG(m, 1, m) OVER(ORDER BY d) = m, 0, 1) AS n FROM @t),
c2 AS(SELECT m, SUM(n) OVER(ORDER BY d) AS n FROM c1) 
    SELECT m, COUNT(*) AS c
    FROM c2
    GROUP BY m, n

输出：

m   c
A   2
B   1
A   3
B   1

这个想法是1在更改消息的行中获取值：

2015-03-01  A   0
2015-03-02  A   0
2015-03-03  B   1
2015-03-04  A   1
2015-03-05  A   0
2015-03-06  A   0
2015-03-07  B   1

第二步只是当前行值+所有先前值的总和：

2015-03-01  A   0
2015-03-02  A   0
2015-03-03  B   1
2015-03-04  A   2
2015-03-05  A   2
2015-03-06  A   2
2015-03-07  B   3

通过这种方式，您可以按消息列和计算列获得分组集。