如何解决通过随时间变化的状态对数据进行分组
我试图将组号分配给数据随时间变化的数据集中不同的行组。在我的示例中，更改的字段是tran_seq，prog_id，deg-id，cur_id和enroll_status。当这些字段中的任何一个与上一行不同时，我需要一个新的分组号。如果字段与上一行相同，则分组号应保持不变。当我尝试ROW_NUMBER（），RANK（）或DENSE_RANK（）时，我得到了同一组的增加值（例如，示例中的前2行）。我觉得我需要对start_date进行排序，因为它是时间数据。
+----+----------+---------+--------+--------+---------------+------------+------------+---------+
| | tran_seq | prog_id | deg_id | cur_id | enroll_status | start_date | end_date | desired |
+----+----------+---------+--------+--------+---------------+------------+------------+---------+
| 1 | 1 | 6 | 9 | 3 | ENRL | 2004-08-22 | 2004-12-11 | 1 |
| 2 | 1 | 6 | 9 | 3 | ENRL | 2006-01-10 | 2006-05-06 | 1 |
| 3 | 1 | 6 | 9 | 59 | ENRL | 2006-08-29 | 2006-12-16 | 2 |
| 4 | 2 | 12 | 23 | 45 | ENRL | 2014-01-21 | 2014-05-16 | 3 |
| 5 | 2 | 12 | 23 | 45 | ENRL | 2014-08-18 | 2014-12-05 | 3 |
| 6 | 2 | 12 | 23 | 45 | LOAP | 2015-01-20 | 2015-05-15 | 4 |
| 7 | 2 | 12 | 23 | 45 | ENRL | 2015-08-25 | 2015-12-11 | 5 |
| 8 | 2 | 12 | 23 | 45 | LOAP | 2016-01-12 | 2016-05-06 | 6 |
| 9 | 2 | 12 | 23 | 45 | ENRL | 2016-05-16 | 2016-08-05 | 7 |
| 10 | 2 | 12 | 23 | 45 | LOAJ | 2016-08-23 | 2016-12-02 | 8 |
| 11 | 2 | 12 | 23 | 45 | ENRL | 2017-01-18 | 2017-05-05 | 9 |
| 12 | 2 | 12 | 23 | 45 | ENRL | 2018-01-17 | 2018-05-11 | 9 |
+----+----------+---------+--------+--------+---------------+------------+------------+---------+
一旦我将数字分组，我想我可以将它们分组以获取最终的目标：带有开始日期和结束日期的不同状态的时间表。对于上面的示例数据，应为：
+---+----------+---------+--------+--------+---------------+------------+------------+
| | tran_seq | prog_id | deg_id | cur_id | enroll_status | start_date | end_date |
+---+----------+---------+--------+--------+---------------+------------+------------+
| 1 | 1 | 6 | 9 | 3 | ENRL | 2004-08-22 | 2006-05-06 |
| 2 | 1 | 6 | 9 | 59 | ENRL | 2004-08-29 | 2006-12-16 |
| 3 | 2 | 12 | 23 | 45 | ENRL | 2014-01-21 | 2014-12-05 |
| 4 | 2 | 12 | 23 | 45 | LOAP | 2015-01-20 | 2015-05-15 |
| 5 | 2 | 12 | 23 | 45 | ENRL | 2015-08-25 | 2015-12-11 |
| 6 | 2 | 12 | 23 | 45 | LOAP | 2016-01-12 | 2016-05-06 |
| 7 | 2 | 12 | 23 | 45 | ENRL | 2016-05-16 | 2016-08-05 |
| 8 | 2 | 12 | 23 | 45 | LOAJ | 2016-08-23 | 2016-12-02 |
| 9 | 2 | 12 | 23 | 45 | ENRL | 2017-01-17 | 2018-05-06 |
+---+----------+---------+--------+--------+---------------+------------+------------+
解决方法
这是一个典型的XY问题，因为您要在中间解决方案中寻求其他解决方案，而不是在询问解决方案本身。
但是，由于您将总体最终目标作为附录进行了补充，因此您可以在没有中间步骤的情况下实现此目标：
declare @t table(tran_seq int,prog_id int,deg_id int,cur_id int,enroll_status varchar(4),start_date date,end_date date,desired int)
insert into @t values
(1,6,9,3,'ENRL','2004-08-22','2004-12-11',1),(1,'2006-01-10','2006-05-06',59,'2006-08-29','2006-12-16',2),(2,12,23,45,'2014-01-21','2014-05-16',3),'2014-08-18','2014-12-05','LOAP','2015-01-20','2015-05-15',4),'2015-08-25','2015-12-11',5),'2016-01-12','2016-05-06',6),'2016-05-16','2016-08-05',7),'LOAJ','2016-08-23','2016-12-02',8),'2017-01-18','2017-05-05',9),'2018-01-17','2018-05-11',9)
;
select tran_seq,prog_id,deg_id,cur_id,enroll_status,min(start_date) as start_date,max(end_date) as end_date
from(select *,row_number() over (order by end_date) - row_number() over (partition by tran_seq,enroll_status order by end_date) as grp
from @t
) AS g
group by tran_seq,grp
order by start_date;
输出
+----------+---------+--------+--------+---------------+------------+------------+
| tran_seq | prog_id | deg_id | cur_id | enroll_status | start_date | end_date |
+----------+---------+--------+--------+---------------+------------+------------+
| 1 | 6 | 9 | 3 | ENRL | 2004-08-22 | 2006-05-06 |
| 1 | 6 | 9 | 59 | ENRL | 2006-08-29 | 2006-12-16 |
| 2 | 12 | 23 | 45 | ENRL | 2014-01-21 | 2014-12-05 |
| 2 | 12 | 23 | 45 | LOAP | 2015-01-20 | 2015-05-15 |
| 2 | 12 | 23 | 45 | ENRL | 2015-08-25 | 2015-12-11 |
| 2 | 12 | 23 | 45 | LOAP | 2016-01-12 | 2016-05-06 |
| 2 | 12 | 23 | 45 | ENRL | 2016-05-16 | 2016-08-05 |
| 2 | 12 | 23 | 45 | LOAJ | 2016-08-23 | 2016-12-02 |
| 2 | 12 | 23 | 45 | ENRL | 2017-01-18 | 2018-05-11 |
+----------+---------+--------+--------+---------------+------------+------------+
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