行集总，周期日期答案

【问题标题】：Row lumping, cycle dates行集总，周期日期
【发布时间】：2019-10-16 17:11:12
【问题描述】：

我想查看lead 类型，如果该类型与该行相同，则合并这些日期以适合一行。

我有下表：

id    start_dt     end_dt    type
 1     1/1/19      2/21/19   cross
 1     2/22/19     6/5/19    cross
 1     6/6/19      8/31/19   cross
 1     9/1/19      10/3/19   AAAA
 1     10/4/19     10/4/19   cross
 1     10/5/19     10/6/19   AAAA
 1     10/7/19     10/10/19  AAAA
 1     10/11/19    12/31/99  cross

预期结果：

 id    start_dt    end_dt    type
 1     1/1/19      8/31/19   cross
 1     9/1/19      10/3/19   AAAA
 1     10/4/19     10/4/19   cross
 1     10/5/19     10/10/19  AAAA
 1     10/11/19    12/31/99  cross

我怎样才能让我的输出看起来像预期的结果？

我已经用lead lag rank 和 case expression 进行了测试，但没有什么值得在这里添加的。我走对了吗？

【问题讨论】：

标签： sql oracle date cycle gaps-and-islands

【解决方案1】：

这是一个gaps-and-islands 问题。通过row_number()分析函数的贡献解决它的一种选择：

select min(start_dt) as startdate, max(end_dt) as enddate, type
  from
  (
   with t(id, start_dt, end_dt,type) as
   (
    select 1, date'2019-01-01', date'2019-02-21', 'cross' from dual union all
    select 1, date'2019-02-22', date'2019-06-05', 'cross'  from dual union all
    select 1, date'2019-06-06', date'2019-08-31', 'cross'  from dual union all  
    select 1, date'2019-09-01', date'2019-10-03', 'AAAA'  from dual union all
    select 1, date'2019-09-04', date'2019-10-04', 'cross'  from dual union all   
    select 1, date'2019-10-05', date'2019-10-06', 'AAAA'  from dual union all
    select 1, date'2019-10-07', date'2019-10-10', 'AAAA'  from dual union all  
    select 1, date'2019-10-11', date'2019-12-31', 'cross'  from dual 
   )
   select type, 
          row_number() over (partition by id, type order by end_dt) as rn1,
          row_number() over (partition by id order by end_dt) as rn2,
          start_dt, end_dt
     from t
   ) tt
 group by type, rn1 - rn2
 order by enddate;

STARTDATE   ENDDATE     TYPE
---------   ---------   -----
01-JAN-19   31-AUG-19   cross
01-SEP-19   03-OCT-19   AAAA
04-SEP-19   04-OCT-19   cross
05-OCT-19   10-OCT-19   AAAA
11-OCT-19   31-DEC-19   cross

Demo

【讨论】：

【解决方案2】：

我实际上认为这是 Oracle 模式匹配功能的一个很好的案例。

with t(id, start_dt, end_dt,type) as
   (
    select 1, date'2019-01-01', date'2019-02-21', 'cross' from dual union all
    select 1, date'2019-02-22', date'2019-06-05', 'cross'  from dual union all
    select 1, date'2019-06-06', date'2019-08-31', 'cross'  from dual union all  
    select 1, date'2019-09-01', date'2019-10-03', 'AAAA'  from dual union all
    select 1, date'2019-09-04', date'2019-10-04', 'cross'  from dual union all   
    select 1, date'2019-10-05', date'2019-10-06', 'AAAA'  from dual union all
    select 1, date'2019-10-07', date'2019-10-10', 'AAAA'  from dual union all  
    select 1, date'2019-10-11', date'2019-12-31', 'cross'  from dual 
   )
SELECT *
FROM t
MATCH_RECOGNIZE(ORDER BY start_dt
                MEASURES a.id AS ID,
                         A.start_dt AS START_DT,
                         NVL(LAST(B.end_dt), A.end_dt) AS END_DT,
                         a.type AS TYPE
                PATTERN (A B*)
                DEFINE B AS start_dt > PREV(start_dt) AND type = PREV(type));

有关该主题的详细入门可以找到here

【讨论】：

另外，如果您需要对所有 ID 执行此操作，您只需在 MATCH_RECOGNIZE 命令的开头添加“PARTITION BY id”即可。（并从 MEASURES 子句中删除“a.id AS ID”）

【解决方案3】：

如果您想查看相邻行以找到可以组合的组，那么我建议 lag() 查找组的开始位置和累积总和：

select id, type, min(start_dt), max(end_dt)
from (select t.*,
             sum(case when prev_end_dt >= start_dt - 1 then 0 else 1 end) over (partition by id, type order by start_dt) as grp
      from (select t.*,
                   lag(end_dt) over (partition by id, type order by start_dt) as prev_end_dt
            from t
           ) t
     ) t
group by id, type, grp
order by id, min(start_dt);

特别是，这将发现type 没有改变但时间范围内存在间隙的情况，如this db<>fiddle 所示，id = 2。

【讨论】：