考虑日期顺序的 SQL 组数据

Posted 2023-03-31

【中文标题】考虑日期顺序的 SQL 组数据

【英文标题】：SQL group data taking date sequence into account

【发布时间】：2020-02-19 22:04:14

【问题描述】：

我正在尝试对可能被不同组破坏的单独行进行分组。如果它们坏了，我希望它们分开分组。

所以我有：

Col1 | Col2
---------------------
| Y  |01/JAN/2012
| Y  |01/FEB/2012
| N  |01/MAR/2012
| Y  |01/APR/2012
| Y  |01/MAY/2012

我想得到结果：

|col1|col2       |GRP
---------------------
| Y  |01/JAN/2012|1
| Y  |01/FEB/2012|1
| N  |01/MAR/2012|2
| Y  |01/APR/2012|3
| Y  |01/MAY/2012|3

我怎样才能做到这一点？

我目前的尝试是这样的：

select
    Col1,
    Col2,
    dense_rank() over (partition by Col1 order by Col2 asc) as grp  
from
    myTABLE
;

但这会将所有的“Y”组合在一起并给我一个这样的序号：

|col1|col2       |GRP
---------------------
| Y  |01/JAN/2012|1
| Y  |01/FEB/2012|2
| N  |01/MAR/2012|1
| Y  |01/APR/2012|3
| Y  |01/MAY/2012|4

【参考方案1】：

这是一种差距和孤岛问题。我建议使用行号的差异来识别“岛”，然后row_number()：

select t.*, dense_rank() over (order by grp) as grp
from (select t.*,
             min(col2) over (partition by col1, seqnum - seqnum_2) as grp
      from (select t.*,
                   row_number() over (order by col2) as seqnum,
                   row_number() over (partition by col1 order by col2) as seqnum_2
            from t
           ) t
     ) t
order by col2;

其实更简单的方法是使用lag()和一个累计和：

select t.*,
       sum(case when col1 = prev_col1 then 0 else 1 end) over (order by col2) as grp
from (select t.*, lag(col1) over (partition by col2) as prev_col1
      from t
     ) t