时间序列数据上的岛屿和差距问题

Question

我有一个示例数据，我一直在尝试获取所需的数据，如下所示。我也能够实施某种岛屿和差距解决方案。这是我达到的最接近预期数据的版本。

DROP TABLE IF EXISTS #data
CREATE TABLE #data(
    factor varchar(50),
    val int,
    [start_date] date, [end_date] date
)
Go
INSERT INTO #data VALUES
('a', 15, '2021-01-01', '2021-01-05'),
('a', 15, '2021-01-08', '2021-01-10'),
('a', 20, '2021-01-11', '2021-01-20'),
('a', 15, '2021-01-21', '2099-01-01'),
('b', 10, '2021-01-01', '2021-01-04'),
('b', 12, '2021-01-05', '2021-01-13'),
('b', 12, '2021-01-17', '2021-01-19'),
('b', 12, '2021-01-20', '2021-01-23'),
('b', 10, '2021-01-24', '2099-01-01');

WITH value_cte As (
    SELECT * ,
    RANK() OVER(PARTITION BY factor ORDER BY [start_date]) - RANK() OVER(PARTITION BY factor, val ORDER BY [start_date]) grp
    FROM #data

)
SELECT factor, val, MIN(start_date) st, MAX(end_date) ed
FROM value_cte
GROUP BY factor, val, grp
ORDER BY factor, st

上述查询的结果：

预期结果：

factor  val st          ed
a       15  2021-01-01  2021-01-05
a       15  2021-01-08  2021-01-10
a       20  2021-01-11  2021-01-20
a       15  2021-01-21  2099-01-01
b       10  2021-01-01  2021-01-04
b       12  2021-01-05  2021-01-13
b       12  2021-01-17  2021-01-23
b       10  2021-01-24  2099-01-01

即使两个连续岛的值相同并且存在间隙，则不应合并该间隙，如果两个岛是连续的，则应将它们合并。不幸的是，我无法在此处更改源（示例数据结构）

Answer 1

您可以使用 lag() 来确定“孤岛”的起点——即没有重叠的地方。然后使用基于日期算术的累计和：

select factor, val, min(start_date), max(end_date)
from (select d.*,
             sum(case when prev_end_date >= dateadd(day, -1, start_date) then 0 else 1 end) over (partition by factor, val order by start_date) as grp
      from (select d.*,
                   lag(end_date) over (partition by factor, val order by start_date) as prev_end_date
            from data d
           ) d
     ) d
group by factor, val, grp
order by factor, min(start_date);

Here 是 SQL Fiddle.

时间序列数据上的岛屿和差距问题

Islands and gaps problem on time-series data

sql

sql-server

gaps-and-islands