时间序列数据上的孤岛和间隙问题

Question

我有一个样本数据，我一直在尝试获取所需的数据，如下所示。我也能够实现某种类型的岛和差距解决方案。这是我接近预期数据的最接近的版本。

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

以上查询的结果：

​

​

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

即使两个连续的岛的值是相同的，并且存在间隙，则不应合并该间隙，如果两个岛是连续的，则应合并它们。不幸的是，我不能在这里更改源代码(示例数据结构)

Answer 1

您可以使用lag()来确定“岛”的起点--也就是没有重叠的地方。然后使用基于日期算法的累积和：

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是一个SQL。