循环组模式中的密集排名

Question

假设我有一张类似的桌子

store	date	is_open
Bay	1/1/2022	true
Bay	1/2/2022	true
Bay	1/3/2022	true
Bay	1/4/2022	false
Bay	1/5/2022	false
Bay	1/6/2022	false
Bay	1/7/2022	true
Bay	1/8/2022	true
Bay	1/9/2022	true
Walmart	1/7/2022	true
Walmart	1/8/2022	false
Walmart	1/9/2022	true

我希望他们使用 partition by 并获得组的排名，例如

store	date	is_open	group
Bay	1/1/2022	true	1
Bay	1/2/2022	true	1
Bay	1/3/2022	true	1
Bay	1/4/2022	false	2
Bay	1/5/2022	false	2
Bay	1/6/2022	false	2
Bay	1/7/2022	true	3
Bay	1/8/2022	true	3
Bay	1/9/2022	true	3
Walmart	1/7/2022	true	1
Walmart	1/8/2022	false	2
Walmart	1/9/2022	true	3

我开始尝试按store 和is_open 进行分区，但真的很困惑在 order by 子句中使用什么，任何帮助将不胜感激。

Answer 1

您可以使用 LAG() 来检测组的开始。

with cte AS (
    SELECT t.*, case when lag(is_open) OVER (PARTITION BY store ORDER BY date) = is_open then 0 else 1 end sflag
    FROM yourTable t
)
SELECT store, date, is_open, sum(sflag) over(PARTITION BY store ORDER BY date) grp
FROM cte
ORDER BY store, date;

db<>fiddle

Answer 2

这实际上是一个空白和孤岛问题。一种方法是使用行数差异法：

WITH cte AS (
    SELECT t.*, ROW_NUMBER() OVER (PARTITION BY store ORDER BY date) rn1,
                ROW_NUMBER() OVER (PARTITION BY store, is_open ORDER BY date) rn2
    FROM yourTable t
),
cte2 AS (
    SELECT t.*, MIN(date) OVER (PARTITION BY store, is_open, rn1 - rn2) AS min_date
    FROM cte t
)

SELECT store, date, is_open,
       DENSE_RANK() OVER (PARTITION BY store ORDER BY rn1 - rn2, min_date) "group"
FROM cte2
ORDER BY store, date;

请注意，我们在此处使用第二个 CTE cte2 来查找每个岛屿的最小日期值。这样做是为了将两个岛与不同的is_open 值（真/假）区分开来，这些值恰好在行号上有相同的差异。保证行数相同的情况下，先上报较早的岛屿。

Demo