如何找到最连续的周

Question

我的数据：

INSERT INTO martin_test (id, user_id, created_at) VALUES
  ('1bb20295-fd7b-4918-a496-313e5babd482', 'abc', '2024-01-04 15:54:51')
, ('08565423-3371-4720-abb3-80c7aef8333d', 'abc', '2024-01-11 15:54:51')
, ('b17443fe-5a4f-4b7c-934d-3d2910a65f44', 'abc', '2024-01-18 15:54:51')
, ('3d267dc3-ee86-44e9-b1fe-fd918a64b77c', 'abc', '2024-02-01 15:54:51')
, ('d28d73d3-bc9c-4192-998a-a6bafce604a5', 'abc', '2024-02-08 15:54:51')
, ('401d38f8-d277-4605-af33-b4a9bd2eef25', 'abc', '2024-02-22 15:54:51')
, ('b804fa29-23af-4d93-a5c9-187767fec3c9', 'abc', '2024-02-29 15:54:51')
;

我的询问：

SELECT *,CASE WHEN gap_weeks > 1 THEN 1 END, avg(CASE WHEN gap_weeks > 1 THEN 7 END) OVER (PARTITION BY user_id ORDER BY week_number) AS grp_avg,
    sum(CASE WHEN gap_weeks > 1 THEN 6 END) OVER (PARTITION BY user_id ORDER BY week_number) AS grp_sum,

           COUNT(CASE WHEN gap_weeks > 1 THEN 1 END) OVER (PARTITION BY user_id ORDER BY week_number) AS grp
    FROM (
      SELECT user_id,
             week_number,
             lag(week_number) OVER (PARTITION BY user_id ORDER BY week_number) AS pre_week_number, 
             week_number - lag(week_number) OVER (PARTITION BY user_id ORDER BY week_number) AS gap_weeks
      FROM (
        select distinct EXTRACT(
              WEEK 
              FROM 
                CAST(created_at AS DATE)
            ) AS week_number ,user_id  from martin_test ) AS a
    ) AS subquery1

为什么abc的grp是'0.1.2.3'？我不知道为什么下一个gap_weeks > 1然后grp将被添加1），并且count＆avg＆sum（case when）over（PARTITION by）如何工作？

我的要求是：
“找到最连续的几周！”

我找到了这篇相关博客文章。

Answer 1

我专注于你宣布的目标：

找到最连续的几周！

可以这样做：

SELECT dense_rank() OVER (ORDER BY count(*) DESC) AS rank
     , user_id
     , count(*) AS consecutive_weeks
     , concat_ws(' - ', min(week_nr), max(week_nr)) AS week_range
FROM (
   SELECT *
        , week_nr - row_number() OVER (PARTITION BY user_id ORDER BY week_nr) AS grp
   FROM  (
      SELECT DISTINCT user_id, extract(week FROM created_at)::int AS week_nr
      FROM   martin_test
      ORDER  BY 1,2
      ) sub
   ) sub1
GROUP  BY user_id, grp
ORDER  BY consecutive_weeks DESC;

小提琴

连续周数将与同一分区上的

row_number()

结果同步增加。如果您从实际周数 (

week_nr

) 中减去该行号，则连续周将获得相同的

grp

数字。然后就可以聚合了。

相关：

• 选择最长的连续序列