Question

我有一个记录应用程序的数据集。它记录时间以及我的小部件是否正常：

CREATE TABLE runs (time int, ok int);

INSERT INTO runs VALUES
(1, NULL),
(2, 1),
(3, 1),
(4, 1),
(5, NULL),
(6, NULL),
(7, 1),
(8, 1),
(9, NULL),
(10, 1)

我想使用窗口函数（我认为）来确定这些“ok”-ness运行的长度。所以最终数据集应如下所示：

time | ok_length
----------------
 2   |   3
 7   |   2
 10  |   1

据我所知：

SELECT
  time,
  ok,
  CASE WHEN
    LAG(ok) OVER (ORDER BY time) IS NOT null
    THEN SUM(ok) OVER (ORDER BY time) END
FROM runs
ORDER BY time

但它完全错了。有人可以帮忙吗？也许我必须在窗口函数的末尾用框架做一些事情，但是当它达到NULL时，该框架必须有条件停止。这是我正在使用的SQL小提琴：http://sqlfiddle.com/#!17/98bf4/3

Answer 1

我认为有一些方法可以简化这一点，但基于值查询的这些类型的计数总是有点冗长。主要内容是：

group_start_cte - 滞后以标记作为不同逻辑分组的开头的行。
group_cte - 累计总和，为所有行提供组ID。
group_cnt - 按逻辑分组ID计算分区。
first_time_for_group - 获取小组开头的时间。

最后我们将group_cnt和first_time_for_group放在一起：

WITH
group_start_cte AS (
SELECT
    TIME,
    ok,
    CASE
      WHEN LAG(ok) OVER (ORDER BY TIME asc) is distinct from ok
      THEN TRUE
    END AS group_start
FROM
    runs
),
group_cte AS (
SELECT
    TIME,
    ok,
    group_start,
    SUM(CASE WHEN group_start THEN 1 ELSE 0 END) OVER (ORDER BY TIME asc) AS grp_id
FROM
    group_start_cte
),
first_time_for_group as (
SELECT
    time,
    grp_id
FROM
    group_cte
WHERE
    group_start IS TRUE
),
group_cnt AS (
SELECT
    grp_id,
    count(*) AS ok_length
FROM
    group_cte
WHERE
    ok IS NOT NULL
GROUP BY
    grp_id
)
SELECT
    TIME,
    ok_length
FROM
    group_cnt
    LEFT JOIN first_time_for_group
    USING (grp_id)
ORDER BY
    time ASC
;

Answer 2

这里有一个不那么详细的解决方案：

select distinct
        min(time) over (partition by gp)
        , sum(ok) over (partition by gp)
from (
        select *
                , time - row_number() over (partition by ok order by time asc) gp
        from runs
        where ok is not null
) rs
order by 1

计算postgresql中运行的长度

2 个答案: