按时间间隔划分

时间:2018-04-14 11:25:14

标签: sql sql-server datetime window-functions

我正在寻找通过datetime值在分区上运行窗口函数的最佳方法。然而,不是按照确切的时间进行分区,而是希望按日期时间进行分区,例如在彼此的15分钟内。

这是我桌子的一小部分。

CREATE TABLE my_table(ID VARCHAR(5), in_time DATETIME)

INSERT INTO my_table (ID, in_time) VALUES
('4844', '2017-04-06 10:15:00.000'),
('5221', '2017-11-24 11:18:00.000'),
('5221', '2017-11-24 11:18:00.000'),
('5221', '2017-11-25 14:23:00.000'),
('8486', '2017-10-10 15:30:00.000'),
('8486', '2017-10-10 15:32:00.000'),
('8486', '2017-10-10 15:46:00.000'), -- new row after updating question
('8486', '2017-10-10 16:00:00.000') -- new row after updating question

以下是我现在使用的查询:

SELECT *, 
    ROW_NUMBER() OVER(PARTITION BY ID, in_time ORDER BY ID, in_time) AS filter_row
FROM my_table

正如预期的那样给了我:

ID      in_time                     filter_row
4844    2017-04-06 10:15:00.000     1
5221    2017-11-24 11:18:00.000     1
5221    2017-11-24 11:18:00.000     2
5221    2017-11-25 14:23:00.000     1
8486    2017-10-10 15:30:00.000     1
8486    2017-10-10 15:32:00.000     1
8486    2017-10-10 15:46:00.000     1
8486    2017-10-10 16:00:00.000     1

我想要达到的目的是:

ID      in_time                     filter_row
4844    2017-04-06 10:15:00.000     1
5221    2017-11-24 11:18:00.000     1
5221    2017-11-24 11:18:00.000     2
5221    2017-11-25 14:23:00.000     1
8486    2017-10-10 15:30:00.000     1
8486    2017-10-10 15:32:00.000     2 -- < notice the 2 here
8486    2017-10-10 15:46:00.000     3 -- < notice the 3 here
8486    2017-10-10 16:00:00.000     4 -- < notice the 4 here

如上所示,ID = 8486的行应该分区在一起,因为它们各自in_timein_time之间的行只有2分钟,14分钟和14分钟。如何有效地做到这一点?

1 个答案:

答案 0 :(得分:2)

以下示例通过基于指定的间隔(以分钟为单位)计算间隔开始时间并按该值进行分区来提供所需的结果。

DECLARE @IntervalMinutes int = 15;
SELECT *, 
    ROW_NUMBER() OVER(
          PARTITION BY ID
        , (DATEADD(minute, (DATEDIFF(minute, '', in_time)/@IntervalMinutes)*@IntervalMinutes, '')
        )
        ORDER BY ID, in_time) AS filter_row
FROM my_table;

修改

上述代码计算固定长度间隔。您可以通过ID标识超出所需间隔的岛屿来解决您更新的问题。以下方法使用NOT EXISTSCROSS APPLY来识别这些岛屿,并确定每个岛屿的间隔开始和结束时间。

DECLARE @IntervalMinutes int = 15;
WITH
    start_intervals AS (
        SELECT DISTINCT
              ID
            , in_time
        FROM dbo.my_table AS a
        WHERE NOT EXISTS(
            SELECT 1
            FROM dbo.my_table AS b
            WHERE
                b.ID = a.ID
                AND b.in_time < a.in_time
                AND b.in_time > DATEADD(minute, -@IntervalMinutes, a.in_time)
            )
        )
    , end_intervals AS (
        SELECT
              ID
            , in_time
        FROM dbo.my_table AS a
        WHERE NOT EXISTS(
            SELECT 1
            FROM dbo.my_table AS b
            WHERE
                b.ID = a.ID
                AND b.in_time > a.in_time
                AND b.in_time < DATEADD(minute, @IntervalMinutes, a.in_time)
            )
    )
    , intervals AS (
        SELECT
              ID
            , start_intervals.in_time AS start_interval
            , end_intervals.in_time AS end_interval
        FROM start_intervals
        CROSS APPLY(
            SELECT TOP(1) in_time
            FROM end_intervals 
            WHERE
                end_intervals.ID = start_intervals.ID
                AND end_intervals.in_time >= start_intervals.in_time
            ) AS end_intervals
        )
SELECT 
      my_table.ID
    , my_table.in_time
    , ROW_NUMBER() OVER(PARTITION BY my_table.ID, intervals.start_interval ORDER BY(intervals.start_interval)) AS filter_row
FROM dbo.my_table
JOIN intervals ON my_table.in_time BETWEEN intervals.start_interval AND intervals.end_interval