Question

我在表格中有这种数据。

我需要以某种方式得到所有行而不是间隔为5分钟的行为了减少它们的数量。

是否有可能以简单的方式使用T-SQL？

ID    AtTime                  Speed  Label 
------------------------------------------------------------------------- 
217   2017-06-06 08:01:01.000   0    Lat: 54.3956 Lon: 86.79349
217   2017-06-06 08:01:23.000   0    Lat: 54.3956 Lon: 86.7935
221   2017-06-06 08:04:20.000   0    Lat: 54.39548 Lon: 86.79372
217   2017-06-06 08:06:24.000   0    Lat: 54.39559 Lon: 86.79347
221   2017-06-06 08:09:21.000   0    Lat: 54.39548 Lon: 86.79372
217   2017-06-06 08:11:25.000   0    Lat: 54.3956 Lon: 86.79346
221   2017-06-06 08:12:21.000   0    Lat: 54.39526 Lon: 86.79405
221   2017-06-06 08:12:30.000   0    Lat: 54.39507 Lon: 86.79451
221   2017-06-06 08:12:36.000   14,4    Lat: 54.39503 Lon: 86.79493
221   2017-06-06 08:12:47.000   10,8    Lat: 54.39518 Lon: 86.79536
221   2017-06-06 08:12:56.000   7,2  Lat: 54.39527 Lon: 86.79578
221   2017-06-06 08:13:06.000   7,2  Lat: 54.39529 Lon: 86.79622
221   2017-06-06 08:14:10.000   0    Lat: 54.39545 Lon: 86.79621

Answer 1

我希望这符合您的需求 - 我们首先创建一个CTE，用于识别数据集中的每个五分钟块。然后，我们使用第二个CTE来选择每个块中最早的行。我不确定ID是否应该用作第二个分区标准，但我已经标记了可以添加的位置：

declare @t table (ID int not null, AtTime datetime not null, Speed decimal(9,4)  not null,
                  Label varchar(29) not null)
insert into @t(ID,AtTime,Speed,Label) values
(217,'2017-06-06T08:01:01.000',0   ,'Lat: 54.3956 Lon: 86.79349 '),
(217,'2017-06-06T08:01:23.000',0   ,'Lat: 54.3956 Lon: 86.7935  '),
(221,'2017-06-06T08:04:20.000',0   ,'Lat: 54.39548 Lon: 86.79372'),
(217,'2017-06-06T08:06:24.000',0   ,'Lat: 54.39559 Lon: 86.79347'),
(221,'2017-06-06T08:09:21.000',0   ,'Lat: 54.39548 Lon: 86.79372'),
(217,'2017-06-06T08:11:25.000',0   ,'Lat: 54.3956 Lon: 86.79346 '),
(221,'2017-06-06T08:12:21.000',0   ,'Lat: 54.39526 Lon: 86.79405'),
(221,'2017-06-06T08:12:30.000',0   ,'Lat: 54.39507 Lon: 86.79451'),
(221,'2017-06-06T08:12:36.000',14.4,'Lat: 54.39503 Lon: 86.79493'),
(221,'2017-06-06T08:12:47.000',10.8,'Lat: 54.39518 Lon: 86.79536'),
(221,'2017-06-06T08:12:56.000',7.2 ,'Lat: 54.39527 Lon: 86.79578'),
(221,'2017-06-06T08:13:06.000',7.2 ,'Lat: 54.39529 Lon: 86.79622'),
(221,'2017-06-06T08:14:10.000',0   ,'Lat: 54.39545 Lon: 86.79621')

;With Times as (
    select distinct u.StartBlock,DATEADD(minute,5,u.StartBlock) as EndBlock
    from @t
        cross apply
        (select DATEADD(minute,((DATEDIFF(minute,0,AtTime)/5)*5),0) as StartBlock) u
), Ordered as (
    select
        *,
        ROW_NUMBER() OVER (PARTITION BY StartBlock /* And ID? */ ORDER BY AtTime) as rn
    from
        @t t
            inner join
        Times tm
            on
                tm.StartBlock <= t.AtTime and
                t.AtTime < tm.EndBlock
)
select *
from Ordered
where rn = 1

结果：

ID          AtTime                  Speed   Label                         StartBlock              EndBlock                rn
----------- ----------------------- ------- ----------------------------- ----------------------- ----------------------- --
217         2017-06-06 08:01:01.000 0.0000  Lat: 54.3956 Lon: 86.79349    2017-06-06 08:00:00.000 2017-06-06 08:05:00.000 1
217         2017-06-06 08:06:24.000 0.0000  Lat: 54.39559 Lon: 86.79347   2017-06-06 08:05:00.000 2017-06-06 08:10:00.000 1
217         2017-06-06 08:11:25.000 0.0000  Lat: 54.3956 Lon: 86.79346    2017-06-06 08:10:00.000 2017-06-06 08:15:00.000 1

请注意，这并不能保证所有行至少相隔5分钟。在病理情况下，您可能有两行，实际上只是间隔时间（例如，如果特定的5分钟间隔只有一行，并且它在该间隔的最后一个可能时刻，并且下一个间隔有一行这恰好发生在起点）。但是，对于正常的数据分布，数据平均相隔五分钟。

T-SQL如何从设置的分钟数中获取每5分钟的行数

1 个答案: