根据与上一个记录和下一个记录的关系选择删除记录

我有一个SQL Server 2014表，其中包含数百万个gps坐标，每个坐标都在特定时间。然而，注册之间的时间间隔并不固定，从1秒到几个小时不等。我只想每4分钟进行一次测量，因此其他记录必须删除。根据与上一个记录和下一个记录的关系选择删除记录

我试过了遍历每条记录的T-SQL中的一个WHILE循环，循环内部有一个带双CROSS APPLY的SELECT语句，如果它位于另外两条不超过4分钟的记录中，它将只返回一条记录分开。然而这个策略太慢了。

这可以通过基于集合的解决方案来完成吗？或者有没有办法加快这个查询？（下面的测试查询只是打印，还没有删除）

SELECT * INTO #myTemp FROM gps ORDER BY TimePoint asc 

declare @Id Uniqueidentifier 
declare @d1 varchar(19) 
declare @d2 varchar(19) 
declare @d3 varchar(19) 


While EXISTS (select * from #myTemp) 
BEGIN 
    select top 1 @Id = ID FROM #myTemp order by TimePoint asc 

    SELECT 
     @d1 = convert(varchar(19), a.justbefore, 121), 
     @d2 = convert(varchar(19), b.tijdstip, 121), 
     @d3 = convert(varchar(19), c.justafter, 121) 
    FROM Gps B CROSS APPLY 
     (
      SELECT top 1 TimePoint as justbefore 
      FROM Gps 
      WHERE (B.TimePoint > TimePoint) AND (B.Id = @Id) 
      ORDER by TimePoint desc 
     ) A 
     CROSS APPLY (
      SELECT top 1 TimePoint as justafter 
      FROM Gps 
      WHERE (Datediff(n,A.justbefore,TimePoint) between -4 AND 0) 
        AND (B.TimePoint < TimePoint) 
      ORDER by TimePoint asc 
     ) C 

    print 'ID=' + Cast(@id as varchar(50)) 
       + '/d1=' + @d1 + '/d2=' + @d2 + '/d3=' + @d3     

    DELETE #myTemp where Id = @id 
END

Sample data: 
    Id  TimePoint   Lat  Lon 
    1  20170725 13:05:27 12,256 24,123 
    2  20170725 13:10:27 12,254 24,120 
    3  20170725 13:10:29 12,253 24,125 
    4  20170725 13:11:55 12,259 24,127 
    5  20170725 13:11:59 12,255 24,123 
    6  20170725 13:14:28 12,254 24,126 
    7  20170725 13:16:52 12,259 24,121 
    8  20170725 13:20:53 12,257 24,125

在这种情况下，记录3,4,5应予删除。记录7应保持为7和8之间的差距超过4分钟。

来源

2017-07-26 Jan

你能发布一些示例数据和预期的结果？ –

我同意样本数据和预期的结果会使这更容易通过。但我会建议做一个搜索差距和岛屿有很多例子。诀窍是你会希望将记录分成4分钟的增量，并能够识别每个组中的第一条记录。 – Matt

看着数字...它看起来像1 & 2住（相隔5分钟）... 3，4，& 5应该去... 6次住宿（4分钟从2）... 7应该去（仅6 2分钟）和8个住宿（6 6分）...

If this is correct, the following will do what you're looking for... 


IF OBJECT_ID('tempdb..#TestData', 'U') IS NOT NULL 
DROP TABLE #TestData; 

CREATE TABLE #TestData (
    Id INT NOT NULL PRIMARY KEY CLUSTERED, 
    TimePoint DATETIME2(0) NOT NULL, 
    Lat DECIMAL(9,3), 
    Lon DECIMAL(9,3) 
    ); 

INSERT #TestData (Id, TimePoint, Lat, Lon) VALUES 
    (1, '20170725 13:05:27', 12.256, 24.123), 
    (2, '20170725 13:10:27', 12.254, 24.120), 
    (3, '20170725 13:10:29', 12.253, 24.125), 
    (4, '20170725 13:11:55', 12.259, 24.127), 
    (5, '20170725 13:11:59', 12.255, 24.123), 
    (6, '20170725 13:14:28', 12.254, 24.126), 
    (7, '20170725 13:16:52', 12.259, 24.121), 
    (8, '20170725 13:20:53', 12.257, 24.125); 

-- SELECT * FROM #TestData td; 

--================================================================================ 

WITH 
    cte_AddLag AS (
     SELECT 
      td.Id, td.TimePoint, td.Lat, td.Lon, 
      MinFromPrev = DATEDIFF(mi, LAG(td.TimePoint, 1) OVER (ORDER BY td.TimePoint), td.TimePoint) 
     FROM 
      #TestData td 
     ), 
    cte_TimeGroup AS (
     SELECT 
      *, 
      TimeGroup = ISNULL(SUM(al.MinFromPrev) OVER (ORDER BY al.TimePoint ROWS UNBOUNDED PRECEDING)/4, 0) 
     FROM 
      cte_AddLag al 
     ) 
SELECT TOP 1 WITH TIES 
    tg.Id, 
    tg.TimePoint, 
    tg.Lat, 
    tg.Lon 
FROM 
    cte_TimeGroup tg 
ORDER BY 
    ROW_NUMBER() OVER (PARTITION BY tg.TimeGroup ORDER BY tg.TimePoint);

结果...

Id   TimePoint     Lat          Lon 
----------- --------------------------- --------------------------------------- --------------------------------------- 
1   2017-07-25 13:05:27   12.256         24.123 
2   2017-07-25 13:10:27   12.254         24.120 
6   2017-07-25 13:14:28   12.254         24.126 
8   2017-07-25 13:20:53   12.257         24.125

HTH，杰森

来源

2017-07-26 02:50:42

理论上，第7点应该留下来，目标是尽可能接近4分钟的距离。通过消除第7点，点6和点8之间的间隔变得超过6分钟。 – Jan

有没有这样做的有效方法。 –

根据与上一个记录和下一个记录的关系选择删除记录

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