Question

我试图将一个数据集放在一起为GIS应用程序提供数据，并寻找一个SQL Guru来帮助我避免使用游标和临时表，因为我们有大约3000万条记录要处理（I＆＃39; m使用SQL Server 2012）。数据集的要点包括VehicleID，PositionTime，纬度，经度和速度。我们需要通过VehicleID和PositionTime排序记录，以便映射Lat / Long点和跟踪运动。如果速度大于或等于0.2，那么它是可接受的记录。但是，如果速度小于0.2，则认为它不移动并需要特殊处理。在下面的示例中，对于记录一和二，我需要平均纬度，长度和速度，并保持最大位置时间并将其放入单个记录中。第三和第四条记录都很好。记录5和6也需要合并为一个记录，因此车辆1将从6个记录变为4个记录。在记录＆＃34;标准化＆＃34;之后，我们需要获得每个车辆的记录和位置时间之间的日期差异。对于VehicleID 2，我们将重新开始相同的过程（我们有大约7,000辆车）。

VehicleID  PositionTime         Latitude    Longitude    Speed
1          11/20/2013 18:09:27  29.54608    -95.04444    0.1
1          11/20/2013 18:47:35  29.54608    -95.04444    0
1          11/20/2013 20:34:45  29.546105   -95.04442    5
1          11/20/2013 20:46:44  29.54607833 -95.04443167 3
1          11/20/2013 21:01:44  29.54606667 -95.04442833 0
1          11/20/2013 21:16:43  29.546095   -95.04443167 0.1
2          11/20/2013 21:31:44  29.54609    -95.04441    5
2          11/20/2013 21:46:44  29.54607667 -95.04443    0

Answer 1

我认为你要做的就是聚合缓慢/不动的车辆

这有帮助吗？

Create or replace view vehicles_view as 
 select VehicleId, 
        Max(PositionTime) as PositionTime, 
        avg(Latitude) as Latitude, 
        Avg(Longitude) as Longitude, 
        avg(Speed) as speed
 from vehicles
 where Speed < 0.2
 group by vehicleId

union

 select VehicleId, 
  PositionTime , 
  Latitude, 
  Longitude, 
  Speed 
 from vehicles
 where Speed >= 0.2

Answer 2

好的，这是一些代码：

DECLARE @t TABLE
    (
      VehicleID INT ,
      PositionTime DATETIME ,
      Latitude DECIMAL(20, 10) ,
      Longitude DECIMAL(20, 10) ,
      Speed DECIMAL(20, 10)
    )

INSERT  INTO @t
VALUES  ( 1, '11/20/2013 18:09:27', 29.54608, -95.04444, 0.1 ),
        ( 1, '11/20/2013 18:47:35', 29.54608, -95.04444, 0 ),
        ( 1, '11/20/2013 20:34:45', 29.546105, -95.04442, 5 ),
        ( 1, '11/20/2013 20:46:44', 29.54607833, -95.04443167, 3 ),
        ( 1, '11/20/2013 21:01:44', 29.54606667, -95.04442833, 0 ),
        ( 1, '11/20/2013 21:16:43', 29.546095, -95.04443167, 0.1 ),
        ( 2, '11/20/2013 21:31:44', 29.54609, -95.04441, 5 ),
        ( 2, '11/20/2013 21:46:44', 29.54607667, -95.04443, 0 );
WITH    cte1
          AS ( SELECT   VehicleID ,
                        PositionTime ,
                        Latitude ,
                        Longitude ,
                        CASE WHEN Speed <= 0.2 THEN 0
                             ELSE Speed
                        END AS Speed
               FROM     @t
             ),
        cte2
          AS ( SELECT   * ,
                        SUM(Speed) OVER ( PARTITION BY VehicleID ORDER BY PositionTime ) AS s
               FROM     cte1
             ),
        cte3
          AS ( SELECT   * ,
                        RANK() OVER ( PARTITION BY VehicleID ORDER BY Speed, s ) AS r
               FROM     cte2
             ),
        cte4
          AS ( SELECT   VehicleID ,
                        MAX(PositionTime) AS PositionTime ,
                        AVG(Latitude) AS Latitude ,
                        AVG(Longitude) AS Longitude ,
                        MAX(Speed) AS Speed
               FROM     cte3
               GROUP BY VehicleID ,
                        r
             )
    SELECT  * ,
            DATEDIFF(ss,
                     LAG(PositionTime) OVER ( PARTITION BY VehicleID ORDER BY PositionTime ),
                     PositionTime) AS DiffInSeconds
    FROM    cte4
    ORDER BY PositionTime

输出：

VehicleID   PositionTime            Latitude         Longitude      Speed           DiffInSeconds
1           2013-11-20 18:47:35.000 29.5460800000   -95.0444400000  0.0000000000    NULL
1           2013-11-20 20:34:45.000 29.5461050000   -95.0444200000  5.0000000000    6430
1           2013-11-20 20:46:44.000 29.5460783300   -95.0444316700  3.0000000000    719
1           2013-11-20 21:16:43.000 29.5460808350   -95.0444300000  0.0000000000    1799
2           2013-11-20 21:31:44.000 29.5460900000   -95.0444100000  5.0000000000    NULL
2           2013-11-20 21:46:44.000 29.5460766700   -95.0444300000  0.0000000000    900

让我们来看看吧。在cte1中，我只是将Speed <=0.2的所有值都转换为0 在cte2我汇总了Speed个值，因此我在Speed = 0

中得到相同的值

Speed           s
0.0000000000    0.0000000000
0.0000000000    0.0000000000
5.0000000000    5.0000000000
3.0000000000    8.0000000000
0.0000000000    8.0000000000
0.0000000000    8.0000000000
5.0000000000    5.0000000000
0.0000000000    5.0000000000

然后在cte3我将排名功能应用于Speed和增量总和s的组合。所以我得到了：

VehicleID   Speed           s               r
1           0.0000000000    0.0000000000    1
1           0.0000000000    0.0000000000    1
1           0.0000000000    8.0000000000    3
1           0.0000000000    8.0000000000    3
1           3.0000000000    8.0000000000    5
1           5.0000000000    5.0000000000    6
2           0.0000000000    5.0000000000    1
2           5.0000000000    5.0000000000    2

请注意，r的{{1}}为Speed，因此您可以VehicleID和r在此处应用分组。我在cte4中执行该操作以获取不同的行，并在日期和Max上应用汇总，如lat和long。{/ p>

最后，从cte4中选择了使用Avg函数计算当前行与行之间的差异。

避免SQL Server中的光标

2 个答案: