在 SQL 表上查找每个参与者的速度

Question

我有一个关于几个人的动作 (people_movement) 的 SQL 表，如下所示：

Name	Time	Position Increment (m)
Brittany	10.20	-75
Brittany	10.30	800
Brittany	10.45	800
Brittany	11.00	900
Alan	10.15	400
Alan	10.16	-700
Alan	1020	800
Casey	9.50	800
Casey	10.01	-400
Casey	10.05	200

注意记录了时间，所以对于第一条记录，开始时间是未知的，但是以一行的时间作为下一行的开始时间。 有成千上万的参与者，行数不同，有些参与者有 4 或 5 行，但许多参与者有 100 行。 我需要计算他们的移动速度（米/分钟），所以表格如下所示：

Name	Start Time	End Time	Position Increment (m)	Increment Speed (m/min)
Brittany	-	10.20	-75	-
Brittany	10.20	10.30	800	80
Brittany	10.30	11.45	800	53.33
Brittany	10.45	11.00	900	60
Alan	-	10.15	400	-
Alan	10.15	10.16	-700	-700
Alan	10.16	10.20	800	200
Casey	-	09.50	800	-
Casey	09.50	10.01	-400	-0.18
Casey	10.01	10.05	200	50

请注意，增量速度只是（位置增量/（开始时间 - 结束时间）），其中结束时间就是每个参与者（姓名）下一次移动的时间。

有没有办法进行这样的计算？

谢谢

Answer 1

使用自联接。

CREATE TABLE speed_test(name varchar(10), time int, position int);
mysql> INSERT INTO speed_test VALUES ('Brittany', 1020, -75);
Query OK, 1 row affected (0,01 sec)

mysql> INSERT INTO speed_test VALUES ('Brittany', 1030, 800);
Query OK, 1 row affected (0,00 sec)

mysql> INSERT INTO speed_test VALUES ('Brittany', 1045, 800);
Query OK, 1 row affected (0,00 sec)

mysql> INSERT INTO speed_test VALUES ('Brittany', 1100, 900);
Query OK, 1 row affected (0,00 sec)

mysql> INSERT INTO speed_test VALUES ('Alan', 1015, 400);
Query OK, 1 row affected (0,00 sec)

mysql> INSERT INTO speed_test VALUES ('Alan', 1016, -700);
Query OK, 1 row affected (0,01 sec)

mysql> INSERT INTO speed_test VALUES ('Alan', 1020, 800);

   SELECT t1.name , t2.time as start_time, t1.time as end_time , 
  t1.position, t1.position/(t1.time-t2.time) 
    FROM speed_test t1 
    LEFT JOIN 
    speed_test t2
    ON t2.name=t1.name 
       AND t2.time<t1.time 
       AND t2.time=(SELECT max(time) FROM  speed_test WHERE t1.name=name AND time<t1.time);

+----------+------------+----------+----------+-------------------------------+
| name     | start_time | end_time | position | t1.position/(t1.time-t2.time) |
+----------+------------+----------+----------+-------------------------------+
| Brittany |       NULL |     1020 |      -75 |                          NULL |
| Brittany |       1020 |     1030 |      800 |                       80.0000 |
| Brittany |       1030 |     1045 |      800 |                       53.3333 |
| Brittany |       1045 |     1100 |      900 |                       16.3636 |
| Alan     |       NULL |     1015 |      400 |                          NULL |
| Alan     |       1015 |     1016 |     -700 |                     -700.0000 |
| Alan     |       1016 |     1020 |      800 |                      200.0000 |
+----------+------------+----------+----------+-------------------------------+

速度的差异是因为时间列中的 int vs time。

Answer 2

由于您的 TimeFormat 是 HH.mm，因此您需要分别捕获小时和分钟。

如果您使用的是SQL Server，请尝试以下操作：

Create table #temp
(
 Name varchar(100), Time numeric(6,2), Position_Increment int
)
INSERT INTO #temp
values
('Brittany',10.20,-75 )
,('Brittany',10.30,800 )
,('Brittany',10.45,800 )
,('Brittany',11.00,900 )
,('Alan',10.15,400 )
,('Alan',10.16,-700)
,('Alan',10.20,800 )
,('Casey',9.50,800 )
,('Casey',10.01,-400)
,('Casey',10.05,200 )



;with cte as 
(
      select Name
            ,Time as EndTime
            ,Position_Increment
            ,Lag(EndTimeHH,1,0) over (partition by Name order by Time ) as StartTimeHH
            ,Lag(EndTimemm,1,0) over (partition by Name order by Time ) as StartTimemm
            ,StartTime
            ,EndTimeHH
            ,EndTimemm
      from
      (
         select *
               ,SUBSTRING(cast(time as varchar),1,CHARINDEX('.',time)-1) as EndTimeHH
               ,SUBSTRING(cast(time as varchar),CHARINDEX('.',time)+1,len(time)) as EndTimemm
               ,Lag(Time,1,0) over (partition by Name order by Time ) as StartTime
         from #temp
      ) as t
)

select Name
      ,CASE WHEN StartTime =0 Then '-' ELSE CAST(StartTime AS varchar) END AS StartTime
      ,EndTime
      ,Position_Increment
      ,CASE WHEN StartTime =0 
            Then '-' 
            ELSE CAST(Position_Increment /ABS(((StartTimeHH*60+StartTimemm) - (EndTimeHH*60+EndTimemm))) AS varchar)
       END AS IncrementSpeed
from cte
order by name
 
Drop table #temp

Answer 3

下面的伪代码解释了要做什么：

select name,
       lag(time) over (partition by name order by time) as start_time,
       time as end_time,
       ( position_increment /
         (time - lag(time) over (partition by name order by time))
       ) as increment_speed
from t;

需要注意的是，日期/时间函数因数据库而异，因此您需要使用适当的逻辑来减去两次。