一次取三个记录的滚动平均值，两个 group by 值

Question

我正在处理的数据没有主键，而是按 TMC 值和日期值分组。我正在尝试做的是取 5 分钟的纪元值，然后将它们组合成一个 15 分钟的纪元，一次平均三个记录。所以 epoch 0, 1, 2 一起平均得到一个新记录，然后是 3,4,5 等等。这是一个示例表。我不知道如何正确格式化：

TMC         DATE    EPOCH   TRAVEL_TIME_ALL_VEHICLES
113N04415   1012016 0       228
113N04415   1012016 1       228
113N04415   1012016 2       222
113N04415   1012016 3       224
113N04415   1012016 4       229
113N04415   1012016 5       223

我认为我对这段代码很接近，因为它适用于第一条记录，但第二条记录失败：

select tmc, date, avg(Travel_TIME_ALL_VEHICLES) AS TRAVEL_avg
from (select I40_2016_ALL.*, 
             row_number() over (order by TMC, date, epoch) as rn 
      from I40_2016_ALL) I40_2016_ALL
where tmc = '113n04415' and date = '1012016'
group by TMC, DATE, floor((rn - 1) / 3)
ORDER BY TMC, DATE

Answer 1

嗯嗯，没试过，但是怎么样：

    select tmc, date, floor((epoch + 1) / 3) as part, 
           avg(Travel_TIME_ALL_VEHICLES) AS TRAVEL_avg
      from I40_2016_ALL
     where tmc = '113n04415' and date = '1012016'
     group by TMC, DATE, floor((epoch + 1) / 3)

或者如果你想要每一行都有总数...

        select tmc, date, floor((epoch + 1) / 3) as part, 
               avg(Travel_TIME_ALL_VEHICLES) 
                over (partition by TMC, DATE, floor((epoch + 1) / 3)) AS TRAVEL_avg
          from I40_2016_ALL
         where tmc = '113n04415' and date = '1012016'

Answer 2

你可以像这样使用整数除法：

SELECT TMC, [DATE], EPOCH, AVG(TRAVEL_TIME_ALL_VEHICLES) AS TRAVEL_avg       
FROM (
   SELECT TMC, [DATE], TRAVEL_TIME_ALL_VEHICLES,
          ROW_NUMBER() OVER (ORDER BY TMC, [DATE], EPOCH) AS rn
   FROM I40_2016_ALL) AS t
GROUP BY TMC, [DATE], (rn -1)/ 3
ORDER BY TMC, [DATE]

此查询将行 1,2,3 放在一组中，将行 4, 5, 6 放在下一组中，等等。

Answer 3

类似：

SELECT AVG(TRAVEL_TIME_ALL_VEHICLES) 
FROM
(
SELECT TRAVEL_TIME_ALL_VEHICLES AS TRAVEL_TIME_ALL_VEHICLES
,(ROW_NUMBER() OVER (Order by EPOCH) + 2)/3 AS GroupID
FROM
YourTable
) AS d
GROUP BY d.GroupID