如何 return 在子查询中找到列 B 的最大值的行上列 A 的值

Question

我有时间序列数据，我正在创建 90 个样本的滚动平均值 window（15 分钟）。然后我想找到每小时最多 15 分钟 window，以便数据按小时分组。但是我还想 return 最长 15 分钟的开始或结束时间戳 window。目前我只是 return 在每个小时开始时输入时间戳。

这是我开始使用的数据示例：

theDate    theTime     value
4/30/2019   22:47:53    0
4/30/2019   22:47:43    0
4/30/2019   22:47:33    0
4/30/2019   22:47:23    0
4/30/2019   22:47:13    0
4/30/2019   22:47:03    0
4/30/2019   22:46:53    0
4/30/2019   22:46:43    0

我已经尝试 row_number() 并在何处进行过滤，但我认为我没有正确地应用它们，因为有几个嵌套的子查询。

select
        min([theDate]) as theDate, 
        min([theTime]) as theTime, 
        max([Value]) as maxValue, 
        max([rolling_avg]) as maxDM,
        timeHour as timeHour

from( select [theDate], [theTime], [Value],
       avg(windowAvg) over(order by theDate DESC, theTime rows between 90 preceding and current row) as rolling_avg,
       datepart(hh,theTime) as timeHour
  from (select [theDate], [theTime], [Value], sum([Value]) as windowAvg
          from [Data].[dbo].[tOne]
          Where ([theDate] > convert(DAte,DATEADD(month, -1, GETDATE())))
          group by theDate, theTime, Value 

          )tOneTemp   
          )tOneTempTwo
          group by theDate, timeHour
          order by theDate DESC, theTime DESC

我正在寻找下面的 table 除了 theTime 始终是我计算的每小时分钟数我希望它是 15 分钟内的值 window计算了 MaxDM。

这是我使用当前代码最终得到的示例：

theDate    theTime   maxValue   maxDM       timeHour
2019-04-30  22:00:04    508.8   660.643956  22
2019-04-30  21:00:03    1071.3  798.206593  21
2019-04-30  20:00:03    1022.2  817.539560  20
2019-04-30  19:00:04    871.4   574.786813  19
2019-04-30  18:00:04    944.0   670.095604  18

我希望看到的是 "theTime" 反映 "maxValue" window.

的开始

这是我希望看到的示例（*注意时间）：

theDate    theTime   maxValue   maxDM       timeHour
2019-04-30  22:10:34    508.8   660.643956  22
2019-04-30  21:45:03    1071.3  798.206593  21
2019-04-30  20:12:03    1022.2  817.539560  20
2019-04-30  19:32:04    871.4   574.786813  19
2019-04-30  18:56:04    944.0   670.095604  18

谢谢 Ali，这里是我对您的代码所做的一些修改，以修复一些错误，但是 "TheTimeOfMaxDM" returns as "NULL" 对于所有行。

                    min([theDate]) as theDate, 
                    min([theTime]) as theTime, 
                    (select top(1) r.[theTime] from RunningAvg15Min r where 
                        r.[theDate]=(select min([theDate]) from RunningAvg15Min) and
                        r.[theTime]>=(select min([theTime]) from RunningAvg15Min)
                        and r.rolling_avg=(select max([rolling_avg]) from RunningAvg15Min)) as TheTimeOfMaxDM,
                    max([KW]) as maxValue, 
                    max([rolling_avg]) as maxDM,
                    timeHour as timeHour
                from RunningAvg15Min
                GROUP BY theDate, timeHour
                ORDER BY thedate DESC, thetime DESC

Answer 1

我在 CTE 中重写了您的查询，因此我可以再次使用子查询。请检查以下内容：- 所以我在您的查询中添加了另一个名为 TheTimeOfMaxDM 的输出。希望这是你需要的。

;with LastMonthData as (
               --get the data for the last month, and sum value if same time stamp is recorded (this is not logical)
               select [theDate], [theTime], [Value], sum([Value]) as windowAvg
                  from [tOne]
                  Where ([theDate] > convert(Date,DATEADD(month, -1, '2019-05-3')))--changed this to match the data provided.
                  group by theDate, theTime, Value 
          ),
RunningAvg15Min as (
                    select [theDate], [theTime], [Value],
                    avg(windowAvg) over(order by theDate DESC, theTime rows between 90 preceding and current row) as rolling_avg,
                    datepart(hh,theTime) as timeHour
                from LastMonthData
              ),DataSetGrouping as (
              select
                    min([theDate]) as theDate, 
                    min([theTime]) as theTime, 
                    min(datepart(hh,theTime)) as timeHour,
                    max([Value]) as maxValue, 
                    max([rolling_avg]) as maxDM
                from RunningAvg15Min
                GROUP BY
                    DATEPART(YEAR, TheDate),
                    DATEPART(MONTH, TheDate),
                    DATEPART(DAY, TheDate),
                    DATEPART(HOUR, TheTime),
                    (DATEPART(MINUTE, TheTime) / 60) ---change the 60 to 15, to change the grouping from per hour to per 15min

                )
                select * ,
                 (select top(1) r.[theTime] from RunningAvg15Min r where 
                        r.[theDate]=theDate
                        and r.[theTime]>=theTime
                        and r.rolling_avg=maxDM
                        order by r.[theDate],r.[theTime]
                        )  [TheTimeOfMaxDM]
                from DataSetGrouping
                ORDER BY thedate DESC, thetime DESC

另一个注意事项，我不知道你是否需要这个，但如果你想按 15 分钟分组，你可以在分组部分使用下面的内容

GROUP BY
    DATEPART(YEAR, TheDate),
    DATEPART(MONTH, TheDate),
    DATEPART(DAY, TheDate),
    DATEPART(HOUR, TheTime),
    (DATEPART(MINUTE, TheTime) / 15)

我会尝试验证一个观点，让我们以下面的输出行为例：-

theDate     theTime             Value   rolling_avg         timeHour
2019-04-07  17:20:49.0000000    398.3   314.499999999997    17

在 2019-04-07 的 17 和 18 之间，最大值 rolling_avg 为 314.499999999997，而该最大值的时间为 17:20:49.0000000，我们可以看到正在替换最后一个cte 的一部分具有以下内容：-

select * from RunningAvg15Min where rolling_avg between 314 and 315
                and thedate='2019-04-07' and theTime between '17:00:00' and '18:00:00'

产出

theDate      theTime            Value   rolling_avg         timeHour
2019-04-07  17:20:49.0000000    398.3   314.499999999997    17

总结一下我的回答，您可以使用我编写的查询中的 CTE 来简化子查询，并在需要时在您的查询中多次引用它们。希望这会有所帮助。

如何 return 在子查询中找到列 B 的最大值的行上列 A 的值

How to return the Value from Column A on the row where the Max of Column B is found in a subquery

tsql

subquery

row-number

max