使用 T-SQL 窗口函数从 1 分钟数据中检索 5 分钟平均值

Question

我有一个数据库表，其中包含一分钟的开盘价、收盘价、最高价、最低价、交易量值。我使用的是 SQL Server 2017，但可以选择 2019 RC。

我正在尝试找到一个高效的 SQL Server 查询，可以将这些汇总到 5 分钟的窗口中，其中：

• Open = 窗口的第一个 Open 值
• 关闭 = 窗口的最后关闭值
• High = 窗口的最大 High 值
• Lo​​w = min 窗口的最小值
• 音量 = 窗口内的平均音量

理想情况下，此查询将考虑数据中的空白，即基于日期计算而不是计算前后行。

例如说我有（这里是 6 分钟的数据）：

|时间 |打开 |关闭 |高 |低 |体积 | |-----------------|------|--------|------|-----|-- ------| | 2019-10-30 09:30 | 5 | 10 | 15 | 1 | 125000 | | 2019-10-30 09:31 | 10 | 15 | 20 | 5 | 100000 | | 2019-10-30 09:32 | 15 | 20 | 25 | 10 | 120000 | | 2019-10-30 09:33 | 20 | 25 | 30 | 15 | 10000 | | 2019-10-30 09:34 | 20 | 22 | 40 | 2 | 13122 | | 2019-10-30 09:35 | 22 | 30 | 35 | 4 | 15000 |未考虑在内，因为这将是下一个 5 分钟窗口的第一行

我正在尝试编写一个可以给我的查询（这是 5 分钟聚合的第一个示例）：

|时间 |打开 |关闭 |高 |低 |体积 | |-----------------|------|--------|------|-----|-- --------| | 2019-10-30 09:30 | 5 | 30 | 40 | 1 | 50224.4 |

有什么建议吗？我用 OVER 子句及其 PARTITION / RANGE 选项将我的头撞到墙上

Answer 1

您希望以 5 分钟为间隔分析数据。您可以使用带有以下分区子句的窗口函数：

partition by datepart(year, t.[time]),
    datepart(month, t.[time]),
    datepart(day, t.[time]),
    datepart(hour, t.[time]),
    (datepart(minute, t.[time]) / 5)

查询：

select *
from (
    select  
        t.time,
        row_number() over(
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
            order by [time]
        ) [rn],
        first_value([open]) over(
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
            order by [time]
        ) [open],
        last_value([close]) over(
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
            order by [time]
        ) [close],
        max([high]) over (
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
        ) [high],
        min([low]) over (
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
        ) [low],
        avg([volume]) over (
            partition by datepart(year, [time]),
                datepart(month, [time]),
                datepart(day, [time]),
                datepart(hour, [time]),
                (datepart(minute, [time]) / 5)
        ) [volume]
    from mytable t
) t
where rn = 1

Answer 2

你可以试试这个。

  SELECT
      MIN([Time]) [Time], 
      Min([Open]) [Open],
      LEAD(Min([Open])) OVER (ORDER BY MIN([Time])) AS [Close],
      Max([High]) [High], 
      Min([Low]) [Low], 
      Avg(Volume) Volume
  FROM SampleData
  GROUP BY DATEADD(Minute, -1* DATEPART(Minute, Time) %5, Time)

sql fiddle

Answer 3

问题的要点是将日期时间值四舍五入到 5 分钟边界（假设数据类型为 datetime）可以使用 DATEADD(MINUTE, DATEDIFF(MINUTE, 0, time) / 5 * 5, 0) 完成。休息是基本的分组/窗口功能：

WITH cte AS (
  SELECT clamped_time
       , [Open]
       , [Close]
       , [High]
       , [Low]
       , [Volume]
       , rn1 = ROW_NUMBER() OVER (PARTITION BY clamped_time ORDER BY [Time])
       , rn2 = ROW_NUMBER() OVER (PARTITION BY clamped_time ORDER BY [Time] DESC)
  FROM t
  CROSS APPLY (
      SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, 0, time) / 5 * 5, 0)
  ) AS x(clamped_time)
)
SELECT clamped_time
     , MIN(CASE WHEN rn1 = 1 THEN [Open] END) AS [Open]
     , MIN(CASE WHEN rn2 = 1 THEN [Close] END) AS [Close]
     , MAX([High]) AS [High]
     , MIN([Low]) AS [Low]
     , AVG([Volume])
FROM cte
GROUP BY clamped_time

Demo on db<>fiddle