使用 T-SQL window 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 table，其中包含证券的一分钟开盘价、收盘价、高价、低价和成交量值。我使用的是 SQL Server 2017，但 2019 RC 是一个选项。

我正在尝试找到一个高效的 SQL 服务器查询，可以将这些查询聚合成 5 分钟 windows，其中：

Open = window
收盘价 = window
High = window
低 = window
交易量 = window

理想情况下，此查询会考虑数据中的差距，即基于日期计算而不是计算前/后行。

例如我有（这里有 6 分钟的数据）：

| Time             | Open | Close | High | Low | Volume |
|------------------|------|-------|------|-----|--------|
| 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  | Not factored in, since this would be the first row of the next 5-minute window

我正在尝试编写一个查询（这是 5 分钟聚合的第一个示例）：

| Time             | Open | Close | High | Low | Volume  |
|------------------|------|-------|------|-----|---------|
| 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 选项撞墙

Answer 1

您想以 5 分钟为间隔分析数据。您可以使用带有以下分区子句的 window 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（假设数据类型是 datetime）可以使用 DATEADD(MINUTE, DATEDIFF(MINUTE, 0, time) / 5 * 5, 0) 来完成。休息是基本 grouping/window 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

使用 T-SQL window 函数从 1 分钟数据中检索 5 分钟平均值

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