在 PostgreSQL 中基于瞬时时间前后 2 分钟创建每小时平均值

Question

我有一个采样频率为 2 分钟的时态数据库，我想将瞬时每小时值提取为每天 00:00、01:00、02、... 23。
所以，我想从平均值中得到平均值：

HH-1:58、HH:00 和 HH:02 = HH 点钟的平均值

或

HH-1:59、HH:01 和 HH:03 = HH 点钟的平均值

样本数据1：

9/28/2007 23:51 -1.68
9/28/2007 23:53 -1.76
9/28/2007 23:55 -1.96
9/28/2007 23:57 -2.02
9/28/2007 23:59 -1.92
9/29/2007 0:01  -1.64
9/29/2007 0:03  -1.76
9/29/2007 0:05  -1.83
9/29/2007 0:07  -1.86
9/29/2007 0:09  -1.94

预期结果：

午夜 00 点：

(-1.92+-1.64+-1.76)/3

样本数据2：

9/28/2007 23:54 -1.44
9/28/2007 23:56 -1.58
9/28/2007 23:58 -2.01
9/29/2007 0:00  -1.52
9/29/2007 0:02  -1.48
9/29/2007 0:04  -1.46

预期结果：

(-2.01+-1.52+-1.48)/3

Answer 1

SELECT  hr, ts, aval
FROM    (
        SELECT  *, ROW_NUMBER() OVER (PARTITION BY hr ORDER BY ts) rn
        FROM    (
                SELECT  *,
                        DATE_TRUNC('hour', ts) AS hr,
                        AVG(value) OVER (ORDER BY ts ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING) AS aval
                FROM    mytable
                ) q
        ) q
WHERE   rn = 1

Answer 2

PostgreSQL 的window functions 使涉及相邻行的任何事情都比以前简单得多。未尝试但应该大致正确：

select
  date_trunc('hour', newest_time) as average_time,
  (oldest_temp + middle_temp + newest_temp) / 3 as average_temp
from (
  select
    date_trunc('hour', sample_time) as average_time,
    lag(sample_time, 2) over w as oldest_time,
    lag(sample_time, 1) over w as middle_time,
    sample_time as newest_time,
    lag(sample_temp, 2) over w as oldest_temp,
    lag(sample_temp, 1) over w as middle_temp,
    sample_temp as newest_temp
  from
    samples
  window
    w as (order by sample_time)
) as s
where
  oldest_time = newest_time - '4 minutes'::interval and
  middle_time = newest_time - '2 minutes'::interval and
  extract(minute from newest_time) in (2, 3);

我已在 where 子句中将此限制为您所描述的场景 - 最新值在 :02 或 :03，前 2 个值在 2 分钟和 4 分钟前。以防万一您丢失了一些数据，否则会产生奇怪的结果，例如在更长的时间间隔内进行平均。