基于measurment与PostgreSQL的10分钟的间隔时间上的平均DB聚集(Average

我有一个时态数据库，我想在下面的时间序列数据的使用进行平均通过对功能的时间序列数据的时间汇总：

Hourly average of HH= (HH-1):51, HH:01, HH:11, HH:21, HH:31, HH:41.

此外，上述的值是瞬时值。

样本数据：

    ambtemp      ddate_ttime
1     -1.42 2007-09-28 23:39:09
2     -1.24 2007-09-28 23:41:09
3     -1.28 2007-09-28 23:43:09
4     -1.28 2007-09-28 23:45:09
5     -1.24 2007-09-28 23:47:09
6     -1.42 2007-09-28 23:49:09
7     -1.68 2007-09-28 23:51:09
8     -1.76 2007-09-28 23:53:09
9     -1.96 2007-09-28 23:55:09
10    -2.02 2007-09-28 23:57:09
11    -1.92 2007-09-28 23:59:09
12    -1.64 2007-09-29 00:01:09
13    -1.76 2007-09-29 00:03:09
14    -1.83 2007-09-29 00:05:09
15    -1.86 2007-09-29 00:07:09
16    -1.94 2007-09-29 00:09:09
17    -1.87 2007-09-29 00:11:09
18    -1.87 2007-09-29 00:13:09
19    -1.80 2007-09-29 00:15:09
20    -1.64 2007-09-29 00:17:09
21    -1.60 2007-09-29 00:19:09
22    -1.90 2007-09-29 00:21:09
23    -2.08 2007-09-29 00:23:09
24    -1.94 2007-09-29 00:25:09
25    -2.12 2007-09-29 00:27:09
26    -1.87 2007-09-29 00:29:09
27    -2.18 2007-09-29 00:31:09
28    -1.98 2007-09-29 00:33:09
29    -1.73 2007-09-29 00:35:09
30    -1.84 2007-09-29 00:37:09
31    -2.04 2007-09-29 00:39:09
32    -1.86 2007-09-29 00:41:09
33    -1.94 2007-09-29 00:43:09
34    -1.77 2007-09-29 00:45:09

和预期的结果：

Hourly mean for 0:00 = average of values of (23:51, 0:01, 0:11, 0:21, 0:31, 0:41)

每小时平均0:00 =（-1.68 + -1.64 -1.87 + + + -1.90 -2.18 + -1.86）/ 6

Answer 1:

select dt, average_temp
from (
    select
        date_trunc('hour', ddate_ttime + interval '10 minutes') dt,
        avg(ambtemp) over(
            partition by date_trunc('hour', ddate_ttime + interval '10 minutes')
        ) average_temp
    from sample
    where extract(minute from ddate_ttime) in (51, 01, 11, 21, 31, 41)
) s
group by 1, 2
order by dt

SQL小提琴

文章来源: Average aggregation on temporal DB based on measurment with 10-minute intervals in PostgreSQL