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時制データベースがあり、次の時系列のデータを使用して、時系列データの平均関数を介して時制集計を実行したいと思います。

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

また、上記の値は瞬時値です。

サンプルデータ：

    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

そして期待される結果：

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

時間平均0:00=（-1.68 + -1.64 + -1.87 + -1.90 + -2.18 + -1.86）/ 6

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1 に答える 1

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 フィドル

于 2013-01-14T10:59:17.373 に答える