Question

我目前正在使用这个：

select avg(tank_level)
from (
    select row_number() over (order by id) as rn, tank_level
    from data_tanksensor
    where sensors_on_site_id = 91
) s
group by (rn + ((Select count(*)/10 From data_tanksensor where sensors_on_site_id = 91)-1))/ (Select count(*)/10 From data_tanksensor where sensors_on_site_id = 91)
;

从表中获得10个平均值。此表还有时间戳，我想获得10个平均tank_level中每个的平均时间戳。这用于创建历史图表。如果有人可以帮我修改此查询，也可以获得非常感谢的平均时间戳。提前谢谢。

表格如下所示

. id sensors_on_site_id tank_level timestamps [PK] bigint integer double precision time without time zone ........... .................. ................ ...................... 12345 91 7.5 2017-03-24 11:16:31.143362 12346 91 7.6 2017-03-24 11:21:31.148639 12347 91 5.4 2017-03-24 11:26:31.155739 12348 91 3.6 2017-03-24 11:31:31.156478 12349 91 8.5 2017-03-24 11:36:31.157303 12350 91 4.2 2017-03-24 11:41:31.172008 例如，如果我只想从中得到平均值，那么我的原始查询将是

select avg(tank_level) from ( select row_number() over (order by id) as rn, tank_level from data_tanksensor where sensors_on_site_id = 91 ) s group by (rn + ((Select count(*)/2 From data_tanksensor where sensors_on_site_id = 91)-1))/ (Select count(*)/2 From data_tanksensor where sensors_on_site_id = 91) ; 粗略的查询缺少可以平衡时间戳的部分，这是我想要弄清楚的。但是我试图获得的预期结果将是

avg timestamp double precision timestamp without time zone ................ ........................... 6.833333 2017-03-24 11:21:31... 5.433333 2017-03-24 11:36:31... 同样，这只是样本数据，平均每行数百个。感谢

Answer 1

您可以将时间戳转换为unix时间戳，将值相加，除以值的数量并转换回时间戳。一个快速工作的例子：

with
    __ts as(
        select unnest(array[
            '2015-11-22 09:31:00', '2015-11-22 09:32:00', '2015-11-23 11:31:00', '2015-11-23 11:32:00',
            '2015-11-23 11:34:00', '2015-11-23 15:28:00', '2015-11-23 15:29:00', '2015-11-24 10:49:00',
            '2015-11-24 10:50:00', '2015-11-24 11:18:00'
        ]::timestamp without time zone[]) as ts
    )
select
    to_timestamp(sum(extract(epoch from ts)) / (select count(1) from __ts))
from
    __ts

Answer 2

select to_timestamp(avg(timestamps)) "timestamps", avg(tank_level) "TankLevel" from ( select row_number() over (order by id) as rn, tank_level, extract(epoch from timestamps) "timestamps" from data_tanksensor where sensors_on_site_id = 91 ) s group by (rn + ((Select count(*)/10 From data_tanksensor where sensors_on_site_id = 91)-1))/ (Select count(*)/10 From data_tanksensor where sensors_on_site_id = 91) order by timestamps asc ; 弄清楚了谢谢大家的例子

如何获得平均时间戳？ PostgreSQL的

2 个答案: