Oracle group by ordered values - oracle

Question

I have a table like this:
[Date] [hour] [valid] [value] [type]
And I need to do the averange for each hour of the 23 previous values and itself (24 values) and only if I have 18 or more valid values.
I do with one hour with this query:
select type, avg(value)
from table
where type = 10
and valid = 1
and ((date = '11/10/2016' and hour <= 22) or
(date = '11/09/2016' and hour > 22))
group by type
having count(*)>=18;
But I need to do it with a date interval for each hour in that interval.
It can be possible?
Thank you.

Answer 1

Sorry but keeping date separately with hour is not good. So I'll merge it into date in with clause. If you don't want that you can use to_date(date || ':' || hour, 'dd/mm/yyyy:hh24') everywhere where I have d and ignore with clause
with dat as (select to_date(date || ':' || hour, 'dd/mm/yyyy:hh24') d, valid, value, type from table)
select d1.d, d1.type, avg(d2.value)
from dat d1 cross join dat d2
where d1.d < (d2.d + 1) and d1.d >= d2.d and d2.valid = 1
group by d1.d, d1.type
having count(*) >= 18;
I hope it will work. If you provided any test data I would test it.

Answer 2

you may write next, I assume column valid contains Y and N values.
select type
,max(avg1)
from (
select type
,avg(value) over (order by trunc(date,'HH') range between interval '24' hour preceding and interval '0' hour following ) as avg1
,count(decode(valid,'Y',1)) over (order by trunc(date,'HH') range between interval '24' hour preceding and interval '0' hour following ) as count1
from table
where type = 10
and valid = 1)
where count1 >= 18
group by type;