How to return all the rows in the yellow census blocks? - sql

Question

Hey the schema is like this: for the whole dataset, we should order by machine_id first, then order by ss2k. after that, for each machine, we should find all the rows with at least consecutively 5 flag = 'census'. In this dataset, the result should be all the yellow rows..
I cannot return the last 4 rows of the yellow blocks by using this:
drop table if exists qz_panel_census_228_rank;
create table qz_panel_census_228_rank as
select t.*
from (select t.*,
count(*) filter (where flag = 'census') over (partition by machine_id, date order by ss2k rows between current row and 4 following) as census_cnt5,
count(*) filter (where flag = 'census') over (partition by machine_id, date) as count_census,
row_number() over (partition by machine_id, date order by ss2k) as seqnum,
count(*) over (partition by machine_id, date) as cnt
from qz_panel_census_228 t
) t
where census_cnt5 = 5
group by 1,2,3,4,5,6,7,8,9,10,11
DISTRIBUTED BY (machine_id);

Answer 1

You were close, but you need to search in both directions:
select t.*
from (select t.*,
case when count(*) filter (where flag = 'census')
over (partition by machine_id, date
order by ss2k
rows between 4 preceding and current row) = 5
or count(*) filter (where flag = 'census')
over (partition by machine_id, date
order by ss2k
rows between current row and 4 following) = 5
then 1
else 0
end as flag
from qz_panel_census_228 t
) t
where flag = 1
Edit:
This approach will not work unless you add an extra count for each possible 5 row window, e.g. 3 preceding and 1 following, 2 preceding and 2 following, etc. This results in ugly code and is not very flexible.
The common way to solve this gaps & islands problem is to assign consecutive rows to a common group first:
select *
from
(
select t2.*,
count(*) over (partition by machine_id, date, grp) as cnt
from
(
select t1.*
from (select t.*,
-- keep the same number for 'census' rows
sum(case when flag = 'census' then 0 else 1 end)
over (partition by machine_id, date
order by ss2k
rows unbounded preceding) as grp
from qz_panel_census_228 t
) t1
where flag = 'census' -- only census rows
) as t2
) t3
where cnt >= 5 -- only groups of at least 5 census rows

Answer 2

Wow, there has to be a better way of doing this, but the only way I could figure out was to create blocks of consecutive 'census' values. This looks awful but might be a catalyst to a better idea.
with q1 as (
select
machine_id, recorded, ss2k, flag, date,
case
when flag = 'census' and
lag (flag) over (order by machine_id, ss2k) != 'census'
then 1
else 0
end as block
from foo
),
q2 as (
select
machine_id, recorded, ss2k, flag, date,
sum (block) over (order by machine_id, ss2k) as group_id,
case when flag = 'census' then 1 else 0 end as census
from q1
),
q3 as (
select
machine_id, recorded, ss2k, flag, date, group_id,
sum (census) over (partition by group_id order by ss2k) as max_count
from q2
),
groups as (
select group_id
from q3
group by group_id
having max (max_count) >= 5
)
select
q2.machine_id, q2.recorded, q2.ss2k, q2.flag, q2.date
from
q2
join groups g on q2.group_id = g.group_id
where
q2.flag = 'census'
If you run each query within the with clauses in isolation, I think you will see how this evolves.