Fun with row_number() - Redshift Postgres - Time sequence and restarting numbering - postgresql

Question

I am looking to number streaks within my data, the goal is to find where at least 3 consecutive streaks are flagged by the np.
Here is a subset of my data:
drop table if exists bi_test;
create table test (id varchar(12),rd date,np decimal);
insert into test
select 'aaabbbccc', '2016-07-25'::date, 0 union all
select 'aaabbbccc', '2016-08-01'::date, 0 union all
select 'aaabbbccc', '2016-08-08'::date, 0 union all
select 'aaabbbccc', '2016-08-15'::date, 0 union all
select 'aaabbbccc', '2016-08-22'::date, 1 union all
select 'aaabbbccc', '2016-08-29'::date, 0 union all
select 'aaabbbccc', '2016-09-05'::date, 1 union all
select 'aaabbbccc', '2016-09-12'::date, 0 union all
select 'aaabbbccc', '2016-09-19'::date, 1;
I am hoping to use row_number() and count(), but it doesn't seem to be giving me the result I want.
select
*
,row_number() over (partition by t.id order by t.rd) all_ctr
,count(t.id) over (partition by t.id) all_count
,row_number() over (partition by t.id,t.np order by t.rd) np_counter
,count(t.id) over (partition by t.id,t.np) np_non_np
from
bi_adhoc.test t
order by
t.rd;
Here are my results, and the desired result:
id rd np all_ctr all_count np_counter np_non_np **Desired**
aaabbbccc 7/25/2016 0 1 9 1 6 **1**
aaabbbccc 8/1/2016 0 2 9 2 6 **2**
aaabbbccc 8/8/2016 0 3 9 3 6 **3**
aaabbbccc 8/15/2016 0 4 9 4 6 **4**
aaabbbccc 8/22/2016 1 5 9 1 3 **1**
aaabbbccc 8/29/2016 0 6 9 5 6 **1**
aaabbbccc 9/5/2016 1 7 9 2 3 **1**
aaabbbccc 9/12/2016 0 8 9 6 6 **1**
aaabbbccc 9/19/2016 1 9 9 3 3 **1**

Answer 1

One way to do this would be to calculate the lag (np) value in a CTE, and then compare the current np and lagged np to detect a streak. This may not be the most optimal way, but seems to work fine.
with source_cte as
(
select
*
,row_number() over (partition by t.id order by t.rd) row_num
,lag(np,1) over (partition by t.id order by t.rd) as prev_np
from
bi_adhoc.test t
)
, streak_cte as
(
select
*,
case when np=prev_np or prev_np is NULL then 1 else 0 end as is_streak
from
source_cte
)
select
*,
case when is_streak=1 then dense_rank() over (partition by id, is_streak order by rd) else 1 end as desired
from
streak_cte
order by
rd;

Answer 2

First, I added some additional data to help fully illustrate the problem...
drop table if exists bi_adhoc.test;
create table bi_adhoc.test (id varchar(12),period date,hit decimal);
insert into bi_adhoc.test
select 'aaabbbccc', '2016-07-25'::date, 0 union all
select 'aaabbbccc', '2016-08-01'::date, 0 union all
select 'aaabbbccc', '2016-08-08'::date, 0 union all
select 'aaabbbccc', '2016-08-15'::date, 1 union all
select 'aaabbbccc', '2016-08-22'::date, 1 union all
select 'aaabbbccc', '2016-08-29'::date, 0 union all
select 'aaabbbccc', '2016-09-05'::date, 0 union all
select 'aaabbbccc', '2016-09-12'::date, 1 union all
select 'aaabbbccc', '2016-09-19'::date, 0 union all
select 'aaabbbccc', '2016-09-26'::date, 1 union all
select 'aaabbbccc', '2016-10-03'::date, 1 union all
select 'aaabbbccc', '2016-10-10'::date, 1 union all
select 'aaabbbccc', '2016-10-17'::date, 1 union all
select 'aaabbbccc', '2016-10-24'::date, 1 union all
select 'aaabbbccc', '2016-10-31'::date, 0 union all
select 'aaabbbccc', '2016-11-07'::date, 0 union all
select 'aaabbbccc', '2016-11-14'::date, 0 union all
select 'aaabbbccc', '2016-11-21'::date, 0 union all
select 'aaabbbccc', '2016-11-28'::date, 0 union all
select 'aaabbbccc', '2016-12-05'::date, 1 union all
select 'aaabbbccc', '2016-12-12'::date, 1;
Then the key was to figure out what a streak was and how to identify each streak so I could partition the data to have something to partition the data.
select
*
,case
when t1.hit = 1 then row_number() over (partition by t1.id,t1.hit_partition order by t1.period)
when t1.hit = 0 then row_number() over (partition by t1.id,t1.miss_partition order by t1.period)
else null
end desired
from
(
select
*
,row_number() over (partition by t.id order by t.id,t.period)
,case
when t.hit = 1 then row_number() over (partition by t.id, t.hit order by t.period)
else null
end hit_counter
,case
when t.hit = 1 then row_number() over (partition by t.id order by t.id,t.period) - row_number() over (partition by t.id, t.hit order by t.period)
else null
end hit_partition
,case
when t.hit = 0 then row_number() over (partition by t.id, t.hit order by t.period)
else null
end miss_counter
,case
when t.hit = 0 then row_number() over (partition by t.id order by t.id,t.period) - row_number() over (partition by t.id, t.hit order by t.period)
else null
end miss_partition
from
bi_adhoc.test t
) t1
order by
t1.id
,t1.period;
The result of this:
id period hit row_number hit_counter hit_partition miss_counter miss_partition desired
aaabbbccc 2016-07-25 0 1 NULL NULL 1 0 1
aaabbbccc 2016-08-01 0 2 NULL NULL 2 0 2
aaabbbccc 2016-08-08 0 3 NULL NULL 3 0 3
aaabbbccc 2016-08-15 1 4 1 3 NULL NULL 1
aaabbbccc 2016-08-22 1 5 2 3 NULL NULL 2
aaabbbccc 2016-08-29 0 6 NULL NULL 4 2 1
aaabbbccc 2016-09-05 0 7 NULL NULL 5 2 2
aaabbbccc 2016-09-12 1 8 3 5 NULL NULL 1
aaabbbccc 2016-09-19 0 9 NULL NULL 6 3 1
aaabbbccc 2016-09-26 1 10 4 6 NULL NULL 1
aaabbbccc 2016-10-03 1 11 5 6 NULL NULL 2
aaabbbccc 2016-10-10 1 12 6 6 NULL NULL 3
aaabbbccc 2016-10-17 1 13 7 6 NULL NULL 4
aaabbbccc 2016-10-24 1 14 8 6 NULL NULL 5
aaabbbccc 2016-10-31 0 15 NULL NULL 7 8 1
aaabbbccc 2016-11-07 0 16 NULL NULL 8 8 2
aaabbbccc 2016-11-14 0 17 NULL NULL 9 8 3
aaabbbccc 2016-11-21 0 18 NULL NULL 10 8 4
aaabbbccc 2016-11-28 0 19 NULL NULL 11 8 5
aaabbbccc 2016-12-05 1 20 9 11 NULL NULL 1
aaabbbccc 2016-12-12 1 21 10 11 NULL NULL 2