Calculating consecutive range of dates with a value in Hive - hive

Question

I want to know if it is possible to calculate the consecutive ranges of a specific value for a group of Id's and return the calculated value(s) of each one.
Given the following data:
+----+----------+--------+
| ID | DATE_KEY | CREDIT |
+----+----------+--------+
| 1 | 8091 | 0.9 |
| 1 | 8092 | 20 |
| 1 | 8095 | 0.22 |
| 1 | 8096 | 0.23 |
| 1 | 8098 | 0.23 |
| 2 | 8095 | 12 |
| 2 | 8096 | 18 |
| 2 | 8097 | 3 |
| 2 | 8098 | 0.25 |
+----+----------+--------+
I want the following output:
+----+-------------------------------+
| ID | RANGE_DAYS_CREDIT_LESS_THAN_1 |
+----+-------------------------------+
| 1 | 1 |
| 1 | 2 |
| 1 | 1 |
| 2 | 1 |
+----+-------------------------------+
In this case, the ranges are the consecutive days with credit less than 1. If there is a gap between date_key column, then the range won't have to take the next value, like in ID 1 between 8096 and 8098 date key.
Is it possible to do this with windowing functions in Hive?
Thanks in advance!

Answer 1

You can do this with a running sum classifying rows into groups, incrementing by 1 every time a credit<1 row is found(in the date_key order). Thereafter it is just a group by.
select id,count(*) as range_days_credit_lt_1
from (select t.*
,sum(case when credit<1 then 0 else 1 end) over(partition by id order by date_key) as grp
from tbl t
) t
where credit<1
group by id

Answer 2

The key is to collapse all the consecutive sequence and compute their length, I struggled to achieve this in a relatively clumsy way:
with t_test as
(
select num,row_number()over(order by num) as rn
from
(
select explode(array(1,3,4,5,6,9,10,15)) as num
)
)
select length(sign)+1 from
(
select explode(continue_sign) as sign
from
(
select split(concat_ws('',collect_list(if(d>1,'v',d))), 'v') as continue_sign
from
(
select t0.num-t1.num as d from t_test t0
join t_test t1 on t0.rn=t1.rn+1
)
)
)
Get the previous number b in the seq for each original a;
Check if a-b == 1, which shows if there is a "gap", marked as 'v';
Merge all a-b to a string, and then split using 'v', and compute length.
To get the ID column out, another string which encode id should be considered.