SQL: Check if n consecutive records are greater than some value

Question

I have a table that contains numbers. I have to find whether there is any case where in n consecutive numbers are greater than some threshold value m. For eg

id      delta         
---------------
1        10  
4        15 
11       22 
23       23  
46       21
57       9

So here, if I want to know if there are 3 consecutive records where value is more than 20 then I should get True. And False when I check for 4 consecutive records. Is that possible? This is on Apache Spark SQL. Thanks.

Answer 1

You can do this using lag:

select t.*
from (select t.*,
             lag(val, 1) over (order by id) as val_1,
             lag(val, 2) over (order by id) as val_2
      from t
     ) t
where val > 20 and val_1 > 20 and val_2 > 20;

This returns the first row that is part of each three-some. If you just want true/false:

select (case when count(*) > 0 then 'true' else 'false' end)
from (select t.*,
             lag(val, 1) over (order by id) as val_1,
             lag(val, 2) over (order by id) as val_2
      from t
     ) t
where val > 20 and val_1 > 20 and val_2 > 20;

EDIT:

I missed the part about not wanting more than 3. So, you can enhance this:

select (case when count(*) > 0 then 'true' else 'false' end)
from (select t.*,
             lag(val, 1) over (order by id) as val_1,
             lag(val, 2) over (order by id) as val_2,
             lag(val, 3) over (order by id) as val_3,
             lead(val, 1) over (order by id) as val_next_1
      from t
     ) t
where (val_3 <= 20 or val_3 is null) and
      (val_2 > 20 and val_1 > 20 and val > 20) and
      (val_next_1 <= 20 or val_next_1 is null);

It is a little tricky because the values can be at the beginning or end of the rows.