SQL group table by "leading rows" without pl/sql - sql

Question

I have this table (short example) with two columns
1 a
2 a
3 a3
4 a
5 a
6 a6
7 a
8 a8
9 a
and I would like to group/partition them into groups separated by those leading "a", ideally to add another column like this, so I can address those groups easily.
1 a 0
2 a 0
3 a3 3
4 a 3
5 a 3
6 a6 6
7 a 6
8 a8 8
9 a 8
problem is that setup of the table is dynamic so I can't use staticly lag or lead functions, any ideas how to do this without pl/sql in postgres version 9.5

Answer 1

Assuming the leading part is a single character. Hence the expression right(data, -1) works to extract the group name. Adapt to your actual prefix.
The solution uses two window functions, which can't be nested. So we need a subquery or a CTE.
SELECT id, data
, COALESCE(first_value(grp) OVER (PARTITION BY grp_nr ORDER BY id), '0') AS grp
FROM (
SELECT *, NULLIF(right(data, -1), '') AS grp
, count(NULLIF(right(data, -1), '')) OVER (ORDER BY id) AS grp_nr
FROM tbl
) sub;
Produces your desired result exactly.
NULLIF(right(data, -1), '') to get the effective group name or NULL if none.
count() only counts non-null values, so we get a higher count for every new group in the subquery.
In the outer query, we take the first grp value per grp_nr as group name and default to '0' with COALESCE for the first group without name (which has a NULL as group name so far).
We could use min() or max() as outer window function as well, since there is only one non-null value per partition anyway. first_value() is probably cheapest since the rows are sorted already.
Note the group name grp is data type text. You may want to cast to integer, if those are clean (and reliably) integer numbers.

Answer 2

This can be achieved by setting rows containing a to a specific value and all the other rows to a different value. Then use a cumulative sum to get the desired number for the rows. The group number is set to the next number when a new value in the val column is encountered and all the proceeding rows with a will have the same group number as the one before and this continues.
I assume that you would need a distinct number for each group and the number doesn't matter.
select id, val, sum(ex) over(order by id) cm_sum
from (select t.*
,case when val = 'a' then 0 else 1 end ex
from t) x
The result for the query above with the data in question, would be
id val cm_sum
--------------
1 a 0
2 a 0
3 a3 1
4 a 1
5 a 1
6 a6 2
7 a 2
8 a8 3
9 a 3

Answer 3

With the given data, you can use a cumulative max:
select . . .,
coalesce(max(substr(col2, 2)) over (order by col1), 0)
If you don't strictly want the maximum, then it gets a bit more difficult. The ANSI solution is to use the IGNORE NULLs option on LAG(). However, Postgres does not (yet) support that. An alternative is:
select . . ., coalesce(substr(reft.col2, 2), 0)
from (select . . .,
max(case when col2 like 'a_%' then col1 end) over (order by col1) as ref_col1
from t
) tt join
t reft
on tt.ref_col1 = reft.col1

Answer 4

You can also try this :
with mytable as (select split_part(t,' ',1)::integer id,split_part(t,' ',2) myvalue
from (select unnest(string_to_array($$1 a;2 a;3 a3;4 a;5 a;6 a6;7 a;8 a8;9 a$$,
';'))t) a)
select id,myvalue,myresult from mytable join (
select COALESCE(NULLIF(substr(myvalue,2),''),'0') myresult,idmin id_down
,COALESCE(lead(idmin) over (order by myvalue),999999999999) id_up
from (
select myvalue,min(id) idmin from mytable group by 1
) a) b
on id between id_down and id_up-1