Window functions and more “local” aggregation

Question

Suppose I have this table:

select * from window_test;

 k | v
---+---
 a | 1
 a | 2
 b | 3
 a | 4

Ultimately I want to get:

Answer 1

EDIT: I've came up with the following query — without window functions at all:

WITH RECURSIVE tree AS (
  SELECT k, v, ''::text as next_k, 0 as next_v, 0 AS level FROM window_test
  UNION ALL
  SELECT c.k, c.v, t.k, t.v + level, t.level + 1
    FROM tree t JOIN window_test c ON c.k = t.k AND c.v + 1 = t.v),
partitions AS (
  SELECT t.k, t.v, t.next_k,
         coalesce(nullif(t.next_v, 0), t.v) AS next_v, t.level
    FROM tree t
   WHERE NOT EXISTS (SELECT 1 FROM tree WHERE next_k = t.k AND next_v = t.v))
SELECT min(k) AS k, v AS min_v, max(next_v) AS max_v
  FROM partitions p
 GROUP BY v
 ORDER BY 2;

I've provided 2 working queries now, I hope one of them will suite you.

SQL Fiddle for this variant.

---

Another way how to achieve this is to use a support sequence.

1. Create a support sequence:

   CREATE SEQUENCE wt_rank START WITH 1;
   

2. The query:

   WITH source AS (
     SELECT k, v,
            coalesce(lag(k) OVER (ORDER BY v), k) AS prev_k
       FROM window_test
       CROSS JOIN (SELECT setval('wt_rank', 1)) AS ri),
   ranking AS (
     SELECT k, v, prev_k,
            CASE WHEN k = prev_k THEN currval('wt_rank')
                 ELSE nextval('wt_rank') END AS rank
       FROM source)
   SELECT r.k, min(s.v) AS min_v, max(s.v) AS max_v
       FROM ranking r
       JOIN source s ON r.v = s.v
      GROUP BY r.rank, r.k
      ORDER BY 2;
   

Answer 2

Would this not do the job for you, without the need for windows, partitions or coalescing. It just uses a traditional SQL trick for finding nearest tuples via a self join, and a min on the difference:

SELECT k, min(v), max(v) FROM (
    SELECT k, v, v + min(d) lim FROM (
        SELECT x.*, y.k n, y.v - x.v d FROM window_test x
        LEFT JOIN window_test y ON x.k <> y.k AND y.v - x.v > 0) 
    z GROUP BY k, v, n)
w GROUP BY k, lim ORDER BY 2;

I think this is probably a more 'relational' solution, but I'm not sure about its efficiency.

Answer 3

This returns your desired result with the sample data. Not sure if it will work for real world data:

select k, 
       min(v) over (partition by group_nr) as min_v,
       max(v) over (partition by group_nr) as max_v
from (
    select *,
           sum(group_flag) over (order by v,k) as group_nr
    from (
    select *,
           case
              when lag(k) over (order by v) = k then null
              else 1
            end as group_flag
    from window_test
    ) t1
) t2
order by min_v;

I left out the DISTINCT though.