Grouped LIMIT in PostgreSQL: show the first N rows for each group?

SqlPostgresql

Sql Problem Overview


I need to take the first N rows for each group, ordered by custom column.

Given the following table:

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

I am using PostgreSQL 8.3.5.

Sql Solutions


Solution 1 - Sql

New solution (PostgreSQL 8.4)

SELECT
  * 
FROM (
  SELECT
    ROW_NUMBER() OVER (PARTITION BY section_id ORDER BY name) AS r,
    t.*
  FROM
    xxx t) x
WHERE
  x.r <= 2;

Solution 2 - Sql

Since v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

It might be faster but, of course, you should test performance specifically on your data and use case.

Solution 3 - Sql

Here's another solution (PostgreSQL <= 8.3).

SELECT
  *
FROM
  xxx a
WHERE (
  SELECT
    COUNT(*)
  FROM
    xxx
  WHERE
    section_id = a.section_id
  AND
    name <= a.name
) <= 2

Solution 4 - Sql

SELECT  x.*
FROM    (
        SELECT  section_id,
                COALESCE
                (
                (
                SELECT  xi
                FROM    xxx xi
                WHERE   xi.section_id = xo.section_id
                ORDER BY
                        name, id
                OFFSET 1 LIMIT 1
                ),
                (
                SELECT  xi
                FROM    xxx xi
                WHERE   xi.section_id = xo.section_id
                ORDER BY 
                        name DESC, id DESC
                LIMIT 1
                )
                ) AS mlast
        FROM    (
                SELECT  DISTINCT section_id
                FROM    xxx
                ) xo
        ) xoo
JOIN    xxx x
ON      x.section_id = xoo.section_id
        AND (x.name, x.id) <= ((mlast).name, (mlast).id)

Solution 5 - Sql

        -- ranking without WINDOW functions
-- EXPLAIN ANALYZE
WITH rnk AS (
        SELECT x1.id
        , COUNT(x2.id) AS rnk
        FROM xxx x1
        LEFT JOIN xxx x2 ON x1.section_id = x2.section_id AND x2.name <= x1.name
        GROUP BY x1.id
        )
SELECT this.*
FROM xxx this
JOIN rnk ON rnk.id = this.id
WHERE rnk.rnk <=2
ORDER BY this.section_id, rnk.rnk
        ;

        -- The same without using a CTE
-- EXPLAIN ANALYZE
SELECT this.*
FROM xxx this
JOIN ( SELECT x1.id
        , COUNT(x2.id) AS rnk
        FROM xxx x1
        LEFT JOIN xxx x2 ON x1.section_id = x2.section_id AND x2.name <= x1.name
        GROUP BY x1.id
        ) rnk
ON rnk.id = this.id
WHERE rnk.rnk <=2
ORDER BY this.section_id, rnk.rnk
        ;

Solution 6 - Sql

A lateral join is the way to go, but you should do a nested query first to improve performance on large tables.

SELECT t_limited.*
FROM (
        SELECT DISTINCT section_id
        FROM t
    ) t_groups
    JOIN LATERAL (
        SELECT *
        FROM t t_all
        WHERE t_all.section_id = t_groups.section_id
        ORDER BY t_all.name
        LIMIT 2
    ) t_limited ON true

Without the nested select distinct, the join lateral runs for every line in the table, even though the section_id is often duplicated. With the nested select distinct, the join lateral runs once and only once for each distinct section_id.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionKouber SaparevView Question on Stackoverflow
Solution 1 - SqlDaveView Answer on Stackoverflow
Solution 2 - SqlposhestView Answer on Stackoverflow
Solution 3 - SqlKouber SaparevView Answer on Stackoverflow
Solution 4 - SqlQuassnoiView Answer on Stackoverflow
Solution 5 - SqlwildplasserView Answer on Stackoverflow
Solution 6 - SqlDavid SkinnerView Answer on Stackoverflow