分组限制在PostgreSQL:显示每组的前N行?

我需要为每个组取前N行，按自定义列排序。

已知下表:

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)

我需要前2行(按名称排序)为每个section_id，即类似的结果:

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

我使用的是PostgreSQL 8.3.5。

当前回答

从v9.3开始，您可以执行横向连接

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;

这样做可能更快，但是当然，您应该针对您的数据和用例专门测试性能。

2016-06-16 14:25:29

其他回答

        -- 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
        ;

2012-12-07 20:53:01

新解决方案(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;

2011-05-19 19:55:29

横向连接是可行的方法，但您应该首先执行嵌套查询，以提高大型表的性能。

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

如果没有嵌套的选择distinct，则连接横向对表中的每一行运行，即使section_id经常是重复的。由于嵌套的选择是不同的，联接横向操作将为每个不同的section_id运行一次且仅运行一次。

2022-03-24 11:36:19

这里是另一个解决方案(PostgreSQL <= 8.3)。

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

2009-07-17 14:41:23

从v9.3开始，您可以执行横向连接

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;

这样做可能更快，但是当然，您应该针对您的数据和用例专门测试性能。

2016-06-16 14:25:29

分组限制在PostgreSQL:显示每组的前N行?

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