我在 Postgres 中有下表,在 a_sno
和 b_sno
两列中有重叠数据。
create table data
( a_sno integer not null,
b_sno integer not null,
PRIMARY KEY (a_sno,b_sno)
);
insert into data (a_sno,b_sno) values
( 4, 5 )
, ( 5, 4 )
, ( 5, 6 )
, ( 6, 5 )
, ( 6, 7 )
, ( 7, 6 )
, ( 9, 10)
, ( 9, 13)
, (10, 9 )
, (13, 9 )
, (10, 13)
, (13, 10)
, (10, 14)
, (14, 10)
, (13, 14)
, (14, 13)
, (11, 15)
, (15, 11);
从前 6 行可以看出,两列中的数据值 4、5、6 和 7 相交/重叠,需要划分到一个组中。第 7-16 行和第 17-18 行也是如此,它们将分别标记为第 2 组和第 3 组。
结果输出应该是这样的:
group | value
------+------
1 | 4
1 | 5
1 | 6
1 | 7
2 | 9
2 | 10
2 | 13
2 | 14
3 | 11
3 | 15
最佳答案
假设所有对都存在于它们的镜像组合中,还有 (4,5)
和 (5,4)
。但以下解决方案在没有镜像复制的情况下也能正常工作。
简单案例
所有连接都可以按单个升序排列,并且像我在 fiddle 中添加的那样复杂是不可能的,我们可以在 rCTE 中使用没有重复的解决方案:
我首先获取每个组的最小 a_sno
,以及最小关联的 b_sno
:
SELECT row_number() OVER (ORDER BY a_sno) AS grp
, a_sno, min(b_sno) AS b_sno
FROM data d
WHERE a_sno < b_sno
AND NOT EXISTS (
SELECT 1 FROM data
WHERE b_sno = d.a_sno
AND a_sno < b_sno
)
GROUP BY a_sno;
这只需要一个查询级别,因为可以在聚合上构建窗口函数:
结果:
grp a_sno b_sno
1 4 5
2 9 10
3 11 15
我避免分支和重复(多行)行 - 长链可能多更昂贵。我在相关子查询中使用 ORDER BY b_sno LIMIT 1
来实现递归 CTE。
性能的关键是匹配索引,它已经由 PK 约束 PRIMARY KEY (a_sno,b_sno)
提供:不是相反 :(b_sno, a_sno )
WITH RECURSIVE t AS (
SELECT row_number() OVER (ORDER BY d.a_sno) AS grp
, a_sno, min(b_sno) AS b_sno -- the smallest one
FROM data d
WHERE a_sno < b_sno
AND NOT EXISTS (
SELECT 1 FROM data
WHERE b_sno = d.a_sno
AND a_sno < b_sno
)
GROUP BY a_sno
)
, cte AS (
SELECT grp, b_sno AS sno FROM t
UNION ALL
SELECT c.grp
, (SELECT b_sno -- correlated subquery
FROM data
WHERE a_sno = c.sno
AND a_sno < b_sno
ORDER BY b_sno
LIMIT 1)
FROM cte c
WHERE c.sno IS NOT NULL
)
SELECT * FROM cte
WHERE sno IS NOT NULL -- eliminate row with NULL
UNION ALL -- no duplicates
SELECT grp, a_sno FROM t
ORDER BY grp, sno;
不太简单的情况
所有节点都可以从根(最小的sno
)开始按升序到达一个或多个分支。
这一次,获取所有更大的sno
并在最后用UNION
去重可能被多次访问的节点:
WITH RECURSIVE t AS (
SELECT rank() OVER (ORDER BY d.a_sno) AS grp
, a_sno, b_sno -- get all rows for smallest a_sno
FROM data d
WHERE a_sno < b_sno
AND NOT EXISTS (
SELECT 1 FROM data
WHERE b_sno = d.a_sno
AND a_sno < b_sno
)
)
, cte AS (
SELECT grp, b_sno AS sno FROM t
UNION ALL
SELECT c.grp, d.b_sno
FROM cte c
JOIN data d ON d.a_sno = c.sno
AND d.a_sno < d.b_sno -- join to all connected rows
)
SELECT grp, sno FROM cte
UNION -- eliminate duplicates
SELECT grp, a_sno FROM t -- add first rows
ORDER BY grp, sno;
与第一个解决方案不同,我们在这里没有得到最后一行 NULL(由相关子查询引起)。
两者都应该表现得很好 - 特别是对于长链/许多分支。结果如愿:
SQL Fiddle (添加行以展示难度)。
无向图
如果存在升序遍历无法从根到达的局部极小值,上述解决方案将不起作用。考虑 Farhęg's solution在这种情况下。
关于SQL分组interescting/overlapping行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29734941/