SQL 分组 interescting/overlapping 行

Question

我在 Postgres 中有以下 table 两列 a_sno 和 b_sno.

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 行数据值 4、5、6 和 7 可以看出，两列 intersects/overlaps 需要划分到一个组中。第 7-16 行和第 17-18 行也将分别标记为第 2 组和第 3 组。

结果输出应如下所示：

group | value
------+------
1     | 4
1     | 5
1     | 6
1     | 7
2     | 9
2     | 10
2     | 13
2     | 14
3     | 11
3     | 15

Answer 1

这是一个开始，可能会提供一些关于方法的想法。递归查询从每条记录的 a_sno 开始，然后尝试沿着 b_sno 的路径，直到到达终点或形成循环。路径由 sno 整数数组表示。

unnest 函数会将数组分成行，因此 sno 值映射到路径数组，例如：

4, {6, 5, 4}

将为数组中的每个值转换为一行：

4, 6
4, 5
4, 4

array_agg 然后通过将值聚合回一个路径来反转操作，但删除重复项和排序。

现在每个a_sno都与一条路径相关联，路径构成了分组。 dense_rank 可用于将分组（集群）映射到数字。

SELECT array_agg(DISTINCT map ORDER BY map) AS cluster
      ,sno
  FROM ( WITH RECURSIVE x(sno, path, cycle) AS (
             SELECT a_sno, ARRAY[a_sno], false FROM  data
             UNION ALL
             SELECT b_sno, path || b_sno, b_sno = ANY(path)
               FROM data, x
               WHERE a_sno = x.sno
                 AND NOT cycle
         )
         SELECT sno, unnest(path) AS map FROM x ORDER BY 1
       ) y
  GROUP BY sno
  ORDER BY 1, 2

输出：

   cluster    | sno 
--------------+-----
 {4,5,6,7}    |   4
 {4,5,6,7}    |   5
 {4,5,6,7}    |   6
 {4,5,6,7}    |   7
 {9,10,13,14} |   9
 {9,10,13,14} |  10
 {9,10,13,14} |  13
 {9,10,13,14} |  14
 {11,15}      |  11
 {11,15}      |  15
(10 rows)

再绕一次排名：

SELECT dense_rank() OVER(order by cluster) AS rank
      ,sno
  FROM (
    SELECT array_agg(DISTINCT map ORDER BY map) AS cluster
          ,sno
      FROM ( WITH RECURSIVE x(sno, path, cycle) AS (
                 SELECT a_sno, ARRAY[a_sno], false FROM  data
                 UNION ALL
                 SELECT b_sno, path || b_sno, b_sno = ANY(path)
                   FROM data, x
                   WHERE a_sno = x.sno
                     AND NOT cycle
             )
             SELECT sno, unnest(path) AS map FROM x ORDER BY 1
           ) y
      GROUP BY sno
      ORDER BY 1, 2
) z

输出：

 rank | sno 
------+-----
    1 |   4
    1 |   5
    1 |   6
    1 |   7
    2 |   9
    2 |  10
    2 |  13
    2 |  14
    3 |  11
    3 |  15
(10 rows)

Answer 2

我想说另外一种方法，可能有用，你可以分两步完成：

1. 每组取 max(sno):

  select q.sno,
         row_number() over(order by q.sno) gn
  from(
    select distinct d.a_sno sno
    from data d
    where not exists (
      select b_sno
      from data
      where b_sno=d.a_sno
      and a_sno>d.a_sno
     )
    )q

结果：

sno gn
7   1
14  2
15  3

2.使用recursive cte查找群组中的所有相关成员：

with recursive cte(sno,gn,path,cycle)as(
  select q.sno,
         row_number() over(order by q.sno) gn,
         array[q.sno],false
  from(
    select distinct d.a_sno sno
    from data d
    where not exists (
      select b_sno
      from data
      where b_sno=d.a_sno
      and a_sno>d.a_sno
     )
    )q
  union all
  select d.a_sno,c.gn,
         d.a_sno || c.path,
         d.a_sno=any(c.path)
  from data d
  join cte c on d.b_sno=c.sno 
  where not cycle
 )
 select distinct gn,sno from cte
 order by gn,sno

结果：

here is the demo 我做了什么。

Answer 3

假设所有对都存在于它们的镜像组合中 (4,5) 和 (5,4)。但是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 中添加的复杂情况是不可能的，我们可以使用这个解决方案而不重复rCTE：

我从每组最小值 a_sno 开始，最小值关联 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 window 函数：

Get the distinct sum of a joined table column

结果：

grp  a_sno  b_sno
1    4      5
2    9      10
3    11     15

我避免分支和重复（倍增）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 来使它在递归 CTE 中运行。

性能的关键是匹配索引，它已经由 PK 约束提供 PRIMARY KEY (a_sno,b_sno)：而不是相反 ~~(b_sno, a_sno)~~:

Is a composite index also good for queries on the first field?

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;

不太简单的案例

所有节点都可以从根（最小sno）的一个或多个分支按升序到达。

这次把全部取大sno，去重可能被多次访问的节点，末尾有UNION：

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。在这种情况下考虑。

SQL 分组 interescting/overlapping 行

SQL grouping interescting/overlapping rows

sql

postgresql

recursive-query

common-table-expression

简单案例

不太简单的案例

无向图