寻找生成森林(WITH RECURSIVE,PostgreSQL 9.5)

2023年6月19日 239次阅读

我有一个任意数量的人的身份表(即别名).每行都有一个以前的名称和一个新名称.在生产中,大约有1M行.例如：

id, old, new
---
1, 'Albert', 'Bob'
2, 'Bob', 'Charles'
3, 'Mary', 'Nancy'
4, 'Charles', 'Albert'
5, 'Lydia', 'Nancy'
6, 'Zoe', 'Zoe'

我想要的是生成用户列表并引用他们各自的身份.这类似于查找连接标识的每个图形中的所有节点,或查找生成林：

User 1: Albert, Bob, Charles (identities: 1,2,4)
User 2: Mary, Nancy, Lydia (identities: 3,5)
User 3: Zoe (identities: 6)

我一直在修补PostgreSQL的WITH RECURSIVE,但它为每个产生了集合和子集.例如：

1,2,4 <-- spanning tree: good
2     <-- subset: discard
3,5   <-- spanning tree: good
4     <-- subset: discard
5     <-- subset: discard
6     <-- spanning tree: good

我需要做什么才能为每个用户生成完整的身份集(即生成树)？

SQLFiddle：http://sqlfiddle.com/#!15/9eaed/4我的最新尝试.这是代码：

WITH RECURSIVE search_graph AS (
   SELECT id
    , id AS min_id
    , ARRAY[id] AS path
    , ARRAY[old,new] AS emails
   FROM   identities

   UNION 

   SELECT identities.id
    , LEAST(identities.id, sg.min_id)
    , (sg.path || identities.id)
    , (sg.emails || identities.old || identities.new)

   FROM search_graph sg
   JOIN identities ON (identities.old = ANY(sg.emails) OR identities.new = ANY(sg.emails))
   WHERE  identities.id <> ALL(sg.path)
)
SELECT array_agg(DISTINCT(p)) from search_graph, unnest(path) p GROUP BY min_id;

结果如下：

1,2,4
2
3,5
4
5
6

最佳答案 我刚才写了一个类似问题的答案：
How to find all connected subgraphs of an undirected graph.在那个问题中,我使用了SQL Server.有关中间CTE的详细说明,请参阅该答案.我将该查询改编为Postgres.

使用Postgres数组功能可以更有效地编写它,而不是将路径连接到文本列.

WITH RECURSIVE
CTE_Idents
AS
(
    SELECT old AS Ident
    FROM identities

    UNION

    SELECT new AS Ident
    FROM identities
)
,CTE_Pairs
AS
(
    SELECT old AS Ident1, new AS Ident2
    FROM identities
    WHERE old <> new

    UNION

    SELECT new AS Ident1, old AS Ident2
    FROM identities
    WHERE old <> new
)
,CTE_Recursive
AS
(
    SELECT
        CTE_Idents.Ident AS AnchorIdent 
        , Ident1
        , Ident2
        , ',' || Ident1 || ',' || Ident2 || ',' AS IdentPath
        , 1 AS Lvl
    FROM 
        CTE_Pairs
        INNER JOIN CTE_Idents ON CTE_Idents.Ident = CTE_Pairs.Ident1

    UNION ALL

    SELECT 
        CTE_Recursive.AnchorIdent 
        , CTE_Pairs.Ident1
        , CTE_Pairs.Ident2
        , CTE_Recursive.IdentPath || CTE_Pairs.Ident2 || ',' AS IdentPath
        , CTE_Recursive.Lvl + 1 AS Lvl
    FROM
        CTE_Pairs
        INNER JOIN CTE_Recursive ON CTE_Recursive.Ident2 = CTE_Pairs.Ident1
    WHERE
        CTE_Recursive.IdentPath NOT LIKE ('%,' || CTE_Pairs.Ident2 || ',%')
)
,CTE_RecursionResult
AS
(
    SELECT AnchorIdent, Ident1, Ident2
    FROM CTE_Recursive
)
,CTE_CleanResult
AS
(
    SELECT AnchorIdent, Ident1 AS Ident
    FROM CTE_RecursionResult

    UNION

    SELECT AnchorIdent, Ident2 AS Ident
    FROM CTE_RecursionResult
)
,CTE_Groups
AS
(
  SELECT
    CTE_Idents.Ident
    ,array_agg(COALESCE(CTE_CleanResult.Ident, CTE_Idents.Ident) 
        ORDER BY COALESCE(CTE_CleanResult.Ident, CTE_Idents.Ident)) AS AllIdents
  FROM
    CTE_Idents
    LEFT JOIN CTE_CleanResult ON CTE_CleanResult.AnchorIdent = CTE_Idents.Ident
  GROUP BY CTE_Idents.Ident
)
SELECT AllIdents
FROM CTE_Groups
GROUP BY AllIdents
;

我在样本数据中添加了一行(7,X,Y).

SQL Fiddle

结果

|          allidents |
|--------------------|
|   Lydia,Mary,Nancy |
| Albert,Bob,Charles |
|                X,Y |
|                Zoe |
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