像这样的聚类是一个迭代过程，步骤数未知。我从来没有找到一个可以在递归查询中完成的解决方案。
我已经六年没有从事crm工作了，但是下面的功能与我们以前生成匹配组的方式类似。对于我们的工作负载来说，一行一行地完成这项工作并没有表现得足够好，而使用java等宿主语言来完成这项工作 HashMap() 以及 HashSet() 反向索引会产生非常混乱的代码。
假设此架构：

\d contact_info 
                 Table "public.contact_info"
      Column      |  Type   | Collation | Nullable | Default 
------------------+---------+-----------+----------+---------
 contact_id_a     | bigint  |           |          | 
 contact_id_b     | bigint  |           |          | 
 ignore_duplicate | boolean |           |          | false
 list_id          | integer |           |          | 496

select * from contact_info ;
 contact_id_a | contact_id_b | ignore_duplicate | list_id 
--------------+--------------+------------------+---------
     16247096 |     16247097 | f                |     496
     16247096 |     16247098 | f                |     496
     16247096 |     16247099 | f                |     496
     16247097 |     16247098 | f                |     496
     16247097 |     16247099 | f                |     496
     16247098 |     16247099 | f                |     496
     16247094 |     16247095 | f                |     496
(7 rows)

此函数创建两个临时表来保存中间集群，然后在不再可能进行集群时返回结果。

create or replace function cluster_contact() 
  returns table (clust_id bigint, contact_id bigint) 
  language plpgsql as $$
declare 
  last_count bigint := 1;
  this_count bigint := 0;
begin
  create temp table contact_match (clust_id bigint, contact_id bigint) on commit drop;
  create index cm_1 on contact_match (contact_id, clust_id);
  create index cm_2 on contact_match using hash (clust_id);
  create temp table contact_hold (clust_id bigint, contact_id bigint) on commit drop;

  with dedup as (
    select distinct least(ci.contact_id_a) as clust_id,
           greatest(ci.contact_id_b) as contact_id
      from contact_info ci
     where not ci.ignore_duplicate
  )
  insert into contact_match
    select d.clust_id, d.clust_id from dedup d
    union
    select d.clust_id, d.contact_id from dedup d;

  while last_count > this_count loop

    if this_count = 0 then 
      select count(distinct cm.clust_id) into last_count from contact_match cm;
    else 
      last_count := this_count;
    end if;

    with new_cid as (
      select cm.contact_id as clust_id_old,
             min(cm.clust_id) as clust_id_new
        from contact_match cm
       group by cm.contact_id
    )
    update contact_match
       set clust_id = nc.clust_id_new
      from new_cid nc
     where contact_match.clust_id = nc.clust_id_old;

    truncate table contact_hold;
    insert into contact_hold 
      select distinct * from contact_match;

    truncate table contact_match;
    insert into contact_match
      select * from contact_hold;

    select count(distinct cm.clust_id) into this_count from contact_match cm;

  end loop;

  return query select * from contact_match order by clust_id, contact_id;
end $$;

我见过的开发人员面临的最大的心理障碍之一就是忽略了 contact_id 对自己。这导致了不相交的处理和不必要的复杂的心理模型的左侧和右侧。

select * from cluster_contact();
 clust_id | contact_id 
----------+------------
 16247094 |   16247094
 16247094 |   16247095
 16247096 |   16247096
 16247096 |   16247097
 16247096 |   16247098
 16247096 |   16247099
(6 rows)

如果您需要澄清此解决方案中的任何步骤，或者它不适合您，请发表意见。
另外，要知道levenshtein在 fuzzystrmatch ，效果很好。
如果你愿意 clust_id 开始于 1 ，更改您的 return query 在此函数中：

return query 
    select dense_rank() over (order by cm.clust_id) as clust_id, 
           cm.contact_id 
      from contact_match cm 
     order by clust_id, contact_id;

它将产生：

select * from cluster_contact();
 clust_id | contact_id 
----------+------------
        1 |   16247094
        1 |   16247095
        2 |   16247096
        2 |   16247097
        2 |   16247098
        2 |   16247099
(6 rows)

可以使用递归cte。这将遍历图形，然后为每一行分配图形中的最小标识符。请注意，您的数据没有每行的唯一标识符，因此首先生成一个：

with recursive d as (
      select row_number() over (order by contact_id_a, contact_id_b) as id, d.*
      from duplicates d
     ),
     cte (id, contact_id_a, contact_id_b, min_id, ids, lev) as (
      select id, contact_id_a, contact_id_b, id as min_id, array[id] as ids, 1 as lev
      from d
      union all
      select d.id, d.contact_id_a, d.contact_id_b, least(d.id, cte.min_id), ids || d.id, lev + 1
      from cte join
           d
           on cte.contact_id_a = d.contact_id_a or cte.contact_id_b = d.contact_id_b
      where d.id <> ALL (cte.ids)
     )
select distinct on (id) cte.*
from cte
order by id, min_id;

列 min_id 包含所需的分组。
下面是一个db<>小提琴来演示代码。