如果我理解正确的话，你似乎希望第二个是第一个的超级集合。那就是：

with k as (
      select k.*, count(*) over (partition by doi) as cnt
      from keywords k
     )
select k.doi, k2.doi
from k join
     k k2
     on k2.keyword = k.keyword
group by k.doi, k2.doi, k.cnt
having count(*) = k.cnt;

如果你想要精确匹配，那么包括 k2.cnt = k.cnt 在 on 条款。
（假设没有重复项。）
编辑：
您可以使用以下方法获得完全相同的关键字：

with k as (
      select k.*, count(*) over (partition by doi) as cnt
      from (select distinct keyword, doi from keywords k) k
     )
select k.doi, k2.doi
from k join
     k k2
     on k2.keyword = k.keyword and k2.cnt = k.cnt
group by k.doi, k2.doi, k.cnt
having count(*) = k.cnt;

或者 listagg() 方法：

select keywords, listagg(doi, ',') within group (order by keyword)
from (select doi,
            listagg(keyword, ',') within group (order by keyword) as keywords
      from (select distinct doi, keyword from keywords) k
      group by doi
     ) d
group by keywords;

这可能受到oracle对字符串长度的限制。

你可以用 listagg 具体如下：

With cte as
(Select t.doi, listagg(keyword,',') within group (order by keyword) as keyword
from your_table t
Group by t.id)
Select distinct t1.doi f_doi, t2.id s_doi, t1.keyword
From cte t1 join cte t2
On t1.adventure = t2.adventure
And t1.doi < t2.doi;