一般来说，这就是@gordonlinoff所说的图行走问题，但是如果任务可以简化为您的示例中那样，使用有限的列并具有排序规则（我假设行的顺序是由您的列定义的），那么任务是完全可解的。使用数据集查看此演示，它会生成所需的结果（请阅读代码中的注解）：

with your_data as ( --your data example
select stack (5,
1, 2,
1, 3,
2, 3,
4, 5,
6, 7 
) as (v1,v2)
) --your data example

select v1, v2, --calculate group Id as a running count of new_grp
       count(new_grp) over(order by v1, v2 rows between unbounded preceding and current row) as gid
from
(
select v1, v2, --calculate new_grp flag
       case when ((not array_contains(prev_tuple,v1) and not array_contains(prev_tuple,v2)) or prev_tuple is null) then true end as new_grp
from
(
select v1, v2, lag(tuple) over (order by v1, v2) prev_tuple --get previous values in array, to simplify code a little bit
  from (select v1, v2, array(v1, v2) as tuple from your_data) s
)s
)s;

结果：

v1      v2      gid
1       2       1
1       3       1
2       3       1
4       5       2
6       7       3