一种方法是使用 self join 作为 distinct 在窗口函数中不支持。

select t1.id,count(distinct t2.port) as cnt
from tbl t1
join tbl t2 on t1.id-t2.id>=0 and t1.id-t2.id<=5 --change this number per requirements
group by t1.id
order by t1.id

这假设id是按顺序排列的。
如果不是，首先得到行号并使用上面的逻辑。就像

with rownums as (select id,port,row_number() over(order by id) as rnum
                 from tbl)
select r1.id,count(distinct r2.port)
from rownums r1
join rownums r2 on r1.rnum-r2.rnum>=0 and r1.rnum-r2.rnum<=5
group by r1.id

我不认为有一个简单的方法。一种方法使用 lag() :

select ( (case when port_5 is not null then 1 else 0 end) +
         (case when port_4 is not null and port_4 not in (port_5) then 1 else 0 end) +
         (case when port_3 is not null and port_3 not in (port_5, port_4) then 1 else 0 end) +
         (case when port_2 is not null and port_2 not in (port_5, port_4, port_3) then 1 else 0 end) +
         (case when port_1 is not null and port_1 not in (port_5, port_4, port_3, port_2) then 1 else 0 end) +
         (case when port is not null and port not in (port_5, port_4, port_3, port_2, port_2) then 1 else 0 end)
       ) as cumulative_distinct_count
from (select t.*,
             lag(port, 5) over (partition by srcport order by id rows) as port_5,
             lag(port, 4) over (partition by srcport order by id rows) as port_4,
             lag(port, 3) over (partition by srcport order by id rows) as port_3,
             lag(port, 2) over (partition by srcport order by id rows) as port_2,
             lag(port, 1) over (partition by srcport order by id rows) as port_1
      from t
     ) t

这是一个复杂的查询，但性能应该还可以。
注： port 以及 srcport 我假设是一样的，但这是借用你的查询。