差距和岛屿 - POSTGRESQL

Question

我有下表

Name PathID PathLength PathStepNum
A      1      5          1
A      1      5          2
B      1      5          3
C      1      5          4
A      1      5          5
Y      2      3          1
Z      2      3          2
Z      2      3          3

我需要得到

PathID   PathName
1        A x2 > B x1 > C x1 > A x1
2        Y x1 > Z x2

所以本质上我需要做的是：对于每个PathID，计算Name连续出现的次数，并按照PathStepNum的顺序将Name + NumConsecOccurrences拼接在一起。

有什么想法吗？ 谢谢！

Answer 1

具有窗口和聚合函数的经典解决方案：

select 
    pathid, 
    string_agg(format('%s x%s', name, count), ' > ' order by min) as path
from (
    select 
        pathid, name, min(pathstepnum), 
        count(*)
    from (
        select 
            pathid, name, pathstepnum, 
            sum(mark) over w as grp
        from (
            select 
                pathid, name, pathstepnum, 
                (name <> lag(name, 1, name) over w)::int as mark
            from my_table
            window w as (partition by pathid order by pathstepnum)
            ) s
        window w as (partition by pathid order by pathstepnum)
        ) s
    group by pathid, name, grp
    ) s
group by pathid

 pathid |           path            
--------+---------------------------
      1 | A x2 > B x1 > C x1 > A x1
      2 | Y x1 > Z x2
(2 rows)    

DbFiddle.

然而，这似乎可以在客户端应用程序（在 python、java 或其他任何使用的）中以更简单和更有效的方式实现。