计算有序行集中的相等、连续的值

Question

我有一个包含两列的 table，例如：

CREATE TABLE actions (
  action_time TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
  "action" text NOT NULL
);

其中包含以下数据：

        action_time         | action 
----------------------------+--------
 2016-12-30 14:12:33.353269 | a
 2016-12-30 14:12:38.536818 | b
 2016-12-30 14:12:43.305001 | a
 2016-12-30 14:12:49.432981 | a
 2016-12-30 14:12:53.536397 | b
 2016-12-30 14:12:57.449101 | b
 2016-12-30 14:13:01.592785 | a
 2016-12-30 14:13:06.192907 | b
 2016-12-30 14:13:11.249181 | b
 2016-12-30 14:13:13.690897 | b
(10 rows)

您可以假定 action_time 列中没有重复值。

如何计算从上一个动作开始连续执行的相同动作的次数？

连续相同动作的次数没有限制，任何一个动作都可以是最后一个。此外，不同操作的多样性没有限制：我只使用了两个来简化示例数据。

对于此示例数据，我希望结果为 3。这是因为最后一个操作是 "b"，并且它连续发生了 3 次。

我认为可以结合 window 函数和 WITH RECURSIVE 子句来实现解决方案，但我不知道该怎么做。

Answer 1

我对经典的间隙和孤岛解决方案进行了一些改动。
注意 ROW_NUMBER 函数如何使用降序 ORDER BY。

select  count(*)

from   (select  

            action
           ,row_number() over (                    order by action_time desc) as rn
           ,row_number() over (partition by action order by action_time desc) as rn_action

        from    mytab
        ) t

group by action
        ,rn - rn_action

having   min(rn) = 1

Answer 2

想到这个：

select count(*)
from t cross join
     (select t2.action
      from t t2
      order by action_time desc
      limit 1
     ) last
where t.action_time >= (select max(t2.action_time)
                        from t t2
                        where t2.action <> last.action
                       );

这应该能够利用 (action_time, action) 上的索引。

Answer 3

这应该可以做到。

SELECT COUNT(*)
FROM actions
WHERE action_time > (
SELECT action_time
  FROM actions 
  WHERE action <> (SELECT action FROM actions ORDER BY action_time DESC LIMIT 1) 
ORDER BY action_time DESC LIMIT 1);

最内层查询

SELECT action FROM actions ORDER BY action_time DESC LIMIT 1

确定最后一个动作。

查询

SELECT action_time
  FROM actions 
  WHERE action <> (SELECT action FROM actions ORDER BY action_time DESC LIMIT 1) 
ORDER BY action_time DESC LIMIT 1

找到具有不同操作的最后一行。

最外层查询查找该行之后的所有行。

Answer 4

改进的解决方案

select  count(*)

from   (select  

            action
           ,row_number() over (                    order by action_time desc) as rn
           ,row_number() over (partition by action order by action_time desc) as rn_action

        from    mytab
        ) t

where   rn = rn_action

计算有序行集中的相等、连续的值

Count equal, consecutive values in an ordered rowset

sql

postgresql

window-functions

gaps-and-islands

postgresql-9.6