Window PostgreSQL 中分组依据的函数

Question

每次操作字段更改为值 1 时，我都需要对每个用户进行计数。如果第一个条目为 1，那也算在内。行乱序但应按 action_date.

顺序计数

换句话说，我认为需要做的是：按 user_id 对行进行分组，按时间戳对它们进行排序，然后计算 action=1 和 action != 前一行的频率。

例子

create table t (
user_id int,
action_date timestamp,
action int
);

Insert into t(user_id, action_date, action)
values
(1, '2017-01-01 00:00:00', 1),
(2, '2017-01-01 00:00:00', 0),
(1, '2017-01-03 00:00:00', 1),
(2, '2017-01-03 00:00:00', 0),
(1, '2017-01-02 00:00:00', 1),
(2, '2017-01-02 00:00:00', 1),
(1, '2017-01-04 00:00:00', 1),
(2, '2017-01-04 00:00:00', 1);

结果应该是

 user_id | count 
---------+-------
       1 |     1
       2 |     2

在回答的帮助下，我可以通过这种方式获得单个帐户的结果，

select user_id, count(*)
from (select user_id, action_date,action,lag(action) over(order by action_date) as prev_action
      from t where user_id=2
     ) t
where (action<>prev_action and action=1) or (action=1 and prev_action is null)
group by user_id;

但我一直在尝试将其扩展到所有用户。

Answer 1

将lag()函数与partition by结合使用：

select user_id, count(*)
from (select t.*,
             lag(action) over (partition by user_id order by action_date) as prev_action
      from t
     ) t
where (action = 1) and (prev_action is distinct from 1)
group by user_id;

Window PostgreSQL 中分组依据的函数

Window function within group by in PostgreSQL

sql

postgresql

group-by

window-functions