Postgresql 中连续相同符号行的总和

Question

假设下表：

date   |  user    |   amount
-----------------------------
01-01  |  John    |       +3
01-02  |  John    |       +2
01-03  |  John    |       +1
01-04  |  John    |       -5
01-05  |  John    |       +1
01-01  |  Jack    |       +2
01-02  |  Jack    |       -1
01-03  |  Jack    |       -6

对于每个用户，按用户对amount进行分组并签名并将总和应用于相同符号的连续金额的方法是什么？ 从而给出这个output：

date   |  user    |   amount
-----------------------------
01-01  |  John    |       +6  <- this is the sum of all consecutive same sign
01-04  |  John    |       -5
01-05  |  John    |       +1
01-01  |  Jack    |       +2
01-02  |  Jack    |       -7  <- this is the sum of all consecutive same sign

我试过使用 window 函数，但我能得到的最接近的是：

select
    sum(amount) over (partition by user, sign(amount) order by date)
from my_table

这不会导致所需的 output。

Answer 1

You can use LAG() window function to check if the previous amount has the same sign as the current amount and create a boolean flag which you can use with SUM() window function to create the consecutive groups with the same sign and then aggregate:

SELECT MIN(date) date,
       "user",
       SUM(amount) amount
FROM (       
  SELECT *, SUM(flag) OVER (PARTITION BY "user" ORDER BY date) grp
  FROM (
    SELECT *, COALESCE(SIGN(amount) <> LAG(SIGN(amount)) OVER (PARTITION BY "user" ORDER BY date), true)::int flag 
    FROM my_table
  ) t  
) t 
GROUP BY "user", grp
ORDER BY "user", date;

请参阅演示。