每天的唯一累积客户

Question

任务：获取每个拒绝原因和每天的唯一累积客户总数。

    Input data sample:
    +---------+--------------+------------+------+
    | Cust_Id |  Decline_Dt  |  Reason    | Days |
    +---------+--------------+------------+------+
    |   A     |  08-09-2020  |  Reason_1  |   0  |
    |   A     |  08-09-2020  |  Reason_1  |   1  | 
    |   A     |  08-09-2020  |  Reason_1  |   2  |
    |   A     |  08-09-2020  |  Reason_1  |   4  |
    |   B     |  08-09-2020  |  Reason_1  |   0  |
    |   B     |  08-09-2020  |  Reason_1  |   2  |
    |   B     |  08-09-2020  |  Reason_1  |   3  |
    |   C     |  08-09-2020  |  Reason_1  |   1  |
    +---------+--------------+------------+------+   
1) Decline_dt - The date on which the payment was declined. (Ignore it for this task)
2) Days - Indicates the # of days after the payment decline happened, the customer interacted with IVR channel. 
3) Reason - Indicates the payment decline reason 
    
    --Expected Output:
    +---------------+-----------+---------------+----------------------------+
    |   Reason      |   Days    | Unique_mtns   | total_cumulative_customers |
    +---------------+-----------+---------------+----------------------------+
    |   Reason_1    |   0       |   2           |           2                |              
    |   Reason_1    |   1       |   2           |           3                |
    |   Reason_1    |   2       |   2           |           3                | 
    |   Reason_1    |   3       |   1           |           3                | 
    |   Reason_1    |   4       |   1           |           3                | 
    +------------------------------------------------------------------------+

我的 Hive 查询：

select a.Reason
        , a.days
        -- , count(distinct a.cust_id) as unique_mtns
        , count(distinct a.cust_id) over (partition by Reason 
                                                order by a.days rows between unbounded preceding and current row) 
                                        as total_cumulative_customers
from table as a  
group by a.reason
        , a.days

输出（不正确）：

+---------------+-----------+----------------------------+
|   Reason      |   Days    | total_cumulative_customers |
+---------------+-----------+----------------------------+
|   Reason_1    |   0       |               2            |              
|   Reason_1    |   1       |               2            |
|   Reason_1    |   2       |               2            | 
|   Reason_1    |   3       |               1            | 
|   Reason_1    |   4       |               1            | 
+--------------------------------------------------------+

理想情况下，我希望在没有 group by 的情况下执行窗口函数。 但是，我得到一个没有 group by 的错误。当我使用 group by 时，我没有得到累积的客户。

Answer 1

如果我没看错，您可以使用子查询来计算每个客户/原因元组的第一天，然后进行条件聚合：

select reason, days, 
    count(distinct cust_id) as unique_mtns,
    sum(sum(case when days = min_days then 1 else 0 end)) 
        over(partition by reason order by days) as total_cumulative_customers
from (
    select reason, cust_id, 
        min(days) over(partition by reason, cust_id) as min_days
    from mytable
) t
group by reason, days

Answer 2

我建议使用row_number() 来枚举行或给定的客户和原因。您的代码在用户 ID 上使用了 count(distinct)，这表明您可能在给定日期有重复项。

这将是：

select reason, days, count(distinct cust_id) as unique_mtns,
       sum(sum(case when seqnum = 1 then 1 else 0 end)) over (partition by reason order by days) as total_cumulative_customers
from (select t.*,
             row_number() over (partition by reason, cust_id order by days) as seqnum
      from t
     ) t
group by reason, days
order by reason, days;