Oracle 中不同的累积计数答案

【问题标题】：Cumulative count distinct in OracleOracle 中不同的累积计数
【发布时间】：2016-03-14 14:33:37
【问题描述】：

我需要数周累计客户数量。有什么好的方法可以在不进行子查询的情况下做到这一点？

我想寻找一些分析函数，但没有发现任何有价值的东西。所有可行的解决方案都使用子查询或总和，这对我的任务来说是错误的。

例如，对于这些表

WITH the_data as (  
   select sysdate - 28 dt, 1 val from dual union all  
   select sysdate - 27 dt, 13 val from dual union all  
   select sysdate - 20 dt, 15 val from dual union all  
   select sysdate - 19 dt, 1 cusval from dual union all  
   select sysdate - 18 dt, 2  from dual union all  
   select sysdate - 17 dt, 3  from dual union all  
   select sysdate - 16 dt, 4  from dual union all  
   select sysdate - 15 dt, 5  from dual union all  
   select sysdate - 14 dt, 6  from dual union all  
   select sysdate - 8 dt, 7  from dual union all  
   select sysdate - 6 dt, 8  from dual union all  
   select sysdate - 3 dt, 9  from dual union all  
   select sysdate - 2 dt, 9  from dual union all  
   select sysdate - 1 dt, 10  from dual)

我想获得关注

+----------------------------------+
| WEEK_NUM    CUM_DISTINCT_COUNT   |
+----------------------------------+
| 8           2                    |
| 9           6                    |
| 10          8                    |
| 11          11                   |
| 12          12                   |
+----------------------------------+

【问题讨论】：

您应该发布您的表结构、示例数据、您尝试过的内容以及期望的结果；没有这些信息，就很难回答。 Here你可以阅读如何提出一个好问题
没有任何数据将很难为您提供帮助，但解决方案可能类似于COUNT(DISTINCT CUST_NUMBER) OVER (PARTITION BY WEEK)

标签： sql oracle count distinct

【解决方案1】：

您如何执行此操作取决于您的计数是否需要来自不同的客户。要显示一些正在运行的聚合选项：

WITH the_data as (
   select sysdate - 28 dt, 1 val from dual union all
   select sysdate - 27 dt, 13 val from dual union all
   select sysdate - 20 dt, 15 val from dual union all
   select sysdate - 19 dt, 1 cusval from dual union all
   select sysdate - 18 dt, 2  from dual union all
   select sysdate - 17 dt, 3  from dual union all
   select sysdate - 16 dt, 4  from dual union all
   select sysdate - 15 dt, 5  from dual union all
   select sysdate - 14 dt, 6  from dual union all
   select sysdate - 8 dt, 7  from dual union all
   select sysdate - 6 dt, 8  from dual union all
   select sysdate - 3 dt, 9  from dual union all
   select sysdate - 2 dt, 9  from dual union all
   select sysdate - 1 dt, 10  from dual)
select distinct trunc(dt,'WW') wk
      ,count(val) over (partition by trunc(dt,'WW')) as wk_cnt
      ,count(distinct val) over (partition by trunc(dt,'WW')) as wk_distinct_cnt
      ,count(val) over (order by trunc(dt,'WW')) as running_wk_cnt
--      ,count(distinct val) over (order by trunc(dt,'WW')) as running_distinct_wk_cnt
from the_data
order by trunc(dt,'WW')

returns
WK,         WK_CNT, WK_DISTINCT_CNT, RUNNING_WK_CNT
12/02/2016, 2,      2,               2
19/02/2016, 3,      3,               5
26/02/2016, 4,      4,               9
04/03/2016, 2,      2,               11
11/03/2016, 3,      2,               14

但是，如果您取消注释该行以执行运行不同计数，您将收到错误，因为完全使用此构造是不可能的。

【讨论】：

【解决方案2】：

SELECT DISTINCT
       TRUNC( date_column, 'WW' ) AS week_number,
       COUNT( customer_id ) OVER ( ORDER BY TRUNC( date_column, 'WW' ) ) AS num_customers
FROM   your_table
ORDER BY week_number

要获得自 2016-01-01 以来不同客户的累积数量，您可以这样做：

SELECT DISTINCT
       last_day_of_week,
       COUNT( is_first_instance_of_customer )
         OVER( ORDER BY last_day_of_week ) AS num_unique_customers
FROM   (
  SELECT TRUNC( date_value, 'WW' ) + 6 AS last_day_of_week,
         CASE ROW_NUMBER()
                OVER ( PARTITION BY customer_id
                       ORDER BY date_value )
         WHEN 1
         THEN 1
         END AS is_first_instance_of_customer
  FROM   your_table
)
ORDER BY last_day_of_week;

【讨论】：

谢谢，但这行不通。这就是我如何获得不同的 week_number 和 num_customers 对。但我需要 2016 年 1 月 1 日到每周最后一天之间的客户数量，加上我的表“your_table”包含 customer_id 的重复项
@Powerfool 更新了不同客户的累积计数。

【解决方案3】：

谢谢大家。这是我要求的有效查询：

with dt as (  
select 'A1' PLANT, '1' CUSTOMERID, date'2016-01-01' CALDAY from dual union all
select 'A1' PLANT, '2' CUSTOMERID, date'2016-01-01' CALDAY from dual union all
select 'A1' PLANT, '3' CUSTOMERID, date'2016-01-01' CALDAY from dual union all
select 'A2' PLANT, '1' CUSTOMERID, date'2016-01-02' CALDAY from dual union all
select 'A2' PLANT, '2' CUSTOMERID, date'2016-01-02' CALDAY from dual union all
select 'A1' PLANT, '1' CUSTOMERID, date'2016-01-08' CALDAY from dual union all
select 'A1' PLANT, '2' CUSTOMERID, date'2016-01-08' CALDAY from dual union all
select 'A2' PLANT, '4' CUSTOMERID, date'2016-01-08' CALDAY from dual union all
select 'A2' PLANT, '4' CUSTOMERID, date'2016-01-15' CALDAY from dual union all
select 'A1' PLANT, '4' CUSTOMERID, date'2016-01-22' CALDAY from dual)


, first_buys_plant as (
  select customerid, plant, calday from (
    select customerid, plant, calday, row_number() over (partition by customerid, plant order by calday) rn
    from dt
  ) where rn = 1
)
select plant, calweek, max(cum_num) cum_num_cust from (
  select 
    plant,
    to_char(calday, 'ww') calweek,
    count(customerid) over(partition by plant order by calday rows between unbounded preceding and current row) cum_num
  from first_buys_plant
) group by plant, calweek
order by 1,2

【讨论】：