sql - oracle SQL选择过去x天滚动期间的不同客户

Question

假设您有一张客户表，日期如下:
[客户表]

+----------+-----------+----------+
| customer | date      | purchase |
+----------+-----------+----------+
| 1        | 1/01/2016 | 12       |
+----------+-----------+----------+
| 1        | 1/12/2016 | 3        |
+----------+-----------+----------+
| 2        | 5/03/2016 | 5        |
+----------+-----------+----------+
| 3        | 1/16/2016 | 6        |
+----------+-----------+----------+
| 3        | 3/22/2016 | 1        |
+----------+-----------+----------+  

我想编写一个查询来计算在过去 10 天内有多少不同的客户购买了一个滚动周期，从每个日历日开始并向后计算 10 天。因此，对于 2016 年的每个独特的日子，最终输出将是一个日历，其中每一天都有一个不同的客户计数，这些客户存在于日历当天的前 10 天，如下所示:
[结果表]

+-----------+------------------+
| date      | unique customers |
+-----------+------------------+
| 1/01/2016 | 112              |
+-----------+------------------+
| 1/02/2016 | 104              |
+-----------+------------------+
| 1/03/2016 | 140              |
+-----------+------------------+
| 1/04/2016 | 133              |
+-----------+------------------+
| ....      | 121              |
+-----------+------------------+  

我想出的一个解决方案是创建一个只有一列的日历表，然后使用不等式连接将日历表连接到客户表。我认为这是非常低效的，并且正在寻求更快的解决方案。所以我的第一步是像这样创建一个日历:
[日历]

+-----------+
| date      |
+-----------+
| 1/01/2016 |
+-----------+
| 1/02/2016 |
+-----------+
| 1/03/2016 |
+-----------+
| 1/04/2016 |
+-----------+
| 1/05/2016 |
+-----------+  

然后对于该日历中的每一天，为了计算每一天之前的不同客户集，我加入了一个不等式，如下所示:

select
count(distinct customer) as unique customers
from calendar c
left join mytable m
on c.date>=m.date and m.date>=c.date-10  

虽然我相信这是正确的，但它运行得非常慢(假设一个日历有 2 年的时间有几百万客户)。是否有 oracle 分析函数可以帮到我？

Answer 1

Is there an oracle analytic function that may help me out here?

不是真的 - 来自 COUNT() documentation :

If you specify DISTINCT, then you can specify only the query_partition_clause of the analytic_clause. The order_by_clause and windowing_clause are not allowed.

您可能需要 DISTINCT 和 windowing_clause，这是不允许的。

更新:

使用按客户分区的非DISTINCT 分析查询组合，然后按天聚合，您可以获得与无效语法相同的效果:

Oracle 设置:

CREATE TABLE table_name ( customer, dt ) AS
  SELECT 1, DATE '2017-01-10' FROM DUAL UNION ALL
  SELECT 1, DATE '2017-01-11' FROM DUAL UNION ALL
  SELECT 1, DATE '2017-01-15' FROM DUAL UNION ALL
  SELECT 1, DATE '2017-01-20' FROM DUAL UNION ALL
  SELECT 2, DATE '2017-01-12' FROM DUAL UNION ALL
  SELECT 2, DATE '2017-01-19' FROM DUAL UNION ALL
  SELECT 3, DATE '2017-01-10' FROM DUAL UNION ALL
  SELECT 3, DATE '2017-01-13' FROM DUAL UNION ALL
  SELECT 3, DATE '2017-01-15' FROM DUAL UNION ALL
  SELECT 3, DATE '2017-01-20' FROM DUAL;

查询:

注意:下面的查询只是查询一个月的数据和前两天的范围来说明原理，但是把参数改成12个月10天也很容易。

SELECT day,
       SUM( has_order_in_range ) AS unique_customers
FROM   (
  SELECT customer,
         day,
         LEAST(
           1,
           COUNT(dt) OVER ( PARTITION BY customer
                            ORDER BY day
                            RANGE BETWEEN INTERVAL '2' DAY PRECEDING
                                      AND INTERVAL '0' DAY FOLLOWING )
         ) AS has_order_in_range
  FROM   table_name t
         PARTITION BY ( customer )
         RIGHT OUTER JOIN
         ( -- Create a calendar for one month
           SELECT DATE '2017-01-01' + LEVEL - 1 AS day
           FROM   DUAL
           CONNECT BY DATE '2017-01-01' + LEVEL - 1 < ADD_MONTHS( DATE '2017-01-01', 1 )
         ) d
         ON ( t.dt = d.day )
)
GROUP BY day
ORDER BY day;

输出:

DAY                 UNIQUE_CUSTOMERS
------------------- ----------------
2017-01-01 00:00:00                0
2017-01-02 00:00:00                0
2017-01-03 00:00:00                0
2017-01-04 00:00:00                0
2017-01-05 00:00:00                0
2017-01-06 00:00:00                0
2017-01-07 00:00:00                0
2017-01-08 00:00:00                0
2017-01-09 00:00:00                0
2017-01-10 00:00:00                2
2017-01-11 00:00:00                2
2017-01-12 00:00:00                3
2017-01-13 00:00:00                3
2017-01-14 00:00:00                2
2017-01-15 00:00:00                2
2017-01-16 00:00:00                2
2017-01-17 00:00:00                2
2017-01-18 00:00:00                0
2017-01-19 00:00:00                1
2017-01-20 00:00:00                3
2017-01-21 00:00:00                3
2017-01-22 00:00:00                2
2017-01-23 00:00:00                0
2017-01-24 00:00:00                0
2017-01-25 00:00:00                0
2017-01-26 00:00:00                0
2017-01-27 00:00:00                0
2017-01-28 00:00:00                0
2017-01-29 00:00:00                0
2017-01-30 00:00:00                0
2017-01-31 00:00:00                0