雪花 SQL 计数与窗口函数不同

Question

我有一个包含每周客户和商店信息的数据集，

问题 - 我必须计算一些特征，例如截至当前周的过去 1、2、3、4、5、6..so 的总独立客户总数。

我在窗口函数上使用 count distinct of customers 列时遇到错误 -

我尝试使用 concat 函数来创建数组，但它也不起作用 - 感谢您的帮助！

SELECT STORE,WEEK,

count(distinct Customers) over (partition by STORE order by WEEK rows between 1 preceding and 1 preceding) as last_1_week_customers,
count(distinct Customers) over (partition by STORE order by WEEK rows between 2 preceding and 2 preceding) as last_2_week_customers
    from TEST_TABLE
    group by STORE,WEEK

错误- SQL 编译错误：distinct 不能与窗口框架或订单一起使用。

我该如何解决这个错误？

输入

CREATE TABLE TEST_TABLE (STORE STRING,WEEK STRING,Customers STRING);


INSERT INTO TEST_TABLE VALUES

('A','1','AA'),
('A','1','DD'),
('A','2','AA'),
('A','2','BB'),
('A','2','CC'),
('A','3','AA'); 

输出

Answer 1

嗯...我认为你在这里根本不需要窗口函数...

首先，我们可以从一个简单的分组开始：

select
    store,
    week,
    count(distinct customers) as cnt
from
    test_table
where
    week >= [this week's number minus 5]
group by
   store, week

这将产生一个简单的表格：

<头>

商店	周	cnt
A	1	2
A	2	3
A	3	1

此时我想请您考虑一下这是否已经足够了。可能您已经可以将这种格式的数据用于您需要的任何目的。但如果不是，那么我们可以进一步修改它以获得“旋转”输出。

在此查询中，将 ${w} 替换为本周的数字：

select
    store,
    count(distinct case when week=${w} then customers else null end) as cnt_now,
    count(distinct case when week=${w-1} then customers else null end) as cnt_minus_1,
    count(distinct case when week=${w-2} then customers else null end) as cnt_minus_2,
    count(distinct case when week=${w-3} then customers else null end) as cnt_minus_3,
    count(distinct case when week=${w-4} then customers else null end) as cnt_minus_4,
    count(distinct case when week=${w-5} then customers else null end) as cnt_minus_5
from
    test_table
where
    week >= {$w-5}
group by
   store

记住 - COUNT() 和 COUNT(DISTINCT) 只计算 NON-NULL 值。

<头>

商店	cnt_now	cnt_minus_1	cnt_minus_2	cnt_minus_3	cnt_minus_4	cnt_minus_5
A	1	2	3	4	5	6
B	9	8	7	6	5	4