在Snowflake sql中，如何使用partition by和order by计算不同的值？

我的数据如下：

| user | eventorder| postal|
|:---- |:---------:| -----:|
| A    | 1         | 60616 |
| A    | 2         | 10000 |
| A    | 3         | 60616 |
| B    | 1         | 20000 |
| B    | 2         | 30000 |
| B    | 3         | 40000 |
| B    | 4         | 30000 |
| B    | 5         | 20000 |

| user | eventorder| postal| travelledStop|
|:---- |:---------:| -----:| ------------:|
| A    | 1         | 60616 |  1    |
| A    | 2         | 10000 |  2    |
| A    | 3         | 60616 |  2    |
| B    | 1         | 20000 |  1    |
| B    | 2         | 30000 |  2    |
| B    | 3         | 40000 |  3    |
| B    | 4         | 30000 |  3    |
| B    | 5         | 20000 |  3    |

我需要解决的问题是：在用户旅行的每个事件顺序之前，有多少不同的站点？

理想结果应如下所示：

| user | eventorder| postal|
|:---- |:---------:| -----:|
| A    | 1         | 60616 |
| A    | 2         | 10000 |
| A    | 3         | 60616 |
| B    | 1         | 20000 |
| B    | 2         | 30000 |
| B    | 3         | 40000 |
| B    | 4         | 30000 |
| B    | 5         | 20000 |

| user | eventorder| postal| travelledStop|
|:---- |:---------:| -----:| ------------:|
| A    | 1         | 60616 |  1    |
| A    | 2         | 10000 |  2    |
| A    | 3         | 60616 |  2    |
| B    | 1         | 20000 |  1    |
| B    | 2         | 30000 |  2    |
| B    | 3         | 40000 |  3    |
| B    | 4         | 30000 |  3    |
| B    | 5         | 20000 |  3    |

以A为例，当事件顺序为1时，它仅行驶60616-1站。 当事件顺序为2时，它已行驶60616次，停10000-2次。 当事件顺序为3时，该用户已行驶的不同站点为60616和10000。-2站

我不允许使用count distinct和按顺序分区。我想在（按用户顺序按事件顺序划分）上做一些类似于count（distinct（postal））的事情，但这是不允许的

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有人知道如何解决这个问题吗？非常感谢

我使用了您提供的样本数据（只是一个样本的子集，但这应该可以扩展）。这里的目标基本上是为每一行生成一个数组，该数组累积了以前事件的所有postals

with _temp as (
select 'A' as usr, 1 as EventOrder, '60616' as Postal
UNION ALL
select 'A' as usr, 2 as EventOrder, '10000' as Postal
UNION ALL
select 'A' as usr, 3 as EventOrder, '60616' as Postal
),
_intermediate as (
select usr
    , eventorder
    , postal
    , array_slice(
          array_agg(postal)
            within group (order by eventorder)
            OVER (Partition by usr)
           , 0, eventorder) as full_array
from _temp
group by usr, eventorder, postal
)
select usr, eventorder, postal, count(distinct f.value)
from _intermediate i, lateral flatten(input => i.full_array) f
group by usr, eventorder, postal

---

也许最简单的方法是使用子查询并计算“1”：

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我喜欢@Daniel Zagales的答案，但这里有一个解决方法，使用

densite_-rank

和

sum

with temp as (
select 'A' as usr, 1 as EventOrder, '60616' as Postal
UNION ALL
select 'A' as usr, 2 as EventOrder, '10000' as Postal
UNION ALL
select 'A' as usr, 3 as EventOrder, '60616' as Postal  
UNION ALL
select 'B' as usr, 1 as EventOrder, '20000' as Postal  
UNION ALL
select 'B' as usr, 2 as EventOrder, '30000' as Postal  
UNION ALL
select 'B' as usr, 3 as EventOrder, '40000' as Postal 
UNION ALL
select 'B' as usr, 4 as EventOrder, '30000' as Postal  
UNION ALL
select 'B' as usr, 5 as EventOrder, '20000' as Postal 
),
temp2 as(
select temp.* ,dense_rank()over(partition by usr,Postal order by EventOrder) rks
from temp 
)
select usr,eventorder,postal,sum(case when rks = 1 then 1 else 0 END)over(partition by usr order by EventOrder) travelledStop
from temp2 
order by usr,EventOrder 

基本上使用

density_-rank

得到第一个出现的停止，而不是总结

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非常好的解决方案！我也试着做同样的事情，但不知道如何为每一行构建数组（考虑的是窗口框架，但不受支持），但是array_slice（）是一个很好的方法。