如何在 Snowflake sql 中使用分区依据和排序依据来计算不同的值？

Question

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 |

我需要解决的问题：在用户旅行的每个事件顺序之前有多少个不同的站点？

理想的结果应该是这样的：

| 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，只走了60616 - 1站。当事件顺序为 2 时，它已行进 60616 和 10000 - 2 个停靠点。当事件顺序为 3 时，此用户经过的不同停靠点为 60616 和 10000。- 2 停靠点。

我不允许将 count distinct 与 partition by order by 一起使用。我想做类似 count(distinct(postal)) over (partition by user order by eventorder) 的事情，但这是不允许的。

有谁知道如何解决这个问题？非常感谢！

Answer 1

我使用了您提供的示例数据（只是 A 的一个子集，但这应该会向外扩展）。这里的目标基本上是为每一行生成一个数组，该数组累积以前事件的所有邮政。

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

Answer 2

也许最简单的方法是使用子查询并计算“1”：

select t.*,
       sum(case when seqnum = 1 then 1 else 0 end) over (partition by usr order by eventorder) as num_postals
from (select t.*,
             row_number() over (partition by usr, postal order by eventorder) as seqnum
      from t
     ) t

Answer 3

我喜欢@Daniel Zagales 的回答，但这是使用 dense_rank 和 sum

的解决方法

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dense_rank先出现停止而不是总结。

db<>fiddle

如何在 Snowflake sql 中使用分区依据和排序依据来计算不同的值？

How to count distinct value with partition by and order by in Snowflake sql?

sql

data-analysis

data-partitioning

data-science

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