Snowflake 查询具有过滤条件的分区

Question

我在 Snowflake table 中有一个基于键值的记录，其中对于给定的 product_id 有几十个键值对记录。请参阅以下示例：

with t1 (product_id, key, value) as 
    (
    select 101, 'grade',                'high' union all
    select 101, 'expense_cost',     'high' union all
    select 101, 'maintenance_cost', 'medium' union all
    select 102, 'grade',                'medium' union all
    select 102, 'expense_cost',     'high' union all
    select 103, 'expense_cost',     'high' union all
    select 103, 'maintenance_cost', 'medium' 
    )
    select * from t1;

鉴于此数据模型，要求是获取 product_id 与键值过滤条件相匹配的数据。

示例 1：获取所有 product_id，其中 key=(grade) 具有 value=(high or medium) & key=(expense_cost)具有价值=（高）

示例 2：获取所有 product_id where key=(grade) has value=(high) & key=(maintenance_cost) has value= （高或中）

我可以使用 Snowflake PIVOT 函数解决此要求，该函数首先将键值数据结构转换为列数据结构，然后使用 [=14= 应用 Filter 条件] 条款。有没有更好的方法在不使用 PIVOT 的情况下解决这个问题，例如使用一些 Window 功能等？

示例 1 的基于 PIVOT 的解决方案：

with t1 (product_id, key, value) as 
(
select 101, 'grade',                'high' union all
select 101, 'expense_cost',     'high' union all
select 101, 'maintenance_cost', 'medium' union all
select 102, 'grade',                'medium' union all
select 102, 'expense_cost',     'high' union all
select 103, 'expense_cost',     'high' union all
select 103, 'maintenance_cost', 'medium' 
)
select * from (
select product_id, key, value 
  from t1 
  where key in ('grade','expense_cost','maintenance_cost')
) pivot(min(value) for key in ('grade','expense_cost','maintenance_cost'))
as p (product_id, grade, expense_cost, maintenance_cost)
where grade in ('high','medium')
and expense_cost in ('high');

注意：在使用 Window 函数的情况下，输出必须仅包含符合条件或通过过滤条件的行。输出不得包含合格分区中的所有记录。

Answer 1

要使用的模式是 QUALIFY 结合 COUNT_IF。如果以后需要，可以旋转它：

"Requirement 1: Fetch all product_id where grade is (high or medium) & expense_cost = high"

with t1 (product_id, key, value) as 
    (
    select 101, 'grade',                'high' union all
    select 101, 'expense_cost',     'high' union all
    select 101, 'maintenance_cost', 'medium' union all
    select 102, 'grade',                'medium' union all
    select 102, 'expense_cost',     'high' union all
    select 103, 'expense_cost',     'high' union all
    select 103, 'maintenance_cost', 'medium' 
    )
select * 
from t1
qualify COUNT_IF(key='grade' AND value IN ('medium', 'high')) 
            OVER(PARTITION BY product_id) > 0
    AND COUNT_IF(key = 'expense_cost' AND value = 'high') 
            OVER(PARTITION BY product_id) > 0;

输出：

Requirement 2: Fetch all product_id where key=(grade) has value=(high) & key=(maintenance_cost) has value=(high or medium)

with t1 (product_id, key, value) as 
    (
    select 101, 'grade',                'high' union all
    select 101, 'expense_cost',     'high' union all
    select 101, 'maintenance_cost', 'medium' union all
    select 102, 'grade',                'medium' union all
    select 102, 'expense_cost',     'high' union all
    select 103, 'expense_cost',     'high' union all
    select 103, 'maintenance_cost', 'medium' 
    )
select * 
from t1
qualify COUNT_IF(key='grade' AND value IN ('high')) 
            OVER(PARTITION BY product_id) > 0
    AND COUNT_IF(key = 'maintenance_cost' AND value IN ('medium','high'))
            OVER(PARTITION BY product_id) > 0
    AND COUNT_IF(key='expense_cost' AND value IN ('high')) 
           OVER(PARTITION BY product_id) = 0  -- explicitly excluding

输出：

Answer 2

我认为 self-documenting 如果您 sub-select 使用 conditional-aggregation 排位赛 product_ids，然后过滤排位赛 product_ids 加上条件where 子句。下面是要求 #2 的演示，但您可以轻松地针对 #1 对其进行修改。

with t (product_id, keys, value) as 

(select 101, 'grade','high' union all
 select 101, 'expense_cost','high' union all
 select 101, 'maintenance_cost','medium' union all
 select 102, 'grade','medium' union all
 select 102, 'expense_cost','high' union all
 select 103, 'expense_cost','high' union all
 select 103, 'maintenance_cost','medium' ),

product_ids as

(select product_id
 from t
 group by product_id
 having sum(case when keys='grade' and value ='high' then 1 end)>0 and
        sum(case when keys='maintenance_cost' and value in ('high', 'medium') then 1 end)>0)
                            
select * 
from t
where product_id in (select product_id from product_ids) and
    ((keys='grade' and value ='high') or (keys='maintenance_cost' and value in ('high', 'medium')))

Answer 3

在下面添加我的解决方案，这是对@Lukasz Szozda 解决方案的改进，该解决方案存在以下两个问题：

一个。对于过滤器标准中的每个属性，他的代码需要添加一个增加雪花计算的额外 Window 函数线性时间。

b。有许多属性需要从结果集中删除，并且为每个属性添加这么多 Window 函数会使 SQL 查询非常臃肿。

这是我改进后的代码：

with t1 (product_id, key, value) as 
(
select 101, 'grade',                'high' union all
select 101, 'expense_cost',     'high' union all
select 101, 'maintenance_cost', 'medium' union all
select 102, 'grade',                'medium' union all
select 102, 'expense_cost',     'high' union all
select 103, 'expense_cost',     'high' union all
select 103, 'maintenance_cost', 'medium' 
)
select * , 
 (CASE 
      WHEN key = 'grade' AND value IN ('medium', 'high') THEN TRUE
      WHEN key = 'expense_cost' AND value = 'high' THEN TRUE
      ELSE FALSE
  END) AS is_allowed
from t1
WHERE key IN ('grade', 'expense_cost')
QUALIFY COUNT_IF(is_allowed=TRUE) OVER(PARTITION BY product_id) = 2;

Snowflake 查询具有过滤条件的分区

Snowflake query a partition with filter condition

snowflake-cloud-data-platform

window-functions

partitioning