Question

我有一个包含客户，产品和类别的购买数据集。

customer     product     category    sales_value
       A     aerosol     air_care             10
       B     aerosol     air_care             12
       C     aerosol     air_care              7
       A     perfume     air_care              8
       A     perfume     air_care              2
       D     perfume     air_care             11
       C      burger         food             13
       D       fries         food              6
       C       fries         food              9

我希望对于每种产品，至少购买一次该产品的客户在该产品上花费的销售价值与在该产品类别上花费的销售价值之间的比率。

另一种说法：让至少购买一次fries的客户，对于所有客户，计算A）在fries上花费的销售价值总和，以及B）销售总和在food上花费的价值。

中间表的格式如下：

product    category  sum_spent_on_product           sum_spent_on_category    ratio
                                                 by_people_buying_product
aerosol    air_care                    29                              39     0.74
perfume    air_care                    21                              31     0.68
 burger        food                    13                              22     0.59
  fries        food                    15                              28     0.53

示例：至少购买过aerosol的人在此产品上总共花费了1800。相同的人在air_care类别（aerosol所属）上总共花费了3600。因此，aerosol的比率为0.5。

我尝试使用left join lateral解决这个问题，并为每个product计算给定的中间结果，但是我不知道如何包含条件{{1} }：

only for customers who bought this specific product

上面的查询列出了每种产品在产品类别上花费的总和，但没有必需的产品购买者条件。

select distinct (product_id) , category , c.sales_category from transactions t left join lateral ( select sum(sales_value) as sales_category from transactions where category = t.category group by category ) c on true ;是正确的方法吗？普通SQL中还有其他解决方案吗？

Answer 1

我将使用窗口功能来计算每个类别中每个客户的总花费：

SELECT
  customer, product, category, sales_value,
  sum(sales_value) OVER (PARTITION BY customer, category) AS tot_cat
FROM transactions;

 customer | product | category | sales_value | tot_cat 
----------+---------+----------+-------------+---------
 A        | aerosol | air_care |       10.00 |   20.00
 A        | perfume | air_care |        8.00 |   20.00
 A        | perfume | air_care |        2.00 |   20.00
 B        | aerosol | air_care |       12.00 |   12.00
 C        | aerosol | air_care |        7.00 |    7.00
 C        | fries   | food     |        9.00 |   22.00
 C        | burger  | food     |       13.00 |   22.00
 D        | perfume | air_care |       11.00 |   11.00
 D        | fries   | food     |        6.00 |    6.00

然后我们只需要总结一下。当客户多次购买同一产品时，只会出现一个问题。在您的示例中，客户A购买了两次香水。为了解决这个问题，让我们同时按客户，产品和类别分组（并汇总sales_value列）：

SELECT
  customer, product, category, SUM(sales_value) AS sales_value,
  SUM(SUM(sales_value)) OVER (PARTITION BY customer, category) AS tot_cat
FROM transactions
GROUP BY customer, product, category

 customer | product | category | sales_value | tot_cat 
----------+---------+----------+-------------+---------
 A        | aerosol | air_care |       10.00 |   20.00
 A        | perfume | air_care |       10.00 |   20.00 <-- this row summarizes rows 2 and 3 of previous result
 B        | aerosol | air_care |       12.00 |   12.00
 C        | aerosol | air_care |        7.00 |    7.00
 C        | burger  | food     |       13.00 |   22.00
 C        | fries   | food     |        9.00 |   22.00
 D        | perfume | air_care |       11.00 |   11.00
 D        | fries   | food     |        6.00 |    6.00

现在，我们只需要对sales_value和tot_cat求和即可得到中间结果表。我使用公用表表达式来获取名称为t的先前结果：

WITH t AS (
  SELECT
    customer, product, category, SUM(sales_value) AS sales_value,
    SUM(SUM(sales_value)) OVER (PARTITION BY customer, category) AS tot_cat
  FROM transactions
  GROUP BY customer, product, category
)
SELECT
  product, category,
  sum(sales_value) AS sales_value, sum(tot_cat) AS tot_cat,
  sum(sales_value) / sum(tot_cat) AS ratio
FROM t
GROUP BY product, category;

 product | category | sales_value | tot_cat |         ratio          
---------+----------+-------------+---------+------------------------
 aerosol | air_care |       29.00 |   39.00 | 0.74358974358974358974
 fries   | food     |       15.00 |   28.00 | 0.53571428571428571429
 burger  | food     |       13.00 |   22.00 | 0.59090909090909090909
 perfume | air_care |       21.00 |   31.00 | 0.67741935483870967742

Answer 2

我希望对于每种产品，至少购买一次该产品的客户在该产品上花费的销售价值与在该产品类别上花费的销售价值之间的比率。

如果我正确理解这一点，则可以按人员和类别汇总销售额，以获取类别的总计。在Postgres中，您可以保留一系列产品并将其用于匹配。因此，查询看起来像：

select p.product, p.category,
       sum(p.sales_value) as product_only_sales, 
       sum(pp.sales_value) as comparable_sales
from purchases p join
     (select customer, category, array_agg(distinct product) as products, sum(sales_value) as sales_value
      from purchases p
      group by customer, category
     ) pp
     on p.customer = pp.customer and p.category = pp.category and p.product = any (pp.products)
group by p.product, p.category;

Here是db <>小提琴。

编辑：

数据允许在产品日期中重复。那丢东西了。解决方案是为每个客户按产品进行预聚合：

select p.product, p.category, sum(p.sales_value) as product_only_sales, sum(pp.sales_value) as comparable_sales
from (select customer, category, product, sum(sales_value) as sales_value
      from purchases p
      group by customer, category, product
     ) p join
     (select customer, category, array_agg(distinct product) as products, sum(sales_value) as sales_value
      from purchases p
      group by customer, category
     ) pp
     on p.customer = pp.customer and p.category = pp.category and p.product = any (pp.products)
group by p.product, p.category

Here是本示例的db <>小提琴。

左连接横向用于条件求和

2 个答案: