2

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

对于每种产品，我想要至少购买该产品一次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：fries以至少购买过一次的客户为例，计算 A) 花费在的销售价值fries的总和和 B) 花费在的销售价值的总和food。

中间表将采用以下形式：

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：人们至少购买aerosol过一次，在该产品上总共花费了 1800。总体而言，相同的人在该air_care类别（aerosol属于）上花费了 3600。因此，比率为aerosol0.5。

我尝试使用left join lateral并计算每个给定的中间结果来解决这个问题product，但我无法理解如何包含条件only for customers who bought this specific product：

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
;

上面的查询列出了每个产品在产品类别上的花费总和，但没有所需的产品购买者条件。

是left join lateral正确的方法吗？普通 SQL 中还有其他解决方案吗？

4

2 回答 2

3

我会使用一个窗口函数来计算每个客户在每个类别中的总花费：

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。当客户多次购买相同的产品时，就会出现问题。在您的示例中，客户A购买了两次香水。为了克服这个问题，让我们同时按客户、产品和类别进行分组（并对sales_value列求和）：

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

现在我们只需将 sales_value 和 tot_cat 相加即可得到中间结果表。我使用一个公用表表达式来获取名称下的先前结果t：

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
于 2019-02-24T20:31:17.240 回答
2

对于每种产品，我想要至少购买该产品一次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。

如果我理解正确，您可以按人员和类别汇总销售额以获得该类别的总数。在 Postgres 中，您可以保留一系列产品并将其用于匹配。所以，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;

这是一个 db<>fiddle。

编辑：

数据允许产品的日期重复。这会让事情变得很糟糕。解决方案是为每个客户按产品预先聚合：

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

这是此示例的 db<>fiddle。

于 2019-02-24T13:29:16.243 回答