BigQuery:如何找到按比例计算的每周平均销售额?

时间:2020-07-29 07:03:50

标签: sql group-by google-bigquery subquery week-number

解决方案: 我们的最终目标是获得具有以下公式的Scaled AverageWeeklySales(AWS):

Scaled AWS = (Number of Weeks article has had a record within the store / maximum Number of Weeks in data) * (AWS /  Number of Weeks)

输入“ table_a”结构:

Store   Article     Date    Week    Year    Sales
xx      xx          xx      xx      xx      xx

试用: 构造查询直到找到AWS(在商店级别),然后卡在那里继续进行。由于我是新手,因此不确定通过BigQuery SQL如何实现此可扩展AWS。需要帮助!

WITH total_weekly_sales AS (
SELECT
  Store,
  Week,
  SUM(Sales) AS TotalWeeklySales
FROM
  table_a
GROUP BY
  Store,
  Week)
SELECT
  Store,
  AVG(TotalWeeklySales) AS AverageWeeklySales
FROM
  total_weekly_sales tws
GROUP BY
  Store

样本数据:

Store   Article     Date            Week        Year    Sales
11      aa          2019-07-01      202001      2020    4.9
11      bb          2019-07-07      202001      2020    22.5
11      cc          2019-07-08      202002      2020    10.4
12      aa          2019-07-01      202001      2020    5.3
12      bb          2019-07-07      202001      2020    20.2

评论查询:

  • AWS -表示平均每周销售额[将根据以下位置的给定数据得出 正如我在“ Tried Out”会话中提到的 问题,以查找商店级别的每周总销售量并获取 平均得出AWS]
  • 周数-商店级别[对于每家商店,多少周数据 当下。 例如。对于商店11,此变量值为2,而对于 商店12,此变量值为1]
  • 数据中的最大周数-例如:从我们的示例数据中, 每个商店的价值是2 [说明: 计算当前的总周数并取最大值 所有商店的周数。在这里,商店11在2周内 总计,而商店12总共有1周。所以,这里的最大值 将是2,它被分配为此变量的值]

所需结果: 缩放后的AWS Formula Part_1结果[周数文章在商店中有一条记录/数据中的最大周数]

Store   part_1_value    
11      1.5 (3/2)
12      1 (2/2)

按比例缩放的AWS公式Part_2结果[AWS /周数]

Store   part_2_value    
11      18.9 (37.8/2)
12      25.5 (25.5/1)

最终输出为: [Part_1 * Part_2]

Store   ScaledAverageWeeklySales    
11      28.35 (1.5*18.9)
12      25.5 (1*25.5)

总的来说,我们的最终结果应该在商店级别。 预先感谢!

1 个答案:

答案 0 :(得分:1)

在您澄清之后,我能够创建一个查询来计算您要寻找的内容。

我已将您提供的数据与WITH语句以及BigQuery中的MAXCOUNT聚合内置方法一起使用。下面是查询,

WITH unique_articles AS (
SELECT Store, COUNT(DISTINCT Article) as uniq_art FROM `test-proj-261014.bq_load_codelab.sales_week`
GROUP BY Store
),
max_weeks_st AS (
#not possible to use DATE_DIFF cos the format of the date, the number of weeks would be 1 for the sample data
#so WEEK column will be used instead
SELECT Store, COUNT(Distinct Week) as max_weeks_st  FROM `test-proj-261014.bq_load_codelab.sales_week` 
group by Store
), 
#below returns a struct. So in order to access the int64 value name_of_struct.max_weeks_data
max_weeks_data AS(
SELECT MAX(a.max_weeks_st) AS max_weeks_data FROM max_weeks_st a
),
sum_sales AS (
SELECT Store, SUM(sales) as sum_sales  FROM `test-proj-261014.bq_load_codelab.sales_week` 
GROUP BY Store
), 
final_data AS(
select a.Store, a.uniq_art, b.max_weeks_st,c.sum_sales, max_weeks_data  FROM  unique_articles a LEFT JOIN sum_sales c USING(Store)
LEFT JOIN max_weeks_st b USING(Store) CROSS JOIN  max_weeks_data 
)
SELECT Store,((uniq_art/max_weeks_data.max_weeks_data)*(sum_sales/ max_weeks_st)) as AWS_result  FROM final_data a

和输出

Row Store   AWS_result
1   11      28.35
2   12      25.5

请注意, max_weeks_data 结构 。因此,要访问其值,必须遵循以下语法: name_of_struct.max_weeks_data 。此外,请注意,您描述的每个变量都是在临时表中计算的,该表始终带有用于联接表的商店ID。