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:
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] 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!
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MAX,COUNT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 是 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。