计算上一期而不是下一期购买的不同客户 Bigquery

Posted 2023-03-31

【中文标题】计算上一期而不是下一期购买的不同客户 Bigquery

【英文标题】：Count distinct customers who bought in previous period and not in next period Bigquery

【发布时间】：2021-09-01 09:13:24

【问题描述】：

我在 bigquery 中有一个数据集，其中包含 order_date: DATE 和 customer_id。

order_date |    CustomerID
2019-01-01 |    111
2019-02-01 |    112
2020-01-01 |    111
2020-02-01 |    113
2021-01-01 |    115
2021-02-01 |    119

我尝试在上一年的月份和当年的相同月份之间计算不同的 customer_id。比如从2019-01-01到2020-01-01，然后从2019-02-01到2020-02-01，然后谁没有在明年同期2020-01-01到2021- 01-01，然后 2020-02-01 到 2021-02-01。 我期望的输出

order_date| count distinct CustomerID|who not buy in the next period
2020-01-01| 5191                     |250
2020-02-01| 4859                     |500
2020-03-01| 3567                     |349
..........| ....                     |......

并且下一个期间不应包括上一个。

我尝试了下面的代码，但它以另一种方式工作

with customers as (
                  select distinct date_trunc(date(order_date),month) as dates,    
                         CUSTOMER_WID
                  from t 
                  where date(order_date) between '2018-01-01' and current_date()-1
            )
            
            select 
            dates,
            customers_previous,
            customers_next_period
            from 
            (
            select dates,
            count(CUSTOMER_WID) as customers_previous,
            count(case when customer_wid_next is null then 1 end) as customers_next_period,
            from (
            select prev.dates,
            prev.CUSTOMER_WID,
            next.dates as next_dates,
            next.CUSTOMER_WID as customer_wid_next
            from customers as prev
            left join  customers
            as next on next.dates=date_add(prev.dates,interval 1 year) 
            and prev.CUSTOMER_WID=next.CUSTOMER_WID 
            ) as t2
            group by dates
            )
            order by 1,2

提前致谢。

【问题讨论】：

是否有任何我们可以看到的代码代表您尝试过的代码，或者您正在寻求帮助而没有花任何时间解决问题？

【参考方案1】：

如果我理解正确，您是在尝试在时间窗口上计算值，为此我建议使用窗口函数 - docs here 和 here 是一篇很好的文章，解释了它是如何工作的。

也就是说，我的建议是：

SELECT DISTINCT 
    periods,
    COUNT(DISTINCT CustomerID) OVER 12mos AS count_customers_last_12_mos

FROM (
    SELECT 
        order_date,
        FORMAT_DATE('%Y%m', order_date) AS periods,
        customer_id

    FROM dataset
    )

WINDOW 12mos AS ( # window of last 12 months without current month
    PARTITION BY periods ORDER BY periods DESC
    ROWS BETWEEN 12 PRECEEDING AND 1 PRECEEDING 
    )

我相信您可以从中构建一些自定义项来改进您想要的聚合。

【参考方案2】：

您可以使用unnest(generate_date_array()) 生成句点。然后使用joins 引入前 12 个月和未来 12 个月的客户。最后，汇总并统计客户数：

select period,
       count(distinct c_prev.customer_wid),
       count(distinct c_next.customer_wid)
from unnest(generate_date_array(date '2020-01-01', date '2021-01-01', interval '1 month')) period join
     customers c_prev
     on c_prev.order_date <= period and
        c_prev.order_date > date_add(period, interval -12 month) left join
     customers c_next
     on c_next.customer_wid = c_prev.customer_wid and
        c_next.order_date > period and
        c_next.order_date <= date_add(period, interval 12 month)
group by period;