问题描述
我正在尝试从下表中进行
| user_id | touch | Date | Purchase Amount
| 1 | Impression| 2020-09-12 |0
| 1 | Impression| 2020-10-12 |0
| 1 | Purchase | 2020-10-13 |125$
| 1 | Email | 2020-10-14 |0
| 1 | Impression| 2020-10-15 |0
| 1 | Purchase | 2020-10-30 |122
| 2 | Impression| 2020-10-15 |0
| 2 | Impression| 2020-10-16 |0
| 2 | Email | 2020-10-17 |0
到
| user_id | path | Number of days between First Touch and Purchase | Purchase Amount
| 1 | Impression,Impression,Purchase | 2020-10-13(Purchase) - 2020-09-12 (Impression) |125$
| 1 | Email,Purchase | 2020-10-30(Purchase) - 2020-10-14(Email) | 122$
| 2 | Impression,Email | 2020-12-31 (Fixed date) - 2020-10-15(Impression) | 0$
本质上，每次在逗号分隔的字符串中遇到“购买”时，我都试图为表中的每个唯一用户创建一个新行。
此外，请计算每个唯一用户的首次接触和首次购买之间的差异。创建新行时，我们对同一个用户执行相同的操作，如上例所示。
从我收集的小数据来看，我需要混合使用交叉连接和字符串 agg，但我尝试在字符串 agg 中使用 case 语句，但无法获得所需的结果。
在 sql (Bigquery) 中是否有更好的方法来做到这一点。
谢谢
解决方法
意味着如果有购买联系，您需要分行的解决方案。
使用以下查询：
Select user_id,Aggregation function according to your requirement,Sum(purchase_amount)
From
(Select t.*,Sum(case when touch = 'Purchase' then 1 else 0 end) over (partition by user_id order by date) as sm
From t) t
Group by user_id,sm
我们可以将此视为一个缺口和岛屿问题，每个岛屿都以购买结束。我们如何定义组？通过计算我们提前购买的数量（包括当前行） - 因此在查询中使用降序排序。
select user_id,string_agg(touch order by date),min(date) as first_date,max(date) as max_date,date_diff(max(date),min(date)) as cnt_days
from (
select t.*,countif(touch = 'Purchase') over(partition by user_id order by date desc) as grp
from mytable t
) t
group by user_id,grp
以下是 BigQuery 标准 SQL
#standardSQL
select user_id,string_agg(touch order by date) path,min(date),day) days,sum(amount) amount
from (
select user_id,touch,date,amount,countif(touch = 'Purchase') over win grp
from `project.dataset.table`
window win as (partition by user_id order by date rows between unbounded preceding and 1 preceding)
)
group by user_id,grp
如果适用于您问题中的样本数据 - 输出为
另一个变化，如果触摸中没有购买，我们从我们设置的固定窗口计算天数。如何将其添加到上面的查询中？
select user_id,date_diff(if(countif(touch = 'Purchase') = 0,'2020-12-31',max(date)),grp
带输出
您可以为每一行创建一个值，该值对应于 table.touch = 'Purchase' 的实例数，然后可用于分组：
with r as (select row_number() over(order by t1.user_id) rid,t1.* from table t1)
select t3.user_id,group_concat(t3.touch),sum(t3.amount),date_diff(max(t3.date),min(t3.date))
from (select
(select sum(r1.touch = 'Purchase' AND r1.rid < r2.rid) from r r1) c1,r2.* from r r2
) t3
group by t3.c1;