SQL / HIVE：如何计算购买天数

Question

SQL / Hive：我希望计算访客购买的天数。 这是我的数据的样子

date    visitor orders
1-Jan   A   0  
1-Jan   B   0  
4-Jan   B   1  
5-Jan   A   0  
12-Jan  A   1

这是我期望的结果：

Days to purchase    count of visitors
0   0
1   0 
2   0
3   1
4   0
5   0
.   .
.   .
.   .
11  1

有什么帮助吗？

Answer 1

如果我对您的理解正确：那么您需要做的就是找到每个访客+订单组合的最低日期

select visitor,orders,min(date) as min.date from table group by visitor,orders

这应该给出类似：

visitor orders min.date
  A         0  1-Jan 
  B         0  1-Jan
  B         1  4-Jan
  A         1  12-Jan

该表（我们称其为tbl）可以自行加入

select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
from (select visitor,min.date as first.visit from tbl where orders=0) A 
inner join (select visitor,min.date as purchase.date from tbl where orders=1) B
on A.visitor=B.visitor

现在，将此查询与外部查询一起包装以计算具有相同datediffs的访问者：

 select days.to.purchase,count(visitors) as visitors from 
 (select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
    from (select visitor,min.date as first.visit from tbl where orders=0) A 
    inner join (select visitor,min.date as purchase.date from tbl where orders=1) B
    on A.visitor=B.visitor
) joined
group by days.to.purchase order by days.to.purchase

希望我正确理解你。 我不确定这是否是正确的解决方案，但您并没有给我太多的开始:)

完整的解决方案可能是：

 select days.to.purchase,count(visitors) as visitors from 
 (select A.visitor,datediff(day,purchase.date,first.visit) as days.to.purchase 
    from 
(select visitor,min.date as first.visit from 
(select visitor,orders,min(date) as min.date from table group by visitor,orders) tbl where orders=0) A 
    inner join 
(select visitor,min.date as purchase.date from 
(select visitor,orders,min(date) as min.date from table group by visitor,orders) tbl where orders=1) B
    on A.visitor=B.visitor
) joined
group by days.to.purchase order by days.to.purchase