Question

SQL / Hive：我希望计算访问者购买的天数。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

这是我期待的结果：

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

任何帮助？

Answer 1

如果我理解正确的话：你需要做的是找到每个访客+订单组合的最小日期

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

这应该是这样的：

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

这个表（让我们称之为tbl）可以自己加入来提供

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，使用外部查询包装此查询以计算具有相同日期的访问者：

 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

希望我能正确理解你。我不确定这是否是正确的解决方案，但你并没有给我很多开始：）

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

SQL / HIVE：如何计算购买天数

1 个答案: