可变行数之间的天数差异

Question

我在 Teradata 数据集中遇到这种情况：

ID                DATE1                         
134311784   14-2-2019   (first visit)           
134311784   7-3-2019    (second visit)          
134311784   22-3-2019   (third visit)           
234567891   10-2-2019   (first visit)           
234567891   17-2-2019   (second visit)          
234567891   22-2-2019   (third visit)           
234567891   18-3-2019   (fourth visit)  

我希望这样：

ID                DATE1          DIFFERENCE     
134311784   14-2-2019              0        NEW VISITOR
134311784   7-3-2019               21       --> (second -first ) = 21
134311784   22-3-2019              15       --> (third - second ) = 15
234567891   10-2-2019              0        NEW VISITOR
234567891   17-2-2019              7        
234567891   22-2-2019              5        ..
234567891   18-3-2019              24       ..

我该如何做到这一点？
我尝试使用子查询，但它没有给出想要的结果。

SELECT   t1.visit, t1.date1,T2.date1, abs( t2.date2-t1.date1) as difference
FROM data_want T1
left join data_want T2
on T1.visit=T2.visit
WHERE t2.date1 > t1.date1

Answer 1

无需自连接，只需一个有序解析/窗口聚合函数

SELECT   t1.ID, t1.date1,
   t1.date1 - LAG(t1.date1,1,t1.date1) OVER (PARTITION BY t1.ID ORDER BY t1.date1) difference
FROM data_want T1
ORDER BY t1.ID, t1.date1

或COALESCE(t1.date1 - LAG(t1.date1) OVER (PARTITION BY t1.ID ORDER BY T1.date1),0) 将给出相同的结果（差异 = 0 首次访问）。如果没有默认子句或 COALESCE，首次访问将为 NULL。

Answer 2

我猜你可以尝试这样做：

SELECT   t1.visit, t1.date1,T2.date1, DATEDIFF(t2.date1, t1.date1) t as difference
FROM data_want T1
left join data_want T2
on T1.visit=T2.visit
WHERE t2.date1 > t1.date1