使用 Impala 获取连续行程的计数

Question

样本数据

touristid|day
ABC|1
ABC|1
ABC|2
ABC|4
ABC|5
ABC|6
ABC|8
ABC|10

输出应该是

touristid|trip
ABC|4

4 后面的逻辑是连续天数不同的连续天数 sqq 1,1,2 是第 1，然后 4,5,6 是第 2，然后 8 是第 3，10 是第 4 我想要这个输出使用 impala 查询

Answer 1

使用 lag() 函数获取前一天，如果 day-prev_day>1 则计算 new_trip_flag，然后 count(new_trip_flag)。

演示：

with table1 as (
select 'ABC' as touristid, 1  as day union all
select 'ABC' as touristid, 1  as day union all
select 'ABC' as touristid, 2  as day union all
select 'ABC' as touristid, 4  as day union all
select 'ABC' as touristid, 5  as day union all
select 'ABC' as touristid, 6  as day union all
select 'ABC' as touristid, 8  as day union all
select 'ABC' as touristid, 10 as day 
)

select touristid, count(new_trip_flag) trip_cnt
  from 
       ( -- calculate new_trip_flag
         select touristid,
                case when (day-prev_day) > 1 or prev_day is NULL then true end  new_trip_flag
           from       
                ( -- get prev_day
                  select touristid, day, 
                         lag(day) over(partition by touristid order by day) prev_day
                    from table1
                )s
        )s
 group by touristid;

结果：

touristid   trip_cnt    
ABC         4   

这同样适用于 Hive。