使用Impala获取连续旅行的次数

本文介绍了使用Impala获取连续旅行的次数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

样本数据

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

输出应为

touristid|trip ABC|4

4后的逻辑是连续天数，连续天数sqq 1,1,2为第一，然后4,5,6为第二，然后8为第三，而10为第四 我想要使​​用impala查询的输出

Logic behind 4 is count of consecutive days distinct consecutive days sqq 1,1,2 is 1st then 4,5,6 is 2nd then 8 is 3rd and 10 is 4th I want this output using impala query

推荐答案

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

Get previous day using lag() function, calculate new_trip_flag if the day-prev_day>1, then 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

在Hive中也一样.