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;
1条答案
按热度按时间n6lpvg4x1#
使用lag()函数获取前一天,如果前一天>1,则计算新的\u trip \u标志,然后计数(新的\u trip \u标志)。
演示:
结果:
同样的方法也适用于Hive。