hive 统计用户在同一地点停留时长

需求

1. 对同一个用户，在同一个位置，连续的多条记录进行合并
2. 合并原则：开始时间取最早的，停留时长加和

字段

userID, locationID, time, duration

样例数据

user_a  location_a  2022-02-03 08:00:00 60
user_a  location_a  2022-02-03 09:00:00 60
user_a  location_a  2022-02-03 11:00:00 60
user_a  location_a  2022-02-03 12:00:00 60

结果数据

user_a  location_a  2022-02-03 08:00:00 120
user_a  location_a  2022-02-03 11:00:00 120

数据准备

create table temp_dura_0303
(
    user_id     string, --'用户id',
    location_id string, -- '位置id'
    time_        string, -- '时间'
    duration    int     --'持续时间'
) row format delimited fields terminated by '\t'
    stored as orc
    tblproperties ("orc.compress" = "snappy");

insert into temp_dura_0303 values
("user_a", "location_a","2022-02-03 08:00:00",60),
("user_a", "location_a","2022-02-03 09:00:00",60),
("user_a", "location_a","2022-02-03 11:00:00",60),
("user_a", "location_a","2022-02-03 12:00:00",60),
("user_a", "location_b","2022-02-03 10:00:00",60),
("user_a", "location_c","2022-02-03 08:00:00",60),
("user_a", "location_c","2022-02-03 09:00:00",60),
("user_a", "location_c","2022-02-03 10:00:00",60),
("user_b", "location_a","2022-02-03 15:00:00",60),
("user_b", "location_a","2022-02-03 16:00:00",60),
("user_b", "location_a","2022-02-03 18:00:00",60);
select * from temp_dura_0303;

  解题思路

本题主要是将连续数据进行合并，就是将连续的数据分成一个组里面，所有的问题都能解决，因此本题实质上是构造分组条件，如何去分组的问题。根据需求得知本题还是连续性问题，那么问题转换为如何判断连续问题，如果连续则相邻两行的差值相同。我们先求出相邻两行的时间差值观察结果。

select user_id,
       location_id,
       time_,
       duration,
       int((unix_timestamp(time_) -
            unix_timestamp(lag(time_, 1, time_) over (partition by user_id,duration order by time_))) / 3600) time_cz
from temp_dura_0303;

可以看出，同一个用户id及位置id里面连续的差值一定是小于等于1的，而在出现不连续的分界点时，其差值是大于1的，因此我们将出现差值大于1的地方视为一个转折点，一个事件，此时被标记为1,否则标记为0，根据重分组算法，我们进行如下分组。

select user_id,
       location_id,
       time_,
       duration,
       sum(case when time_cz > 1 then 1 else 0 end) over (partition by user_id,location_id order by time_) grep_flag
from (select user_id,
             location_id,
             time_,
             duration,
             int((unix_timestamp(time_) -
                  unix_timestamp(lag(time_, 1, time_) over (partition by user_id,location_id order by time_))) /
                 3600) time_cz
      from temp_dura_0303) t;

我们可以看到上述新加的列中已经依据我们的想法将数据进行区分。我们将这种大于某参照量的标记为1进行累加的算法称为重分组算法，用SQL表示为sum(if(XXX>YYY,1,0)) over(partition by XXX order by XXX).我们按照user_id,location_id及grp_id进行分组，将需要的数据分到一个组里，然后求取最小时间值及对duration时间进行求和即为最终结果,sql如下：

select user_id, location_id, min(time_) start_time, sum(duration) sum_duration
from (select user_id,
             location_id,
             time_,
             duration,
             sum(case when time_cz > 1 then 1 else 0 end)
                 over (partition by user_id,location_id order by time_) grep_flag
      from (select user_id,
                   location_id,
                   time_,
                   duration,
                   int((unix_timestamp(time_) -
                        unix_timestamp(lag(time_, 1, time_) over (partition by user_id,location_id order by time_))) /
                       3600) time_cz
            from temp_dura_0303) t) t
group by user_id, location_id, grep_flag;