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问题描述

现在我们有一张用户访问区域的记录表，有三个字段：user_id表示用户ID，area表示用户访问的区域，visit_time是访问时间。求用户对某一区域访问的开始和结束时间，结果如右表所示。

分析问题

101访问了上午和晚上分别访问了A，中文访问了B
102 早上9点访问了C，晚上19点也访问了C，总共访问了两次。
我们需要对同一用户访问相同地点的不同次做区分，然后聚合就好了
那我们定一个规则，如果用户换区域或者相邻两条记录间隔超过3小时，那么不算相同访问了。

准备数据

-- 删除表
drop table if exists visit;
-- 建表
CREATE TABLE visit (
  user_id bigint,
  area string,
  report_time string
);

-- 插入数据
INSERT overwrite table visit  
VALUES
  (101, 'A','2023-01-01 09:00:00'),
  (101, 'A','2023-01-01 10:00:00'),
  (101, 'B','2023-01-01 14:00:00'),
  (101, 'B','2023-01-01 15:00:00'),
  (101, 'A','2023-01-01 19:00:00'),
  (101, 'A','2023-01-01 20:00:00'),
  (102, 'C','2023-01-01 09:00:00'),
  (102, 'C','2023-01-01 10:00:00'),
  (102, 'C','2023-01-01 11:00:00'),
  (102, 'C','2023-01-01 19:00:00'),
  (102, 'C','2023-01-01 20:00:00');  
  
-- 查看数据
select * from visit

方案1：适用于换场+长间隔

1、取出每个用户的上一条访问记录的时间地点

select user_id
,report_time,area
,lag(area,1) over( partition by user_id order by report_time) as last_area
,lag(report_time,1) over( partition by user_id order by report_time) as last_time
from visit 

2、设置一个标识列，开始为0，其他为0

select t1.*
,case 
when last_area is null then 1  -- 没有访问
when last_area  !=area then 1  -- 换地方
when (unix_timestamp(report_time)-
        unix_timestamp(last_time))/(60*60)>=3 then 1  -- 超过3小时
when last_area = area then 0 end as num
from (
  select user_id
  ,report_time,area
  ,lag(area,1) over( partition by user_id order by report_time) as last_area
  ,lag(report_time,1) over( partition by user_id order by report_time) as last_time
  from visit 
) t1

3、使用sum over累计求和，给每次访问打标，相同访问标记一样

select t1.*
,sum(case 
when last_area is null then 1  -- 没有访问
when last_area  !=area then 1  -- 换地方
when (unix_timestamp(report_time)-
        unix_timestamp(last_time))/(60*60)>=3 then 1  -- 超过3小时
when last_area = area then 0 end)
over(partition by user_id order by report_time) as num
from (
  select user_id
  ,report_time,area
  ,lag(area,1) over( partition by user_id order by report_time) as last_area
  ,lag(report_time,1) over( partition by user_id order by report_time) as last_time
  from visit 
) t1

4、找到每个用户每个区域的首次末次访问时间

with t1 as (
  -- 上次访问记录
  select user_id
  ,report_time,area
  ,lag(area,1) over( partition by user_id order by report_time) as last_area
  ,lag(report_time,1) over( partition by user_id order by report_time) as last_time
  from visit 
)
,t2 as (
  -- 判断访问
     select *
    ,case 
    when last_area is null then 1  -- 没有访问
    when last_area  !=area then 1  -- 换地方
    when (unix_timestamp(report_time)-
            unix_timestamp(last_time))/(60*60)>=3 then 1  -- 超过3小时
    when last_area = area then 0 end as num
    from t1
)
,t3 as  (
  -- sum over 打标
   select user_id,area,report_time
   ,sum(num) over(partition by user_id order by report_time) as num
    from t2
  
)
-- 找到每个用户每个区域的起止时间
select user_id,area
,num 
,min(report_time) start_time
,max(report_time) end_time
from t3
group by user_id,area,num
-- order by user_id,start_time

实现了功能，感觉有点繁琐，抛转引玉吧！有好的方案，欢迎评论。

方案2：仅适用于换场

1、标记相同场地的不同访问。按用户编号减去按用户和场地编号，就可以找出换场次序，但是无法标记长时间间隔的，比如102上午、下午分别来了一次。

select user_id
,report_time,area
,row_number()over(partition by user_id order by report_time) as rn1 -- 按用户编码
,row_number()over(partition by user_id,area order by report_time) as rn2 -- 按用户和场地编码
-- rn1 -rn2就可以区分相同场的不同次
,row_number()over(partition by user_id order by report_time)-
row_number()over(partition by user_id,area order by report_time) as rn3 
from visit 
order by user_id,report_time

2、按照用户、场地和编码聚合，就可以找出起止时间了。A 是正常的，但C是不对的。

select user_id
,area
,rn3
,min(report_time) as start_time
,max(report_time) as end_time
from (
    select user_id
    ,report_time,area
    ,row_number()over(partition by user_id order by report_time) as rn1 -- 按用户编码
    ,row_number()over(partition by user_id,area order by report_time) as rn2 -- 按用户和场地编码
    -- rn1 -rn2就可以区分相同场的不同次
    ,row_number()over(partition by user_id order by report_time)-
    row_number()over(partition by user_id,area order by report_time) as rn3 
    from visit 
    order by user_id,report_time
) t1
group by  user_id
,area
,rn3