Hive-SQL查询连续活跃登录用户
2022/1/19 2:08:32
本文主要是介绍Hive-SQL查询连续活跃登录用户,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
连续活跃登陆的用户指至少连续2天都活跃登录的用户
解决类似场景的问题
创建数据
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CREATE TABLE test5active(
dt string,
user_id string,
age int )
ROW format delimited fields terminated BY ',' ;
INSERT INTO TABLE test5active VALUES
( '2019-02-11' , 'user_1' ,23),( '2019-02-11' , 'user_2' ,19),
( '2019-02-11' , 'user_3' ,39),( '2019-02-11' , 'user_1' ,23),
( '2019-02-11' , 'user_3' ,39),( '2019-02-11' , 'user_1' ,23),
( '2019-02-12' , 'user_2' ,19),( '2019-02-13' , 'user_1' ,23),
( '2019-02-15' , 'user_2' ,19),( '2019-02-16' , 'user_2' ,19);
|
思路一:
1、因为每天用户登录次数可能不止一次,所以需要先将用户每天的登录日期去重。
2、再用row_number() over(partition by _ order by _)函数将用户id分组,按照登陆时间进行排序。
3、计算登录日期减去第二步骤得到的结果值,用户连续登陆情况下,每次相减的结果都相同。
4、按照id和日期分组并求和,筛选大于等于2的即为连续活跃登陆的用户。
第一步:用户登录日期去重
1 |
select DISTINCT dt,user_id from test5active;
|
第二步:用row_number() over()函数计数
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select
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from
(
select DISTINCT dt,user_id from test5active
)t1;
|
第三步:日期减去计数值得到结果
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select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from
(
select DISTINCT dt,user_id from test5active
)t1)t2;
|
第四步:根据id和结果分组并计算总和,大于等于2的即为连续登陆的用户,得到 用户id,开始日期,结束日期,连续登录天数
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select
t3.user_id, min (t3.dt), max (t3.dt), count (1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from
(
select DISTINCT dt,user_id from test5active
)t1
)t2
)t3 group by t3.user_id,t3.dis having count (1)>1;
|
用户id 开始日期 结束日期 连续登录天数
最后:连续登陆的用户
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select distinct t4.user_id
from
(
select
t3.user_id, min (t3.dt), max (t3.dt), count (1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from
(
select
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from
(
select DISTINCT dt,user_id from test5active
)t1
)t2
)t3 group by t3.user_id,t3.dis having count (1)>1
)t4;
|
思路二:使用lag(向后)或者lead(向前)
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select
user_id,t1.dt,
lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id
from
(
select DISTINCT dt,user_id from test5active
)t1;
|
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select
distinct t2.user_id
from
(
select
user_id,t1.dt,
lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id
from
(
select DISTINCT dt,user_id from test5active
)t1
)t2 where datediff(last_date_id,t2.dt)=1;
|
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