HiveSQL练习题：计算近一个月活跃、连续活跃、沉默用户需求

2024-07-25 15:42:32 浏览数 (1)

set hive.exec.mode.local.auto=true; 开启hive的local模式

一、新的需求建表

1.1 建表语句：

代码语言：javascript复制
create table tmp.test(
   deviceid   string,
   dt         string
)row format delimited 
fields terminated by ','

1.2 插入语句：

代码语言：javascript复制
INSERT INTO tmp.test VALUES
	('deviceid1','2022-11-01'),
	('deviceid1','2022-11-01'),
	('deviceid1','2022-11-02'),
	('deviceid1','2022-11-02'),
	('deviceid1','2022-11-03'),
	('deviceid1','2022-11-05'),
	('deviceid1','2022-11-05'),
	('deviceid1','2022-11-05'),
	('deviceid1','2022-11-06'),
	('deviceid1','2022-11-07'),
	('deviceid1','2022-11-07'),
	('deviceid1','2022-11-08'),
	('deviceid1','2022-11-09'),
	('deviceid1','2022-11-10'),
	('deviceid2','2022-11-01'),
	('deviceid2','2022-11-01'),
	('deviceid2','2022-11-02'),
	('deviceid2','2022-11-02'),
	('deviceid2','2022-11-03'),
	('deviceid2','2022-11-05'),
	('deviceid2','2022-11-05'),
	('deviceid2','2022-11-05'),
	('deviceid2','2022-11-06'),
	('deviceid2','2022-11-07'),
	('deviceid2','2022-11-07'),
	('deviceid2','2022-11-08'),
	('deviceid2','2022-11-09'),
	('deviceid2','2022-11-10');

SELECT * FROM tmp.test

二、需求一

2.1 最近一个月内，有过连续活跃2天的用户数

代码语言：javascript复制
SELECT deviceid,count(*)
from (
	SELECT deviceid, dt_sub,count(*) count_n
	from (
		SELECT deviceid , dt
				,date_sub(dt,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt)) as dt_sub
				,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt) as rn
		from (
					select deviceid,dt
					from tmp.test 
					where dt >= date_sub('2022-11-15',30)
					group by deviceid,dt
		)t1
	)t2
	group by deviceid,dt_sub
	having count(*) >= 2
)t3	
group by deviceid

2.2 实现思路解析

2.2.1 求出连续登录的天数

本小节利用row_number()开窗计算出每个设备deviceid的行号，再利用日期dt减去行号，求出相同的dt_sub即为连续登录的日期。

代码语言：javascript复制
SELECT deviceid , dt
		,date_sub(dt,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt)) as dt_sub
		,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt) as rn
from (
			select deviceid,dt
			from tmp.test 
			where dt >= date_sub('2022-11-15',30)
			group by deviceid,dt
)t1

Step 1:子查询从表 tmp.test 中选择 dt 大于等于 '2022-11-15' 减去 30天的数据，并按 deviceid 和 dt 进行分组。返回的结果将包含 deviceid 和 dt 列的值，这里是为了筛选出近一个月30天的数据。

Step 2:在外层查询中，使用窗口函数ROW_NUMBER()将每个 deviceid 分组内的数据按照 dt 进行排序，并为每行分配一个行号，即 rn 列。Step 2查询结果如下：

代码语言：javascript复制
deviceid	dt		dt_sub		rn
deviceid1	2022-11-01	2022-10-31	1
deviceid1	2022-11-02	2022-10-31	2
deviceid1	2022-11-03	2022-10-31	3
deviceid1	2022-11-05	2022-11-01	4
deviceid1	2022-11-06	2022-11-01	5
deviceid1	2022-11-07	2022-11-01	6
deviceid1	2022-11-08	2022-11-01	7
deviceid1	2022-11-09	2022-11-01	8
deviceid1	2022-11-10	2022-11-01	9
deviceid2	2022-11-01	2022-10-31	1
deviceid2	2022-11-02	2022-10-31	2
deviceid2	2022-11-03	2022-10-31	3
deviceid2	2022-11-05	2022-11-01	4
deviceid2	2022-11-06	2022-11-01	5
deviceid2	2022-11-07	2022-11-01	6
deviceid2	2022-11-08	2022-11-01	7
deviceid2	2022-11-09	2022-11-01	8
deviceid2	2022-11-10	2022-11-01	9
2.2.2 汇总求和

本小节求出连续登录大于2天的数量，并进行汇总计算。

代码语言：javascript复制
SELECT deviceid, dt_sub,count(*) count_n
	from (
		SELECT deviceid , dt
				,date_sub(dt,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt)) as dt_sub
				,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt) as rn
		from (
					select deviceid,dt
					from tmp.test 
					where dt >= date_sub('2022-11-15',30)
					group by deviceid,dt
		)t1
	)t2
	group by deviceid,dt_sub
	having count(*) >= 2

Step 3：对结果进行分组，按照 deviceid 和 dt_sub 进行分组，并计算了每个分组中的行数，即 count_n 列。

代码语言：javascript复制
deviceid	dt_sub	count_n
deviceid1	2022-10-31	3
deviceid1	2022-11-01	6
deviceid2	2022-10-31	3
deviceid2	2022-11-01	6

Step 4：对满足条件的分组再次进行了 GROUP BY deviceid，统计每个 deviceid 对应的分组数,即为需求用户数。

代码语言：javascript复制
SELECT deviceid,count(*)
from (
	SELECT deviceid, dt_sub,count(*) count_n
	from (
		SELECT deviceid , dt
				,date_sub(dt,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt)) as dt_sub
				,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt) as rn
		from (
					select deviceid,dt
					from tmp.test 
					where dt >= date_sub('2022-11-15',30)
					group by deviceid,dt
		)t1
	)t2
	group by deviceid,dt_sub
	having count(*) >= 2
)t3	
group by deviceid
代码语言：javascript复制
deviceid	count(*)
deviceid1	2
deviceid2	2

三、需求二

2.1 统计最近一个月内，每个用户的活跃总天数

2.1.1 Step 1
代码语言：javascript复制
select deviceid,dt
from tmp.test 
where dt >= date_sub('2022-11-15',30)

这行语句可以理解为统计最近 11 月 15 日前的 30 天内的用户数据。它使用了 date_sub() 函数来计算日期，并使用 >= 运算符筛选出符合条件的数据，即日期大于等于最近的 11 月 15 日前的 30 天。这样就可以获取最近一个月内的用户数据。

代码语言：javascript复制
deviceid	dt
deviceid1	2022-11-01
deviceid1	2022-11-01
deviceid1	2022-11-02
deviceid1	2022-11-02
deviceid1	2022-11-03
deviceid1	2022-11-05
deviceid1	2022-11-05
deviceid1	2022-11-05
deviceid1	2022-11-06
deviceid1	2022-11-07
deviceid1	2022-11-07
deviceid1	2022-11-08
deviceid1	2022-11-09
deviceid1	2022-11-10
deviceid2	2022-11-01
deviceid2	2022-11-01
deviceid2	2022-11-02
deviceid2	2022-11-02
deviceid2	2022-11-03
deviceid2	2022-11-05
deviceid2	2022-11-05
deviceid2	2022-11-05
deviceid2	2022-11-06
deviceid2	2022-11-07
deviceid2	2022-11-07
deviceid2	2022-11-08
deviceid2	2022-11-09
deviceid2	2022-11-10
2.1.2 Step 2
代码语言：javascript复制
SELECT deviceid, COUNT(DISTINCT dt) AS active_days
	from (
		select deviceid,dt
		from tmp.test 
		where dt >= date_sub('2022-11-15',30)
	)t1
GROUP BY deviceid;

按照 deviceid 进行分组，并使用 COUNT(DISTINCT dt) 函数计算每个设备的唯一日期数，即活跃天数。结果将返回每个设备和其对应的活跃天数。

代码语言：javascript复制
deviceid	active_days
deviceid1	9
deviceid2	9

四、需求三

4.1 最近一个月内，每个用户的最大连续登陆天数

代码语言：javascript复制
select deviceid,max(count_n) max_days
from (
	SELECT deviceid, dt_sub,count(*) count_n
	from (
		SELECT deviceid , dt
				,date_sub(dt,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt)) as dt_sub
				,ROW_NUMBER() over(PARTITION  by deviceid ORDER by dt) as rn
		from (
					select deviceid,dt
					from tmp.test 
					where dt >= date_sub('2022-11-15',30)
					group by deviceid,dt
		)t1
	)t2
	group by deviceid,dt_sub
)t3
group by deviceid

这个需求比较简单，在需求一的基础上，增加对deviceid的分组和对count_n的求最大值即可

代码语言：javascript复制
deviceid	max_days
deviceid1	6
deviceid2	6

五、需求四

5.1 最近一个月内，连续活跃[1-3]天的人数，[4-6]天的人数，[7 天的人数

代码语言：javascript复制
SELECT x1,count(x1)
from (
	SELECT deviceid,dt_sub,active_days,
			case when  active_days >= 1 and active_days <= 3 then '[1-3]'
	          	 when active_days >= 4 and active_days <= 6 then '[4-6]'
	          	 when active_days >= 7 then '[7 ]'  end  x1
	from (
		SELECT deviceid,dt_sub,count(*) as active_days
		FROM (
		    SELECT deviceid, dt,
		   			ROW_NUMBER() OVER (PARTITION BY deviceid ORDER BY dt) as rn,
		           date_sub(dt, ROW_NUMBER() OVER (PARTITION BY deviceid ORDER BY dt))  AS dt_sub
		    FROM (
		        SELECT deviceid, dt
		        FROM tmp.test 
		        WHERE dt >= date_sub('2022-11-15', 30)
		        GROUP BY deviceid, dt
		    ) t1
		) t2
		group by deviceid,dt_sub
	)t3
)t4
group by x1
Step 1

仍然是先计算出连续活跃的天数，在内层查询：

代码语言：javascript复制
deviceid	dt_sub	active_days
deviceid1	2022-10-31	3
deviceid1	2022-11-01	6
deviceid2	2022-10-31	3
deviceid2	2022-11-01	6
Step 2

使用case when 进行如下结果样例操作，可以进行行转列：

代码语言：javascript复制
/**
 * 区间         人数
 * [1-3]    2
 * [4-6]    3
 * [7       4
 */
/**
 *  deviceid1	[1-3]
	deviceid1	[4-6]
	deviceid2	[1-3]
	deviceid2	[4-6]
 */

六、需求五

6.1 最近30天内，沉默天数超过3天的有多少人，超过5天有多少人

6.1.1 更换表

更换满足本需求案例的新的数据和表。

代码语言：javascript复制
CREATE TABLE tmp.testdt (
  deviceid VARCHAR(255),
  dt DATE
);

INSERT INTO tmp.testdt (deviceid, dt) VALUES
('deviceid1', '2023-06-01'),
('deviceid1', '2023-06-02'),
('deviceid1', '2023-06-03'),
('deviceid1', '2023-06-29'),
('deviceid1', '2023-06-30');

6.2 需求实现

代码语言：javascript复制
select	count(DISTINCT if(dt_diff > 3,deviceid,null)) as `超过3天`,
		count(DISTINCT if(dt_diff > 5,deviceid,null)) as `超过5天`
from (
	select deviceid ,dt,pre_date,datediff(date_sub(dt,1),pre_date) as dt_diff
	from (
		SELECT deviceid,dt, LAG(dt,1,dt) OVER (partition by deviceid ORDER BY dt) AS pre_date
		from (
			select deviceid,dt
			from (
				select deviceid,dt
				from tmp.testdt where dt >= date_sub('2023-06-30',30)
				group by deviceid,dt
			)t1
		)t2
	)t3
)t4

6.3 代码技术点

代码语言：javascript复制
SELECT deviceid,dt, LAG(dt,1,dt) OVER (partition by deviceid ORDER BY dt) AS pre_date
		from (
			select deviceid,dt
			from (
				select deviceid,dt
				from tmp.testdt where dt >= date_sub('2023-06-30',30)
				group by deviceid,dt
			)t1
		)t2

Step 1:

以上查询用lag开窗，求出了给定date类型列中每个日期的前一个出现的日期，仅在本列中求出。查询结果如下：

代码语言：javascript复制
deviceid	dt	pre_date
deviceid1	2023-06-01	2023-06-01
deviceid1	2023-06-02	2023-06-01
deviceid1	2023-06-03	2023-06-02
deviceid1	2023-06-29	2023-06-03
deviceid1	2023-06-30	2023-06-29

Step 2:

使用datediff计算出dt和pre_date的间隔日期，即为沉默的天数。

代码语言：javascript复制
deviceid	dt	pre_date dt_diff
deviceid1	2023-06-01	2023-06-01	-1
deviceid1	2023-06-02	2023-06-01	0
deviceid1	2023-06-03	2023-06-02	0
deviceid1	2023-06-29	2023-06-03	25
deviceid1	2023-06-30	2023-06-29	0

最后通过count(DISTINCT if(dt_diff > 3,deviceid,null)) as 超过3天``给定查找的格式字段，即可求出。

代码语言：javascript复制
超过3天	超过5天
1	1
date dt 登录 函数 数据

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