大数据面试SQL040 计算次日留存率

一、题目

现有用户登录记录表，已经按照用户日期进行去重处理。以用户登录的最早日期作为新增日期，请计算次日留存率是多少。

+----------+-------------+
| user_id  | login_date  |
+----------+-------------+
| aaa      | 2023-12-01  |
| bbb      | 2023-12-01  |
| bbb      | 2023-12-02  |
| ccc      | 2023-12-02  |
| bbb      | 2023-12-03  |
| ccc      | 2023-12-03  |
| ddd      | 2023-12-03  |
| ccc      | 2023-12-04  |
| ddd      | 2023-12-04  |
+----------+-------------+

二、分析

维度	评分
题目难度	⭐️⭐️⭐️⭐️
题目清晰度	⭐️⭐️⭐️⭐️⭐️
业务常见度	⭐️⭐️⭐️⭐️⭐️

三、SQL

指标定义：

次日留存用户：新增用户第二天登录（活跃）的用户；

次日留存率：t+1日留存用户数/t日新增用户；

1.根据登录日志，使用开窗函数计算出用户的最小登录时间作为新增日期first_day,然后计算当天日期和新增日期的时间差。

select
  user_id,
  login_date,
  min(login_date)over(partition by user_id order by login_date asc) as first_day,
  datediff(login_date,min(login_date)over(partition by user_id order by login_date asc)) as date_diff
from t_login_040

查询结果

2.我们根据first_day进行分组，date_diff=0的为当天新增用户，date_diff=1的为次日登录的用户

select
first_day,
count(case when  date_diff = 0 then user_id end) as new_cnt,
count(case when date_diff =1 then user_id end) as next_act_cnt
from(
select
  user_id,
  login_date,
  min(login_date)over(partition by user_id order by login_date asc) as first_day,
  datediff(login_date,min(login_date)over(partition by user_id order by login_date asc)) as date_diff
from t_login_040
)t
group by first_day
order by first_day asc

查询结果

3.用次日留存数/新增用户数据即为留存率，因为新增可能为0，所以需要先判断。

select
first_day,
concat(if(count(case when  date_diff = 0 then user_id end) =0,0,count(case when date_diff =1 then user_id end) /count(case when  date_diff = 0 then user_id end))*100,'%') as next_act_per
from(
select
  user_id,
  login_date,
  min(login_date)over(partition by user_id order by login_date asc) as first_day,
  datediff(login_date,min(login_date)over(partition by user_id order by login_date asc)) as date_diff
from t_login_040
)t
group by first_day
order by first_day asc

查询结果

四、建表语句和数据插入

create table t_login_040
(
user_id string COMMENT '用户ID',
login_date string COMMENT '登录日期'
) COMMENT '用户登录记录表'
stored as orc
;

insert into t_login_040(user_id,login_date)
values
('aaa','2023-12-01'),
('bbb','2023-12-01'),
('bbb','2023-12-02'),
('ccc','2023-12-02'),
('bbb','2023-12-03'),
('ccc','2023-12-03'),
('ddd','2023-12-03'),
('ccc','2023-12-04'),
('ddd','2023-12-04');