SQL速算N日留存

news2024/9/28 13:16:00

之前才哥发布了《用SQL进行用户留存率计算》

链接：https://mp.weixin.qq.com/s/QJ8JUO00bVJe_K6sx_ttaw

简化数据后得到如下结构的数据：

由于用户和登录日期被设置为主键所以不需要再进行去重，下面看看如何快速求七日留存。

数据下载地址：https://gitcode.net/as604049322/blog_data/-/blob/master/role_login.sql

首先我们求每个用户的安装日期和第几天仍然再登录：

select 
	role_id,
	install_date,
	datediff(b.event_date,install_date) day_diff
from (
	select
		role_id,min(event_date) install_date
	from role_login group by role_id
) a join role_login b using(role_id);

然后我们一口气求得七日留存：

with cte as (
	select 
		role_id,
		install_date,
		datediff(b.event_date,install_date) day_diff
	from (
		select
			role_id,min(event_date) install_date
		from role_login group by role_id
	) a join role_login b using(role_id)
)
select 
	install_date,
	sum(day_diff=0) 新增用户数,
	sum(day_diff=1) 次日留存,
	sum(day_diff=2) 2日留存,
	sum(day_diff=3) 3日留存,
	sum(day_diff=4) 4日留存,
	sum(day_diff=5) 5日留存,
	sum(day_diff=6) 6日留存,
	sum(day_diff=7) 7日留存,
	concat(round(sum(day_diff=1)*100/sum(day_diff=0),2),"%") 次日留存率,
	concat(round(sum(day_diff=2)*100/sum(day_diff=0),2),"%") 2日留存率,
	concat(round(sum(day_diff=3)*100/sum(day_diff=0),2),"%") 3日留存率,
	concat(round(sum(day_diff=4)*100/sum(day_diff=0),2),"%") 4日留存率,
	concat(round(sum(day_diff=5)*100/sum(day_diff=0),2),"%") 5日留存率,
	concat(round(sum(day_diff=6)*100/sum(day_diff=0),2),"%") 6日留存率,
	concat(round(sum(day_diff=7)*100/sum(day_diff=0),2),"%") 7日留存率
from cte where day_diff<=7
group by 1 order by 1;