2023-05-26 14:29 深圳大学 数据分析师
关注
select
count(u2.date) /count(u1.date )
from
(select distinct device_id, date from question_practice_detail) as u1
left join (select distinct device_id, date from question_practice_detail) as u2
on (u1.device_id = u2.device_id) and (date_add(u1.date,interval 1 day)=u2.date)
count(u2.date) /count(u1.date )
from
(select distinct device_id, date from question_practice_detail) as u1
left join (select distinct device_id, date from question_practice_detail) as u2
on (u1.device_id = u2.device_id) and (date_add(u1.date,interval 1 day)=u2.date)
「求助大佬帮看看这道算法题吧！」题目：现在运营想要查看用户在某天刷题后第二天还会再来刷题的平均概率。请你取出相应数据。 示例：question_practice_detail id devic...
https://gw-c.nowcoder.com/api/sparta/jump/link?link=https%3A%2F%2Fwww.nowcoder.com%2Fpractice%2F126083961ae0415fbde061d7ebbde453
全部评论
相关推荐
昨天 17:16
门头沟学院 Java 点赞 评论 收藏
分享
牛客吹哨人：哨哥晚点统一更新到黑名单：能救一个是一个！26届毁意向毁约裁员黑名单https://www.nowcoder.com/discuss/1525833
点赞 评论 收藏
分享
