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://www.nowcoder.com/practice/126083961ae0415fbde061d7ebbde453
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