select
sum(case when datediff(t.py,t.date)=1 then 1 else  0 end)/
count(distinct device_id,date)
from(
select
id,device_id,date,
lead(date,1)over(partition by device_id order by date) as py
from question_practice_detail
) as t
唯一难点在于，不知道那个平均后来才知道，每个人每天的天数，不是全部时间