题解 | #统计每个用户的平均刷题数#

select
"山东大学" as  university,
B.difficult_level,
count(*)/count(distinct A.device_id)
from question_practice_detail as A 
inner join question_detail as B
on A.question_id=B.question_id
and A.device_id in (select device_id from user_profile where university="山东大学")
group by B.difficult_level