题解 | #统计每个用户的平均刷题数#

select a.university,
difficult_level,
count(question_practice_detail.question_id)/count(distinct(question_practice_detail.device_id))
from question_practice_detail 
left join (
    select device_id,university 
    from user_profile) a 
on question_practice_detail.device_id = a.device_id
left join question_detail
on question_practice_detail.question_id = question_detail.question_id
where a.university = "山东大学"
group by a.university,difficult_level;