题解 | #统计每个学校各难度的用户平均刷题数#

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