题解 | #统计每个学校各难度的用户平均刷题数#

select 
    university,
    difficult_level,
    round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from question_practice_detail as qpd

left join user_profile as up
on up.device_id=qpd.device_id

left join question_detail as qd
on qd.question_id=qpd.question_id

group by university, difficult_level