题解 | #统计每个用户的平均刷题数#

select 
university,difficult_level, count(qpd.device_id)/count(distinct qpd.device_id) as avg_answer_cnt
from 
question_practice_detail qpd left join question_detail qd on qpd.question_id =qd.question_id
left join
user_profile up on qpd.device_id = up.device_id
group by university,difficult_level
having university='山东大学'