题解 | #统计每个用户的平均刷题数#

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 where up.university='山东大学' group by university , difficult_level;