题解 | #统计每个学校各难度的用户平均刷题数#

select university, difficult_level, count(1)/count(distinct question_practice_detail.device_id) as avg_answer_cnt from user_profile inner join question_practice_detail on
user_profile.device_id = question_practice_detail.device_id inner join question_detail
on question_practice_detail.question_id	= question_detail.question_id group by 
university,difficult_level