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) as avg_answer_cnt
from question_practice_detail
inner join user_profile 
on question_practice_detail.device_id = user_profile.device_id
inner join question_detail
on question_practice_detail.question_id = question_detail.question_id
group by university,question_detail.difficult_level