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