题解 | 统计每个学校各难度的用户平均刷题数
select
up.university,
qd.difficult_level,
round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from
question_practice_detail as qpd
inner join user_profile as up on qpd.device_id = up.device_id
inner join question_detail as qd on qpd.question_id = qd.question_id
group by
up.university,
qd.difficult_level
