统计每个用户的平均刷题数

count(distinct up.device_id)

select
  university,
  difficult_level,
  count(qpd.question_id) / count(distinct up.device_id) avg_answer_cnt
from
  user_profile up
  inner join question_practice_detail qpd on up.device_id = qpd.device_id
  inner join question_detail qd on qpd.question_id = qd.question_id
where
  university = "山东大学"
group by
  difficult_level