题解 | #统计每个学校的答过题的用户的平均答题数#

select a.ee, a.cc / b.ff as avg_answer_cnt

from ( select user_profile.university as ee, count(question_id) as cc from user_profile, question_practice_detail where user_profile.device_id = question_practice_detail.device_id group by user_profile.university ) a join ( select user_profile.university as dd , count( distinct user_profile.device_id) as ff from user_profile, question_practice_detail where user_profile.device_id = question_practice_detail.device_id group by user_profile.university order by user_profile.university ) b on a.ee= b.dd

order by b.dd