题解 | #统计每个学校的答过题的用户的平均答题数#

SELECT up.university, 
    COUNT(qpd.question_id) / COUNT(DISTINCT(qpd.device_id))
    AS avg_answer_cnt
FROM user_profile AS up JOIN question_practice_detail AS qpd
ON up.device_id = qpd.device_id
GROUP BY up.university