题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    a. university, c. difficult_level, count(b. question_id)/count(distinct a. device_id) avg_answer_cnt
from
    user_profile a
JOIN
    question_practice_detail b on a. device_id = b. device_id
JOIN
    question_detail c on b. question_id = c. question_id
GROUP BY a. university, c. difficult_level