题解 | #统计每个学校各难度的用户平均刷题数#
题意明确：
计算每个学校用户不同难度下的用户平均答题题目数
问题分解：
- 限定条件：无；
- 每个学校：按学校分组
group by university - 不同难度：按难度分组
group by difficult_level - 平均答题数：总答题数除以总人数
count(qpd.question_id) / count(distinct qpd.device_id) - 来自上面信息三个表，需要联表，up与qpd用device_id连接，qd与qpd用question_id连接。
细节问题：
- 表头重命名：as
- 平均值精度：保留4位小数round(x, 4)
完整代码：
|-------------------------------|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| 1 2 3 4 5 6 7 8 9 10 11 12 13 | select ``university, ``difficult_level, ``round(count(qpd.question_id) / count(distinct qpd.device_id),``4``) as avg_answer_cnt from question_practice_detail as qpd left join user_profile as up on up.device_id=qpd.device_id left join question_detail as qd on qd.question_id=qpd.question_id group by university, difficult_level |