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题解 | 统计每个用户的平均刷题数

select university,difficult_level,round(count(question_id)/count(distinct device_id),4) as avg_answer_cnt from (select a.device_id,a.university,b.ques...

2025-11-21

1 26

题解 | 统计每个学校的答过题的用户的平均答题数

select university,count(question_id)/count(distinct device_id) as avg_answer_cnt from (select a.device_id,a.university,b.question_id from user_profile...

2025-11-21

1 22