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题解 | #浙大不同难度题目的正确率#

select difficult_level, sum(result='right')/count(*) as correct_rate from (select user_profile.device_id, question_id, result, university from user_pr...

2024-07-15

0 118

题解 | #统计每个学校各难度的用户平均刷题数#

select university, difficult_level, count(*) / count(distinct device_id) as avg_answer_cnt from ( select c.questio...

2024-07-14

0 119

题解 | #统计每个学校的答过题的用户的平均答题数#

select university, avg(num) as avg_answer_cnt from ( select university, count(*) as num from ...

2024-07-14

0 107