SQL20. 分组排序练习题
描述
id | device_id | gender | age | university | gpa | active_days_within_30 | question_cnt | answer_cnt |
1 | 2138 | male | 21 | 北京大学 | 3.4 | 7 | 2 | 12 |
2 | 3214 | male | | 复旦大学 | 4.0 | 15 | 5 | 25 |
3 | 6543 | female | 20 | 北京大学 | 3.2 | 12 | 3 | 30 |
4 | 2315 | female | 23 | 浙江大学 | 3.6 | 5 | 1 | 2 |
5 | 5432 | male | 25 | 山东大学 | 3.8 | 20 | 15 | 70 |
6 | 2131 | male | 28 | 山东大学 | 3.3 | 15 | 7 | 13 |
7 | 4321 | female | 26 | 复旦大学 | 3.6 | 9 | 6 | 52 |
university | avg_question_cnt |
浙江大学 | 1.0000 |
北京大学 | 2.5000 |
复旦大学 | 5.5000 |
山东大学 | 11.0000 |
示例1
输入:
drop table if exists user_profile; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52);
输出:
浙江大学|1.0000 北京大学|2.5000 复旦大学|5.5000 山东大学|11.0000
Mysql 解法, 执行用时: 36ms, 内存消耗: 6388KB, 提交时间: 2021-11-30
select university, avg(question_cnt) as avg_question_cnt from user_profile group by university order by avg_question_cnt
Mysql 解法, 执行用时: 36ms, 内存消耗: 6392KB, 提交时间: 2021-12-27
SELECT university, AVG(question_cnt) as avg_question_cnt FROM user_profile GROUP BY university order BY avg_question_cnt asc
Mysql 解法, 执行用时: 36ms, 内存消耗: 6396KB, 提交时间: 2022-01-25
select university,avg(question_cnt) as avg_question_cnt from user_profile group by university order by avg(question_cnt);
Mysql 解法, 执行用时: 36ms, 内存消耗: 6396KB, 提交时间: 2021-12-12
SELECT university,avg(question_cnt) avg_question_cnt from user_profile GROUP by university ORDER by 2
Mysql 解法, 执行用时: 36ms, 内存消耗: 6404KB, 提交时间: 2022-01-25
select university,avg(question_cnt) as avg_question_cnt from user_profile group by university ORDER by avg_question_cnt asc;