SQL33. 找出每个学校GPA最低的同学
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 | 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 |
device_id | university | gpa |
6543 | 北京大学 | 3.2000 |
4321 | 复旦大学 | 3.6000 |
2131 | 山东大学 | 3.3000 |
2315 | 浙江大学 | 3.6000 |
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:
6543|北京大学|3.200 4321|复旦大学|3.600 2131|山东大学|3.300 2315|浙江大学|3.600
Mysql 解法, 执行用时: 36ms, 内存消耗: 6380KB, 提交时间: 2022-01-03
SELECT device_id,university,gpa FROM ( SELECT *,ROW_NUMBER() over (PARTITION BY university ORDER BY gpa) AS rn FROM user_profile ) AS temp WHERE temp.rn=1
Mysql 解法, 执行用时: 36ms, 内存消耗: 6384KB, 提交时间: 2022-01-25
select a.device_id,a.university,a.gpa from user_profile a inner join (select university,min(gpa) gpa from user_profile group by university) b on a.university=b.university and a.gpa=b.gpa order by a.university # select university,min(gpa) gpa from user_profile group by university
Mysql 解法, 执行用时: 36ms, 内存消耗: 6384KB, 提交时间: 2022-01-03
select device_id,university,gpa from( select *,row_number()over(partition by university order by gpa)as rank_1 from user_profile ) as test where rank_1=1
Mysql 解法, 执行用时: 36ms, 内存消耗: 6400KB, 提交时间: 2022-01-22
select device_id,university,gpa from user_profile where (university,gpa) in (select university,min(gpa) from user_profile group by university) order by university
Mysql 解法, 执行用时: 36ms, 内存消耗: 6420KB, 提交时间: 2022-01-22
SELECT device_id,university,gpa from user_profile WHERE (university,gpa) in ( select university,min(gpa) from user_profile group by university) ORDER BY university