SQL类别高难度试卷得分的截断平均值

id	exam_id	tag	difficulty	duration	release_time
1	9001	SQL	hard	60	2020-01-01 10:00:00
2	9002	算法	medium	80	2020-08-02 10:00:00

示例数据：exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）

id	uid	exam_id	start_time	submit_time	score
1	1001	9001	2020-01-02 09:01:01	2020-01-02 09:21:01	80
2	1001	9001	2021-05-02 10:01:01	2021-05-02 10:30:01	81
3	1001	9001	2021-06-02 19:01:01	2021-06-02 19:31:01	84
4	1001	9002	2021-09-05 19:01:01	2021-09-05 19:40:01	89
5	1001	9001	2021-09-02 12:01:01	(NULL)	(NULL)
6	1001	9002	2021-09-01 12:01:01	(NULL)	(NULL)
7	1002	9002	2021-02-02 19:01:01	2021-02-02 19:30:01	87
8	1002	9001	2021-05-05 18:01:01	2021-05-05 18:59:02	90
9	1003	9001	2021-09-07 12:01:01	2021-09-07 10:31:01	50
10	1004	9001	2021-09-06 10:01:01	(NULL)	(NULL)

根据输入你的查询结果如下：

tag	difficulty	clip_avg_score
SQL	hard	81.7

从examination_info表可知，试卷9001为高难度SQL试卷，该试卷被作答的得分有[80,81,84,90,50]，去除最高分和最低分后为[80,81,84]，平均分为81.6666667，保留一位小数后为81.7

输入描述

输入数据中至少有3个有效分数

示例1

输入:

drop table if exists examination_info;
CREATE TABLE  examination_info (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    exam_id int UNIQUE NOT NULL COMMENT '试卷ID',
    tag varchar(32) COMMENT '类别标签',
    difficulty varchar(8) COMMENT '难度',
    duration int NOT NULL COMMENT '时长',
    release_time datetime COMMENT '发布时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;

drop table if exists exam_record;
CREATE TABLE exam_record (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid int NOT NULL COMMENT '用户ID',
    exam_id int NOT NULL COMMENT '试卷ID',
    start_time datetime NOT NULL COMMENT '开始时间',
    submit_time datetime COMMENT '提交时间',
    score tinyint COMMENT '得分'
)CHARACTER SET utf8 COLLATE utf8_general_ci;

INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES
  (9001, 'SQL', 'hard', 60, '2020-01-01 10:00:00'),
  (9002, '算法', 'medium', 80, '2020-08-02 10:00:00');

INSERT INTO exam_record(uid,exam_id,start_time,submit_time,score) VALUES
(1001, 9001, '2020-01-02 09:01:01', '2020-01-02 09:21:01', 80),
(1001, 9001, '2021-05-02 10:01:01', '2021-05-02 10:30:01', 81),
(1001, 9001, '2021-06-02 19:01:01', '2021-06-02 19:31:01', 84),
(1001, 9002, '2021-09-05 19:01:01', '2021-09-05 19:40:01', 89),
(1001, 9001, '2021-09-02 12:01:01', null, null),
(1001, 9002, '2021-09-01 12:01:01', null, null),
(1002, 9002, '2021-02-02 19:01:01', '2021-02-02 19:30:01', 87),
(1002, 9001, '2021-05-05 18:01:01', '2021-05-05 18:59:02', 90),
(1003, 9001, '2021-02-06 12:01:01', null, null),
(1003, 9001, '2021-09-07 10:01:01', '2021-09-07 10:31:01', 50);

输出:

SQL|hard|81.7

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

Mysql 解法, 执行用时: 36ms, 内存消耗: 6392KB, 提交时间: 2022-01-25

select tag,difficulty,
round((sum(score)-min(score)-max(score))/
      (count(score)-2),1) as avg_score
from examination_info 
join exam_record using(exam_id)
where tag='SQL' and difficulty='hard';

Mysql 解法, 执行用时: 36ms, 内存消耗: 6444KB, 提交时间: 2021-12-18

select t.tag, t.difficulty, round(avg(t.score), 1) from(
select tag, difficulty, score,
         row_number() over(partition by tag order by score) rank1,
         row_number() over(partition by tag order by score desc) rank2 
     from examination_info join exam_record 
     on examination_info.exam_id = exam_record.exam_id and score is not null 
         and tag = 'SQL' and difficulty = 'hard'
    ) t
    where t.rank1 <> 1 and t.rank2 <> 1 group by t.tag

Mysql 解法, 执行用时: 36ms, 内存消耗: 6460KB, 提交时间: 2021-12-01

SELECT tag,difficulty,ROUND((SUM(score)-max(score)-MIN(score))/(COUNT(score)-2),1) AS clip_avg_socre
FROM exam_record as er
join examination_info AS ei
WHERE er.exam_id=ei.exam_id and difficulty='hard' and tag='SQL'

Mysql 解法, 执行用时: 36ms, 内存消耗: 6464KB, 提交时间: 2021-12-31

select tag,difficulty,
round((sum(score)-min(score)-max(score))/
      (count(score)-2),1)as clip_avg_score
from examination_info as ei 
join exam_record as er
on ei.exam_id=er.exam_id
where ei.tag='SQL' and difficulty='hard'
;

Mysql 解法, 执行用时: 36ms, 内存消耗: 6508KB, 提交时间: 2022-01-01

select tag,
difficulty,
round((sum(score)-max(score)-min(score))/(count(score)-2),1)as clip_avg_score
from examination_info ei
left join exam_record er
on ei.exam_id = er.exam_id
where difficulty = 'hard' and submit_time is not null and tag = 'SQL'

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