统计每个用户的平均刷题数

用户信息表：user_profile

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	NULL	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	male	28	复旦大学	3.6	9	6	52

第一行表示:id为1的用户的常用信息为使用的设备id为2138，性别为男，年龄21岁，北京大学，gpa为3.4，在过去的30天里面活跃了7天，发帖数量为2，回答数量为12

最后一行表示:id为7的用户的常用信息为使用的设备id为432，性别为男，年龄28岁，复旦大学，gpa为3.6，在过去的30天里面活跃了9天，发帖数量为6，回答数量为52

题库练习明细表：question_practice_detail

id	device_id	question_id	result
1	2138	111	wrong
2	3214	112	wrong
3	3214	113	wrong
4	6534	111	right
5	2315	115	right
6	2315	116	right
7	2315	117	wrong
8	5432	117	wrong
9	5432	112	wrong
10	2131	113	right
11	5432	113	wrong
12	2315	115	right
13	2315	116	right
14	2315	117	wrong
15	5432	117	wrong
16	5432	112	wrong
17	2131	113	right
18	5432	113	wrong
19	2315	117	wrong
20	5432	117	wrong
21	5432	112	wrong
22	2131	113	right
23	5432	113	wrong

第一行表示:id为1的用户的常用信息为使用的设备id为2138，在question_id为111的题目上，回答错误

......

最后一行表示:id为23的用户的常用信息为使用的设备id为5432，在question_id为113的题目上，回答错误

表：question_detail

id	question_id	difficult_level
1	111	hard
2	112	medium
3	113	easy
4	115	easy
5	116	medium
6	117	easy

第一行表示: 题目id为111的难度为hard

....

第一行表示: 题目id为117的难度为easy

请你写一个SQL查询，计算山东、不同难度的用户平均答题量，根据示例，你的查询应返回以下结果(结果在小数点位数保留4位，4位之后四舍五入)：

university	difficult_level	avg_answer_cnt
山东大学	easy	4.5000
山东大学	medium	3.0000

山东大学有设备id为5432和2131这2个用户，这2个用户总共在question_practice_detail表下有12条答题记录，且答题题目是112，113，117，且数目分别为3，6，3，从question_detail可以知道题目难度分别为medium,easy,easy，所以，easy共有9个，故easy的题目平均答题量= 9(easy=9)/2 (device_id=3214 or device_id=5432) =4.5000，medium共有3个，medium的答题只有device_id=5432的用户，故medium的题目平均答题量= 3(medium=9)/1 ( device_id=5432) =3.0000

示例1

输入:

drop table if exists `user_profile`;
drop table if  exists `question_practice_detail`;
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 
);
CREATE TABLE `question_practice_detail` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`question_id`int NOT NULL,
`result` varchar(32) NOT NULL
);
CREATE TABLE `question_detail` (
`id` int NOT NULL,
`question_id`int NOT NULL,
`difficult_level` varchar(32) NOT NULL
);

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);
INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong');
INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong');
INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong');
INSERT INTO question_practice_detail VALUES(4,6543,111,'right');
INSERT INTO question_practice_detail VALUES(5,2315,115,'right');
INSERT INTO question_practice_detail VALUES(6,2315,116,'right');
INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong');
INSERT INTO question_practice_detail VALUES(8,5432,117,'wrong');
INSERT INTO question_practice_detail VALUES(9,5432,112,'wrong');
INSERT INTO question_practice_detail VALUES(10,2131,113,'right');
INSERT INTO question_practice_detail VALUES(11,5432,113,'wrong');
INSERT INTO question_practice_detail VALUES(12,2315,115,'right');
INSERT INTO question_practice_detail VALUES(13,2315,116,'right');
INSERT INTO question_practice_detail VALUES(14,2315,117,'wrong');
INSERT INTO question_practice_detail VALUES(15,5432,117,'wrong');
INSERT INTO question_practice_detail VALUES(16,5432,112,'wrong');
INSERT INTO question_practice_detail VALUES(17,2131,113,'right');
INSERT INTO question_practice_detail VALUES(18,5432,113,'wrong');
INSERT INTO question_practice_detail VALUES(19,2315,117,'wrong');
INSERT INTO question_practice_detail VALUES(20,5432,117,'wrong');
INSERT INTO question_practice_detail VALUES(21,5432,112,'wrong');
INSERT INTO question_practice_detail VALUES(22,2131,113,'right');
INSERT INTO question_practice_detail VALUES(23,5432,113,'wrong');
INSERT INTO question_detail VALUES(1,111,'hard');
INSERT INTO question_detail VALUES(2,112,'medium');
INSERT INTO question_detail VALUES(3,113,'easy');
INSERT INTO question_detail VALUES(4,115,'easy');
INSERT INTO question_detail VALUES(5,116,'medium');
INSERT INTO question_detail VALUES(6,117,'easy');

输出:

山东大学|easy|4.5000
山东大学|medium|3.0000

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: 6588KB, 提交时间: 2021-12-31

select university ,difficult_level,count(qpd.question_id)/count(DISTINCT qpd.device_id) as avg_answer_cnt
from user_profile as u
inner join question_practice_detail as qpd
on u.device_id = qpd.device_id
inner join question_detail as qd
on qpd.question_id = qd.question_id
where university = '山东大学'
group by difficult_level

Mysql 解法, 执行用时: 37ms, 内存消耗: 6380KB, 提交时间: 2022-01-22

select
    a.university,
    c.difficult_level,
    COUNT(b.question_id) / COUNT(DISTINCT b.device_id) avg_answer_cnt
from
    user_profile a
    JOIN question_practice_detail b on a.device_id = b.device_id
    JOIN question_detail c on b.question_id = c.question_id
WHERE
    a.university = '山东大学'
GROUP BY
    c.difficult_level

Mysql 解法, 执行用时: 37ms, 内存消耗: 6384KB, 提交时间: 2022-01-27

SELECT up.university,qd.difficult_level,COUNT(qpd.result) / COUNT(DISTINCT qpd.device_id)
FROM question_practice_detail AS qpd
LEFT JOIN user_profile AS up
ON qpd.device_id=up.device_id
LEFT JOIN question_detail AS qd
ON qpd.question_id=qd.question_id
WHERE up.university='山东大学'
GROUP BY qd.difficult_level

Mysql 解法, 执行用时: 37ms, 内存消耗: 6384KB, 提交时间: 2021-12-21

select u.university, q2.difficult_level,
       count(q1.question_id) / count(distinct(q1.device_id))
from user_profile as u
inner join question_practice_detail as q1
on u.device_id=q1.device_id
inner join question_detail as q2
on q1.question_id=q2.question_id
group by u.university, q2.difficult_level
having u.university='山东大学'

Mysql 解法, 执行用时: 37ms, 内存消耗: 6388KB, 提交时间: 2022-01-27

SELECT u.university,qd.difficult_level,
count(qpd.question_id)/count(DISTINCT qpd.device_id) as avg_answer_cnt
from question_practice_detail qpd inner join user_profile u
ON qpd.device_id = u.device_id
INNER join question_detail as qd
on qpd.question_id = qd.question_id
where university = '山东大学'
GROUP BY qd.difficult_level

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