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统计每个用户的平均刷题数

[编程题]统计每个用户的平均刷题数
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题目:运营想要查看参加了答题的山东大学的用户在不同难度下的平均答题题目数,请取出相应数据
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
....
最后一行表示: 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
select
    t.university,
    d.difficult_level,
    count(d.difficult_level) / count(distinct t.user_device_id) as 'avg_answer_cnt'
from
    question_detail d
    inner join (
        select
            u.device_id AS user_device_id,
            u.university,
            q.question_id
        from
            user_profile u
            inner join question_practice_detail q on u.device_id = q.device_id
    ) t on d.question_id = t.question_id
where t.university = '山东大学'
group by
    t.university,
    d.difficult_level;

发表于 2025-06-20 21:03:31 回复(0)
select
        u.university,
        d.difficult_level,
        count(q.question_id)/count(distinct u.device_id) as co
from user_profile u
left join question_practice_detail q
on u.device_id = q.device_id
left join question_detail d
on q.question_id  = d.question_id
where university = '山东大学'
group by u.university,d.difficult_level
发表于 2025-06-20 18:18:28 回复(0)
select u.university, 
qd.difficult_level, 
round(count(qpd.question_id)/count(distinct qpd.device_id),4) 
as avg_answer_cnt 
from question_practice_detail as qpd 
left join question_detail as qd on qpd.question_id = qd.question_id
inner join user_profile as u on qpd.device_id = u.device_id and u.university = '山东大学'
group by qd.difficult_level
逻辑就是构造表格然后选数据
发表于 2025-06-16 21:41:25 回复(0)
SELECT
    u.university,
    qd.difficult_level,
    ROUND(COUNT(qp.question_id)/COUNT(DISTINCT qp.device_id),4) avg_answer_cnt
FROM
    question_practice_detail qp,
    user_profile u,
    question_detail qd
WHERE
    qp.device_id = u.device_id
    AND qp.question_id = qd.question_id and university="山东大学"
GROUP BY
    university, difficult_level
发表于 2025-06-11 18:31:55 回复(0)
SELECT
    up.university,
    qd.difficult_level,
    COUNT(qpd.question_id)/COUNT(DISTINCT(up.device_id)) AS avg_answer_cnt
FROM
    user_profile up
    INNER JOIN question_practice_detail qpd ON up.device_id = qpd.device_id
    INNER JOIN question_detail qd ON qd.question_id = qpd.question_id
GROUP BY
    up.university,
    qd.difficult_level
HAVING up.university = "山东大学"
发表于 2025-06-05 16:33:56 回复(0)
select a.university,c.difficult_level,
round(count(b.question_id)/count(distinct(b.device_id)),4) as avg_anwser_cnt
from user_profile as a
inner join qusetion_practice_detail as b
on a.device_id = b.device_id
inner join question_detail as c
on b.device_id = c.device_id
where a.university = '山东大学'
group by c.difficult_level;
为啥我一直报错呢
发表于 2025-05-31 10:30:29 回复(0)
select t3.university ,t4.difficult_level,
    count(t3.question_id)/count(distinct t3.device_id) as avg_answer_cnt
from (
    select t1.university,t2.question_id,t1.device_id
    from user_profile t1 right join question_practice_detail t2
    on t1.device_id=t2.device_id
    where t1.university='山东大学'
    )t3
inner join question_detail t4
on t3.question_id=t4.question_id
group by t3.university,t4.difficult_level;
我先是获取了我想要的数据,再通过子查询继续获取。
同时练习连接和子查询
发表于 2025-05-28 11:47:14 回复(0)
写错的代码如下
select 
   university,
   difficult_level,
   count(qpd.question_id)/count(qpd.device_id) as avg_answer_cnt
from question_practice_detail qpd

join user_profile up on qpd.device_id=up.device_id
join question_detail qd on qpd.question_id=qd.question_id

where university='山东大学'
group by difficult_level;
错误原因如下:
没有对题库练习明细表question_practice_detail中的device_id去重;
注意,去重后,用左/内连接都可以运行成功。
发表于 2025-04-04 15:21:15 回复(0)
select university,difficult_level,
round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from question_practice_detail as qpd
    inner join user_profile as up
        on up.device_id = qpd.device_id and up.university = '山东大学'
    inner join question_detail as qd
        on qd.question_id = qpd.question_id
group by difficult_level
发表于 2025-04-03 18:17:59 回复(0)
SELECT
    university,
    difficult_level,
    round(count(q.question_id)/count(distinct(q.device_id)),4) as avg_answer_cnt
FROM user_profile as u
WHERE u.university = "山东大学"
JOIN question_practice_detail as q
ON q.device_id = u.device_id
JOIN question_detail as qd
ON q.question_id = qd.question_id
GROUP BY difficult_level;
这样为什么报错
发表于 2025-03-27 11:07:35 回复(1)
SELECT
    us.university,
    qd.difficult_level,
    ROUND(COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id),4) AS avg_answer_cnt
FROM user_profile us
JOIN question_practice_detail qpd USING device_id
JOIN question_detail qd USING question_id
GROUP BY qb.difficult_level 
HAVING us.university = '山东大学'
那位大神帮我看看出啥问题了

发表于 2025-03-22 22:39:02 回复(0)
select university,difficult_level,
round(count(B.question_id)/count(distinct B.device_id),4) as 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
group by
university,difficult_level
having
university="山东大学";

回答平均值=回答问题的总数量/唯一设备ID的总数量
避免字段数据来源模糊,一定在记得在索取字段前限定所在表的名字。

发表于 2025-03-20 17:13:19 回复(0)
题目中说要筛选答过题的人数,如果直接用JOIN的话可以,但是如果用LEFT JOIN ,是不是应该在添加一个,question_id不等于null(虽然这个题里的数据所有人都答过题)
SELECT 
    university,
	difficult_level,
	count(*)/count(distinct(u.device_id)) avg_answer_cnt
FROM user_profile u LEFT JOIN question_practice_detail qpd
ON u.device_id = qpd.device_id
JOIN question_detail qd 
ON qpd.question_id = qd.question_id
WHERE university='山东大学'
AND qpd.question_id IS NOT NULL
GROUP BY difficult_level



发表于 2025-03-15 13:50:05 回复(0)
从上一题中,直接加入where
university = '山东大学'
条件,这样有什么不妥吗?
select
    university,
    difficult_level,
    round(
        count(qpd.question_id) / count(distinct qpd.device_id),
        4
    ) avg_answer_cnt
from
    question_practice_detail qpd
    join user_profile up on up.device_id = qpd.device_id
    join question_detail qd on qd.question_id = qpd.question_id
where
    university = '山东大学'
group by
    university,
    difficult_level
发表于 2025-03-12 18:59:29 回复(0)
SELECT
university,
difficult_level,
ROUND(COUNT(qpd.question_id) / COUNT(DISTINCT qpd.device_id), 4) AS avg_answer_cnt
FROM
question_practice_detail qpd
LEFT JOIN user_profile up USING(device_id)
LEFT JOIN question_detail qd USING(question_id)
WHERE university = '山东大学'
GROUP BY
university,
difficult_level;

为啥用Having会报错呢? 只能用where吗
发表于 2025-03-12 16:22:18 回复(0)