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分组计算练习题

求求看看我好吗

avg(active_days_within_30)as avg_active_days,

avg(question_cnt)as avg_question_cnt

这两句有使用avg函数的必要吗

如果单看答案的需求是不是没必要写

查看原题

加油！小菜鸡

求指点，为啥语句和题解一样，但显示答案错误，是哪个细节错了，经常犯这些错，解不出答案😫

SELECT gender,university,COUNT(gender),
avg(active_days_within_30)as avg_active_days,
avg(question_cnt)as avg_question_cnt
FROM user_profile
group by university;

发表于 2021-08-27 01:56:33 回复(27)

Scorpio-Xin

uu们，为什么我写的和题解的答案一样，可是还是提醒我出错，我实在没找到哪里出错了，大家能帮我看看是哪里出错了可以嘛？谢谢！

发表于 2022-06-01 22:24:14 回复(15)

Mysql

牛客477988881号

疑惑，为啥count(distinct divice_id)就不对了呢?

select gender,university, count(distinct device_id) user_num, avg(active_days_within_30) avg_active_day, avg(question_cnt) avg_question_cnt
from user_profile
group by university,gender

发表于 2022-01-18 15:04:21 回复(18)

菠萝冰棒🍍

SELECT 
    gender,
    university,
    COUNT(*) AS user_num,
    ROUND(AVG(active_days_within_30), 1) AS avg_active_day,
    ROUND(AVG(question_cnt ), 1) AS avg_question_cnt
FROM 
    user_profile
GROUP BY 
    gender,
    university
ORDER BY 
    gender ASC,
    university ASC

注意：结果保留1位小数，1位小数之后的四舍五入,查询出来的结果按照gender、university升序排列

所以后面一定要加order by，不然不成功

发表于 2024-12-04 15:39:46 回复(11)

Mysql

有滋有味小熊鞋

SELECT gender,university,count(gender) as user_num,
avg(active_days_within_30) as avg_active_days,avg(question_cnt) as avg_question_cnt
from user_profile
GROUP by university,gender;

发表于 2021-10-03 16:40:12 回复(5)

Mysql

酸菜鱼土豆大侠

【场景】：每个学校不同性别的用户

【分类】：分组计算

分析思路

select 查询结果 [性别;学校;count(设备ID) as 用户数;avg(30天内活跃记录) as 平均活跃天数;avg(发帖记录) as 平均发帖数]
from 从哪张表中查找数据 [user_profile]
group by 分组 [学校;性别]

求解代码

方法一：

select
    gender,
    university,
    count(device_id) as user_num,
    avg(active_days_within_30) as avg_active_day,
    avg(question_cnt) as avg_question_cnt
from user_profile
group by university, gender

发表于 2022-11-23 19:19:53 回复(4)

Mysql

月光死磕英语

select gender,university,count(id) user_num,avg(active_days_within_30) avg_active_day,avg(question_cnt) avg_question_cnt from user_profile

group by university,gender

求大佬解答，为什么第三个user_num 可以是id,也可以是性别(gender) 我不能理解

发表于 2021-12-04 14:58:59 回复(10)

键盘盘盘盘

select
gender
,university
,count(gender) as uer_num
,avg(active_days_within_30) as avg_active_day
,avg(question_cnt) as avg_question_cnt
FROM user_profile

group by gender,university

发表于 2021-11-03 20:10:24 回复(4)

会飞的包子123

SELECT gender,university,COUNT(gender) user_num,
AVG(active_days_within_30) avg_active_days,
AVG(question_cnt) avg_question_cnt
FROM user_profile
GROUP BY university,gender

发表于 2021-08-29 16:07:32 回复(3)

Mysql

冷凡社长

select gender,university
,count(device_id) as user_num
,round(avg(active_days_within_30),1) as avg_active_day
,round(avg(question_cnt),1) as avg_question_cnt
from user_profile
group by gender,university

这个题的考察点比较多。
多列分组。
聚合函数。
使用round控制小数点。

发表于 2022-07-14 15:01:23 回复(8)

Mysql

霸总王子奇

select gender,
    university,
    count(device_id) as user_num,
    round(avg(active_days_within_30),1) as avg_active_day,
    round(avg(question_cnt),1)  as avg_question_cnt 
from 
    user_profile 
group by 
    gender,university 
order by
    gender,university;

注意：输出按照题目要求：

1.对性别和学校分组 group by gender,university

2.查询出来的结果按照gender、university升序排列 order by gender,university

发表于 2024-12-05 12:15:00 回复(1)

Mysql

牛客747227703号

select gender,university,
count(gender) as user_num,
avg(active_days_within_30) as avg_active_days,
avg(question_cnt) as avg_question_cnt
from user_profile
group by gender,university

发表于 2021-08-26 14:37:59 回复(0)

JiangJJ

select gender,university,count(gender) as user_num,
ROUND(avg(active_days_within_30),1),
ROUND(avg(question_cnt),1)
from user_profile
group by gender,university

发表于 2022-05-24 01:11:50 回复(1)

Mysql

这昵称有毒

select gender, university, count(gender), avg(active_days_within_30), avg(question_cnt) from user_profile group by gender,university;

发表于 2021-08-25 18:52:02 回复(1)

牛客462217447号

想问一下group by时候gender和university的先后顺序有影响吗，结果是没区别的，但是应该要按顺序吧

发表于 2022-08-13 16:34:51 回复(3)

Mysql

牛客270645913

SELECT gender, university, count(gender) as user_num, 
ROUND(avg(active_days_within_30), 1) as avg_active_day,
ROUND(avg(question_cnt), 1) as avg_question_cnt
FROM user_profile
GROUP BY gender, university

发表于 2021-11-27 22:04:45 回复(0)

coderLYC

user_num不是每个学校不同性别的用户量么? 为什么可以count(gender), 这样不是只会计算男生和女生的数量了么?

不知道该怎么同时满足两个字段进行count, 求教

发表于 2022-07-01 15:55:30 回复(5)

Mysql

牛客655660060号

select gender,university,COUNT(gender),avg(active_days_within_30),avg(question_cnt) from user_profile group by gender,university;

发表于 2021-08-26 19:04:19 回复(0)

Mysql

滚筒洗衣机🌸

select gender,university,count(*) user_num,round(avg(active_days_within_30),1) avg_active_day ,
round(avg(question_cnt),1) avg_question_cnt
from user_profile
group by gender,university

发表于 2022-03-12 17:23:22 回复(0)

Mysql

牛客316872201号

select
l.gender,
l.university,
count(gender) as user_num ,
avg(l.active_days_within_30) as avg_active_day,
avg(l.question_cnt) as avg_question_cnt
FROM user_profile l

group by l.university ,l.gender

题目示例表格中那个字段：“avg(l.question_cnt) as avg_question-cnt”的最后这个“-cnt”有问题，应该是下划线“_cnt”

发表于 2021-10-25 10:08:31 回复(0)

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问题信息

SQL

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分组计算练习题

分析思路

求解代码

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