HiveSQL题——聚合函数(sum/count/max/min/avg)

目录

一、窗口函数的知识点

1.1 窗户函数的定义

1.2 窗户函数的语法

1.3 窗口函数分类

聚合函数

排序函数

前后函数 

头尾函数

1.4 聚合函数

二、实际案例

2.1 每个用户累积访问次数

0 问题描述

1 数据准备

2 数据分析

3 小结

2.2 各直播间最大的同时在线人数

0 问题描述

1 数据准备

2 数据分析

3 小结

2.3 历史至今每个小时内同时在线人数

0 问题描述

1 数据准备

2 数据分析

3 小结

2.4 某个时间段、每个小时内同时在线人数

0 问题描述

1 数据准备

2 数据分析

3 小结

2.5 学生各学科的成绩

0 问题描述

1 数据准备

2 数据分析

3 小结


一、窗口函数的知识点

1.1 窗户函数的定义

        窗口函数可以拆分为【窗口+函数】。窗口函数官网指路:

LanguageManual WindowingAndAnalytics - Apache Hive - Apache Software Foundationicon-default.png?t=N7T8https://cwiki.apache.org/confluence/display/Hive/LanguageManual%20WindowingAndAnalytics

  • 窗口:限定函数的计算范围(窗口函数:针对分组后的数据,从逻辑角度指定计算的范围,并没有从物理上真正的切分,只有group by 是物理分组,真正意义上的分组)
  • 函数:计算逻辑
  •  窗口函数的位置:跟sql里面聚合函数的位置一样,from -> join -> on -> where -> group by->select 后面的普通字段,窗口函数 -> having -> order by  -> lmit 。 窗口函数不能跟聚合函数同时出现。聚合函数包括count、sum、 min、max、avg。
  • sql 执行顺序:from -> join -> on -> where -> group by->select 后面的普通字段,聚合函数-> having -> order by -> limit

1.2 窗户函数的语法

      <窗口函数>window_name  over ( [partition by 字段...]  [order by 字段...]  [窗口子句] )

  • window_name:给窗口指定一个别名。
  • over:用来指定函数执行的窗口范围,如果后面括号中什么都不写,即over() ,意味着窗口包含满足where 条件的所有行,窗口函数基于所有行进行计算。
  • 符号[] 代表:可选项;  | : 代表二选一
  •  partition by 子句: 窗口按照哪些字段进行分组,窗口函数在不同的分组上分别执行。分组间互相独立。
  • order by 子句:每个partition内部按照哪些字段进行排序,如果没有partition ,那就直接按照最大的窗口排序,且默认是按照升序(asc)排列。
  • 窗口子句:显示声明范围(不写窗口子句的话,会有默认值)。常用的窗口子句如下:
    rows between unbounded preceding and  unbounded following; -- 上无边界到下无边界(一般用于求 总和)
    rows between unbounded preceding and current row;  --上无边界到当前记录(累计值)
    rows between 1 preceding and current row; --从上一行到当前行
    rows between 1 preceding and 1 following; --从上一行到下一行
    rows between current row and 1 following; --从当前行到下一行

       ps: over()里面有order by子句,但没有窗口子句时 ,即: <窗口函数> over ( partition by 字段... order by 字段... )此时窗口子句是有默认值的 -->  rows between unbounded preceding and current row (上无边界到当前行)。

      此时窗口函数语法:<窗口函数> over ( partition by 字段... order by 字段... ) 等价于

     <窗口函数> over ( partition by 字段... order by 字段... rows between unbounded preceding and current row)
      需要注意有个特殊情况:当order by 后面跟的某个字段是有重复行的时候, <窗口函数> over ( partition by 字段... order by 字段... )  不写窗口子句的情况下,窗口子句的默认值是:range between unbounded preceding and current row(上无边界到当前相同行的最后一行)。

    因此,遇到order by 后面跟的某个字段出现重复行,且需要计算【上无边界到当前行】,那就需要手动指定窗口子句 rows between unbounded preceding and current row ,偷懒省略窗口子句会出问题~

      ps: 窗口函数的执行顺序是在where之后,所以如果where子句需要用窗口函数作为条件,需要多一层查询,在子查询外面进行。

     【例如】求出登录记录出现间断的用户Id

select
    id
from (
         select
             id,
             login_date,
             lead(login_date, 1, '9999-12-31')
                  over (partition by id order by login_date) next_login_date
             --窗口函数 lead(向后取n行)
             --lead(column1,n,default)over(partition by column2 order by column3) 查询当前行的后边第n行数据,如果没有就为null
         from (--用户在同一天可能登录多次,需要去重
                  select
                      id,
                      date_format(`date`, 'yyyy-MM-dd') as login_date
                  from user_log
                  group by id, date_format(`date`, 'yyyy-MM-dd')
              ) tmp1
     ) tmp2
where  datediff(next_login_date, login_date) >=2
group by id;

1.3 窗口函数分类

      哪些函数可以是窗口函数呢?(放在over关键字前面的)

  • 聚合函数

sum(column) over (partition by .. order by .. 窗口子句);
count(column) over (partition by .. order by .. 窗口子句);
max(column) over  (partition by .. order by .. 窗口子句);
min(column) over (partition by .. order by .. 窗口子句);
avg(column) over (partition by .. order by .. 窗口子句);

     ps : 高级聚合函数

             collect_list 收集并形成list集合,结果不去重;

             collect_set 收集并形成set集合,结果去重; 

      举例:

--每个月的入职人数以及姓名
 
select 
month(replace(hiredate,'/','-')),
    count(*) as cnt,
    collect_list(name) as name_list
from employee
group by month(replace(hiredate,'/','-'));
 
 
/*
输出结果
month  cn  name_list
4	    2	["宋青书","周芷若"]
6	    1	["黄蓉"]
7	    1	["郭靖"]
8	    2	["张无忌","杨过"]
9	    2	["赵敏","小龙女"]
*/
  • 排序函数

--  顺序排序——1、2、3
row_number() over(partition by .. order by .. )
 
--  并列排序,跳过重复序号——1、1、3(横向加)
rank() over(partition by .. order by .. )
 
-- 并列排序,不跳过重复序号——1、1、2(纵向加)
dense_rank()  over(partition by .. order by .. )
  • 前后函数 

-- 取得column列的前n行,如果存在则返回,如果不存在,返回默认值default
lag(column,n,default) over(partition by order by) as lag_test
-- 取得column列的后n行,如果存在则返回,如果不存在,返回默认值default
lead(column,n,default) over(partition by order by) as lead_test
  • 头尾函数

---当前窗口column列的第一个数值,如果有null值,则跳过
first_value(column,true) over (partition by ..order by.. 窗口子句) 
 
---当前窗口column列的第一个数值,如果有null值,不跳过
first_value(column,false) over (partition by ..order by.. 窗口子句)
 
--- 当前窗口column列的最后一个数值,如果有null值,则跳过
last_value(column,true) over (partition by ..order by.. 窗口子句) 
 
--- 当前窗口column列的最后一个数值,如果有null值,不跳过
last_value(column,false) over (partition by ..order by.. 窗口子句) 
 

1.4 聚合函数

       sum() /count() /max() /min() /avg()  函数,一般用于开窗求累积汇总值。

sum(column) over (partition by .. order by .. 窗口子句);
count(column) over (partition by .. order by .. 窗口子句);
max(column) over  (partition by .. order by .. 窗口子句);
min(column) over (partition by .. order by .. 窗口子句);
avg(column) over (partition by .. order by .. 窗口子句);

二、实际案例

2.1 每个用户累积访问次数

0 问题描述

    统计每个用户累积访问次数

1 数据准备

create table if not exists table6
(
    userid         string comment '用户id',
    visitdate      string comment '访问时间',
    visitcount     int comment '访问次数'
)
    comment '用户访问次数';

2 数据分析

select
    userid,
    visit_date,
    vc1,
     --再求出用户历史至今的累积访问次数
    sum(vc1) over (partition by userid order by visit_date ) as vc2
from (   --先求出用户每个月的累积访问次数
         select
             userid,
             date_format(visitdate, 'yyyy-MM') as visit_date,
             sum(visitcount)  as vc1
         from table6
         group by userid, date_format(visitdate, 'yyyy-MM')
     ) tmp1;

3 小结

2.2 各直播间最大的同时在线人数

0 问题描述

   根据直播间的用户访问记录,统计各直播间最大的同时在线人数。

1 数据准备

create table if not exists table7
(
    room_id      int comment '直播间id',
    user_id      int comment '用户id',
    login_time   string comment '用户进入直播间时间',
    logout_time  string comment '用户离开直播间时间'
)
    comment '直播间的用户访问记录';
INSERT overwrite table table7
VALUES (1,100,'2021-12-01 19:00:00', '2021-12-01 19:28:00'),
       (1,100,'2021-12-01 19:30:00', '2021-12-01 19:53:00'),
       (2,100,'2021-12-01 21:01:00', '2021-12-01 22:00:00'),
       (1,101,'2021-12-01 19:05:00', '2021-12-01 20:55:00'),
       (2,101,'2021-12-01 21:05:00', '2021-12-01 21:58:00'),
       (1,102,'2021-12-01 19:10:00', '2021-12-01 19:25:00'),
       (2,102,'2021-12-01 19:55:00', '2021-12-01 21:00:00'),
       (3,102,'2021-12-01 21:05:00', '2021-12-01 22:05:00'),
       (1,104,'2021-12-01 19:00:00', '2021-12-01 20:59:00'),
       (2,104,'2021-12-01 21:57:00', '2021-12-01 22:56:00'),
       (2,105,'2021-12-01 19:10:00', '2021-12-01 19:18:00'),
       (3,106,'2021-12-01 19:01:00', '2021-12-01 21:10:00');

2 数据分析

select
    room_id,
    max(num)
from (
         select
             room_id,
             sum(flag) over (partition by room_id order by dt) as num
         from (
                  select
                      room_id,
                      user_id,
                      login_time as dt,
                      --对登入该直播间的人,标记 1
                      1          as flag
                  from table7
                  union
                  select
                      room_id,
                      user_id,
                      logout_time as dt,
                       --对退出该直播间的人,标记 -1
                      -1          as flag
                  from table7
              ) tmp1
     ) tmp2
--求出直播间最大的同时在线人数
group by room_id;

3 小结

    该题的关键点在于:对每个用户进入/退出直播间的行为进行打标签,再利用sum()over聚合函数计算最终的数值。

2.3 历史至今每个小时内同时在线人数

       由案例2.2 引申出来的案例 2.3和 案例2.4

0 问题描述

    根据直播间用户访问记录,不限制时间段,统计历史至今的各直播间​​​每个小时内的同时在线人数

1 数据准备

create table if not exists table7
(
    room_id      int comment '直播间id',
    user_id      int comment '用户id',
    login_time   string comment '用户进入直播间时间',
    logout_time  string comment '用户离开直播间时间'
)
    comment '直播间的用户访问记录';
INSERT overwrite table table7
VALUES (1,100,'2021-12-01 19:00:00', '2021-12-01 19:28:00'),
       (1,100,'2021-12-01 19:30:00', '2021-12-01 19:53:00'),
       (2,100,'2021-12-01 21:01:00', '2021-12-01 22:00:00'),
       (1,101,'2021-12-01 19:05:00', '2021-12-01 20:55:00'),
       (2,101,'2021-12-01 21:05:00', '2021-12-01 21:58:00'),
       (1,102,'2021-12-01 19:10:00', '2021-12-01 19:25:00'),
       (2,102,'2021-12-01 19:55:00', '2021-12-01 21:00:00'),
       (3,102,'2021-12-01 21:05:00', '2021-12-01 22:05:00'),
       (1,104,'2021-12-01 19:00:00', '2021-12-01 20:59:00'),
       (2,104,'2021-12-01 21:57:00', '2021-12-01 22:56:00'),
       (2,105,'2021-12-01 19:10:00', '2021-12-01 19:18:00'),
       (3,106,'2021-12-01 19:01:00', '2021-12-01 21:10:00');

2 数据分析

   完整代码如下:

with temp_data as (
    select
        room_id,
        user_id,
        login_time,
        logout_time,
        hour(login_time) as min_time,
        --  hour('2021-12-01 19:30:00') = 19
        hour(logout_time) as max_time,
        length(space(hour(logout_time) - hour(login_time))) as lg,
        split(space(hour(logout_time) - hour(login_time)), '') as dis
    from table7
)

select
    room_id,
    on_time,
    count(1) as cnt
from (
         select distinct
             room_id,
             user_id,
             min_time,
             max_time,
             dis,
             dis_index,
             (min_time + dis_index) as on_time
         from temp_data lateral view posexplode(dis) n as dis_index,dis_data
         order by user_id,
                  min_time,
                  max_time,
                  dis,
                  dis_index
     ) tmp1
group by room_id, on_time
order by room_id, on_time;

     代码拆解分析:

--以一条数据为例,
 room_id  user_id     login_time               logout_time
  1         100    '2021-12-01 19:00:00'     '2021-12-01 21:28:00'
(1)上述数据取时间hour(login_time) as min_time 、hour(logout_time)as max_time
        1(room_id),100(user_id),19(min_time),21(max_time)
(2)split(space(hour(logout_time) - hour(login_time)), '') 的结果:
     根据[21-19]=2,利用space函数生成长度是2的空格字符串,再用split拆分
        1(room_id),100(user_id),19(min_time),21(max_time),['','','']
(3)用posexplode经过转换增加行(列转行,炸裂),通过下角标index来获取 on_time时间,
     根据数组['','',''],得到index的取值是0,1,2
     炸裂得出下面三行数据(一行变三行)
        1(room_id),100(user_id),19(min_time),19 = 19+0 (on_time = min_time+index)
        1(room_id),100(user_id),19(min_time),20 = 19+1 (on_time = min_time+index)
        1(room_id),100(user_id),19(min_time),21 = 19+2 (on_time = min_time+index)
     炸裂的目的:将用户在线的时间段[19-21] 拆分成具体的小时,19,20,21;
(4)根据room_id,on_time进行分组,求出每个直播间分时段的在线人数 

3 小结

    上述代码中用到的函数有:

一、字符串函数
 1、空格字符串函数:space
 语法:space(int n)
 返回值:string
 说明:返回值是n的空格字符串
 举例:select length (space(10)) --> 10
 一般space函数和split函数结合使用:select split(space(3),'');  -->   ["","","",""]

 
 2、split函数(分割字符串)
 语法:split(string str,string pat)
 返回值:array
 说明:按照pat字符串分割str,会返回分割后的字符串数组
 举例:select split ('abcdf','c') from test; -> ["ab","df"]

 3、repeat:重复字符串
 语法:repeat(string A, int n)
 返回值:string
 说明:将字符串A重复n遍。
 举例:select repeat('123', 3); -> 123123123
 一般repeat函数和split函数结合使用:select split(repeat(',',4),',');  -->  
  ["","","","",""]


二、炸裂函数
 explode 
    语法:lateral view explode(split(a,',')) tmp  as new_column
    返回值:string
    说明:按照分隔符切割字符串,并将数组中内容炸裂成多行字符串
    举例:select student_score from test lateral view explode(split(student_score,',')) 
tmp as student_score
 
posexplode
    语法:lateral view posexploed(split(a,',')) tmp as pos,item 
    返回值:string
    说明:按照分隔符切割字符串,并将数组中内容炸裂成多行字符串(炸裂具备瞎下角标 0,1,2,3)
    举例:select student_name, student_score from test
   lateral view posexplode(split(student_name,',')) tmp1 as student_name_index,student_name
   lateral view posexplode(split(student_score,',')) tmp2 as student_score_index,student_score
   where student_score_index = student_name_index
 
 

2.4 某个时间段、每个小时内同时在线人数

0 问题描述

    根据直播间用户访问记录,统计某个时间段的各直播间​​​每个小时内的同时在线人数,假设时间段是['2021-12-01 19:00:00', '2021-12-01 23:00:00']

1 数据准备

​create table if not exists table7
(
    room_id      int comment '直播间id',
    user_id      int comment '用户id',
    login_time   string comment '用户进入直播间时间',
    logout_time  string comment '用户离开直播间时间'
)
    comment '直播间的用户访问记录';
INSERT overwrite table table7
VALUES (1,100,'2021-12-01 19:00:00', '2021-12-01 19:28:00'),
       (1,100,'2021-12-01 19:30:00', '2021-12-01 19:53:00'),
       (2,100,'2021-12-01 21:01:00', '2021-12-01 22:00:00'),
       (1,101,'2021-12-01 19:05:00', '2021-12-01 20:55:00'),
       (2,101,'2021-12-01 21:05:00', '2021-12-01 21:58:00'),
       (1,102,'2021-12-01 19:10:00', '2021-12-01 19:25:00'),
       (2,102,'2021-12-01 19:55:00', '2021-12-01 21:00:00'),
       (3,102,'2021-12-01 21:05:00', '2021-12-01 22:05:00'),
       (1,104,'2021-12-01 19:00:00', '2021-12-01 20:59:00'),
       (2,104,'2021-12-01 21:57:00', '2021-12-01 22:56:00'),
       (2,105,'2021-12-01 19:10:00', '2021-12-01 19:18:00'),
       (3,106,'2021-12-01 19:01:00', '2021-12-01 21:10:00');

​

2 数据分析

   完整代码如下:

with temp_data1 as (
    select
        room_id,
        user_id,
        login_time,
        logout_time,
        hour(login_time) as min_time,
        hour(logout_time)  as max_time,
        split(space(hour(logout_time) - hour(login_time)), '') as dis
    from table7
    where login_time >= '2021-12-01 19:00:00'
      and login_time <= '2021-12-01 21:00:00'
)

select
    room_id,
    on_time,
    count(1) as cnt
from (select distinct
          room_id,
          user_id,
          min_time,
          max_time,
          dis_index,
          (min_time + dis_index) as on_time
      from temp_data1 lateral view posexplode(dis) n1 as dis_index, dis_data
      order by user_id,
               min_time,
               max_time,
               dis_index) tmp
group by room_id, on_time
order by room_id, on_time;

3 小结

    解题思路与2.3一致,只需要限制下时间区间

2.5 学生各学科的成绩

0 问题描述

    基于不同的窗口限定范围(窗口边界),统计各学生的学科成绩。

1 数据准备

create table if not exists table9
(
    name    string comment '学生名称',
    subject string comment '学科',
    score   int comment '分数'
)
    comment '学生分数';
INSERT overwrite table table9
VALUES ('a','数学',12),
       ('b','数学',19),
       ('c','数学',17),
       ('d','数学',24),
       ('a','英语',77),
       ('c','英语',11),
       ('d','英语',34),
       ('a','语文',61);

2 数据分析

select
    name,
    subject,
    score,
    --1.全局聚合
    sum(score) over () as sum1,
    --2.根据学科分组,组内全局聚合
    sum(score) over (partition by subject) as sum2,
    --3.根据学科分组,根据分数排序,计算由起点到当前行的累积值
    sum(score) over (partition by subject order by score)  as sum3,
    --4.根据学科分组,根据分数排序,计算由起点到当前行的累积值 (sum3跟sum4的结果是一样的)
    sum(score) over (partition by subject order by score rows between unbounded preceding and current row ) as sum4,
    --5.根据学科分组,根据分数排序,计算上一行到当前行的累积值
    sum(score) over (partition by subject order by score rows between 1 preceding and current row ) as sum5,
    --6.根据学科分组,根据分数排序,计算上一行到下一行的累积值
    sum(score) over (partition by subject order by score rows between 1 preceding and 1 following)  as sum6,
    --7.根据学科分组,根据分数排序,计算当前行到后面所有行的累积值
    sum(score) over (partition by subject order by score rows between current row and unbounded following ) as sum7
from table9;

3 小结

  窗口函数 = 窗口+ 函数,解题时需要梳理清楚函数的计算范围。

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