hive分析函数篇：函数高级

墨蓝 2022-05-10 09:54 269阅读 0赞

*  hive与关系型数据库
 *  hive函数

# part1: hive对比关系型数据库 #

hive: 基于Hadoop的一个数据仓库工具，将结构化的数据文件映射成一张表，并提供类sql语句的查询功能, 使用Hql作为查询接口，使用HDFS存储，使用mapreduce计算

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th>mysql</th> 
   <th>hive</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>存储位置</td> 
   <td>本地文件系统</td> 
   <td>hdfs文件系统</td> 
  </tr> 
  <tr> 
   <td>表文件存储形式</td> 
   <td>数据文件,索引文件</td> 
   <td>metadata表(使用关系型数据库), 真实数据文件</td> 
  </tr> 
  <tr> 
   <td>数据格式</td> 
   <td>系统定义</td> 
   <td>用户自定义(serde,fileformat,compress)</td> 
  </tr> 
  <tr> 
   <td>作业引擎</td> 
   <td>MyISAM</td> 
   <td>mapreduce</td> 
  </tr> 
  <tr> 
   <td>使用场景</td> 
   <td>online 事务处理(实时查询)</td> 
   <td>online 分析(高延迟,读多写少,数据挖掘)</td> 
  </tr> 
  <tr> 
   <td>使用的脚本语言</td> 
   <td>sql</td> 
   <td>hql</td> 
  </tr> 
  <tr> 
   <td>可扩展性</td> 
   <td>弱</td> 
   <td>强</td> 
  </tr> 
 </tbody> 
</table>

# part2: hive分析函数 #

##### 1，窗口函数+ 聚合函数+ 普通字段 #####

hive (db)> select * from stu;
    OK
    stu.class_id	stu.stu_id	stu.score
    1	1	90
    1	2	94
    1	3	87
    1	4	89
    2	4	96
    2	2	59
    2	1	75
    3	2	67
    3	1	87

现有数据如上：（如何用一条sql求出： 每班总分，平均分，最高分学生信息 ？）  
1，hql实现

select * from (
    select *, 
    max(score)over(partition by class_id) max,
    sum(score)over(partition by class_id) sum, 
    avg(score)over( partition by class_id) avg
    from stu )tmp
    where tmp.score=tmp.max;
    
    Total jobs = 1
    OK
    tmp.class_id	tmp.stu_id	tmp.score	tmp.max	tmp.sum	tmp.avg
    1	2	94	94	360	90.0
    2	4	96	96	230	76.66666666666667
    3	1	87	87	154	77.0
    Time taken: 3.266 seconds, Fetched: 3 row(s)

2， mysql实现

select stu.class_id, stu.stu_id, stu.score ,tab_max.max ,tab_sum.sum, tab_avg.avg from stu  
    
    inner join (select class_id ,sum(score)sum  from stu  group by class_id) tab_sum 
    on tab_sum.class_id=stu.class_id
    
    inner join (select class_id ,avg(score)avg from stu group by class_id) tab_avg 
    on tab_avg.class_id=stu.class_id
    
    inner join (select class_id,max(score)max from stu group by class_id ) tab_max
    on stu.class_id=tab_max.class_id
    
    where stu.score=tab_max.max;
    
    Total jobs = 10
    OK
    stu.class_id	stu.stu_id	stu.score	tab_max.max	tab_sum.sum	tab_avg.avg
    1	2	94	94	360	90.0
    2	4	96	96	230	76.66666666666667
    3	1	87	87	154	77.0
    Time taken: 119.113 seconds, Fetched: 3 row(s)

##### 2， 一条语句 + 多个group by = 多条件分组(grouping sets, with cube, with rollup) #####

<table> 
 <thead> 
  <tr> 
   <th></th> 
   <th></th> 
   <th align="left">with cube</th> 
   <th align="left">grouping sets</th> 
   <th align="left">with rollup</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>group 组合数</td> 
   <td>1个条件(A)</td> 
   <td align="left">2(2^1)种 (0,1)</td> 
   <td align="left">1(2^1-1)种 (0)</td> 
   <td align="left">2(1+1)种 (0,1)</td> 
  </tr> 
  <tr> 
   <td>group 组合数</td> 
   <td>2个条件(A,B)</td> 
   <td align="left">4(2^2)种(00,01,10,11)</td> 
   <td align="left">3(2^2-1)种 (01, 10,11)</td> 
   <td align="left">3(2+1)种 (00,01,11)</td> 
  </tr> 
  <tr> 
   <td>group 组合数</td> 
   <td>3个条件(A,B,C)</td> 
   <td align="left">8(2^3)种 (000,001,010, 011 ,100, 101, 110, 111)</td> 
   <td align="left">7种(2^3-1)–(000,001,011,111)</td> 
   <td align="left">4(3+1)种(000,001,011,111)</td> 
  </tr> 
  <tr> 
   <td>group 组合数</td> 
   <td>4个条件(A,B,C,D)</td> 
   <td align="left">16(2^4)种 (0000,0001,0010, 0100 ,1000, 1001, 1010, 1100…)</td> 
   <td align="left">15种(2^4-1)–(0000,0001,0010, 0100 ,1000, 1001, 1010, 1100…)</td> 
   <td align="left">5(4+1)种(0000,0001,0011,0111, 1111)</td> 
  </tr> 
 </tbody> 
</table>

grouping\_\_id ： 根据group by后面声明的顺序字段是否存在于当前group by中的一个二进制位组合数据。【 比如（A,C）的grouping\_\_id(A,C) = grouping(A)+grouping(B)+grouping ( C ) ：二进制=101 也就是5】

hive (db)> select * from orders;
    OK
    orders.id	orders.produce_name	orders.price	orders.customerid
    1	phone	7000.0	1
    2	book	34.0	1
    3	shooe	300.0	1
    4	cup	90.0	5
    5	book	34.0	2
    6	shooe	300.0	2
    7	cup	90.0	2
    8	cup	90.0	3
    9	cup	90.0	3
    10	book	78.0	4
    11	cup	90.0	5
    #三个条件： grouping sets(...)
    select customerid, price, produce_name,grouping__id gid from orders 
    group by customerid,price,produce_name
    grouping sets (
    customerid,
    price,
    produce_name,
    
    (customerid,price),
    (customerid,produce_name),
    (price,customerid),
    (price,produce_name),
    (customerid,price),
    (customerid,produce_name),
    (customerid,price,produce_name))
    order by gid;
    #三个条件： with cube/ with rollup
    select customerid, price, produce_name,grouping__id gid from orders 
    group by customerid,price,produce_name
    with cube 
    order by gid;

##### 3， 分区内求最值（top k问题）：LAG,LEAD,FIRST\_VALUE, LAST\_VALUE #####

hive (db)> select * from urls;
    OK
    urls.id	urls.visit_time	urls.url
    USER1	2016-10-12 01:10:00	url1 
    USER1	2016-10-12 01:15:10	url2          
    USER1	2016-10-12 01:16:40	url3          
    USER1	2016-10-12 02:13:00	url4          
    USER1	2016-10-12 03:14:30	url5          
    USER2	2016-11-12 01:10:00	url1          
    USER2	2016-11-12 01:15:10	url2          
    USER2	2016-11-12 01:16:40	url3          
    USER3	2016-11-12 02:13:00	url2          
    USER3	2016-11-12 03:14:30	url3

有数据如上， 要求出： 每个用户的页面的访问次序，每个页面的访问时长

select tmp3.* ,first_value(tmp3.times)over(partition by tmp3.id order by tmp3.times desc) maxtime from (
    select tmp2.id,tmp2.starts,tmp2.ends, nvl(tmp2.url1,'起始url')url1,tmp2.url2, 
    nvl(tmp2.ends-tmp2.starts ,0)times from (
    select tmp.id,tmp.starts, 
    lead(tmp.starts,1,null)over(partition by tmp.id order by tmp.starts )ends,
    lag(tmp.url)over(partition by tmp.id order by tmp.starts) url1,
    tmp.url as url2 
    from
    (select id,url , 
    unix_timestamp(visit_time,'yyyy-MM-dd HH:mm:ss') starts 
    from urls )tmp
    )tmp2
    )tmp3;
    
    Total jobs = 2
    OK
    tmp3.id	tmp3.starts	tmp3.ends	tmp3.url1	tmp3.url2	tmp3.times	maxtime
    USER1	1476209580	1476213270	url3          	url4          	3690	3690
    USER1	1476206200	1476209580	url2          	url3          	3380	3690
    USER1	1476205800	1476206110	起始url	url1 	310	3690
    USER1	1476206110	1476206200	url1 	url2          	90	3690
    USER1	1476213270	NULL	url4          	url5          	0	3690
    USER2	1478884200	1478884510	起始url	url1          	310	310
    USER2	1478884510	1478884600	url1          	url2          	90	310
    USER2	1478884600	NULL	url2          	url3          	0	310
    USER3	1478887980	1478891670	起始url	url2          	3690	3690
    USER3	1478891670	NULL	url2          	url3 	0	3690

##### 4，分段/排名函数NTILE( n ), ROW\_NUMBER, RANK, DENSE\_RANK #####

hive (db)> select * from orders;
    OK
    orders.id	orders.produce_name	orders.price	orders.customerid
    1	phone	7000.0	1
    2	book	90.0	1
    3	shooe	300.0	1
    4	cup	90.0	1
    8	book2	34.0	2
    6	shooe	300.0	2
    7	cup	34.0	2
    13	video	30.0	5
    11	cup	90.0	5
    14	pc	9000.0	1
    15	chair	200.0	5
    16	chair2	200.0	5

数据如上， 要求出每个用户的商品购买总数，价格最高的由哪些？ 价格最高的前1/3商品是哪些？

select * from (
    
    select produce_name, price, customerid,
    count(*)over(partition by customerid ) cnt,
    rank()over(partition by customerid order by price desc )rk,
    ntile(3)over(partition by customerid order by price desc) nt
    from orders
    
    )tmp 
    where tmp.rk=1
    or tmp.nt=1
    
    OK
    tmp.produce_name	tmp.price	tmp.customerid	tmp.cnt	tmp.rk	tmp.nt
    pc	9000.0	1	5	1	1
    phone	7000.0	1	5	2	1
    shooe	300.0	2	3	1	1
    chair	200.0	5	4	1	1
    chair2	200.0	5	4	1	1

# part3: 其他hive常用函数 #

![在这里插入图片描述][70]  
![在这里插入图片描述][70 1]  
![在这里插入图片描述][70 2]  
![在这里插入图片描述][70 3]

[70]: /images/20220510/2f1ece98fa5344e3b0b93a24f272b037.png
[70 1]: /images/20220510/949825aad445484d80e79138f672d1e9.png
[70 2]: /images/20220510/5fef64b8439341548d64fbd1b6f06cc4.png
[70 3]: /images/20220510/ca4a8469a4c34bf18896b3f4e5cd2ced.png