当前位置:   article > 正文

hive中map相关函数总结_hive map函数

hive map函数

hive官方函数解释

hive官网函数大全地址: hive官网函数大全地址

Return TypeNameDescription
mapmap(key1, value1, key2, value2, …)Creates a map with the given key/value pairs.
arraymap_values(Map<K.V>)Returns an unordered array containing the values of the input map.
arraymap_keys(Map<K.V>)Returns an unordered array containing the keys of the input map.
map<string,string>str_to_map(text[, delimiter1, delimiter2])Splits text into key-value pairs using two delimiters. Delimiter1 separates text into K-V pairs, and Delimiter2 splits each K-V pair. Default delimiters are ‘,’ for delimiter1 and ‘:’ for delimiter2.
Tkey,Tvalueexplode(MAP<Tkey,Tvalue> m)Explodes a map to multiple rows. Returns a row-set with a two columns (key,value) , one row for each key-value pair from the input map. (As of Hive 0.8.0.).

示例

1、map(key1, value1, key2, value2, …)

SELECT map('name', '张三', 'age', 20, 'gender', '男') AS student;
---结果:
student	
{"age":"20","gender":"男","name":"张三"}
  • 1
  • 2
  • 3
  • 4

2、map_values(Map<K.V>)

SELECT map_keys(map('name', '张三', 'age', 20, 'gender', '男')) AS keys;
---结果:
keys
["name","age","gender"]
  • 1
  • 2
  • 3
  • 4

3、map_values(Map<K.V>)

SELECT map_values(map('name', '张三', 'age', 20, 'gender', '男')) AS values;
---结果:
values	
["张三","20","男"]
  • 1
  • 2
  • 3
  • 4

4、str_to_map(str, delimiter1, delimiter2)
str_to_map 函数用于将一个字符串转换为 Map 对象。具体来说,str_to_map 函数会将一个由键值对组成的字符串解析成一个 Map 对象,其中键和值之间使用指定的分隔符进行分隔。其中,str 是要转换的字符串,delimiter1 是键值对之间的分隔符,delimiter2 是键和值之间的分隔符。默认情况下,delimiter1 的值是 ‘,’,delimiter2 的值是 ‘:’。

SELECT str_to_map('name:张三,age:20,gender:男', ',', ':') AS student;
---结果:
student	
{"age":"20","gender":"男","name":"张三"}

SELECT str_to_map('name=张三,age=20,gender=男', ',', '=') AS student;
---结果:
student	
{"age":"20","gender":"男","name":"张三"}
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9

5、explode (map)

select explode(map('A',10,'B',20,'C',30));
select explode(map('A',10,'B',20,'C',30)) as (key,value);
select tf.* from (select 0) t lateral view explode(map('A',10,'B',20,'C',30)) tf;
select tf.* from (select 0) t lateral view explode(map('A',10,'B',20,'C',30)) tf as key,value;
---上述四个结果均为:
key     value
A       10	
B       20	
C       30
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9

实战

给出一组学生数据,有名字,课程,等级,分数等字段,现在求每门课的情况,包含平均成绩,及这门课包含哪些学生及学生的等级

with stud as
( select  'zhang3' as name ,'优' as grade  ,'math' as course ,'88' as score  
  union all 
  select  'li4' as name ,'良' as grade  ,'math' as course ,'72' as score
    union all 
  select  'zhao6' as name ,'差' as grade  ,'math' as course ,'44' as score
    union all 
  select  'wang5' as name ,'优' as grade  ,'chinese' as course ,'80' as score
    union all 
  select  'zhao6' as name ,'优' as grade  ,'chinese' as course ,'55' as score
    union all 
  select  'tian7' as name ,'优' as grade  ,'chinese' as course ,'75' as score
)

--sql1
select course, collect_set(concat(name,':',grade)) as collect , avg(score) from stud group by course;
---结果:
course             collect                                             avg(score)	
math        ["li4:良","zhao6:差","zhang3:优"]                           68.0
chinese     ["wang5:优","tian7:优","zhao6:优"]                          70.0
----sql2
select course, concat_ws(',',collect_set(concat(name,':',grade))) as strings , avg(score) from stud group by course;
---结果:
course                      strings                                        avg(score)
math             li4:良,zhao6:差,zhang3:优                                  68.0
chinese          wang5:优,tian7:优,zhao6:优                                 70.0
----sql3
select course, str_to_map(concat_ws(',',collect_set(concat(name,':',grade))),',',':') as maps , avg(score) from stud group by course;
---结果:
course                               maps                              avg(score)	
math                 {"li4":"良","zhang3":"优","zhao6":"差"}             68.0
chinese              {"tian7":"优","wang5":"优","zhao6":"优"}            70.0
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25
  • 26
  • 27
  • 28
  • 29
  • 30
  • 31
  • 32

注意:
第一种sql,collect 字段的类型是array;第二种sql,strings字段的类型是string;第三种sql,maps字段的类型是map;
问题来了,能否在第二种的基础上,实现第一种和第三种的结果,且字段类型是string;
下面实现第二种转化为第三种,实际上就是map格式转换成json字符串;

with stud as
( select  'zhang3' as name ,'优' as grade  ,'math' as course ,'88' as score  
  union all 
  select  'li4' as name ,'良' as grade  ,'math' as course ,'72' as score
    union all 
  select  'zhao6' as name ,'差' as grade  ,'math' as course ,'44' as score
    union all 
  select  'wang5' as name ,'优' as grade  ,'chinese' as course ,'80' as score
    union all 
  select  'zhao6' as name ,'优' as grade  ,'chinese' as course ,'55' as score
    union all 
  select  'tian7' as name ,'优' as grade  ,'chinese' as course ,'75' as score
)

select 
course
,concat('{"',string2,'"}') as string3
from  
(select 
course
,regexp_replace(string1,'\\,','\\"\\,\\"') as string2
from  
(
select 
    course,
    concat_ws(',', collect_list(concat_ws('":"', k,v) ) ) as string1
from (
select course, str_to_map(concat_ws(',',collect_set(concat(name,':',grade))),',',':') as maps , avg(score) 
from stud group by course
)test_map_1
lateral view outer explode(maps) kv as k,v
group by course
) tt 
) tm 

---结果:
course                               string3                            	
math                 {"li4":"良","zhang3":"优","zhao6":"差"}           
chinese              {"tian7":"优","wang5":"优","zhao6":"优"}        
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25
  • 26
  • 27
  • 28
  • 29
  • 30
  • 31
  • 32
  • 33
  • 34
  • 35
  • 36
  • 37
  • 38
  • 39
声明:本文内容由网友自发贡献,不代表【wpsshop博客】立场,版权归原作者所有,本站不承担相应法律责任。如您发现有侵权的内容,请联系我们。转载请注明出处:https://www.wpsshop.cn/w/盐析白兔/article/detail/86369
推荐阅读
相关标签
  

闽ICP备14008679号