如何将字符串解析为 HIVE 中的映射数组
How to parse an string to an array of maps in HIVE
我有一个从系统日志中提取的配置单元 table。数据以一种奇怪的格式(映射数组)编码,其中数组的每个元素都包含 field_name
和 value
。列类型为 STRING。就像下面的例子:
select 1 as user_id, '[{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}]' as user_info
union all
select 2 as user_id, '[{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}]' as user_info;
它创建了这样的东西:
user_id
user_info
1
[{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}]
2
[{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}]
请注意,数组大小并不总是相同。我正在尝试将地图数组转换为简单地图。然后,这就是我期望的结果:
user_id
user_info
1
{"name":"Bob", "gender":"M"}
2
{"name":"Ana", "gender":"F", "age":22}
我计划通过 3 个步骤实现:(1) 解析字符串列以创建映射数组,(2) 分解数组(使用横向视图),(3) 收集字段列表,以及按 user_id
分组
我正在努力完成第一步:解析字符串列以创建映射数组。任何帮助将不胜感激:D
查看代码中的注释。要转换为映射的字符串数组由此 split(user_info, '(?<=\}) *, *(?=\{)')
生成。然后分解,每个元素转化为map。
with mydata as
(select 1 as user_id, '[{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}]' as user_info
union all
select 2 as user_id, '[{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}]' as user_info
)
select user_id,
--build new map
str_to_map(concat('name:', name, nvl(concat(',','gender:', gender),''), nvl(concat(',','age:', age),'') )) as user_info
from
(
select user_id,
--get name, gender, age, aggregate by user_id
max(case when user_info['field'] = 'name' then user_info['value'] end) name,
max(case when user_info['field'] = 'gender' then user_info['value'] end) gender,
max(case when user_info['field'] = 'age' then user_info['value'] end) age
from
(
select s.user_id,
--remove {} and ", convert to map
str_to_map(regexp_replace(e.element,'^\{| *"|\}$','')) as user_info
from
(
select user_id, regexp_replace(user_info, '^\[|\]$','') as user_info -- remove []
from mydata
)s lateral view outer explode(split(user_info, '(?<=\}) *, *(?=\{)'))e as element --split by comma between }{ with optional spaces in between
) s
group by user_id
)s
结果:
user_id user_info
1 {"name":"Bob","gender":"M"}
2 {"name":"Ana","gender":"F","age":"22"}
我有一个从系统日志中提取的配置单元 table。数据以一种奇怪的格式(映射数组)编码,其中数组的每个元素都包含 field_name
和 value
。列类型为 STRING。就像下面的例子:
select 1 as user_id, '[{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}]' as user_info
union all
select 2 as user_id, '[{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}]' as user_info;
它创建了这样的东西:
user_id | user_info |
---|---|
1 | [{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}] |
2 | [{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}] |
请注意,数组大小并不总是相同。我正在尝试将地图数组转换为简单地图。然后,这就是我期望的结果:
user_id | user_info |
---|---|
1 | {"name":"Bob", "gender":"M"} |
2 | {"name":"Ana", "gender":"F", "age":22} |
我计划通过 3 个步骤实现:(1) 解析字符串列以创建映射数组,(2) 分解数组(使用横向视图),(3) 收集字段列表,以及按 user_id
我正在努力完成第一步:解析字符串列以创建映射数组。任何帮助将不胜感激:D
查看代码中的注释。要转换为映射的字符串数组由此 split(user_info, '(?<=\}) *, *(?=\{)')
生成。然后分解,每个元素转化为map。
with mydata as
(select 1 as user_id, '[{"field":"name", "value":"Bob"}, {"field":"gender", "value":"M"}]' as user_info
union all
select 2 as user_id, '[{"field":"gender", "value":"F"}, {"field":"age", "value":22}, {"field":"name", "value":"Ana"}]' as user_info
)
select user_id,
--build new map
str_to_map(concat('name:', name, nvl(concat(',','gender:', gender),''), nvl(concat(',','age:', age),'') )) as user_info
from
(
select user_id,
--get name, gender, age, aggregate by user_id
max(case when user_info['field'] = 'name' then user_info['value'] end) name,
max(case when user_info['field'] = 'gender' then user_info['value'] end) gender,
max(case when user_info['field'] = 'age' then user_info['value'] end) age
from
(
select s.user_id,
--remove {} and ", convert to map
str_to_map(regexp_replace(e.element,'^\{| *"|\}$','')) as user_info
from
(
select user_id, regexp_replace(user_info, '^\[|\]$','') as user_info -- remove []
from mydata
)s lateral view outer explode(split(user_info, '(?<=\}) *, *(?=\{)'))e as element --split by comma between }{ with optional spaces in between
) s
group by user_id
)s
结果:
user_id user_info
1 {"name":"Bob","gender":"M"}
2 {"name":"Ana","gender":"F","age":"22"}