处理Hive中的数据倾斜的方法

目录

• 1 groupby(大表分组-局部聚合+全局聚合)
• 2 join（大中表Join - 加salt + 小表膨胀）
• 3 双大表Join - 抽样取倾斜key+BroadJoin
• 4 小结

1 groupby(大表分组-局部聚合+全局聚合)

示例1：

select label,sum(cnt) as all from 
(
    select rd,label,sum(1) as cnt from 
    (
        select id,label,round(rand(),2) as rd,value from tmp1
    ) as tmp
    group by rd,label
) as tmp
group by label;

示例2：

select 
	split(new_source,'\\_')[0] as source 
	,sum(cnt) as cnt 
from  
(select  
	concat(source,'_', rand()*100) as  new_source
	,count(1) as cnt 
from  test_table 
where day ='2022-01-01'
group by 
	concat(source,'_', rand()*100)
)tt 
group by 
	split(new_source,'\\_')[0]

2 join（大中表Join - 加salt + 小表膨胀）

示例1：

select label,sum(value) as all from 
(
    select rd,label,sum(value) as cnt from
    (
        select tmp1.rd as rd,tmp1.label as label,tmp1.value*tmp2.value as value 
        from 
        (
            select id,round(rand(),1) as rd,label,value from tmp1
        ) as tmp1
        join
        (
            select id,rd,label,value from tmp2
            lateral view explode(split('0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9',',')) mytable as rd
        ) as tmp2
        on tmp1.rd = tmp2.rd and tmp1.label = tmp2.lab编程客栈el
    ) as tmp1
    group by rd,label
) as tmp1
group by label;

示例2：

select 
	source
	,source_name
	,sum(cnt) as cnt 
from  
(select 
	t1.source 
	,new_source
	,nvl(source_name,'未知') as source_name 
	,count(imei) as cnt 
from  
(select  
	编程imei
	,sphpource 
	,concat(cast(rand()*10 as int ),'_',source ) as new_source
from  test_table_1
where day ='2022-01-01'
) t1 
inner join 
(
select 
	source_name 
	,concat(preflix,'_',source) as new_source
from  test_table_1
where day ='2022-01-01'
lateral view explode(split('0,1,2,3,4,5,6,7,8,9,10',','))b as preflix 
) t2 
on t1.new_source =t2.new_source
group by 
t1.source 
,new_source
,nvl(soujsrce_name,'未知')
) tta  
group by 
	source
	,source_name

3 双大表Join - 抽样取倾斜key+BroadJoin

##优化前：
create table test.tmp_table_test_all as 
select  
imei 
,lable_id 
,nvl(label_name,'未知')
from tmp_table_1  t1  
left join 
(select  
lable_id
,label_name
from  tmp_table_2 
where day ='2024-01-01') t2 
on t1.lable_id =t2.lable_id
where t1.day ='2024-01-01'
;
 
## 优化后 ：
create table test.tmp_table_test_all_new  as 
 
 
with tmp_table_test_1 as 
(select  
lable_id 
,count(1) as cnt 
from tmp_table_1  t1 
tablesample(5 percent) --抽样取5%的数据，减少table scan的量
group by lable_id
order by cnt desc 
limit 100
) 
 
 
select  
	imei 
	,lable_id 
	,nvl(label_name,'未知') as  label_name
from tmp_table_1  t1 
left join  tmp_table_test_1  t2
on tjavascript1.lable_id =t2.lable_id
left join 
(select  
	lable_id
	,label_name
from  tmp_table_2 
where day ='2024-01-01') t3
on t1.lable_id =t3.lable_id
where t1.day ='2024-01-01' and  t2.lable_id is null 
 
union all  
 
select  
	imei 
	,lable_id 
	,nvl(label_name,'未知') as  label_name 
from tmp_table_1  t1 
inner  join 
(select  
	lable_id
from  tmp_table_test_1  t1 
left   join   tmp_table_2  t2 
on t1.lable_id =t2.lable_id
where t2.day ='2024-01-01') t3
on t1.lable_id =t3.lable_id
where t1.day ='2024-01-01' 
;

4 小结

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