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Hive对同一表中其他数组列的排序数组列

sql

我在蜂巢中有一张桌子,两列分别为col1 array<int>col2 array<double>。输出如下图

col1                col2
[1,2,3,4,5]         [0.43,0.01,0.45,0.22,0.001]

我想按升序对该col2进行排序,并且col1还应相应地更改其索引,例如

col1                col2
[5,2,4,3,1]        [0.001,0.01,0.22,0.43,0.45]

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2021-04-07

共1个答案

小编典典

分解两个数组,进行排序,然后再次聚合数组。sort在子查询中使用之前collect_list对数组进行排序:

with your_data as(
select array(1,2,3,4,5) as col1,array(0.43,0.01,0.45,0.22,0.001)as col2
)

select original_col1,original_col2, collect_list(c1_x) as new_col1, collect_list(c2_x) as new_col2
from
(
select d.col1 as original_col1,d.col2 as original_col2, c1.x as c1_x, c2.x as c2_x, c1.i as c1_i  
 from your_data d
      lateral view posexplode(col1) c1 as i,x
      lateral view posexplode(col2) c2 as i,x
where c1.i=c2.i 
distribute by original_col1,original_col2
sort by c2_x
)s
group by original_col1,original_col2;

结果:

OK
original_col1   original_col2                   new_col1        new_col2
[1,2,3,4,5]     [0.43,0.01,0.45,0.22,0.001]     [5,2,4,1,3]     [0.001,0.01,0.22,0.43,0.45]
Time taken: 34.642 seconds, Fetched: 1 row(s)

编辑:同一脚本的简化版本,您可以不用第二次posexplode,而是按位置使用直接引用 d.col2[c1.i] as c2_x

with your_data as(
select array(1,2,3,4,5) as col1,array(0.43,0.01,0.45,0.22,0.001)as col2
)

select original_col1,original_col2, collect_list(c1_x) as new_col1, collect_list(c2_x) as new_col2
from
(
select d.col1 as original_col1,d.col2 as original_col2, c1.x as c1_x, d.col2[c1.i] as c2_x, c1.i as c1_i  
 from your_data d
      lateral view posexplode(col1) c1 as i,x
distribute by original_col1,original_col2
sort by c2_x
)s
group by original_col1,original_col2;
2021-04-07