0

说我有一个结构

{1001, {{id=1001, count=20, key=a}, {id=1001, count=30, key=b}}}
{1002, {{id=1002, count=40, key=a}, {id=1001, count=50, key=b}}}

我希望它把它变成

{id=1001, a=20, b=30}
{id=1002, a=40, b=50}

我可以使用哪些 Pig 命令来执行此操作?

4

2 回答 2

1

看起来您正在旋转,类似于Pivoting in Pig。但是你已经有一袋元组了。进行内部连接将是昂贵的,因为它会导致额外的 Map Reduce Jobs。要快速做到这一点,您需要在嵌套的 foreach 中进行过滤。修改后的代码将类似于:

inpt = load '..../pig/bag_pivot.txt' as (id : int, b:bag{tuple:(id : int, count : int, key : chararray)});

result = foreach inpt {
    col1 = filter b by key == 'a';
    col2 = filter b by key == 'b';
    generate id, flatten(col1.count) as a, flatten(col2.count) as b;
};

样本输入数据:

1001    {(1001,20,a),(1001,30,b)}
1002    {(1002,40,a),(1001,50,b)}

输出:

(1001,20,30)
(1002,40,50)
于 2012-08-01T19:49:25.950 回答
1

不确定你的起始关系的格式是什么,但对我来说它看起来像 (int, bag:{tuple:(int,int,chararray)})?如果是这样,这应该工作:

flattened = FOREACH x GENERATE $0 AS id, flatten($1) AS (idx:int, count:int, key:chararray);
a = FILTER flattened BY key == 'a';
b = FILTER flattened BY key == 'b';
joined = JOIN a BY id, b BY id;
result = FOREACH joined GENERATE a::id AS id, a::count AS a, b::count AS b;
于 2012-07-31T22:52:02.470 回答