hadoop - 如何使用 Apache Pig 获取类似 GROUP BY 的 SQL？

Question

我有以下名为 movieUserTagFltr 的输入:

(260,{(260,starwars),(260,George Lucas),(260,sci-fi),(260,cult classic),(260,Science Fiction),(260,classic),(260,supernatural powers),(260,nerdy),(260,Science Fiction),(260,critically acclaimed),(260,Science Fiction),(260,action),(260,script),(260,"imaginary world),(260,space),(260,Science Fiction),(260,"space epic),(260,Syfy),(260,series),(260,classic sci-fi),(260,space adventure),(260,jedi),(260,awesome soundtrack),(260,awesome),(260,coming of age)})
(858,{(858,Katso Sanna!)})
(924,{(924,slow),(924,boring)})
(1256,{(1256,Marx Brothers)})

它遵循模式:(movieId:int, tags:bag{(movieId:int, tag:cararray),...})

基本上，第一个数字代表电影 ID，随后的包包含与该电影相关的所有关键字。我想以这样的方式对这些关键字进行分组，以便输出如下内容:

(260,{(1,starwars),(1,George Lucas),(1,sci-fi),(1,cult classic),(4,Science Fiction),(1,classic),(1,supernatural powers),(1,nerdy),(1,critically acclaimed),(1,action),(1,script),(1,"imaginary world),(1,space),(1,"space epic),(1,Syfy),(1,series),(1,classic sci-fi),(1,space adventure),(1,jedi),(1,awesome soundtrack),(1,awesome),(1,coming of age)})
(858,{(1,Katso Sanna!)})
(924,{(1,slow),(1,boring)})
(1256,{(1,Marx Brothers)})

请注意，对于 ID 为 260 的电影，标签 Science Fiction 出现了 4 次。使用 GROUP BY 和 COUNT，我设法使用以下脚本计算每部电影的不同关键字:

sum = FOREACH group_data { 
    unique_tags = DISTINCT movieUserTagFltr.tags::tag;
    GENERATE group, COUNT(unique_tags) as tag;
};

但这只会返回一个全局计数，我想要一个本地计数。所以我当时的逻辑是:

result = iterate over each tuple of group_data {
    generate a tuple with $0, and a bag with {
        foreach distinct tag that group_data has on it's $1 variable do {
            generate a tuple like: (tag_name, count of how many times that tag appeared on $1)
        }
    }
}

Answer 1

您可以展平原始输入，以便每个 movieID 和 tag 都是它们自己的记录。然后按 movieID 和 tag 分组以获得每个组合的计数。最后，按 movieID 分组，这样您就可以得到每部电影的一袋标签和计数。

假设您从 movieUserTagFltr 和您描述的架构开始:

A = FOREACH movieUserTagFltr GENERATE FLATTEN(tags) AS (movieID, tag);
B = GROUP A BY (movieID, tag);
C = FOREACH B GENERATE
    FLATTEN(group) AS (movieID, tag),
    COUNT(A) AS movie_tag_count;
D = GROUP C BY movieID;

您的最终架构是:

D: {group: int,C: {(movieID: int,tag: chararray,movie_tag_count: long)}}