pig-无法转储数据

e7arh2l6  于 2021-05-29  发布在  Hadoop
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我有两个数据集,一个是电影,另一个是收视率
电影数据看起来像

MovieID#Title#Genre
1#Toy Story (1995)#Animation|Children's|Comedy
2#Jumanji (1995)#Adventure|Children's|Fantasy
3#Grumpier Old Men (1995)#Comedy|Romance

评级数据看起来像

UserID#MovieID#Ratings#RatingsTimestamp
1#1193#5#978300760
1#661#3#978302109
1#914#3#978301968

我的剧本如下

1) movies_data = LOAD '/user/admin/MoviesDataset/movies_new.dat' USING PigStorage('#') AS (movieid:int,
    moviename:chararray,moviegenere:chararray);

    2) ratings_data = LOAD '/user/admin/RatingsDataset/ratings_new.dat' USING PigStorage('#') AS (Userid:int,
    movieid:int,ratings:int,timestamp:long);

    3) moviedata_ratingsdata_join = JOIN movies_data BY movieid, ratings_data BY movieid;

    4) moviedata_ratingsdata_join_group = GROUP moviedata_ratingsdata_join BY movies_data.movieid;

    5) moviedata_ratingsdata_averagerating = FOREACH moviedata_ratingsdata_join_group GENERATE group,
    AVG(moviedata_ratingsdata_join.ratings) AS Averageratings, (moviedata_ratingsdata_join.Userid) AS userid;

    6) DUMP moviedata_ratingsdata_averagerating;

我得到这个错误

2017-03-25 06:46:50,332 [main] ERROR org.apache.pig.tools.pigstats.PigStats - ERROR 0: org.apache.pig.backend.executionengine.ExecException: ERROR 0: Exception while executing (Name: moviedata_ratingsdata_join_group: Local Rearrange[tuple]{int}(false) - scope-95 Operator Key: scope-95): org.apache.pig.backend.executionengine.ExecException: ERROR 0: Exception while executing (Name: moviedata_ratingsdata_averagerating: New For Each(false,false)[bag] - scope-83 Operator Key: scope-83): org.apache.pig.backend.executionengine.ExecException: ERROR 0: Scalar has more than one row in the output. 1st : (1,Toy Story (1995),Animation|Children's|Comedy), 2nd :(2,Jumanji (1995),Adventure|Children's|Fantasy) (common cause: "JOIN" then "FOREACH ... GENERATE foo.bar" should be "foo::bar" )

如果删除第6行,脚本将成功执行
为什么我不能转储第5行中生成的关系?

snvhrwxg

snvhrwxg1#

使用消歧运算符( :: )在 JOIN , COGROUP , CROSS ,或 FLATTEN 操作员。
关系 movies_data 以及 ratings_data 两者都有一列 movieid . 形成关系时 moviedata_ratingsdata_join_group ,使用 :: 运算符来标识哪个列 movieid 用于 GROUP .
所以你的 4) 看起来像,

4) moviedata_ratingsdata_join_group = GROUP moviedata_ratingsdata_join BY movies_data::movieid;

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