hadoop - Pig- 无法转储数据

Question

我有两个数据集，一个用于电影，另一个用于评级

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

评分数据看起来像

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

我的脚本如下

    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;

我收到这个错误

 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，脚本执行成功

为什么我不能 DUMP 在第 5 行生成的关系？

Answer 1

使用 disambiguate operator ( :: ) 识别 JOIN 之后的字段名称, COGROUP , CROSS , 或 FLATTEN运营商。

关系 movies_data和 ratings_data两者都有一列 movieid .形成关系时moviedata_ratingsdata_join_group , 使用 ::运算符来识别哪个列 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;