我有一种复杂的要求
1）1）对于Pinterest
twitter handle , pinterest_post , pinterest_likes.
handle "what" , 7
JavaPairRDD<String ,Pinterest> PintRDD
2）for Instagram
Twitter handle , instargam_post , instagram_likes
handle "hello" , 10
handle2 "hi" ,20
JavaPairRDD<String ,Pinterest> instRDD
3）本体论
twitter handle , categories , sub_categories
handle , Products , MakeUp
handle , Products, MakeUp
handle2 , Services , Face
JavaPairRDD<String ,ontologies1> ontologiesPair
最终输出应为
如果找到值，则应从相应的对象打印，否则应打印空白值。
编辑 - 根据Umberto的代码
public class Combine3PairRDD {
public static void main(String[] args) {
CommonUtils generateSparkContext = new CommonUtils();
JavaSparkContext sc = generateSparkContext.createSparkContext();
JavaPairRDD<String, Pinterest> pintRDD = sc
.parallelizePairs(Arrays.asList(new Tuple2<String, Pinterest>("handle", new Pinterest("what", 7))));
JavaPairRDD<String, Instagram> instRDD = sc
.parallelizePairs(Arrays.asList(new Tuple2<String, Instagram>("handle", new Instagram("hello", 10)),
new Tuple2<String, Instagram>("handle2", new Instagram("Hi", 20))));
JavaPairRDD<String, Ontologies> ontologiesPair = sc.parallelizePairs(
Arrays.asList(new Tuple2<String, Ontologies>("handle", new Ontologies("marketing", "MakeUp")),
new Tuple2<String, Ontologies>("handle2", new Ontologies("Service", "Face")),
new Tuple2<String, Ontologies>("handle", new Ontologies("products", "MakeUp"))));
JavaPairRDD<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>> grouped = ontologiesPair
.cogroup(instRDD, pintRDD);
System.out.println("size of cogreop"+grouped.count());
grouped.foreach(new functionn());
JavaPairRDD<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>> groupedWithDuplicated = grouped
.flatMapToPair(new PairFlatMapFunction<Tuple2<String,Tuple3<Iterable<Ontologies>,Iterable<Instagram>,Iterable<Pinterest>>>,String,Tuple3<Iterable<Ontologies>,Iterable<Instagram>,Iterable<Pinterest>>>() {
private static final long serialVersionUID = 853578182309543660L;
@Override
public Iterable<Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>> call(
Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>> entry)
throws Exception {
List<Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>> withDuplicate = new ArrayList<Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>>();
String key = entry._1();
List<Ontologies>listOntologies = Lists.newArrayList(entry._2()._1());
List<Instagram>listInstagram = Lists.newArrayList(entry._2()._2());
List<Pinterest>listPinterest = Lists.newArrayList(entry._2()._3());
Set<Ontologies> setOntologies = new HashSet<Ontologies>(listOntologies);
Set<Instagram> setInstagram = new HashSet<Instagram>(listInstagram);
Set<Pinterest> setPinterest = new HashSet<Pinterest>(listPinterest);
if(setOntologies.size() < listOntologies.size()){
/* There are duplicates */
withDuplicate
.add(new Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>(
key, new Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>(
entry._2()._1(), entry._2()._2(), entry._2()._3())));
}
if(setInstagram.size() < listInstagram.size()){
/* There are duplicates */
withDuplicate
.add(new Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>(
key, new Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>(
entry._2()._1(), entry._2()._2(), entry._2()._3())));
}
if(setPinterest.size() < listPinterest.size()){
/* There are duplicates */
withDuplicate
.add(new Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>(
key, new Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>(
entry._2()._1(), entry._2()._2(), entry._2()._3())));
}
withDuplicate
.add(new Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>(
key, new Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>(
entry._2()._1(), entry._2()._2(), entry._2()._3())));
return withDuplicate;
}
});
List<Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>>> mapResult2 = groupedWithDuplicated
.collect();
for (Tuple2<String, Tuple3<Iterable<Ontologies>, Iterable<Instagram>, Iterable<Pinterest>>> entry : mapResult2) {
Ontologies ontologies = new Ontologies("", "");
Pinterest pinterest = new Pinterest("", -1);
Instagram instagram = new Instagram("", -1);
if (entry._2()._1().iterator().hasNext()) {
ontologies = entry._2()._1().iterator().next();
}
if (entry._2()._2().iterator().hasNext()) {
instagram = entry._2()._2().iterator().next();
}
if (entry._2()._3().iterator().hasNext()) {
pinterest = entry._2()._3().iterator().next();
}
System.out.println(entry._1() + " " + pinterest.getPinterest_post() + " " + " " + pinterest.getPinterest_likes() + " "
+ instagram.getInstagram_post() + " " + instagram.getInstagram_likes() + " " + ontologies.getCategories() + " "
+ ontologies.getSub_categories());
}
}
}
The three wrappers are as follows
public class Pinterest implements Serializable{
private static final long serialVersionUID = 1226764093455880169L;
public String twitterHandle;
public String pinterest_post ;
public int pinterest_likes;
Pinterest(String pinterest_post,int pinterest_likes){
this.pinterest_post=pinterest_post;
this.pinterest_likes=pinterest_likes;
}
public int hashCode(){
return (int)
pinterest_post.hashCode() *
pinterest_likes;
}
public boolean equals(Object o) {
if(o == null) return false;
Pinterest other = (Pinterest) o;
if(this.twitterHandle != other.twitterHandle) return false;
if(! this.pinterest_post.equals(other.pinterest_post)) return false;
if(this.pinterest_likes != other.pinterest_likes) return false;
return true;
}
}
public class Ontologies implements Serializable{
private static final long serialVersionUID = 1996294848173720136L;
public String twitterHandle;
public String categories ;
public String sub_categories ;
Ontologies(String categories,String sub_categories){
this.categories=categories;
this.sub_categories=sub_categories;
}
public int hashCode(){
return (int)
categories.hashCode() *
sub_categories.hashCode();
}
public boolean equals(Object o) {
if(o == null) return false;
Ontologies other = (Ontologies) o;
if(this.twitterHandle != other.twitterHandle) return false;
if(! this.categories.equals(other.categories)) return false;
if(! this.sub_categories.equals(other.sub_categories)) return false;
return true;
}
}
public class Instagram implements Serializable {
private static final long serialVersionUID = 7351892713578143761L;
public String twitterHandle;
public String instagram_post ;
public int instagram_likes;
Instagram(String instagram_post,int instagram_likes){
this.instagram_post=instagram_post;
this.instagram_likes=instagram_likes;
}
public int hashCode(){
return (int)
instagram_post.hashCode() *
instagram_likes;
}
public boolean equals(Object o) {
if(o == null) return false;
Instagram other = (Instagram) o;
if(this.twitterHandle != other.twitterHandle) return false;
if(! this.instagram_post.equals(other.instagram_post)) return false;
if(this.instagram_likes != other.instagram_likes) return false;
return true;
}
}
上面的代码在构造函数的值中可以正常工作，如代码
new Tuple2<String, Ontologies>("handle", new Ontologies("Products", "MakeUp")),
new Tuple2<String, Ontologies>("handle2", new Ontologies("Service", "Face")),
new Tuple2<String, Ontologies>("handle", new Ontologies("Products", "MakeUp")))
工作正常并打印
handle what 7 hello 10 Products MakeUp
handle what 7 hello 10 Products MakeUp
handle2 -1 Hi 20 Service Face
但是当我改变构造函数时
new Tuple2<String, Ontologies>("handle", new Ontologies("Marketing", "MakeUp"))
new Tuple2<String, Ontologies>("handle2", new Ontologies("Service", "Face")),
new Tuple2<String, Ontologies>("handle", new Ontologies("Products", "MakeUp")))
new Tuple2<String, Ontologies>("handle", new Ontologies("Products", "MakeUp")))
我想要为键创建两行，即句柄和句柄2即
handle what 7 hello 10 Marketing MakeUp
handle what 7 hello 10 Products MakeUp
handle what 7 hello 10 Products MakeUp
handle2 -1 Hi 20 Service Face
，我该如何实现呢？
答案 0 :(得分：0)