RDD联接：将两个不同的RDD对联接之后，结果RDD密钥值和顺序已更改？

Question

我有两个RDD可以说

 RDD1 : [(1,a),(2,b),(3,c)]    
 RDD2 : [(1,d),(2,e),(3,f)]

现在正在使用join加入这些RDD

 RDD3 = RDD1.join(RDD2);

我已经用以下代码显示了RDD3中的元素

 for(Tuple2<Integer,Tuple2<String,String>> tuple : RDD3.collect()) 
                      System.out.println(tuple._1()+":"+tuple._2()._1()+","+tuple._2()._2());

我已经看到了奇怪的结果

5:b,e
4:a,d 
6:c,f

正如我所期望的那样

1:a,d
1:b,e 
1:c,f

有什么办法可以像上面那样获得所需的输出吗？ 否则会错误地解释RDD行为？ 请建议

编辑：

其实我正在读取这样的数据

JavaDoubleRDD data1 = sc.parallelizeDoubles(Arrays.asList(45.25,22.15,33.24));
JavaDoubleRDD data2 = sc.parallelizeDoubles(Arrays.asList(23.45,19.35,12.45));

接着

JavaPairRDD<Double,Double> lat1 = data1.cartesian(data1);
JavaRDD<Double> lat2 = lat1.map(new Function<Tuple2<Double,Double>,Double>() {
    @Override
    public Double call(Tuple2<Double,Double> t) {
        return Math.pow(t._1()-t._2(),2);
    }
});
 //flag and flag1 are static variables initially equal to 1
JavaPairRDD<Integer,Double> lat3 = lat2.mapToPair(new PairFunction<Double,Integer,Double>() {
    @Override
     public Tuple2<Integer,Double> call(Double d) {
        return new Tuple2<Integer,Double>(flag++,d);
    }
});
System.out.println("Latitude values display");  
    for(Tuple2<?,?> tuple : lat3.collect()) {
                  System.out.println(tuple._1()+":"+tuple._2());
    } 
JavaPairRDD<Double,Double> long1 = data2.cartesian(data2);
JavaRDD<Double> long2 = long1.map(new Function<Tuple2<Double,Double>,Double>() {
        @Override
        public Double call(Tuple2<Double,Double> t) {
                return Math.pow(t._1()-t._2(),2);
    }
});
    JavaPairRDD<Integer,Double> long3 = long2.mapToPair(new PairFunction<Double,Integer,Double>() {
        @Override
        public Tuple2<Integer,Double> call(Double d ) {
                return new Tuple2<Integer,Double>(flag1++,d);
        }
});
System.out.println("Longitude values display"); 
    for(Tuple2<?,?> tuple : long3.collect()) {
                  System.out.println(tuple._1()+":"+tuple._2());
    }
System.out.println("latitude and longitude values join");
JavaPairRDD<Integer,Tuple2<Double,Double>> weightmatrix1 = lat3.join(long3);
System.out.println("Weightmatrix1 Display");
    for(Tuple2<?,Tuple2<?,?>> tuple : weightmatrix1.collect()) {
                  System.out.println(tuple._1()+":"+tuple._2()._1()+","+tuple._2()._2());
    }   

因此，正在做的是基于经度和纬度值计算权重矩阵

Answer 1

当我做：

scala> val rdd1 = sc.parallelize(Array((1,"a"),(2,"b"),(3,"c")))
scala> val rdd2 = sc.parallelize(Array((1,"d"),(2,"e"),(3,"f")))
scala> val rdd3 = rdd1.join(rdd2)
scala> rdd3.toArray.foreach(println(_))

我一直得到：

(1,(a,d))
(2,(b,e))
(3,(c,f))

Answer 2

这就是我尝试的预期结果：

val data1 = sc.parallelize(Array((1,"a"),(2,"b"),(3,"c")))
val data2 = sc.parallelize(Array((1,"d"),(2,"e"),(3,"f")))
val data3 = data1.join(data2)
data3.collect().map(tuple => tuple._1 + ":"+tuple._2._1+","+tuple._2._2).foreach(println(_))

获得：

1:a,d
2:b,e
3:c,f

这就是scala。 我想在Java中应该是相同的输出。