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在java中构造嵌套Tuple2_Java_Apache Spark_Spark Graphx - Fatal编程技术网

在java中构造嵌套Tuple2

在java中构造嵌套Tuple2,java,apache-spark,spark-graphx,Java,Apache Spark,Spark Graphx,我尝试使用嵌套的Tuple2,以便为图中的每个顶点保存三个值:id、weight和label 目前我有: List<Tuple2<Object,Integer>> vertices = Lists.newArrayList( new Tuple2<Object,Integer>(1l,0), new Tuple2<Object,Integer>(2l,Integer.MAX_VALUE), ne

我尝试使用嵌套的Tuple2,以便为图中的每个顶点保存三个值:id、weight和label

目前我有:

List<Tuple2<Object,Integer>> vertices = Lists.newArrayList(
         new Tuple2<Object,Integer>(1l,0),
         new Tuple2<Object,Integer>(2l,Integer.MAX_VALUE),
         new Tuple2<Object,Integer>(3l,Integer.MAX_VALUE),
         new Tuple2<Object,Integer>(4l,Integer.MAX_VALUE),
 );
List vertices=Lists.newArrayList(
新元组2(1l,0),
新的元组2(2l,整数.MAX_值),
新的元组2(3l,整数.MAX_值),
新的元组2(4l,整数.MAX_值),
);
我试图构建嵌套的Tuple2,如下所示:

List<Tuple2<Object,Tuple2<Integer,String>>> vertices = Lists.newArrayList(
         new Tuple2<Object,Tuple2<Integer,String>>(1l,(0,"A")),
         new Tuple2<Object,Tuple2<Integer,String>>(2l,(Integer.MAX_VALUE,"B")),
         new Tuple2<Object,Tuple2<Integer,String>>(3l,(Integer.MAX_VALUE,"C")),
         new Tuple2<Object,Tuple2<Integer,String>>(4l,(Integer.MAX_VALUE,"D"))
);
List vertices=Lists.newArrayList(
新元组2(1l,(0,“A”),
新的元组2(2l,(整数最大值,“B”),
新的元组2(3l,(整数.MAX_值,“C”),
新的元组2(4l,(Integer.MAX_值,“D”))
);
但这给了我错误

你能帮我找出问题所在吗


谢谢

由于未初始化内部元组,因此出现错误:

List<Tuple2<Object,Tuple2<Integer,String>>> vertices = Lists.newArrayList(
     new Tuple2<Object,Tuple2<Integer,String>>(
       1l, new Tuple2<Integer,String>(0,"A")
     ),
     new Tuple2<Object,Tuple2<Integer,String>>(
       2l, new (Integer.MAX_VALUE,"B")
     )
);
List vertices=Lists.newArrayList(
新元组2(
1l,新元组2(0,“A”)
),
新元组2(
2l,新(整数最大值,“B”)
)
);

非常感谢@user9717204