Spark Java - 为什么Tuple2不能成为map中函数的参数？

Question

我是Spark新手。我写了下面的Java代码

JavaPairRDD<Detection, Detection> allCombinations = currentLevelNodes.cartesian(nextLevelNodes); 

allCombinations.map(new Function<Tuple2<Detection, Detection>, Segment>(){ 
    public Segment call(Tuple2<Detection, Detection> combination){ 
    Segment segment = new Segment(); 
    Detection a = combination._1(); 
    Detection b = combination._2(); 
    segment.distance = Math.sqrt(Math.pow((a.x)-(b.x), 2) + Math.pow((a.y)-(b.y), 2)); 

    return segment; 
    } 
}); 

的IDE（Eclipse的霓虹灯）显示了以下消息

Multiple markers at this line 
- The method map(Function<Tuple2<Detection,Detection>,R>) in the type 
AbstractJavaRDDLike<Tuple2<Detection,Detection>,JavaPairRDD<Detection,Detection>> is not applicable for the arguments (new 
Function<Tuple2<Detection,Detection>,Segment>(){}) 
- The type new Function<Tuple2<Detection,Detection>,Segment>(){} must implement the inherited abstract method 
Function<Tuple2<Detection,Detection>,Segment>.apply(Tuple2<Detection,Detection>) 

我该如何解决这个问题？

Answer 1

我使用了Lambda表达式并解决了这个问题。

这是我的示例代码

JavaPairRDD<Detection,Detection> allCombinations = currentLevelNodes.cartesian(nextLevelNodes); 

JavaRDD<Segment> segmentRDD = allCombinations.map(tuple -> new Segment(tuple._1, tuple._2)); 

和我添加的构造

public Segment(Detection a, Detection b){ 
    this.distance = Math.sqrt(Math.pow(a.x-b.x, 2)+Math.pow(a.y-b.y, 2)); 
} 

和它的作品。