Question

我已经查看了来自互联网的标签传播算法源代码，并且在自定义算法的一部分中使用了它。我使用的代码如下：


    def run[VD, ED: ClassTag](graph: Graph[VD, ED], maxSteps: Int): Graph[VertexId, ED] = {
      require(maxSteps > 0, s"Maximum of steps must be greater than 0, but got ${maxSteps}")

      val lpaGraph: Graph[VD, ED] = graph.mapVertices { case (y,x) => x}
      def sendMessage(e: EdgeTriplet[VertexId, ED]): Iterator[(VertexId, Map[VertexId, Long])] = {
        Iterator((e.srcId, Map(e.dstAttr -> 1L)), (e.dstId, Map(e.srcAttr -> 1L)))
      }
      def mergeMessage(count1: Map[VertexId, Long], count2: Map[VertexId, Long])
      : Map[VertexId, Long] = {
        // Mimics the optimization of breakOut, not present in Scala 2.13, while working in 2.12
        val map = mutable.Map[VertexId, Long]()
        (count1.keySet ++ count2.keySet).map { i =>
          val count1Val = count1.getOrElse(i, 0L)
          val count2Val = count2.getOrElse(i, 0L)
          map.put(i, count1Val + count2Val)
        }
        map
      }
      def vertexProgram(vid: VertexId, attr: Long, message: Map[VertexId, Long]): VertexId = {
        if (message.isEmpty) attr else message.maxBy(_._2)._1
      }
      val initialMessage = Map[VertexId, Long]()
      Pregel(lpaGraph, initialMessage, maxIterations = maxSteps)(
        vprog = vertexProgram,
        sendMsg = sendMessage,
        mergeMsg = mergeMessage)
    }

在原始源代码的开头，每个节点都获得与该代码相同的ID作为初始标签（属性）：

val lpaGraph: Graph[VD, ED] = graph.mapVertices { case (y,x) => y}

但是在我的自定义算法中，我希望节点通过使用以下代码将其当前标签（属性）作为其初始标签：

val lpaGraph: Graph[VD, ED] = graph.mapVertices { case (y,x) => x}

但是当我使用上面的代码时，出现一些错误，例如vprog = vertexProgram中的错误，错误是类型不匹配：

Error:(176, 17) type mismatch;
 found   : org.apache.spark.graphx.VertexId (which expands to)  Long
 required: VD
    vprog = vertexProgram,

有人可以帮助我解决这个问题吗？我已经使用GraphLoader加载边缘列表以制作图形，而RDD如下所示：

(1,2)
(2,2)
(3,8) 
...

为什么我无法在Spark GraphX中正确运行Pregel程序？

0 个答案: