如何并行化Spark scala计算？

Question

当我使用spark API运行类似的代码时，它在许多不同的（分布式）作业中运行，并且成功运行。当我运行它时，我的代码（应该做与Spark代码相同的事情），我得到一个堆栈溢出错误。知道为什么吗？

代码如下：

import java.util.Arrays
        import org.apache.spark.mllib.linalg.{Vectors, Vector}
        import org.apache.spark.mllib.linalg._
        import org.apache.spark.mllib.linalg.distributed.RowMatrix
        import org.apache.spark.rdd.RDD
        import org.apache.spark.api.java.JavaRDD
        import breeze.linalg.{axpy => brzAxpy, inv, svd => brzSvd, DenseMatrix => BDM, DenseVector => BDV,
          MatrixSingularException, SparseVector => BSV, CSCMatrix => BSM, Matrix => BM}

        val EPSILON = {
            var eps = 1.0
            while ((1.0 + (eps / 2.0)) != 1.0) {
              eps /= 2.0
            }
            eps
          }

        def dot(x: Vector, y: Vector): Double = {
            require(x.size == y.size,
              "BLAS.dot(x: Vector, y:Vector) was given Vectors with non-matching sizes:" +
              " x.size = " + x.size + ", y.size = " + y.size)
            (x, y) match {
              case (dx: DenseVector, dy: DenseVector) =>
                dot(dx, dy)
              case (sx: SparseVector, dy: DenseVector) =>
                dot(sx, dy)
              case (dx: DenseVector, sy: SparseVector) =>
                dot(sy, dx)
              case (sx: SparseVector, sy: SparseVector) =>
                dot(sx, sy)
              case _ =>
                throw new IllegalArgumentException(s"dot doesn't support (${x.getClass}, ${y.getClass}).")
            }
         }

         def fastSquaredDistance(
              v1: Vector,
              norm1: Double,
              v2: Vector,
              norm2: Double,
              precision: Double = 1e-6): Double = {
            val n = v1.size
            require(v2.size == n)
            require(norm1 >= 0.0 && norm2 >= 0.0)
            val sumSquaredNorm = norm1 * norm1 + norm2 * norm2
            val normDiff = norm1 - norm2
            var sqDist = 0.0
            /*
             * The relative error is
             * <pre>
             * EPSILON * ( \|a\|_2^2 + \|b\\_2^2 + 2 |a^T b|) / ( \|a - b\|_2^2 ),
             * </pre>
             * which is bounded by
             * <pre>
             * 2.0 * EPSILON * ( \|a\|_2^2 + \|b\|_2^2 ) / ( (\|a\|_2 - \|b\|_2)^2 ).
             * </pre>
             * The bound doesn't need the inner product, so we can use it as a sufficient condition to
             * check quickly whether the inner product approach is accurate.
             */
            val precisionBound1 = 2.0 * EPSILON * sumSquaredNorm / (normDiff * normDiff + EPSILON)
            if (precisionBound1 < precision) {
              sqDist = sumSquaredNorm - 2.0 * dot(v1, v2)
            } else if (v1.isInstanceOf[SparseVector] || v2.isInstanceOf[SparseVector]) {
              val dotValue = dot(v1, v2)
              sqDist = math.max(sumSquaredNorm - 2.0 * dotValue, 0.0)
              val precisionBound2 = EPSILON * (sumSquaredNorm + 2.0 * math.abs(dotValue)) /
                (sqDist + EPSILON)
              if (precisionBound2 > precision) {
                sqDist = Vectors.sqdist(v1, v2)
              }
            } else {
              sqDist = Vectors.sqdist(v1, v2)
            }
            sqDist
        }

        def findClosest(
              centers: TraversableOnce[Vector],
              point: Vector): (Int, Double) = {
            var bestDistance = Double.PositiveInfinity
            var bestIndex = 0
            var i = 0
            centers.foreach { center =>
              // Since `\|a - b\| \geq |\|a\| - \|b\||`, we can use this lower bound to avoid unnecessary
              // distance computation.
              var lowerBoundOfSqDist = Vectors.norm(center, 2.0) - Vectors.norm(point, 2.0)
              lowerBoundOfSqDist = lowerBoundOfSqDist * lowerBoundOfSqDist
              if (lowerBoundOfSqDist < bestDistance) {
                val distance: Double = fastSquaredDistance(center, Vectors.norm(center, 2.0), point, Vectors.norm(point, 2.0))
                if (distance < bestDistance) {
                  bestDistance = distance
                  bestIndex = i
                }
              }
              i += 1
            }
            (bestIndex, bestDistance)
        }

         def pointCost(
              centers: TraversableOnce[Vector],
              point: Vector): Double =
            findClosest(centers, point)._2



        def clusterCentersIter: Iterable[Vector] =
            clusterCenters.map(p => p)


        def computeCostZep(indata: RDD[Vector]): Double = {
            val bcCenters = indata.context.broadcast(clusterCenters)
            indata.map(p => pointCost(bcCenters.value, p)).sum()
          }

        computeCostZep(projectedData)

我相信我正在使用与spark相同的所有并行化工作，但它对我不起作用。任何关于使我的代码分发/帮助了解为什么在我的代码中发生内存溢出的建议都将是非常有帮助的

val clusters = KMeans.train(projectedData, numClusters, numIterations)

val clusterCenters = clusters.clusterCenters




// Evaluate clustering by computing Within Set Sum of Squared Errors
val WSSSE = clusters.computeCost(projectedData)
println("Within Set Sum of Squared Errors = " + WSSSE)

Answer 1

正在发生的事情似乎很简单：您在这里递归调用dot方法：

def dot(x: Vector, y: Vector): Double = {
        require(x.size == y.size,
          "BLAS.dot(x: Vector, y:Vector) was given Vectors with non-matching sizes:" +
          " x.size = " + x.size + ", y.size = " + y.size)
        (x, y) match {
          case (dx: DenseVector, dy: DenseVector) =>
            dot(dx, dy)
          case (sx: SparseVector, dy: DenseVector) =>
            dot(sx, dy)
          case (dx: DenseVector, sy: SparseVector) =>
            dot(sy, dx)
          case (sx: SparseVector, sy: SparseVector) =>
            dot(sx, sy)
          case _ =>
            throw new IllegalArgumentException(s"dot doesn't support (${x.getClass}, ${y.getClass}).")
        }
     }

对dot的后续递归调用与前一个调用具有相同的参数-因此递归永远不会有结论。

stacktrace还告诉您--注意位置在dot方法处：