使用Scala将Matrix中的Matrix转换为RowMatrix

Question

我真的想将我的org.apache.spark.mllib.linalg.Matrix转换为org.apache.spark.mllib.linalg.distributed.RowMatrix

我可以这样做:

val xx = X.computeGramianMatrix()  //xx is type org.apache.spark.mllib.linalg.Matrix
val xxs = xx.toString()
val xxr = xxs.split("\n").map(row => row.replace("   "," ").replace("  "," ").replace("  "," ").replace("  "," ").replace(" ",",").split(","))
val xxp = sc.parallelize(xxr)
val xxd = xxp.map(ar => Vectors.dense(ar.map(elm => elm.toDouble)))
val xxrm: RowMatrix = new RowMatrix(xxd)

然而,这真的很糟糕,而且完全是黑客攻击.有人能告诉我一个更好的方法吗？

注意我使用的是Spark 1.3.0版

Answer 1

我建议您将Matrix转换为RDD [Vector],您可以自动将其转换为RowMatrix.

让我们考虑以下示例:

import org.apache.spark.rdd._
import org.apache.spark.mllib.linalg._


val denseData = Seq(
  Vectors.dense(0.0, 1.0, 2.0),
  Vectors.dense(3.0, 4.0, 5.0),
  Vectors.dense(6.0, 7.0, 8.0),
  Vectors.dense(9.0, 0.0, 1.0)
)

val dm: Matrix = Matrices.dense(3, 2, Array(1.0, 3.0, 5.0, 2.0, 4.0, 6.0))

您需要定义一个方法将Matrix转换为RDD [Vector]

def matrixToRDD(m: Matrix): RDD[Vector] = {
   val columns = m.toArray.grouped(m.numRows)
   val rows = columns.toSeq.transpose // Skip this if you want a column-major RDD.
   val vectors = rows.map(row => new DenseVector(row.toArray))
   sc.parallelize(vectors)
}

现在您可以在Matrix上应用转换:

 import org.apache.spark.mllib.linalg.distributed.RowMatrix
 val rows = matrixToRDD(dm)
 val mat = new RowMatrix(rows)

我希望这可以帮助你!