转换矩阵，以RowMatrix在使用Scala的Apache星火 [英] Convert Matrix to RowMatrix in Apache Spark using Scala

问题描述

我真的很想我的org.apache.spark.mllib.linalg.Matrix转换为org.apache.spark.mllib.linalg.distributed.RowMatrix

I'd really like to convert my org.apache.spark.mllib.linalg.Matrix to org.apache.spark.mllib.linalg.distributed.RowMatrix

我能做到这样：

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)

不过，这确实是粗暴和总的黑客。有人能告诉我一个更好的办法？

However, that is really gross and a total hack. Can someone show me a better way?

请注意我用的星火版本1.3.0

Note I am using Spark version 1.3.0

推荐答案

我建议您将矩阵转换为RDD [矢量]，它可以自动转换为RowMatrix。

I suggest that you convert your Matrix to an RDD[Vector] which you can automatically convert to a RowMatrix.

让我们看看下面的例子：

Let's consider the following example :

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，你的矩阵转换为RDD [矢量]

You'll need to define a method to convert your Matrix to an 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)
}

，现在你可以适用于你的矩阵转换：

and now you can apply the conversion on your Matrix :

 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！

I hope that this can help!

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