如何在Scala中使用Spark声明稀疏向量? [英] How to declare a sparse Vector in Spark with Scala?

问题描述

我正在尝试创建稀疏向量(

I'm trying to create a sparse Vector (the mllib.linalg.Vectors class, not the default one) but I can't understand how to use Seq. I have a small test file with three numbers/line, which I convert to an rdd, split the text in doubles and then group the lines by their first column.

测试文件

1 2 4
1 3 5    
1 4 8    
2 7 5    
2 8 4    
2 9 10

代码

val data = sc.textFile("/home/savvas/DWDM/test.txt")
val data2 = data.map(s => Vectors.dense(s.split(' ').map(_.toDouble)))
val grouped = data2.groupBy( _(0) )

这将导致grouped具有这些值

(2.0,CompactBuffer([2.0,7.0,5.0], [2.0,8.0,4.0], [2.0,9.0,10.0]))
(1.0,CompactBuffer([1.0,2.0,4.0], [1.0,3.0,5.0], [1.0,4.0,8.0]))

但是我似乎无法弄清楚下一步.我需要获取grouped的每一行并为其创建一个向量，以便新RDD的每一行都具有一个向量，该向量在第二个值指定的索引中具有CompactBuffer的第三个值.简而言之，我的意思是我想要这样的示例中的数据.

But I can't seem to figure out the next step. I need to take each line of grouped and create a vector for it, so that each line of the new RDD has a vector with the third value of the CompactBuffer in the index specified by the second value. In short, what I mean is that I want my data in the example like this.

[0, 0, 0, 0, 0, 0, 5.0, 4.0, 10.0, 0]
[0, 4.0, 5.0, 8.0, 0, 0, 0, 0, 0, 0]

我知道我需要使用稀疏向量，并且有三种构造方法.我已经尝试过使用带有tuple2(index，value)的Seq，但是我不明白如何创建这样的Seq.

I know I need to use a sparse vector, and that there are three ways to construct it. I've tried using a Seq with a tuple2(index, value) , but I cannot understand how to create such a Seq.

推荐答案

一种可能的解决方案如下.首先，让我们将数据转换为预期的类型:

One possible solution is something like below. First lets convert data to expected types:

import org.apache.spark.rdd.RDD

val pairs: RDD[(Double, (Int, Double))] = data.map(_.split(" ") match {
  case Array(label, idx, value) => (label.toDouble, (idx.toInt, value.toDouble))
})

下一步查找最大索引(向量的大小):

next find a maximum index (size of the vectors):

val nCols = pairs.map{case (_, (i, _)) => i}.max + 1

分组并转换:

import org.apache.spark.mllib.linalg.SparseVector

def makeVector(xs: Iterable[(Int, Double)]) = {
  val (indices, values) = xs.toArray.sortBy(_._1).unzip
  new SparseVector(nCols, indices.toArray, values.toArray)
}

val transformed: RDD[(Double, SparseVector)] = pairs
  .groupByKey
  .mapValues(makeVector)

您可以使用CoordinateMatrix:

import org.apache.spark.mllib.linalg.distributed.{CoordinateMatrix, MatrixEntry}

val entries: RDD[MatrixEntry] = data.map(_.split(" ") match {
  case Array(label, idx, value) => 
    MatrixEntry(label.toInt, idx.toInt, value.toDouble)
})

val transformed: RDD[(Double, SparseVector)] = new CoordinateMatrix(entries)
  .toIndexedRowMatrix
  .rows
  .map(row => (row.index.toDouble, row.vector))

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