Matrix Transpose on RowMatrix in Spark

Question

Suppose I have a RowMatrix.

1. How can I transpose it. The API documentation does not seem to have a transpose method.
2. The Matrix has the transpose() me

Answer 1

For very large and sparse matrix, (like the one you get from text feature extraction), the best and easiest way is:

def transposeRowMatrix(m: RowMatrix): RowMatrix = {
  val indexedRM = new IndexedRowMatrix(m.rows.zipWithIndex.map({
    case (row, idx) => new IndexedRow(idx, row)}))
  val transposed = indexedRM.toCoordinateMatrix().transpose.toIndexedRowMatrix()
  new RowMatrix(transposed.rows
    .map(idxRow => (idxRow.index, idxRow.vector))
    .sortByKey().map(_._2))      
}

For not so sparse matrix, you can use BlockMatrix as the bridge as mentioned by aletapool's answer above.

However aletapool's answer misses a very important point: When you start from RowMaxtrix -> IndexedRowMatrix -> BlockMatrix -> transpose -> BlockMatrix -> IndexedRowMatrix -> RowMatrix, in the last step (IndexedRowMatrix -> RowMatrix), you have to do a sort. Because by default, converting from IndexedRowMatrix to RowMatrix, the index is simply dropped and the order will be messed up.

val data = Array(
  MllibVectors.sparse(5, Seq((1, 1.0), (3, 7.0))),
  MllibVectors.dense(2.0, 0.0, 3.0, 4.0, 5.0),
  MllibVectors.dense(4.0, 0.0, 0.0, 6.0, 7.0),
  MllibVectors.sparse(5, Seq((2, 2.0), (3, 7.0))))

val dataRDD = sc.parallelize(data, 4)

val testMat: RowMatrix = new RowMatrix(dataRDD)
testMat.rows.collect().map(_.toDense).foreach(println)

[0.0,1.0,0.0,7.0,0.0]
[2.0,0.0,3.0,4.0,5.0]
[4.0,0.0,0.0,6.0,7.0]
[0.0,0.0,2.0,7.0,0.0]

transposeRowMatrix(testMat).
  rows.collect().map(_.toDense).foreach(println)

[0.0,2.0,4.0,0.0]
[1.0,0.0,0.0,0.0]
[0.0,3.0,0.0,2.0]
[7.0,4.0,6.0,7.0]
[0.0,5.0,7.0,0.0]

Answer 2

If anybody interested, I've implemented the distributed version @javadba had proposed.

  def transposeRowMatrix(m: RowMatrix): RowMatrix = {
    val transposedRowsRDD = m.rows.zipWithIndex.map{case (row, rowIndex) => rowToTransposedTriplet(row, rowIndex)}
      .flatMap(x => x) // now we have triplets (newRowIndex, (newColIndex, value))
      .groupByKey
      .sortByKey().map(_._2) // sort rows and remove row indexes
      .map(buildRow) // restore order of elements in each row and remove column indexes
    new RowMatrix(transposedRowsRDD)
  }


  def rowToTransposedTriplet(row: Vector, rowIndex: Long): Array[(Long, (Long, Double))] = {
    val indexedRow = row.toArray.zipWithIndex
    indexedRow.map{case (value, colIndex) => (colIndex.toLong, (rowIndex, value))}
  }

  def buildRow(rowWithIndexes: Iterable[(Long, Double)]): Vector = {
    val resArr = new Array[Double](rowWithIndexes.size)
    rowWithIndexes.foreach{case (index, value) =>
        resArr(index.toInt) = value
    }
    Vectors.dense(resArr)
  } 

Answer 3

You can use BlockMatrix, which can be created from an IndexedRowMatrix:

BlockMatrix matA = (new IndexedRowMatrix(...).toBlockMatrix().cache();
matA.validate();

BlockMatrix matB = matA.transpose();

Then, can be easily put back as IndexedRowMatrix. This is described in the spark documentation.

Answer 4

This is a variant of the previous solution but working for sparse row matrix and keeping the transposed sparse too when needed:

  def transpose(X: RowMatrix): RowMatrix = {
    val m = X.numRows ().toInt
    val n = X.numCols ().toInt

    val transposed = X.rows.zipWithIndex.flatMap {
      case (sp: SparseVector, i: Long) => sp.indices.zip (sp.values).map {case (j, value) => (i, j, value)}
      case (dp: DenseVector, i: Long) => Range (0, n).toArray.zip (dp.values).map {case (j, value) => (i, j, value)}
    }.sortBy (t => t._1).groupBy (t => t._2).map {case (i, g) =>
      val (indices, values) = g.map {case (i, j, value) => (i.toInt, value)}.unzip
      if (indices.size == m) {
        (i, Vectors.dense (values.toArray) )
      } else {
        (i, Vectors.sparse (m, indices.toArray, values.toArray))
      }
    }.sortBy(t => t._1).map (t => t._2)

    new RowMatrix (transposed)
  }

Hope this help!

Answer 5

Getting the transpose of RowMatrix in Java:

public static RowMatrix transposeRM(JavaSparkContext jsc, RowMatrix mat){
List<Vector> newList=new ArrayList<Vector>();
List<Vector> vs = mat.rows().toJavaRDD().collect();
double [][] tmp=new double[(int)mat.numCols()][(int)mat.numRows()] ;

for(int i=0; i < vs.size(); i++){
    double[] rr=vs.get(i).toArray();
    for(int j=0; j < mat.numCols(); j++){
        tmp[j][i]=rr[j];
    }
}

for(int i=0; i < mat.numCols();i++)
    newList.add(Vectors.dense(tmp[i]));

JavaRDD<Vector> rows2 = jsc.parallelize(newList);
RowMatrix newmat = new RowMatrix(rows2.rdd());
return (newmat);
}

Answer 6

You are correct: there is no

 RowMatrix.transpose()

method. You will need to do this operation manually.

Here is the non-distributed/local matrix versions:

def transpose(m: Array[Array[Double]]): Array[Array[Double]] = {
    (for {
      c <- m(0).indices
    } yield m.map(_(c)) ).toArray
}

The distributed version would be along the following lines:

    origMatRdd.rows.zipWithIndex.map{ case (rvect, i) =>
        rvect.zipWithIndex.map{ case (ax, j) => ((j,(i,ax))
    }.groupByKey
    .sortBy{ case (i, ax) => i }
    .foldByKey(new DenseVector(origMatRdd.numRows())) { case (dv, (ix,ax))  =>
              dv(ix) = ax
     }

Caveat: I have not tested the above: it will have bugs. But the basic approach is valid - and similar to work I had done in the past for a small LinAlg library for spark.