Convert dataframe into Spark mllib matrix in Scala

Question

I have a Spark dataframe named df as input:

+---------------+---+---+---+---+
|Main_CustomerID| A1| A2| A3| A4|
+---------------+---+---+---+---+
|            101|  1|  0|  2|  1|
|            102|  0|  3|  1|  1|
|            103|  2|  1|  0|  0|
+---------------+---+---+---+---+

I need to collect the values of A1 , A2 , A3 , A4 into a mllib matrix such as,

dm: org.apache.spark.mllib.linalg.Matrix =
1.0  0.0  2.0  1.0
0.0  3.0  1.0  1.0
2.0  1.0  0.0  0.0

How can I achieve this in Scala?

Answer 1

You can do it as follows, first get all columns that should be included in the matrix:

import org.apache.spark.sql.functions._

val matrixColumns = df.columns.filter(_.startsWith("A")).map(col(_))

Then convert the dataframe to an RDD[Vector] . Since the vector need to contain doubles this conversion need to be done here too.

import org.apache.spark.mllib.linalg.Vectors
import org.apache.spark.mllib.linalg.distributed.{IndexedRow, IndexedRowMatrix}

val rdd = df.select(array(matrixColumns:_*).as("arr")).as[Array[Int]].rdd
  .zipWithIndex()
  .map{ case(arr, index) => IndexedRow(index, Vectors.dense(arr.map(_.toDouble)))}

Then convert the rdd to an IndexedRowMatrix which can be converted, if required, to a local Matrix:

val dm = new IndexedRowMatrix(rdd).toBlockMatrix().toLocalMatrix()

---

For smaller matrices that can be collected to the driver there is an easier alternative:

val matrixColumns = df.columns.filter(_.startsWith("A")).map(col(_))

val arr = df.select(array(matrixColumns:_*).as("arr")).as[Array[Int]]
  .collect()
  .flatten
  .map(_.toDouble)

val rows = df.count().toInt
val cols = matrixColumns.length

// It's necessary to reverse cols and rows here and then transpose
val dm = Matrices.dense(cols, rows, arr).transpose()