CODEWITHSUNDEEP
scalascala-breeze
Armand Grillet
Armand Grillet

Reputation: 3399

Creating a DenseMatrix from a Transpose

I started using Breeze since a few weeks and I am not able to do something that seems simple. I want to transform a Transpose into a DenseMatrix, for example:

val matrix = DenseMatrix((1.0, 3.5), (3.0, 2.0)) // DenseMatrix
val meanCols = mean(matrix(::, *)) // Transpose
val meanColsDM = meanCols.toDenseMatrix // Error: value toDenseMatrix is not a member of breeze.linalg.Transpose

I thought about creating a loop to transform the Transpose into an array to then create the DenseMatrix (1 row, 2 cols using the matrix from the example) but I wonder if there is a simpler way to obtain the same thing.

I need to do this to then concatene the mean of the columns with other matrices, I did not put the code in the example as it is not the source of the problem.

Upvotes: 2

Views: 1178

Answers (1)

Travis Brown
Travis Brown

Reputation: 139028

meanCols is a Transpose[DenseVector[Double]], which is just a wrapper for a DenseVector[Double]. If you want the result in a matrix with one row and two columns, you can transpose it again with .t to get a DenseVector[Double] and then convert that to a matrix with .toDenseVector:

scala> import breeze.linalg._, breeze.stats.mean
import breeze.linalg._
import breeze.stats.mean

scala> val matrix = DenseMatrix((1.0, 3.5), (3.0, 2.0))
matrix: breeze.linalg.DenseMatrix[Double] =
1.0  3.5
3.0  2.0

scala> val meanCols = mean(matrix(::, *))
meanCols: breeze.linalg.Transpose[breeze.linalg.DenseVector[Double]] = ...

scala> val meanColsDM = meanCols.t.toDenseMatrix
meanColsDM: breeze.linalg.DenseMatrix[Double] = 2.0  2.75

Upvotes: 3

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