Question

有没有办法将breeze矢量转换为行/列大小为1的breeze矩阵？我是Scala的新手，发现在过去编写无缝处理向量和矩阵的函数（主要是在Matlab中）非常有用。例如，我想在下面的代码中使用func作为subsetMatrix或subsetVector的输入。

val dummyMatrix = DenseMatrix.eye[Double](3)
val subsetMatrix = dummyMatrix(::,0 to 2)
val subsetVector = dummyMatrix(::,1)


def func(X1: DenseMatrix[Double]): Int = {
// Some operation on X1
}

Answer 1

使用asDenseMatrix

scala> import breeze.linalg._
import breeze.linalg._

scala> val dv = DenseVector(1, 2, 3)
dv: breeze.linalg.DenseVector[Int] = DenseVector(1, 2, 3)

scala> dv.asDenseMatrix
res0: breeze.linalg.DenseMatrix[Int] = 1  2  3

scala> (res0.rows, res0.cols)
res1: (Int, Int) = (1,3)

scala>

Answer 2

更好的解决方案通常是首先创建subsetVector时使用范围切片（如果您希望它是Nx1矩阵类型而不是length-N向量类型）。例如：

val subsetSingleColumnMatrix = dummyMatrix(::, 0 until 1)

例如：

scala> val z = DenseMatrix.rand[Double](3, 3)
z: breeze.linalg.DenseMatrix[Double] =
0.9523399908830603    0.5140714248369589   0.23363266806760596
0.27656335841627455   0.2143774018347031   0.3116275714282011
0.020435608706944386  0.13954770594758292  0.5493312961226657

scala> z(::, 0 until 1)
res47: breeze.linalg.DenseMatrix[Double] =
0.9523399908830603
0.27656335841627455
0.020435608706944386

微风矢量到矩阵转换

2 个答案: