矢量通过Scala微风中的矩阵乘法

时间:2016-12-15 21:50:25

标签: scala breeze

在这里,我试图在从列表创建每个2x2矩阵之后将其复用2x2矩阵:

val x1 = List[Double](1.0,2.0);
val x2 = List[List[Double]](List(1,2) , List(3,4));
val dv1 = DenseVector[Double]((x1.toArray):_*) 
val dv2 = DenseMatrix(((x2).toArray):_*) 
val h = dv1 :* dv2

但乘法操作会抛出错误:

 \Main.scala:50: could not find implicit value for parameter op: breeze.linalg.operators.OpMulScalar.Impl2[breeze.linalg.DenseVector[Double],breeze.linalg.DenseMatrix[Double],That]
[error] val h = dv1 :* dv2

尺寸是否正确所以不应抛出此错误?

对于这部分错误:DenseMatrix[Double],That] That应该是Matrix的返回类型,如果是,如何设置?

1 个答案:

答案 0 :(得分:2)

dv1 * dv2不是有效的操作ncol(dv1) = 1 != nrow(dv2) = 2。切换订单并使用*运算符可以提供您想要的结果:

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

scala> :pa
// Entering paste mode (ctrl-D to finish)

val x1 = List[Double](1.0,2.0);
val x2 = List[List[Double]](List(1,2) , List(3,4));
val dv1 = DenseVector[Double]((x1.toArray):_*)
val dv2 = DenseMatrix(((x2).toArray):_*)

// Exiting paste mode, now interpreting.

x1: List[Double] = List(1.0, 2.0)
x2: List[List[Double]] = List(List(1.0, 2.0), List(3.0, 4.0))
dv1: breeze.linalg.DenseVector[Double] = DenseVector(1.0, 2.0)
dv2: breeze.linalg.DenseMatrix[Double] =
1.0  2.0
3.0  4.0

scala> dv1 * dv2
java.lang.IllegalArgumentException: requirement failed: b.rows == 1 (2 != 1)
  at breeze.linalg.operators.DenseMatrixMultiplyStuff$$anon$36.apply(DenseMatrixOps.scala:100)
  at breeze.linalg.operators.DenseMatrixMultiplyStuff$$anon$36.apply(DenseMatrixOps.scala:98)
  at breeze.linalg.ImmutableNumericOps$class.$times(NumericOps.scala:135)
  at breeze.linalg.DenseVector.$times(DenseVector.scala:51)
  ... 43 elided

scala> dv2 * dv1
res2: breeze.linalg.DenseVector[Double] = DenseVector(5.0, 11.0)