Question

我试图找出scala协方差和逆变。可能会在两个不同的概念之间混淆。处理以下代码：

sealed trait Algorithm[ +T <: Model, P <: Model, R <: AnyVal] {
  def name: String
  def train(trainingData: DenseMatrix[Double]): T
  def predict(row: DenseVector[R], mlModel : P): R
}

然后我有两种声明为：

的算法类型

case class LibLinear() extends Algorithm[Active_Linear, Active_Linear, Double] {
  override val name = "libLinear"
  override def train(trainingData: DenseMatrix[Double]): Active_Linear = {
    ........
  }
  override def predict(row: DenseVector[Double], model: Active_Linear): Double = {
    ..........
  }
}


case class SVM() extends Algorithm[Volume_SVM, Volume_SVM, Double] {
  override val name = "libSVM"
  override def train(trainingData: DenseMatrix[Double]): Volume_SVM = {
    ..........
  }
  override def predict(row: DenseVector[Double], model: Volume_SVM): Double = {
    ...........
  }
}

Active_Linear和Volume_SVM都是Model的子类型。

现在我不能这样做：

val algorithm: Algorithm[Model, Model, Double] =  SVM()

SVM是Algorithm的子类型，Volume_SVM是Model的子类型。我们使用Algorithm和covariant符号声明contravariant。

Answer 1

那是因为算法只在涉及T时是协变的。 P和R被定义为不变量;您必须根据需要在每个+或-之前添加，然后相应地修改代码。根据你的最终作业，我做了一些假设，这就是我的想法：

sealed trait Algorithm[ +T <: Model, +P <: Model, +R <: AnyVal] {
  def name: String
  def train(trainingData: DenseMatrix[Double]): T
  def predict[U >: R, V >: P](row: DenseVector[U], mlModel : V): U
}

case class LibLinear() extends Algorithm[Active_Linear, Active_Linear, Double] {
  override val name = "libLinear"
  override def train(trainingData: DenseMatrix[Double]): Active_Linear = {
    ...
  }

  override def predict[U >: Double, V >: Active_Linear](row: DenseVector[U], model: V): U = {
    ...
  }
}


case class SVM() extends Algorithm[Volume_SVM, Volume_SVM, Double] {
  override val name = "libSVM"
  override def train(trainingData: DenseMatrix[Double]): Volume_SVM = {
      ...
  }
  override def predict[U >: Double, V >: Volume_SVM](row: DenseVector[U], model: V): U = {
      ...
  }
}

你的上一个作业然后工作正常。

Scala协方差赋值为超类型

1 个答案: