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考虑以下定义 DenseVector 和数组之间的双向隐式转换的（工作）片段：

import scala.reflect.ClassTag
import org.apache.spark.mllib.linalg.Vectors
import breeze.linalg.{DenseVector => BDV}

  implicit def bdvToSeq[T](vect: BDV[T])(implicit ct: ClassTag[T]): Seq[T] = vect.toArray.toSeq

  implicit def arrayToVect(darr: Array[Double]): BDV[Double]
   = new BDV(darr)
  implicit def bdvToArray[T](vect: BDV[T])(implicit ct: ClassTag[T]): Array[T] = vect.toArray

  def add(v1: BDV[Double], v2: BDV[Double]) =
    v1.zip(v2).map { x => x._1 + x._2}

让我们试一试：

scala> arr.allVisitableIndicesActive   // Array to DenseVector implicit conversion
res4: Boolean = true

scala> val d = new BDV(Array(77.7, 88.8, 99.9))
d: breeze.linalg.DenseVector[Double] = DenseVector(77.7, 88.8, 99.9)

现在另一个方向：

scala> def myArrMethod(arr: Array[Double]) = println(arr.mkString(","))
myArrMethod: (arr: Array[Double])Unit

scala> myArrMethod(d)   // DenseVector to array conversion
77.7,88.8,99.9

以下方法也有效：注意 darr。toSeq

  def norm(darr: Array[Double]): Double = {
    Math.sqrt(darr.toSeq.foldLeft(0.0) { case (sum, dval) => sum + Math.pow(dval, 2)})
  }

但是，如果我们省略 .toSeq ，那么我们就有隐式冲突：

scala>   def norm(darr: Array[Double]): Double = {
        Math.sqrt(darr.foldLeft(0.0) { case (sum, dval) => sum + Math.pow(dval, 2)})
       }
<console>:34: error: type mismatch;
 found   : darr.type (with underlying type Array[Double])
 required: ?{def foldLeft(x$1: ? >: Double(0.0)): ?}
Note that implicit conversions are not applicable because they are ambiguous:
 both method doubleArrayOps in object Predef of type (xs: Array[Double])scala.collection.mutable.ArrayOps[Double]
 and method arrayToVect of type (darr: Array[Double])breeze.linalg.DenseVector[Double]
 are possible conversion functions from darr.type to ?{def foldLeft(x$1: ? >: Double(0.0)): ?}
           Math.sqrt(darr.foldLeft(0.0) { case (sum, dval) => sum + Math.pow(dval, 2)})
                     ^
<console>:34: error: value foldLeft is not a member of Array[Double]
           Math.sqrt(darr.foldLeft(0.0) { case (sum, dval) => sum + Math.pow(dval, 2)})

但是，我不希望明确放置“.toSeq”。

所以我的问题是：如何设置适当的优先级，以便Predef.toSeq是明显的赢家 - 总是在上面定义的自定义低优先级隐式之前尝试？

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1 回答 1

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您可以将所有隐式转换提取到对象中，并仅在需要时导入它们

trait BDVImplicits {
  implicit def bdvToSeq[T](vect: BDV[T])(implicit ct: ClassTag[T]): Seq[T] =
    vect.toArray.toSeq

  implicit def arrayToVect(darr: Array[Double]): BDV[Double] =
    new BDV(darr)

  implicit def bdvToArray[T](vect: BDV[T])(implicit ct: ClassTag[T]): Array[T] =
    vect.toArray
}

object BDVImplicits extends BDVImplicits

例如，如果您将“规范”方法放在 Utils 中，并且不导入隐式，那么您会很好。您只需要在调用 Util 方法的类中导入 BDVImplicits._ 。

我认为这是解决此问题的最简单方法。

于 2015-01-26T23:16:41.033 回答