Question

这是我迄今为止所做的尝试：

implicit val doubleEq = TolerantNumerics.tolerantDoubleEquality(0.1)

implicit val listEq = new Equivalence[List[Double]] {
  override def areEquivalent(a: List[Double], b: List[Double]): Boolean = {
    (a, b) match {
      case (Nil, Nil) => true
      case (x :: xs, y :: ys) => x === y && areEquivalent(xs, ys)
      case _ => false
    }
  }
}

第一个断言成功但第二个断言失败：

assert(1.0 === 1.01)

assert(List(1.0) === List(1.01))

有没有办法让集合使用我为其元素定义的含义？

Answer 1

就我而言，我重新定义了areEqual方法，提供new Equality[List[Double]]这是Equivalence[List[Double]]的子类，考虑到areEqual将Any作为第二个implicit val listEq = new Equality[List[Double]] { def areEqual(a: List[Double], b: Any): Boolean = { def areEqualRec(a: List[Double], b: List[Double]): Boolean = { (a, b) match { case (Nil, Nil) => true case (x :: xs, y :: ys) => x === y && areEquivalent(xs, ys) case _ => false } } b match { case daList: List[Double] => areEqualRec(a, daList) case _ => false } } }类型参数。

$('#val1').val()

Answer 2

Equality类仅在导入TypeCheckedTripleEquals时使用：

提供返回Boolean的===和！==运算符，将相等性确定委托给Equality类型类，并要求比较两个值的类型为子类型/超类型关系。

这是我用来解决这个问题的基础测试类：

import org.scalactic.{Equivalence, TolerantNumerics, TypeCheckedTripleEquals}
import org.scalatest.FunSuite

abstract class UnitSpec extends FunSuite with TypeCheckedTripleEquals {
  implicit val doubleEq = TolerantNumerics.tolerantDoubleEquality(0.001)

  implicit val listEq = new Equivalence[List[Double]] {
    override def areEquivalent(a: List[Double], b: List[Double]): Boolean = {
      (a, b) match {
        case (Nil, Nil) => true
        case (x :: xs, y :: ys) => x === y && areEquivalent(xs, ys)
        case _ => false
      }
    }
  }
}

如何比较使用Scalatest Equality类型的Double列表？

2 个答案: