Question

我想检查两个Traversable是否包含相同的元素而不考虑它们的顺序。

所以，我自己尝试了，并编写了以下简单示例：

implicit val l = 3
implicit def equality(implicit l: Int) = new Equality[String] {
  override def areEqual(a: String, b: Any): Boolean = (a, b) match {
    case (_, b: Int) => a.toInt == b
  }
}

"Test" should "check how equality works" in {
    List("1") should contain theSameElementsAs Vector(1) //Fine
    List("1", "2") should contain theSameElementsAs Vector(1, 2) //Fine
    List("1", "2", "3") should contain theSameElementsAs Vector(1, 2, 3) //Fine
    List("1", "2", "2") should contain theSameElementsAs Vector(1, 2, 2) //Error
    List("2", "1") should contain theSameElementsAs Vector(1, 2) //Error
}

正如the documentation所说：

＆＃34; contain theSameElementsAs＆＃34;语法让你断言那两个聚合包含相同的对象

它不应该考虑重复和顺序。那有什么不对？

Answer 1

我认为根本问题在于equality关系并不像人们所期望的那样具有反思性。那是x == y然后是y == x。

Scalacheck使用提供的equality函数尝试在目标集合中count the number of repetitions。

遇到equality(2,2)，tryEquality fails internally with a ClassCastException时，将比较默认为false。

底线：Equality需要反思（如同旧的数学要求）才能获得预期的结果。

Answer 2

在左侧使用Int列表而不是字符串列表。

 "Test" should "check how equality works" in {
    List(1) should contain theSameElementsAs Vector(1) //Fine
    List(1, 2) should contain theSameElementsAs Vector(1, 2) //Fine
    List(1, 2, 3) should contain theSameElementsAs Vector(1, 2, 3) //Fine
    List(1, 2, 2) should contain theSameElementsAs Vector(1, 2, 2) //Fine
    List(2, 1) should contain theSameElementsAs Vector(1, 2) //Fine
  }

输出：

[info] Test
[info] - should check how equality works
[info] Run completed in 354 milliseconds.
[info] Total number of tests run: 1
[info] Suites: completed 1, aborted 0
[info] Tests: succeeded 1, failed 0, canceled 0, ignored 0, pending 0
[info] All tests passed.

如何比较两个Traversable包含完全相同的元素？

2 个答案: