相关疑难解决方法(0)

我想做这样的事情:

class Foo extends Ordered[Foo] {
   val x
   val y
   val z
   .
   .
   .
   .
   def compare(that: Foo) = {
      val c0 = this.length compareTo that.length          // primary comparison
      lazy val c1 = this.x compareTo that.x               // secondary comparison
      lazy val c2 = this.y.size compareTo that.y.size     // tertiary comparison
      lazy val c3 = this.z.head compareTo that.z.head     // final tie breaker
      if (c0 != 0) c0 else if (c1 != 0) c1 else if (c2 != 0) c2 else if …

鉴于两个相同的元组,我如何按字典顺序比较它们？看起来这应该像下面的代码片段一样简单,但事实并非如此.任何简单的例子如何做到这一点？

var x = (1,2,3) < (1,2,4)

如果他们列出,我可以定义一个递归函数,比较列表的头部直到找到差异或列表的结尾,但我不认为我可以为元组做到这一点.

这是我的代码示例:

 case class Person(name:String,tel:String){
        def equals(that:Person):Boolean = that.name == this.name && this.tel == that.tel}

 val persons = Array(Person("peter","139"),Person("peter","139"),Person("john","111"))
 sc.parallelize(persons).distinct.collect

它回来了

 res34: Array[Person] = Array(Person(john,111), Person(peter,139), Person(peter,139))

为什么distinct不起作用？我希望结果为Person("john",111),Person("peter",139)

有没有办法为两个不同的类定义相同的隐式排序？

我尝试沿着以下几行做一些事情,但它没有检测到排序.

abstract class Common
case class A extends Common
case class B extends Common

implicit val KeyOrdering = new Ordering[Common] {
    override def compare(x: Common, y: Common): Int = {
      x.toString.compareTo(y.toString)
   }
}