Scala 按多个字段排序

Question

这个问题可能很愚蠢，但我找不到例子，也无法弄清楚。

我想按姓氏、名字和中间名的顺序比较两个 Person 类。这是一种脑死亡的方法:

def compare(that: Person): Int = {
  val last: Int = lastName.compare(that.lastName)
  if (last != 0) last
  else {
    val first: Int = firstName.compare(that.firstName)
    if (first != 0) first
    else middleName.compare(that.middleName)
}

我知道有一些更聪明的方法可以做到这一点(可能使用Ordering)，但我无法具体说明。

托德

一旦我意识到如何在订购中访问正确的内容，我就明白了这一点。

def compare(that: Person): Int = {
  Ordering.Tuple3(Ordering.String, Ordering.String, Ordering.String).compare(
     (lastName, firstName, middleName),
     (that.lastName, that.firstName, that.middleName))
}

我很确定我可以使用更少的显式内容，但这很有效并且相当紧凑。

Answer 1

选项 1:排序

使用 sortBy 方法，这可以非常简单:

case class Person(first: String, middle: String, last: String)
val personList = List(Person("john", "a", "smith"), Person("steve", "x", "scott"), Person("bill", "w", "smith"))
personList.sortBy{ case Person(f,m,l) => (l,f,m) } 

选项 2:延长订购[人]

通过扩展 Ordered[Person]，该类将知道如何对自身进行排序，因此我们得到诸如 sorted、min 和最大免费:

case class Person(first: String, middle: String, last: String) extends Ordered[Person] {
  def compare(that: Person): Int =
    (last compare that.last) match {
      case 0 => 
        (first compare that.first) match {
          case 0 => middle compare that.middle
          case c => c
        }
      case c => c
    }
}

val personList = List(Person("john", "a", "smith"), Person("steve", "x", "scott"), Person("bill", "w", "smith"))
personList.sorted
personList.min
personList.max

选项 3:隐式排序

如果您使用隐式Ordering，那么您将获得sorted、min等，而无需将该特定顺序与原始类绑定(bind)。这种解耦可能很方便，也可能很烦人，具体取决于您的具体情况。

case class Person(first: String, middle: String, last: String)

implicit val ord = new Ordering[Person] { 
  def compare(self: Person, that: Person): Int =
    (self.last compare that.last) match {
      case 0 => 
        (self.first compare that.first) match {
          case 0 => self.middle compare that.middle
          case c => c
        }
      case c => c
    }
}

val personList = List(Person("john", "a", "smith"), Person("steve", "x", "scott"), Person("bill", "w", "smith"))
personList.sorted
personList.min
personList.max