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Слава Иванов
Asked:2020-12-12 06:46:49 +0000 UTC2020-12-12 06:46:49 +0000 UTC

在Scala中按条件从另外两个创建列表

  • 772

有一个类：

case class Person (age: Int, name: String)

有两个列表：

val classRoom1 = List(Person(14, "Jimm"),Person(14, "Jack"), Person(13, "Ricky"))
val classRoom2 = List(Person(14, "Jimm"),Person(14, "Jack"), Person(13, "Ricky"), Person(13,"Susan"), Person(16, "Jack"))

有必要从第二个列表中选择名称字段，选择那些不在第一个列表中的对象并从中创建第三个列表。

val classRoom3 = List(Person(13,"Susan"))

如何以实用的风格做到这一点？

scala

2 个回答

  1. Best Answer

    让我在额头提供一个简单的（就代码而言）解决方案：

    val classRoom1 = List(Person(14, "Jimm"),Person(14, "Jack"), Person(13, "Ricky"))
val classRoom2 = List(Person(14, "Jimm"),Person(14, "Jack"), Person(13, "Ricky"), Person(13,"Susan"), Person(16, "Jack"))

val classRoom3 = classRoom2.filterNot(p => classRoom1.exists(_.name == p.name))

println(classRoom3)    //List(Person(13,Susan))

    困难就在这里n*m。
    如果它不适合您，您可以将其包装在地图中并实现线性复杂性。但是代码会更加混乱。

    • 2

  2. 在小列表的情况下，此选项适用： classRoom2.diff(classRoom2.intersect(classRoom1))

    对于大型列表，最好执行以下操作：

    abstract class Keyable[B] {
  def uniq: B
}

case class Person(age: Int, name: String) extends Keyable[(Int, String)] {
  override def uniq = (age, name)
}

implicit class SeqExtKey[C](val seq: List[Keyable[C]]) extends AnyVal {

  def diffE(anotherSeq: List[Keyable[C]]) = {
    val firstMap = seq.groupBy(_.uniq)
    val secondMap = anotherSeq.groupBy(_.uniq)

    secondMap.keySet.diff(firstMap.keySet).flatMap(secondMap.get).flatten.toList
  }
}

val classRoom1 = List(Person(14, "Jimm"), Person(14, "Jack"), Person(13, "Ricky"))
val classRoom2 = List(Person(14, "Jimm"), Person(14, "Jack"), Person(13, "Ricky"), Person(13, "Susan"), Person(16, "Jack"))

val diff = classRoom1.diffE(classRoom2)

println(diff)

    https://scalafiddle.io/sf/Me1c4S5/0

    • 0

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