压平列表的列表中阶

Question

元素的单一列表我要压平嵌套列表结构

MyGrp(List(MyGrp(List(TypeA(2))), MyGrp(List(TypeB(ABC), TypeC(20.0))))) 

到List(Type(A),TypeB(ABC),TypeC(20.0))

trait Msg { 

    def toCustString(flag:Boolean): String 
    } 

    trait Ele[T] extends Msg { 

    val value: T 

    override def toCustString(flag:Boolean): String = s"${value}" 

    } 

    trait Grp extends Msg { 

    val list: Seq[Msg] 

    override def toCustString(flag: Boolean = false): String = { 

     val sep = if (flag) "\n" else "!" 
     test((builder: StringBuilder, elem: Msg) => builder.append(s"$sep${elem.toCustString(false)}$sep")) 

    } 

    def test(acc: (StringBuilder, Msg) => StringBuilder): String = { 
     list.foldLeft(StringBuilder.newBuilder)(acc).toString() 
    } 


    } 

case class MyMessage(list:Seq[Msg]) extends Grp 
case class TypeA(value: Int) extends Ele[Int] 
case class TypeB(value: String) extends Ele[String] 
case class TypeC(value: Float) extends Ele[Float] 
case class MyGrp (list:Seq[Msg]) extends Grp 

object Demo extends App{ 
    val grp1 = MyGrp(Seq(TypeA(2))) 
    val grp2 = MyGrp(Seq(TypeB("ABC"), TypeC(20))) 
    val s=MyGrp(Seq(grp1,grp2)) 
} 

我已经尝试使用

1. s.list.flatten但它说'错误：（51，10）没有隐式视图 avai来自Msg => scala.collection.GenTraversableOnce [B]。
   s.list.flatten”

2. s.list.map(x=>x.toCustString())但是这给一个字符串形式，我想作一个列表

Answer 1

您可以用flatMap模式匹配，包装非Grp s转换为单 - 元素列表：

val g = MyGrp(List(MyGrp(List(TypeA(2))), MyGrp(List(TypeB("ABC"), TypeC(20.0f))))) 

val out = g.list.flatMap { 
    case grp: Grp => grp.list 
    case msg => List(msg) 
}