如何处理将来的流以创建具有List属性的类的实例

我有一个方法,它以未来为参数,并且在其中也有未来。我想从这个方法创建一个列表,这样它就可以从将来传入的值中获取值。

案例类

case class Color (colorName: String)
case class Shade (shadeName: String)
case class ColoShade (color: Color, shades: List[Shade])

方法

val shadesFuture: Future[Seq[Shade]] = {
    val larSource: Future[Seq[LoanApplicationRegister]] =
      getMySource(ctx, id)
        .map(_.utf8String)
        .map(_.trim)
        .map(s => ShadeParser(s))
        .collect {
          case Right(shade) => shade
        }
        .runWith(Sink.seq)
}

//call the method
myMethod(shadesFuture)

//method definition
def myMethod(shadesFuture: Future[Seq][Shade]])
  :Future[ColorShade] {
    colorFuture
      .map(_.utf8String)
      .map(_.trim)
      .map { s =>
        ColorParser(s)
      }
      .collect {
        case Right(c) => c
      }
      .map { c =>  
         //How can I make an instance of ColorSade here? 
         //I tried 
         val colorShade = ColorShade(c, shadesFuture) 
         //but this doesn't work  

         //I would like to change this to instead be someOtherMethod(colorShade.c)
         someOtherMethod(c) 
      }
}

问题

我怎样才能正确返回 ColorShade 从 myMethod 这样 shades 属性是传入参数的输出 shadesFuture 是吗?