元组不是很容易扁平化的东西--Scala3提供了连接它们的方法：

val xs: Array[((String, String, String),(String, String, String))]

xs.map { case (t1, t2) =>
  t1 ++ t2
} // Array[(String, String, String, String, String, String)]

但在Scala2中，您必须手动组合它们

val xs: Array[((String, String, String),(String, String, String))]

xs.map { case ((a, b, c), (d, e, f)) =>
  (a, b, c, d, e, f)
} // Array[(String, String, String, String, String, String)]

一旦你有了元组的数组...您仍然不能“只是”展平它，因为将数组中的所有元组组合成一个元组的直观、内置、默认的方法是什么？你可以这样做，例如，用foldLeft/foldRight/Reduce，但你仍然需要知道怎么做：

val xs2: Array[(String, String, String, String, String, String)]

xs2.foldLeft(initialValue) { (tuple1, tuple) =>
  // your definition how to combine them
}

例如，如果您想串联相应位置上的值

val xs: Array[((String, String, String),(String, String, String))]

xs.map { case ((a, b, c), (d, e, f)) =>
  (a, b, c, d, e, f)
}.foldLeft(("", "", "", "", "", "")) {
  case ((a1,b1,c1,d1,e1,f1), (a2,b2,c2,d2,e2,f2)) =>
    (a1+a2, b1+b2, c1+c2, d1+d2, e1+e2, f1+f2)
} // (String, String, String, String, String, String)

它还可以通过以下一步完成：

val xs: Array[((String, String, String),(String, String, String))]

xs.foldLeft(("", "", "", "", "", "")) {
  case ((a1,b1,c1,d1,e1,f1), ((a2,b2,c2),(d2,e2,f2))) =>
    (a1+a2, b1+b2, c1+c2, d1+d2, e1+e2, f1+f2)
} // (String, String, String, String, String, String)

但确切的实现取决于您对将元组的元组数组“扁平”为元组的理解。

您可以使用ProductIterator，但由于它返回需要与您拥有的类型匹配的任何内容：

val x=Seq((("A","B","C"),("D","E","F")),(("A1","B1","C1"),("D1","E1","F1")))
x.flatMap(_.productIterator
    .flatMap{
        case ys:(String, String, String)=>
          ys.productIterator.map{ 
               case xs:String => xs 
  }})

res0: Seq[String] = List(A, B, C, D, E, F, A1, B1, C1, D1, E1, F1)