Question

我试图通过定义大小为2 * 3的空矩阵来转置大小为3 * 2的矩阵，我如何创建一个空矩阵？我在注释的代码段中遗漏了一些东西！

type Row = List[Int]
type Matrix = List[Row]

val m:Matrix = List(1 :: 2 :: Nil, 3 :: 4 :: Nil, 5 :: 6 :: Nil)

def transpose(m:Matrix):Matrix = {

  val rows = m.size
  val cols = m.head.size
  val trans= List(List())(rows(cols)) // Int doesn't take parameter

  for (i <- 0 until  cols) {
    for (j <- 0 until  rows) {
      trans(i)(j) = this (j)(i)
    }
  }
  return trans
}

Answer 1

当需要通过index访问元素时，Vector或Array比List更有效率。这是解决方案的Vector版本。

type Row = Vector[Int]
type Matrix = Vector[Row]
val m:Matrix = Vector(Vector(1,2), Vector(3,4), Vector(5,6))

def transpose(mat:Matrix) = {
  def size[A](v: Vector[A]): Int = { var x =0; for(i<-v) x+=1; x}
  for (i <-Range(0,size(mat(0)))) yield for(j <-Range(0,size(mat))) yield mat(j)(i)
}

在REPL中测试：

scala> transpose(m)
res12: scala.collection.immutable.IndexedSeq[scala.collection.immutable.IndexedSeq[Int]] = Vector(Vector(1, 3, 5), Vector(2,
 4, 6))

Scala中的矩阵乘法

1 个答案: