I am trying to transpose a matrix of size 3*2 by defining a empty matrix of size 2*3, how can i create an empty matrix?? I am missing something in the commented piece of code!!

```
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
}
```

scala
answered 5 days ago jwvh #1

An empty matrix has zero rows and zero columns, which isn't going to help you. You might build a matrix populated with some default value like `-1`

but a `List`

is immutable so you can't replace values without rebuilding a fresh `List`

.

Here's one way to get what you want.

```
type Row = List[Int]
type Matrix = List[Row]
def transpose(m:Matrix):Matrix = {
val rLen = m.head.length
val cLen = m.length
List.tabulate(rLen)(cIdx =>
List.tabulate(cLen)(rIdx =>
m(rIdx)(cIdx)
)
)
}
```

This uses the "inbuilt" methods `head`

, `length`

, and `tabulate`

.

OK, so here's a solution that does no `List`

indexing and uses only `head`

, `tail`

, `reverse`

, pre-pending (no appending), `flatten`

, and `length`

(once).

```
type Row = List[Int]
type Matrix = List[Row]
def transpose(m:Matrix):Matrix = {
def loop(data :List[Int], result :Matrix, reserve :Matrix) :Matrix = {
if (data.isEmpty) result
else result match {
case hd::Nil => loop(data.tail, ((data.head::hd)::reserve).reverse, Nil)
case hd::tl => loop(data.tail, tl, (data.head::hd)::reserve)
case _ => Nil
}
}
loop(m.reverse.flatten, List.fill(m.head.length)(Nil), Nil)
}
```

answered 5 days ago RAGHHURAAMM #2

When it is necessary to access elements by `index`

, `Vector`

or `Array`

is more efficient than `List`

s.
Here is the `Vector`

version of solution.

```
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)
}
```

*Test in REPL:*

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