Initializing a 2D (multi-dimensional) array in Scala

Question

It's easy to initialize a 2D array (or, in fact, any multidimensional array) in Java by putting something like that:

int[][] x = new int[][] {
        { 3, 5, 7, },
        { 0, 4, 9, },
        { 1, 8, 6, },
};

It's easy to read, it resembles a 2D matrix, etc, etc.

But how do I do that in Scala?

The best I could come up with looks, well, much less concise:

val x = Array(
    Array(3, 5, 7),
    Array(0, 4, 9),
    Array(1, 8, 6)
)

The problems I see here:

• It repeats "Array" over and over again (like there could be anything else besides Array)
• It requires to omit trailing , in every Array invocation

• If I screw up and insert something besides Array() in the middle of array, it will go okay with compiler, but type of x would silently become Array[Any] instead of Array[Array[Int]]:

  val x = Array(
      Array(3, 5, 7),
      Array(0, 4), 9, // <= OK with compiler, silently ruins x
      Array(1, 8, 6)
  )
  

  There is a guard against it, to specify the type directly, but it looks even more overkill than in Java:

  val x: Array[Array[Int]] = Array(
      Array(3, 5, 7),
      Array(0, 4), 9, // <= this one would trigger a compiler error
      Array(1, 8, 6)
  )
  

  This last example needs Array even 3 times more than I have to say int[][] in Java.

Is there any clear way around this?

Answer 1

Personally I'd suck it up and type out (or cut and paste) "Array" a few times for clarity's sake. Include the type annotation for safety, of course. But if you're really running out of e-ink, a quick easy hack would be simply to provide an alias for Array, for example:

val > = Array

val x: Array[Array[Int]] = >(
  >(3, 5, 7),
  >(0, 4, 9),
  >(1, 8, 6)
)

You could also provide a type alias for Array if you want to shorten the annotation:

type >[T] = Array[T]

val x: >[>[Int]] = ...

Answer 2

I suggest to use Scala 2.10 and macros:

object MatrixMacro {

  import language.experimental.macros

  import scala.reflect.macros.Context
  import scala.util.Try

  implicit class MatrixContext(sc: StringContext) {
    def matrix(): Array[Array[Int]] = macro matrixImpl
  }

  def matrixImpl(c: Context)(): c.Expr[Array[Array[Int]]] = {
    import c.universe.{ Try => _, _ }

    val matrix = Try {
      c.prefix.tree match {
        case Apply(_, List(Apply(_, List(Literal(Constant(raw: String)))))) =>

          def toArrayAST(c: List[TermTree]) =
            Apply(Select(Select(Ident("scala"), newTermName("Array")), newTermName("apply")), c)

          val matrix = raw split "\n" map (_.trim) filter (_.nonEmpty) map {
            _ split "," map (_.trim.toInt)
          }
          if (matrix.map(_.length).distinct.size != 1)
            c.abort(c.enclosingPosition, "rows of matrix do not have the same length")

          val matrixAST = matrix map (_ map (i => Literal(Constant(i)))) map (i => toArrayAST(i.toList))

          toArrayAST(matrixAST.toList)
      }
    }

    c.Expr(matrix getOrElse c.abort(c.enclosingPosition, "not a matrix of Int"))
  }

}

Usage with:

scala> import MatrixMacro._
import MatrixMacro._

scala> matrix"1"
res86: Array[Array[Int]] = Array(Array(1))

scala> matrix"1,2,3"
res87: Array[Array[Int]] = Array(Array(1, 2, 3))

scala> matrix"""
     |   1, 2, 3
     |   4, 5, 6
     |   7, 8, 9
     | """
res88: Array[Array[Int]] = Array(Array(1, 2, 3), Array(4, 5, 6), Array(7, 8, 9))

scala> matrix"""
     |   1, 2
     |   1
     | """
<console>:57: error: rows of matrix do not have the same length
matrix"""
^

scala> matrix"a"
<console>:57: error: not a matrix of Int
              matrix"a"
              ^

I don't think you will get it shorter. ;)

Answer 3

If using a mere List of List (which in itself cannot guarantee that every sub list is of the same size) is not a problem for you, and you are only concerned with easy syntax and avoiding errors at creation-time, scala has many ways to create nice syntax constructs.

One such possibility would be a simple helper:

object Matrix {
  def apply[X]( elements: Tuple3[X, X, X]* ): List[List[X]] = {
    elements.toList.map(_.productIterator.toList.asInstanceOf[List[X]] )
  }
  // Here you might add other overloads for Tuple4, Tuple5 etc if you need "matrixes" of those sizes
}

val x = Matrix(
  (3, 5, 7),
  (0, 4, 9),
  (1, 8, 6)
)

About your concerns:

It repeats "List" over and over again (like there could be anything else besides List)

Not the case here.

It requires to omit trailing , in every List invocation

Unfortunately that is still true here, not much you can do given scala's syntactic rules.

If I screw up and insert something besides List() in the middle of array, it will go okay with compiler, but type of x would silently become List[Any] instead of List[List[Int]]:

val x = List(
  List(3, 5, 7),
  List(0, 4), 9, // <= OK with compiler, silently ruins x
  List(1, 8, 6)
)

The equivalent code now faile to compile:

scala> val x = Matrix(
     |   (3, 5, 7),
     |   (0, 4), 9,
     |   (1, 8, 6)
     | )
<console>:10: error: type mismatch;
 found   : (Int, Int)
 required: (?, ?, ?)
         (0, 4), 9,

And finally if you want to explicitly specify the type of elements (say that you want to protect against the possibility of inadvertently mixing Ints and Doubles), you only have to specify Matrix[Int] instead of the ugly List[List[Int]]:

val x = Matrix[Int](
  (3, 5, 7),
  (0, 4, 9),
  (1, 8, 6)
)

EDIT: I see that you replaced List with Array in your question. To use arrays all you have to use is to replace List with Array and toList with toArray in my code above.

Answer 4

Since I'm also in disgust with this trailing comma issue (i.e. I cannot simply exchange the last line with any other) I sometimes use either a fluent API or the constructor syntax trick to get the syntax I like. An example using the constructor syntax would be:

trait Matrix {
  // ... and the beast
  private val buffer = ArrayBuffer[Array[Int]]()
  def >(vals: Int*) = buffer += vals.toArray
  def build: Array[Array[Int]] = buffer.toArray
}

Which allows:

// beauty ... 
val m = new Matrix {
  >(1, 2, 3)
  >(4, 5, 6)
  >(7, 8, 9)
} build

Unfortunately, this relies on mutable data although it is only used temporarily during the construction. In cases where I want maximal beauty for the construction syntax I would prefer this solution.

In case build is too long/verbose you might want to replace it by an empty apply function.

Answer 5

I don't know if this is the easy way, but I've included some code below for converting nested tuples into '2D' arrays.

First, you need some boiler plate for getting the size of the tuples as well as converting the tuples into [Array[Array[Double]]. The series of steps I used were:

1. Figure out the number of rows and columns in the tuple
2. Turn the nested tuple into a one row Array
3. Reshape the array based on the size of the original tuple.

The code for that is:

object Matrix {

    /**
     * Returns the size of a series of nested tuples. 
     */
    def productSize(t: Product): (Int, Int) = {
        val a = t.productArity
        val one = t.productElement(0)
        if (one.isInstanceOf[Product]) {
            val b = one.asInstanceOf[Product].productArity
            (a,  b)
        }
        else {
            (1, a)
        }
    }

    /**
     * Flattens out a nested tuple and returns the contents as an iterator. 
     */
    def flattenProduct(t: Product): Iterator[Any] = t.productIterator.flatMap {
        case p: Product => flattenProduct(p)
        case x => Iterator(x)
    }

    /**
     * Convert a nested tuple to a flattened row-oriented array.
     * Usage is:
     * {{{
     *  val t = ((1, 2, 3), (4, 5, 6))
     *  val a = Matrix.toArray(t)
     *  // a: Array[Double] = Array(1, 2, 3, 4, 5, 6)
     * }}}
     *
     * @param t The tuple to convert to an array
     */
    def toArray(t: Product): Array[Double] = flattenProduct(t).map(v =>
        v match {
            case c: Char => c.toDouble
            case b: Byte => b.toDouble
            case sh: Short => sh.toDouble
            case i: Int => i.toDouble
            case l: Long => l.toDouble
            case f: Float => f.toDouble
            case d: Double => d
            case s: String => s.toDouble
            case _ => Double.NaN
        }

    ).toArray[Double]

    def rowArrayTo2DArray[@specialized(Int, Long, Float, Double) A: Numeric](m: Int, n: Int,
            rowArray: Array[A]) = {
        require(rowArray.size == m * n)
        val numeric = implicitly[Numeric[A]]
        val newArray = Array.ofDim[Double](m, n)
        for (i <- 0 until m; j <- 0 until n) {
            val idx = i * n + j
            newArray(i)(j) = numeric.toDouble(rowArray(idx))
        }
        newArray
    }

    /**
     * Factory method for turning tuples into 2D arrays
     */
    def apply(data: Product): Array[Array[Double]] = {
        def size = productSize(data)
        def array = toArray(data)
        rowArrayTo2DArray(size._1, size._2, array)
    }

}

Now to use this, you could just do the following:

val a  = Matrix((1, 2, 3))
// a: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))

val b = Matrix(((1, 2, 3), (4, 5, 6), (7, 8, 9)))
// b: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0), 
//                                 Array(4.0, 5.0, 6.0), 
//                                 Array(7.0, 8.0, 9.0))

val c = Matrix((1L, 2F, "3"))    // Correctly handles mixed types
// c: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))

val d = Matrix((1L, 2F, new java.util.Date())) // Non-numeric types convert to NaN
// d: Array[Array[Double]] = Array(Array(1.0, 2.0, NaN))

Alternatively, if you could just call the rowArrayTo2DArray directly using the size of the array you want and a 1D array of values:

val e = Matrix.rowArrayTo2DArray(1, 3, Array(1, 2, 3))
// e: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0))

val f = Matrix.rowArrayTo2DArray(3, 1, Array(1, 2, 3))
// f: Array[Array[Double]] = Array(Array(1.0), Array(2.0), Array(3.0))

val g = Matrix.rowArrayTo2DArray(3, 3, Array(1, 2, 3, 4, 5, 6, 7, 8, 9))
// g: Array[Array[Double]] = Array(Array(1.0, 2.0, 3.0), 
//                                 Array(4.0, 5.0, 6.0), 
//                                 Array(7.0, 8.0, 9.0))

Answer 6

glancing through the answers, i did not find what to me seems the most obvious & simple way to do it. instead of Array, you can use a tuple.
would look something like that:

scala> val x = {(
     | (3,5,7),
     | (0,4,9),
     | (1,8,6)
     | )}

x: ((Int, Int, Int), (Int, Int, Int), (Int, Int, Int)) = ((3,5,7),(0,4,9),(1,8,6))

seems clean & elegant?
i think so :)