Create and populate two-dimensional array in Scala

Question

What's the recommended way of creating a pre-populated two-dimensional array in Scala? I've got the following code:

val map = for {
    x <- (1 to size).toList
} yield for {
        y <- (1 to size).toList
    } yield (x, y)

How do I make an array instead of list? Replacing .toList with .toArray doesn't compile. And is there a more concise or readable way of doing this than the nested for expressions?

Daniel C. Sobral · Accepted Answer · 2010-03-23 02:22:51Z

8

On Scala 2.7, use Array.range:

for {
  x <- Array.range(1, 3)
} yield for {
  y <- Array.range(1, 3)
} yield (x, y)

On Scala 2.8, use Array.tabulate:

Array.tabulate(3,3)((x, y) => (x, y))

answered Mar 23, 2010 at 2:22

Daniel C. Sobral

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6 Comments

asteinlein Over a year ago

Array.tabulate seems sweet, and looks like Array.fromFunction suggested by Rex Kerr in another answer. What's the difference (apart from the fact that I see Array.fromFunction is deprecated in 2.8)? And how do they actually work?

Daniel C. Sobral Over a year ago

@asteinlein I'm not sure about fromFunction -- I had never noticed it before -- but tabulate (which is present on List on Scala 2.7) works pretty much like the for-comprehension you wrote, but 0 until size instead of 1 to size. On Scala 2.7, however, it is one-dimensional only, so you'd need to nest them. On Scala 2.8, you can provide multiple dimensions (up to 5, I think).

asteinlein Over a year ago

Thanks. For anyone else interested, this works nicely on 2.7: List.tabulate(3, x => List.tabulate(3, y => (x, y)).toArray).toArray

jk. Over a year ago

Another straightforward way... val 2d_array:Array[Array[MyClass]] = Array.fill(height) { Array.fill(width) { new MyClass } }

Daniel C. Sobral Over a year ago

@jk. If you think that's the same thing, you didn't understand what this is doing.

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Rex Kerr · Accepted Answer · 2010-03-22 23:02:31Z

2

Among other ways, you can use use Array.range and map:

scala> Array.range(0,3).map(i => Array.range(0,3).map(j => (i,j)))
res0: Array[Array[(Int, Int)]] = Array(Array((0,0), (0,1), (0,2)), Array((1,0), (1,1), (1,2)), Array((2,0), (2,1), (2,2)))

Or you can use Array.fromFunction:

scala> Array.fromFunction((i,j) => (i,j))(3,3)                    
res1: Array[Array[(Int, Int)]] = Array(Array((0,0), (0,1), (0,2)), Array((1,0), (1,1), (1,2)), Array((2,0), (2,1), (2,2)))

Scala 2.8 gives you even more options--look through the Array object. (Actually, that's good advice for 2.7, also....)

answered Mar 22, 2010 at 23:02

Rex Kerr

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