2d scala array iteration

Question

I have a 2d array of type boolean (not important) It is easy to iterate over the array in non-functional style. How to do it FP style?

var matrix = Array.ofDim[Boolean](5, 5)

for ex, I would like to iterate through all the rows for a given column and return a list of int that would match a specific function. Example: for column 3, iterate through rows 1 to 5 to return 4, 5 if the cell at (4, 3), (5, 3) match a specif function. Thx v much

def getChildren(nodeId: Int) : List[Int] = {
    info("getChildren("+nodeId+")")

    var list = List[Int]()
    val nodeIndex = id2indexMap(nodeId)

    for (rowIndex <- 0 until matrix.size) {
      val elem = matrix(rowIndex)(nodeIndex)
      if (elem) {
        println("Row Index = " + rowIndex)
        list = rowIndex :: list
      }
    }

    list
  }

Didier Dupont · Accepted Answer · 2011-09-12 11:38:18Z

4

What about

(1 to 5) filter {i => predicate(matrix(i)(3))}

where predicate is your function?

Note that initialized with (5,5) indexes goes from 0 to 4.

Update: based on your example

def getChildren(nodeId: Int) : List[Int] = {
  info("getChildren("+nodeId+")")
  val nodeIndex = id2indexMap(nodeId)

  val result = (0  until matrix.size).filter(matrix(_)(nodeIndex)).toList
  result.forEach(println)
  result
}

You may move the print in the fiter if you want too, and reverse the list if you want it exactly as in your example

edited Sep 12, 2011 at 11:38

answered Sep 12, 2011 at 10:26

Didier Dupont

29.6k7 gold badges73 silver badges90 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

jts Over a year ago

I think your sol would return an array if the initial structure is a 2d array?

jts Over a year ago

Ah, another difference is I was thinking about filtering directly on the 2d array, but by doing so, I am loosing the row index: val children = matrix.filter(row => row(nodeIndex)) which returns 2 1d arrays - but I do not know which one

Didier Dupont Over a year ago

Not really, the structure I do fitering on is a Range, not an Array (as it was the indexes you wanted in the output). The result type is an IndexedSeq[Int], and it happens to be a Vector. Just do .toList if you want a List, but Vector is usually a better structure than List.

jts Over a year ago

This is the non-FP version of what I'd like to achieve (edit main)

Luigi Plinge · Accepted Answer · 2011-09-12 19:25:30Z

If you're not comfortable with filters and zips, you can stick with the for-comprehension but use it in a more functional way:

for {
  rowIndex <- matrix.indices
  if matrix(rowIndex)(nodeIndex)
} yield { 
  println("Row Index = " + rowIndex)
  rowIndex
}

yield builds a new collection from the results of the for-comprehension, so this expression evaluates to the collection you want to return. seq.indices is a method equivalent to 0 until seq.size. The curly braces allow you to span multiple lines without semicolons, but you can make it in-line if you want:

for (rowIndex <- matrix.indices; if matrix(rowIndex)(nodeIndex)) yield rowIndex

Should probably also mention that normally if you're iterating through an Array you won't need to refer to the indices at all. You'd do something like

for {
  row  <- matrix 
  elem <- row
} yield f(elem)

but your use-case is a bit unusual in that it requires the indices of the elements, which you shouldn't normally be concerned with (using array indices is essentially a quick and dirty hack to pair a data element with a number). If you want to capture and use the notion of position you might be better off using a Map[Int, Boolean] or a case class with such a field.

ziggystar · Accepted Answer · 2011-09-12 11:16:42Z

1

def findIndices[A](aa: Array[Array[A]], pred: A => Boolean): Array[Array[Int]] =
  aa.map(row => 
    row.zipWithIndex.collect{ 
      case (v,i) if pred(v) => i 
  }
)

You can refactor it to be a bit more nicer by extracting the function that finds the indices in a single row only:

def findIndices2[A](xs: Array[A], pred: A => Boolean): Array[Int] =
  xs.zipWithIndex.collect{ 
    case (v,i) if pred(v) => i 
  }

And then write

matrix.map(row  => findIndices2(row, pred))

edited Sep 12, 2011 at 11:16

answered Sep 12, 2011 at 11:04

ziggystar

28.7k9 gold badges76 silver badges127 bronze badges

Collectives™ on Stack Overflow

2d scala array iteration

3 Answers 3

4 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related