Swift method on generic 2 dimensional Array

Question

I would like to create an extension method on the Array type when the array is a 2 dimensional array, ex. [[Int]], but the type should be generic. I'm trying to do something like this:

extension Array [where Element : ???] {

  public func transposed() -> ??? {
     // ...
  }
}

I did. It is illuminating. But the answer is simpler. Also my question is simpler to understand and brief. — andrewz, May 04 '18 at 14:29

score 0 · Answer 1 · answered May 04 '18 at 14:27

This worked in Swift 4.

extension Array where Element: Collection {

    public func transposed() -> [[Element.Iterator.Element]] {
        var result : [[Element.Iterator.Element]] = [[]]
        for row in self {
            for (y,column) in row.enumerated() {
                while (result.count <= y) {
                    result.append([])
                }
                result[y].append(column)
            }
        }
        return result
    }
}

Then also unit testing passed:

func testTransposed() {
    XCTAssertEqual([[1]].transposed(), [[1]])
    XCTAssertEqual([[1,2,3],[4,5]].transposed(), [[1,4],[2,5],[3]])
    XCTAssertEqual([[4,1],[5,2],[3]].transposed(),[[4,5,3],[1,2]])
    XCTAssertEqual([[1,2,3]].transposed(), [[1],[2],[3]])
    XCTAssertEqual([[1],[2],[3]].transposed(),[[1,2,3]])
}

score 0 · Answer 2 · answered May 04 '18 at 16:52

You could do this:

extension Array
{
   func transposed<T>() -> [[T]] where Element == Array<T>
   {
      return self[0].indices.map{ i in self.map{$0[i]} }
   }
}

let a = [ [1,2,3],
          [4,5,6],
          [7,8,9] ]
for r in a.transposed() {print(r)}

// [1, 4, 7]
// [2, 5, 8]
// [3, 6, 9]
// note that the matrix cannot be sparse

Swift method on generic 2 dimensional Array

2 Answers2