Split large Array in Array of two elements

Question

I have large list of objects and I need to split them in a group of two elements for UI propouse.

Example:

[0, 1, 2, 3, 4, 5, 6]

Becomes

Answer 1

You can use oisdk's awesome SwiftSequence framework. There is the chunk function which does exactly what you want:

[1, 2, 3, 4, 5].chunk(2)

[[1, 2], [3, 4], [5]]

Also there's a LOT more functions for sequences, you should definitely check it out.

You can have a look at his implementation of chunk here (It uses generators)

Answer 2

Maybe not the most efficient solution, but the most direct solution:

func toPairs(numbers:[Int])->[[Int]]
{
    var pairs:[[Int]]=[]
    var pair:[Int]=[]
    for var index=0;index<numbers.count;index++ {
        pair.append(numbers[index])
        if pair.count == 2 || index==numbers.count-1 {
            pairs.append(pair)
            pair=[]
        }
    }
    return pairs
}

var numbers=[0,1,2,3,4,5]

var pairs=toPairs(numbers)

print(pairs)

Output on my laptop:

[[0, 1], [2, 3], [4, 5]]
Program ended with exit code: 0

Answer 3

If you're looking for efficiency, you could have a method that would generate each array of 2 elements lazily, so you'd only store 2 elements at a time in memory:

public struct ChunkGen<G : GeneratorType> : GeneratorType {

  private var g: G
  private let n: Int
  private var c: [G.Element]

  public mutating func next() -> [G.Element]? {
    var i = n
    return g.next().map {
      c = [$0]
      while --i > 0, let next = g.next() { c.append(next) }
      return c
    }
  }

  private init(g: G, n: Int) {
    self.g = g
    self.n = n
    self.c = []
    self.c.reserveCapacity(n)
  }
}

public struct ChunkSeq<S : SequenceType> : SequenceType {

  private let seq: S
  private let n: Int

  public func generate() -> ChunkGen<S.Generator> {
    return ChunkGen(g: seq.generate(), n: n)
  }
}

public extension SequenceType {
  func chunk(n: Int) -> ChunkSeq<Self> {
    return ChunkSeq(seq: self, n: n)
  }
}

var g = [1, 2, 3, 4, 5].chunk(2).generate()

g.next() // [1, 2]
g.next() // [3, 4]
g.next() // [5]
g.next() // nil

This method works on any SequenceType, not just Arrays.

For Swift 1, without the protocol extension, you've got:

public struct ChunkGen<T> : GeneratorType {

  private var (st, en): (Int, Int)
  private let n: Int
  private let c: [T]

  public mutating func next() -> ArraySlice<T>? {
    (st, en) = (en, en + n)
    return st < c.endIndex ? c[st..<min(en, c.endIndex)] : nil
  }

  private init(c: [T], n: Int) {
    self.c = c
    self.n = n
    self.st = 0 - n
    self.en = 0
  }
}

public struct ChunkSeq<T> : SequenceType {

  private let c: [T]
  private let n: Int

  public func generate() -> ChunkGen<T> {
    return ChunkGen(c: c, n: n)
  }
}

func chunk<T>(ar: [T], #n: Int) -> ChunkSeq<T> {
  return ChunkSeq(c: ar, n: n)
}

For Swift 3:

public struct ChunkIterator<I: IteratorProtocol> : IteratorProtocol {

  fileprivate var i: I
  fileprivate let n: Int

  public mutating func next() -> [I.Element]? {
    guard let head = i.next() else { return nil }
    var build = [head]
    build.reserveCapacity(n)
    for _ in (1..<n) {
      guard let x = i.next() else { break }
      build.append(x)
    }
    return build
  }

}

public struct ChunkSeq<S: Sequence> : Sequence {

  fileprivate let seq: S
  fileprivate let n: Int

  public func makeIterator() -> ChunkIterator<S.Iterator> {
    return ChunkIterator(i: seq.makeIterator(), n: n)
  }
}

public extension Sequence {
  func chunk(_ n: Int) -> ChunkSeq<Self> {
    return ChunkSeq(seq: self, n: n)
  }
}

var g = [1, 2, 3, 4, 5].chunk(2).makeIterator()

g.next() // [1, 2]
g.next() // [3, 4]
g.next() // [5]
g.next() // nil

Answer 4

Swift 2 Gist

let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

extension Array {
    func splitBy(subSize: Int) -> [[Element]] {
        return 0.stride(to: self.count, by: subSize).map { startIndex in
            let endIndex = startIndex.advancedBy(subSize, limit: self.count)
            return Array(self[startIndex ..< endIndex])
        }
    }
}

let chunks = arr.splitBy(5)

print(chunks) // [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12]]

Answer 5

If you want an array of subslices, you can use the split function to generate it using a closure that captures a state variable and increments it as it passes over each element, splitting only on every nth element. As an extension of Sliceable (Swift 2.0 only, would need to be a free function in 1.2):

extension Sliceable {
    func splitEvery(n: Index.Distance) -> [SubSlice] {
        var i: Index.Distance = 0
        return split(self) { _ in ++i % n == 0 }
    }
}

Subslices are very efficient in so much as they usually share internal storage with the original sliceable entity. So no new memory will be allocated for storing the elements - only memory for tracking the subslices' pointers into the original array.

Note, this will work on anything sliceable, like strings:

"Hello, I must be going"
    .characters
    .splitEvery(3)
    .map(String.init)

returns ["He", "lo", " I", "mu", "t ", "e ", "oi", "g"].

If you want to lazily split the array up (i.e. generate a sequence that only serves up subslices on demand) you could write it using anyGenerator:

extension Sliceable {
    func lazilySplitEvery(n: Index.Distance) -> AnySequence<SubSlice> {

        return AnySequence { () -> AnyGenerator<SubSlice> in
            var i: Index = self.startIndex
            return anyGenerator {
                guard i != self.endIndex else { return nil }
                let j = advance(i, n, self.endIndex)
                let r = i..<j
                i = j
                return self[r]
            }
        }
    }
}


for x in [1,2,3,4,5,6,7].lazilySplitEvery(3) {
    print(x)
}
// prints [1, 2, 3]
//        [4, 5, 6]
//        [7]

Answer 6

The shortest solution (Swift 4), I have seen so far, is from a Gist:

extension Array {

    func chunks(chunkSize: Int) -> [[Element]] {
        return stride(from: 0, to: self.count, by: chunkSize).map {
            Array(self[$0..<Swift.min($0 + chunkSize, self.count)])
        }
    }

}