Splitting an Array into Sub-Arrays in Swift

Question

Problem

Given an array of values how can I split it into sub-arrays made of elements that are equal?

Example

Given this array

let numbers = [1, 1, 1, 3, 3, 4]

I want this output

[[1,1,1], [3, 3], [4]]

What I am NOT looking for

A possible way of solving this would be creating some sort of index to indicate the occurrences of each element like this.

let indexes = [1:3, 3:2, 4:1]

And finally use the index to rebuild the output array.

let subsequences = indexes.sort { $0.0.0 < $0.1.0 }.reduce([Int]()) { (res, elm) -> [Int] in
    return res + [Int](count: elm.1, repeatedValue: elm.0)
}

However with this solution I am losing the original values. Of course in this case it's not a big problem (an Int value is still and Inteven if recreated) but I would like to apply this solution to more complex data structures like this

struct Starship: Equatable {
    let name: String
    let warpSpeed: Int
}

func ==(left:Starship, right:Starship) -> Bool {
    return left.warpSpeed == right.warpSpeed
}

Final considerations

The function I am looking for would be some kind of reverse of flatten(), infact

let subsequences: [[Int]] = [[1,1,1], [3, 3], [4]]
print(Array(subsequences.flatten())) // [1, 1, 1, 3, 3, 4]

I hope I made myself clear, let me know should you need further details.

Answer 1

 // extract unique numbers using a set, then
 // map sub-arrays of the original arrays with a filter on each distinct number

 let numbers = [1, 1, 1, 3, 3, 4]

 let numberGroups = Set(numbers).map{ value in return numbers.filter{$0==value} }

 print(numberGroups)

[EDIT] changed to use Set Initializer as suggested by Hamish

[EDIT2] Swift 4 added an initializer to Dictionary that will do this more efficiently:

 let numberGroups = Array(Dictionary(grouping:numbers){$0}.values)

For a list of objects to be grouped by one of their properties:

 let objectGroups = Array(Dictionary(grouping:objects){$0.property}.values)