Removing duplicate elements from an array in Swift

Question

I might have an array that looks like the following:

[1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]

Or, reall

Answer 1

An alternate (if not optimal) solution from here using immutable types rather than variables:

func deleteDuplicates<S: ExtensibleCollectionType where S.Generator.Element: Equatable>(seq:S)-> S {
    let s = reduce(seq, S()){
        ac, x in contains(ac,x) ? ac : ac + [x]
    }
    return s
}

Included to contrast Jean-Pillippe's imperative approach with a functional approach.

As a bonus this function works with strings as well as arrays!

Edit: This answer was written in 2014 for Swift 1.0 (before Set was available in Swift). It doesn't require Hashable conformance & runs in quadratic time.

Answer 2

Swift 4.x:

extension Sequence where Iterator.Element: Hashable {
  func unique() -> [Iterator.Element] {
    return Array(Set<Iterator.Element>(self))
  }

  func uniqueOrdered() -> [Iterator.Element] {
    return reduce([Iterator.Element]()) { $0.contains($1) ? $0 : $0 + [$1] }
  }
}

usage:

["Ljubljana", "London", "Los Angeles", "Ljubljana"].unique()

or

["Ljubljana", "London", "Los Angeles", "Ljubljana"].uniqueOrdered()

Answer 3

1. First add all the elements of an array to NSOrderedSet.
2. This will remove all the duplicates in your array.
3. Again convert this orderedset to an array.

Done....

Example

let array = [1,1,1,1,2,2,2,2,4,6,8]

let orderedSet : NSOrderedSet = NSOrderedSet(array: array)

let arrayWithoutDuplicates : NSArray = orderedSet.array as NSArray

output of arrayWithoutDuplicates - [1,2,4,6,8]

Answer 4

You can convert to a Set and back to an Array again quite easily:

let unique = Array(Set(originals))

This is not guaranteed to maintain the original order of the array.

Answer 5

Many answers available here, but I missed this simple extension, suitable for Swift 2 and up:

extension Array where Element:Equatable {
    func removeDuplicates() -> [Element] {
        var result = [Element]()

        for value in self {
            if result.contains(value) == false {
                result.append(value)
            }
        }

        return result
    }
}

Makes it super simple. Can be called like this:

let arrayOfInts = [2, 2, 4, 4]
print(arrayOfInts.removeDuplicates()) // Prints: [2, 4]

Filtering based on properties

To filter an array based on properties, you can use this method:

extension Array {

    func filterDuplicates(@noescape includeElement: (lhs:Element, rhs:Element) -> Bool) -> [Element]{
        var results = [Element]()

        forEach { (element) in
            let existingElements = results.filter {
                return includeElement(lhs: element, rhs: $0)
            }
            if existingElements.count == 0 {
                results.append(element)
            }
        }

        return results
    }
}

Which you can call as followed:

let filteredElements = myElements.filterDuplicates { $0.PropertyOne == $1.PropertyOne && $0.PropertyTwo == $1.PropertyTwo }

Answer 6

swift 2

with uniq function answer:

func uniq<S: SequenceType, E: Hashable where E==S.Generator.Element>(source: S) -> [E] {
    var seen: [E:Bool] = [:]
    return source.filter({ (v) -> Bool in
        return seen.updateValue(true, forKey: v) == nil
    })
}

use:

var test = [1,2,3,4,5,6,7,8,9,9,9,9,9,9]
print(uniq(test)) //1,2,3,4,5,6,7,8,9