Find Duplicate Elements In Array Using Swift

问题:

How to find Duplicate Elements in Array? I have array of phone numbers so in the phone numbers i should start searching from the right side to the left side and find similar 6 integers. then i should print them out.

回答1:

To find duplicates, you could build cross reference by phone number, then filter that down to duplicates only. For example, consider:

let contacts = [
    Contact(name: "Rob",     phone: "555-1111"),
    Contact(name: "Richard", phone: "555-2222"),
    Contact(name: "Rachel",  phone: "555-1111"),
    Contact(name: "Loren",   phone: "555-2222"),
    Contact(name: "Mary",    phone: "555-3333"),
    Contact(name: "Susie",   phone: "555-2222")
]

In Swift 4, you can build the cross reference dictionary with:

let crossReference = Dictionary(grouping: contacts, by: { $0.phone })

Or

let crossReference = contacts.reduce(into: [String: [Contact]]()) {
    $0[$1.phone, default: []].append($1)
}

Then, to find the duplicates:

let duplicates = crossReference
    .filter { $1.count > 1 }                 // filter down to only those with multiple contacts
    .sorted { $0.1.count > $1.1.count }      // if you want, sort in descending order by number of duplicates

Clearly use whatever model types make sense for you, but the above uses the following Contact type:

struct Contact {
    let name: String
    let phone: String
}

There are many, many ways to implement this, so I wouldn't focus on the implementation details of the above, but rather focus on the concept: Build cross reference original array by some key (e.g. phone number) and then filter results down to just those keys with duplicate values.

---

It sounds like you want to flatten this structure that reflects the duplicates, into a single array of contacts (I'm not sure why you'd want to do that, as you lose the structure identifying which are duplicates of each other), but if you want to do that, you can flatMap it:

let flattenedDuplicates = crossReference
    .filter { $1.count > 1 }                 // filter down to only those with multiple contacts
    .flatMap { $0.1 }                        // flatten it down to just array of contacts that are duplicates of something else

For Swift 2 or 3 renditions, see previous renditions of this answer.

回答2:

Feeling ~clever~. Given an array of Ints

let x = [1, 1, 2, 3, 4, 5, 5]
let duplicates = Array(Set(x.filter({ (i: Int) in x.filter({ $0 == i }).count > 1})))
// [1, 5]

Please note, this is horrendously efficient for everyone involved, including the compiler, and you.

I'm just showing off.

Edit: lol someone downvoted this, which leads me to reiterate, just in case: please DO NOT USE THIS in production or anywhere else.

回答3:

You could implement it using "Merge sort", but you need to make one modification, during the merge step you should ignore the duplicates.

The easiest way to find duplicate elements is if the phone number is just a 6-digit number and has type Int, you could sort the array of phone numbers and than filter that to find duplicates.

var phoneNumbers = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]

func findDuplicates(sortedArray array: [Int]) -> [Int]
{
    var duplicates: [Int] = []

    var prevItem: Int = 0
    var addedItem: Int = 0

    for item in array
    {
        if(prevItem == item && addedItem != item)
        {
            duplicates.append(item)
            addedItem = item
        }

        prevItem = item
    }

    return duplicates
}

func sortPhoneNumbers(phoneNumbers: [Int]) -> [Int]
{
    return phoneNumbers.sorted({ return $0<$1 })
}

sortPhoneNumbers(phoneNumbers)
findDuplicates(sortPhoneNumbers(phoneNumbers))

In addition, you could implement the findDuplicates method in different ways:

Using Set (Swift 1.2+):

func findDuplicates(array: [Int]) -> [Int]
{
    var duplicates = Set<Int>()
    var prevItem = 0       

    for item in array
    {
        if(prevItem == item)
        {
            duplicates.insert(item)
        }

        prevItem = item
    }

    return Array(duplicates)
}

And so on.

回答4:

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, based on the contacts example of Rob:

let filteredContacts = myContacts.filterDuplicates { $0.name == $1.name && $0.phone == $1.phone }

回答5:

Antoine's solution in Swift 3+ syntax

extension Array {

    func filterDuplicates(includeElement: @escaping (_ lhs: Element, _ rhs: Element) -> Bool) -> [Element] {

        var results = [Element]()

        forEach { (element) in

            let existingElements = results.filter {
                return includeElement(element, $0)
            }

            if existingElements.count == 0 {
                results.append(element)
            }
        }
        return results
    }
}

回答6:

Same as in @tikhop's answer, but as Array extension (Swift 3):

extension Array where Element: Comparable & Hashable {

   public var duplicates: [Element] {

      let sortedElements = sorted { $0 < $1 }
      var duplicatedElements = Set<Element>()

      var previousElement: Element?
      for element in sortedElements {
         if previousElement == element {
            duplicatedElements.insert(element)
         }
         previousElement = element
      }

      return Array(duplicatedElements)
   }

}

回答7:

I've found a way by using reduce, here is the code(Swift 4):

let testNumbers = [1,1,2,3,4,5,2]
let nondupicate = testNumbers.reduce(into: [Int]()) {
    if !$0.contains($1) {
        $0.append($1)
    } else {
        print("Found dupicate: \($1)")
    }
}

As a side effect, it returns an array has no dupicated elements.

You can easily modify it for counting duplicated elements numbers, checking string arrays etc.

回答8:

Based on Rob's answer, an array extension to just finds duplicates is:

extension Array where Element: Hashable {
    func duplicates() -> Array {
        let groups = Dictionary(grouping: self, by: {$0})
        let duplicateGroups = groups.filter {$1.count > 1}
        let duplicates = Array(duplicateGroups.keys)
        return duplicates
    }
}

回答9:

A very simple answer which preserves all duplicates

let originalNums = [5, 3, 2, 3 , 7 , 5,3]
var nums = Array(originalNums)

let numSet = Set(nums)

for num in numSet {
  if let index = nums.index(of: num) {
     nums.remove(at: index)
  }
}

output

[3, 5, 3]

回答10:

I also had a similar problem and have overcome in the following way. (Xcode 8.3.2)

let a = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]
var b = a // copy-on-write so that "a" won't be modified

while let c = b.popLast() {
  b.forEach() {
    if $0 == c {
      Swift.print("Duplication: \(c)")
    }
  }
}

//  Duplication: 456789
//  Duplication: 123456

The point is that the number of comparison. It would be smaller than others.

Assume that the number of items in the array is N. In each loop, the number will be decrementing by one. So, the total number will be (N-1) + (N-2) + (N-3) + ... + 2 + 1 = N * (N-1) / 2 When N = 10, that will be 9 + 8 + ... = 45

In contrast, that of some algorithms might be N * N. When N = 10 that will be 100.

In spite of that, taking into account of the cost of deep-copy or shallow-copy, I agree that @Patrick Perini's brilliant way would be better than this in some situations even the number of that would be N * N.

EDIT:

Alternative way with IteratorProtocol

let a = [123456, 234567, 345678, 123456, 456789, 135790, 456789, 142638]
var i = a.makeIterator()

while let c = i.next() {
  var j = i
  while let d = j.next() {
    if c == d {
      Swift.print("Duplication: \(c)")
    }
  }
}

//  Duplication: 123456
//  Duplication: 456789

That looks more complex, but uses the same idea as before. This does not have unnecessary memory allocations or copies.

My concern is efficiency, i.e. quicker UI response, longer battery life, smaller memory footprint, etc. Avoiding unnecessary memory allocations and/or memory copies which are automatically done by Swift in the behind scene would be crucial if we are providing competitive products. (-;