How to remove an element from an array in Swift

Question

How can I unset/remove an element from an array in Apple\'s new language Swift?

Here\'s some code:

let animals = [\"cats\", \"dogs\", \"chimps\", \"m         


        

Answer 1

I came up with the following extension that takes care of removing elements from an Array, assuming the elements in the Array implement Equatable:

extension Array where Element: Equatable {
  
  mutating func removeEqualItems(_ item: Element) {
    self = self.filter { (currentItem: Element) -> Bool in
      return currentItem != item
    }
  }

  mutating func removeFirstEqualItem(_ item: Element) {
    guard var currentItem = self.first else { return }
    var index = 0
    while currentItem != item {
      index += 1
      currentItem = self[index]
    }
    self.remove(at: index)
  }
  
}
  

Usage:

var test1 = [1, 2, 1, 2]
test1.removeEqualItems(2) // [1, 1]

var test2 = [1, 2, 1, 2]
test2.removeFirstEqualItem(2) // [1, 1, 2]

Answer 2

Few Operation relates to Array in Swift

Create Array

var stringArray = ["One", "Two", "Three", "Four"]

Add Object in Array

stringArray = stringArray + ["Five"]

Get Value from Index object

let x = stringArray[1]

Append Object

stringArray.append("At last position")

Insert Object at Index

stringArray.insert("Going", at: 1)

Remove Object

stringArray.remove(at: 3)

Concat Object value

var string = "Concate Two object of Array \(stringArray[1]) + \(stringArray[2])"

Answer 3

You could do that. First make sure Dog really exists in the array, then remove it. Add the for statement if you believe Dog may happens more than once on your array.

var animals = ["Dog", "Cat", "Mouse", "Dog"]
let animalToRemove = "Dog"

for object in animals {
    if object == animalToRemove {
        animals.remove(at: animals.firstIndex(of: animalToRemove)!)
    }
}

If you are sure Dog exits in the array and happened only once just do that:

animals.remove(at: animals.firstIndex(of: animalToRemove)!)

If you have both, strings and numbers

var array = [12, 23, "Dog", 78, 23]
let numberToRemove = 23
let animalToRemove = "Dog"

for object in array {

    if object is Int {
        // this will deal with integer. You can change to Float, Bool, etc...
        if object == numberToRemove {
        array.remove(at: array.firstIndex(of: numberToRemove)!)
        }
    }
    if object is String {
        // this will deal with strings
        if object == animalToRemove {
        array.remove(at: array.firstIndex(of: animalToRemove)!)
        }
    }
}

Answer 4

As of Xcode 10+, and according to the WWDC 2018 session 223, "Embracing Algorithms," a good method going forward will be mutating func removeAll(where predicate: (Element) throws -> Bool) rethrows

Apple's example:

var phrase = "The rain in Spain stays mainly in the plain."
let vowels: Set<Character> = ["a", "e", "i", "o", "u"]

phrase.removeAll(where: { vowels.contains($0) })
// phrase == "Th rn n Spn stys mnly n th pln."

see Apple's Documentation

So in the OP's example, removing animals[2], "chimps":

var animals = ["cats", "dogs", "chimps", "moose"]
animals.removeAll(where: { $0 == "chimps" } )
// or animals.removeAll { $0 == "chimps" }

This method may be preferred because it scales well (linear vs quadratic), is readable and clean. Keep in mind that it only works in Xcode 10+, and as of writing this is in Beta.

Answer 5

Remove elements using indexes array:

1. Array of Strings and indexes

   let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
   let indexAnimals = [0, 3, 4]
   let arrayRemainingAnimals = animals
       .enumerated()
       .filter { !indexAnimals.contains($0.offset) }
       .map { $0.element }
   
   print(arrayRemainingAnimals)
   
   //result - ["dogs", "chimps", "cow"]
   

2. Array of Integers and indexes

   var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
   let indexesToRemove = [3, 5, 8, 12]
   
   numbers = numbers
       .enumerated()
       .filter { !indexesToRemove.contains($0.offset) }
       .map { $0.element }
   
   print(numbers)
   
   //result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
   

Remove elements using element value of another array

3. Arrays of integers

   let arrayResult = numbers.filter { element in
       return !indexesToRemove.contains(element)
   }
   print(arrayResult)
   
   //result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
   

4. Arrays of strings

   let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
   let arrayRemoveLetters = ["a", "e", "g", "h"]
   let arrayRemainingLetters = arrayLetters.filter {
       !arrayRemoveLetters.contains($0)
   }
   
   print(arrayRemainingLetters)
   
   //result - ["b", "c", "d", "f", "i"]
   

Answer 6

Swift 5

guard let index = orders.firstIndex(of: videoID) else { return }
orders.remove(at: index)