How to remove an element from an array in Swift

问题:

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", "moose"]

How could the element animals[2] be removed from the array?

回答1:

The let keyword is for declaring constants that can't be changed. If you want to modify a variable you should use var instead, e.g:

var animals = ["cats", "dogs", "chimps", "moose"]

animals.remove(at: 2)  //["cats", "dogs", "moose"]

A non-mutating alternative that will keep the original collection unchanged is to use filter to create a new collection without the elements you want removed, e.g:

let pets = animals.filter { $0 != "chimps" }

回答2:

Given

var animals = ["cats", "dogs", "chimps", "moose"]

Remove first element

animals.removeFirst() // "cats"
print(animals)        // ["dogs", "chimps", "moose"]

Remove last element

animals.removeLast() // "moose"
print(animals)       // ["cats", "dogs", "chimps"]

Remove element at index

animals.remove(at: 2) // "chimps"
print(animals)           // ["cats", "dogs", "moose"]

Remove element of unknown index

For only one element

if let index = animals.index(of: "chimps") {
    animals.remove(at: index)
}
print(animals) // ["cats", "dogs", "moose"]

For multiple elements

var animals = ["cats", "dogs", "chimps", "moose", "chimps"]

animals = animals.filter(){$0 != "chimps"}
print(animals) // ["cats", "dogs", "moose"]

Or alternatively

animals.index(of: "chimps").map { animals.remove(at: $0) }

Notes

• The above methods modify the array in place (except for filter) and return the element that was removed.
• Swift Guide to Map Filter Reduce
• If you don't want to modify the original array, you can use dropFirst or dropLast to create a new array.

Updated to Swift 3

回答3:

The above answers seem to presume that you know the index of the element that you want to delete.

Often you know the reference to the object you want to delete in the array. (You iterated through your array and have found it, e.g.) In such cases it might be easier to work directly with the object reference without also having to pass everywhere its index. Hence, I suggest this solution. It uses the identity operator !==, which you use to test whether two object references both refer to the same object instance.

func delete(element: String) {
    list = list.filter() { $0 !== element }
}

Of course this doesn't just work for Strings.

回答4:

Swift 5: Here is a cool and easy extension to remove elements in an array, without filtering :

   extension Array where Element: Equatable {

    // Remove first collection element that is equal to the given `object`:
    mutating func remove(object: Element) {
        guard let index = firstIndex(of: object) else {return}
        remove(at: index)
    }

}

Usage :

var myArray = ["cat", "barbecue", "pancake", "frog"]
let objectToRemove = "cat"

myArray.remove(object: objectToRemove) // ["barbecue", "pancake", "frog"]

Also works with other types, such as Int since Element is a generic type:

var myArray = [4, 8, 17, 6, 2]
let objectToRemove = 17

myArray.remove(object: objectToRemove) // [4, 8, 6, 2]

回答5:

For Swift4:

list = list.filter{$0 != "your Value"}

回答6:

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", atIndex: 1)

Remove Object

stringArray.removeAtIndex(3)

Concat Object value

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

回答7:

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.removeAtIndex(animals.indexOf(animalToRemove)!)
    }
}

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

animals.removeAtIndex(animals.indexOf(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.removeAtIndex(array.indexOf(numberToRemove)!)
        }
    }
    if object is String
    {
        // this will deal with strings
        if object == animalToRemove
        {
        array.removeAtIndex(array.indexOf(animalToRemove)!)
        }
    }
}

回答8:

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.

回答9:

If you don't know the index of the element that you want to remove, and the element is conform the Equatable protocol, you can do:

animals.removeAtIndex(animals.indexOf("dogs")!)

See Equatable protocol answer:How do I do indexOfObject or a proper containsObject

回答10:

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"]
   

回答11:

Regarding @Suragch's Alternative to "Remove element of unknown index":

There is a more powerful version of "indexOf(element)" that will match on a predicate instead of the object itself. It goes by the same name but it called by myObjects.indexOf{$0.property = valueToMatch}. It returns the index of the first matching item found in myObjects array.

If the element is an object/struct, you may want to remove that element based on a value of one of its properties. Eg, you have a Car class having car.color property, and you want to remove the "red" car from your carsArray.

if let validIndex = (carsArray.indexOf{$0.color == UIColor.redColor()}) {
  carsArray.removeAtIndex(validIndex)
}

Foreseeably, you could rework this to remove "all" red cars by embedding the above if statement within a repeat/while loop, and attaching an else block to set a flag to "break" out of the loop.

回答12:

If you have array of custom Objects, you can search by specific property like this:

    if let index = doctorsInArea.indexOf({$0.id == doctor.id}){
        doctorsInArea.removeAtIndex(index)
    }

or if you want to search by name for example

    if let index = doctorsInArea.indexOf({$0.name == doctor.name}){
        doctorsInArea.removeAtIndex(index)
    }

回答13:

Swift 5

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

回答14:

This should do it (not tested):

animals[2..3] = []

Edit: and you need to make it a var, not a let, otherwise it's an immutable constant.

回答15:

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.removeAtIndex(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]

回答16:

extension to remove String object

extension Array {
    mutating func delete(element: String) {
        self = self.filter() { $0 as! String != element }
    }
}

回答17:

I use this extension, almost same as Varun's, but this one (below) is all-purpose:

 extension Array where Element: Equatable  {
        mutating func delete(element: Iterator.Element) {
                self = self.filter{$0 != element }
        }
    }

回答18:

To remove elements from an array, use the remove(at:) , removeLast() and removeAll()

yourArray = [1,2,3,4]

Remove the value at 2 position

yourArray.remove(at: 2 )

Remove the last value from array

yourArray.removeLast()

Removes all members from the set

yourArray.removeAll()