Remove Specific Array Element, Equal to String - Swift

Question

Is there no easy way to remove a specific element from an array, if it is equal to a given string? The workarounds are to find the index of the element of the array you wish to remove, and then removeAtIndex, or to create a new array where you append all elements that are not equal to the given string. But is there no quicker way?

Answer 1

You can use filter() to filter your array as follow

var strings = ["Hello","Playground","World"]

strings = strings.filter { $0 != "Hello" }

print(strings)   // "["Playground", "World"]\n"

edit/update:

Xcode 10 • Swift 4.2 or later

You can use the new RangeReplaceableCollection mutating method called removeAll(where:)

var strings = ["Hello","Playground","World"]

strings.removeAll { $0 == "Hello" }

print(strings)   // "["Playground", "World"]\n"

If you need to remove only the first occurrence of an element we ca implement a custom remove method on RangeReplaceableCollection constraining the elements to Equatable:

extension RangeReplaceableCollection where Element: Equatable {
    @discardableResult
    mutating func removeFirst(_ element: Element) -> Element? {
        guard let index = firstIndex(of: element) else { return nil }
        return remove(at: index)
    }
}

---

Or using a predicate for non Equatable elements:

extension RangeReplaceableCollection {
    @discardableResult
    mutating func removeFirst(where predicate: @escaping (Element) throws -> Bool) rethrows -> Element? {
        guard let index = try firstIndex(where: predicate) else { return nil }
        return remove(at: index)
    }
}

---

var strings = ["Hello","Playground","World"]
strings.removeFirst("Hello")
print(strings)   // "["Playground", "World"]\n"
strings.removeFirst { $0 == "Playground" }
print(strings)   // "["World"]\n"

Answer 2

Using filter like suggested above is nice. But if you want to remove only one occurrence of a value or you assume there are no duplicates in the array and you want a faster algorithm, use this:

EDIT: Swift 5 Update

if let index = array.firstIndex(of: "stringToRemove") {
    array.remove(at: index)
} else {
    // not found
}

Thanks @Thomas Mary.

Swift 3 and 4

if let index = array.index(of: "stringToRemove") {
    array.remove(at: index)
} else {
    // not found
}

Answer 3

It's not clear if by quicker you mean in terms of execution time or amount of code.

In the latter case you can easily create a copy using the filter method. For example, given the following array:

let array = ["1", "2", "3", "4", "5"]

you can create a copy with all elements but "2" as:

let filteredArray = array.filter { $0 != "2" }

Answer 4

You'll want to use filter(). If you have a single element (called say obj) to remove, then the filter() predicate will be { $0 != obj }. If you do this repeatedly for a large array this might be a performance issue. If you can defer removing individual objects and want to remove an entire sub-array then use something like:

var stringsToRemove : [String] = ...
var strings : [String] = ...

strings.filter { !contains(stringsToRemove, $0) }

for example:

 1> ["a", "b", "c", "d"].filter { !contains(["b", "c"], $0) }
$R5: [String] = 2 values {
  [0] = "a"
  [1] = "d"
}

Answer 5

You could use filter() in combination with operator overloading to produce an easily repeatable solution:

func -= (inout left: [String], right: String){
    left = left.filter{$0 != right}    
}

var myArrayOfStrings:[String] = ["Hello","Playground","World"]

myArrayOfStrings -= "Hello"

print(myArrayOfStrings)   // "[Playground, World]"

Answer 6

if you need to delete subArray from array then this is a perfect solution using Swift3:

var array = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]            
let subArrayToDelete = ["c", "d", "e", "ee"]
array = array.filter{ !subArrayToDelete.contains($0) }
print(array) // ["a", "b", "f", "g", "h", "i", "j"]

this is better for your performance rather than deleting one by one.

btw even faster solution is (but it will rearrange items in the final array):

array = Array(Set(array).subtracting(subArrayToDelete))

Answer 7

var ra = ["a", "ab", "abc", "a", "ab"]

print(ra)                               // [["a", "ab", "abc", "a", "ab"]

ra.removeAll(where: { $0 == "a" })

print(ra)                               // ["ab", "abc", "ab"]

Answer 8

Simple loop over Array

var array = ["Apple","Banana","Orange"]

for (index,value) in array.enumerated(){
    if value == "Banana"{
    array.remove(at: index)
}

Answer 9

One small point.

Normally you must simply use .filter{$0 != target}

If for some reason you wish to traverse the array. It's a basic of programming that you DON'T edit/delete items from an array while you're traversing it as, depending on the exact nature of the language/OS/etc it may or may not result in unpredictable behavior for the obvious reasons.

The only safe way to traverse is (in some cases) backwards but (always) using a while formulation which re-looks at the array each pass:

    while let found = urls.firstIndex(of: target) {
        urls.remove(at: found)
    }

This is a basic of working with arrays.

Again, always use .filter{$0 != target}

As a footnote, in the incredible case that it is a performance issue (literally 10s of millions of entries), you'd handle the whole thing completely differently using indices and so on, so it is totally irrelevant to this question.