Remove Specific Array Element, Equal to String - Swift

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?

Solution 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"

Solution 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
}

Solution 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" }

Solution 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"
}

Solution 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]"