Swift dictionary get key for value

Swift 3: a more performant approach for the special case of bijective dictionaries

If the reverse dictionary lookup use case covers a bijective dictionary with a one to one relationship between keys and values, an alternative approach to the collection-exhaustive filter operation would be using a quicker short-circuiting approach to find some key, if it exists.

extension Dictionary where Value: Equatable {
    func someKey(forValue val: Value) -> Key? {
        return first(where: { $1 == val })?.key
    }
}

Example usage:

let dict: [Int: String] = [1: "one", 2: "two", 4: "four"]

if let key = dict.someKey(forValue: "two") { 
    print(key)
} // 2

There is, as far as I know, no built-in Swift function to get all dictionary keys for a given value. Here is a possible implementation:

func allKeysForValue<K, V : Equatable>(dict: [K : V], val: V) -> [K] {
    return map(filter(dict) { $1 == val }) { $0.0 }
}

The filter reduces all key-value pairs to those with the given value. The map maps the (filtered) key-value pairs to the keys alone.

Example usage:

let dict = ["a" : 1, "b" : 2, "c" : 1, "d" : 2]
let keys = allKeysForValue(dict, 1)
println(keys) // [a, c]

Update for Swift 2: As of Xcode 7 beta 2, this can now be achieved with an extension method for dictionaries of equatable values (thanks to Airspeed Velocity to make me aware of this in a comment):

extension Dictionary where Value : Equatable {
    func allKeysForValue(val : Value) -> [Key] {
        return self.filter { $1 == val }.map { $0.0 }
    }
}

let dict = ["a" : 1, "b" : 2, "c" : 1, "d" : 2]
let keys = dict.allKeysForValue(1)
print(keys) // [a, c]

Update for Swift 3:

extension Dictionary where Value: Equatable {
    func allKeys(forValue val: Value) -> [Key] {
        return self.filter { $1 == val }.map { $0.0 }
    }
}

let dict = ["a" : 1, "b" : 2, "c" : 1, "d" : 2]
let keys = dict.allKeys(forValue: 1)
print(keys) // [a, c]

You can use allKeys(for:) if you cast to NSDictionary:

let keys = (dict as NSDictionary).allKeys(for: image) as! [UIImage]