Getting the most frequent value of an array

let myArray = [4, 4, 4, 3, 3, 3, 4, 6, 6, 5, 5, 2]

// Create dictionary to map value to count   
var counts = [Int: Int]()

// Count the values with using forEach    
myArray.forEach { counts[$0] = (counts[$0] ?? 0) + 1 }

// Find the most frequent value and its count with max(by:)    
if let (value, count) = counts.max(by: {$0.1 < $1.1}) {
    print("\(value) occurs \(count) times")
}

Output:

4 occurs 4 times

---

Here it is as a function:

func mostFrequent(array: [Int]) -> (value: Int, count: Int)? {
    var counts = [Int: Int]()

    array.forEach { counts[$0] = (counts[$0] ?? 0) + 1 }

    if let (value, count) = counts.max(by: {$0.1 < $1.1}) {
        return (value, count)
    }

    // array was empty
    return nil
}

if let result = mostFrequent(array: [1, 3, 2, 1, 1, 4, 5]) {
    print("\(result.value) occurs \(result.count) times")    
}

1 occurs 3 times

---

Update for Swift 4:

Swift 4 introduces reduce(into:_:) and default values for array look ups which enable you to generate the frequencies in one efficient line. And we might as well make it generic and have it work for any type that is Hashable:

func mostFrequent<T: Hashable>(array: [T]) -> (value: T, count: Int)? {

    let counts = array.reduce(into: [:]) { $0[$1, default: 0] += 1 }

    if let (value, count) = counts.max(by: { $0.1 < $1.1 }) {
        return (value, count)
    }

    // array was empty
    return nil
}

if let result = mostFrequent(array: ["a", "b", "a", "c", "a", "b"]) {
    print("\(result.value) occurs \(result.count) times")
}

a occurs 3 times

You can also use the NSCountedSet, here's the code

let nums = [4, 4, 4, 3, 3, 3, 4, 6, 6, 5, 5, 2]
let countedSet = NSCountedSet(array: nums)
let mostFrequent = countedSet.max { countedSet.count(for: $0) < countedSet.count(for: $1) }

Thanks to @Ben Morrow for the smart suggestions in the comments below.