Count the occurences a particular integer has in an array

Question

How to count the number of times a particular number occurs in an array. I cant find a method for this for swift. can someone guide me please?

Thanks :)

Answer 1

Xcode 9 or later • Swift 4 or later

In Swift 4 you can use the new Dictionary method reduce(into:) as follow:

extension Sequence where Element: Hashable {
    var frequency: [Element: Int] {
        return reduce(into: [:]) { $0[$1, default: 0] += 1 }
    }
    func frequency(of element: Element) -> Int {
        return frequency[element] ?? 0
    }
}

---

Playground testing:

let numbers = [0, 0, 1, 1, 1, 2, 3, 4, 5, 5]
print(numbers.frequency) // [2: 1, 4: 1, 5: 2, 3: 1, 1: 3, 0: 2]

print(numbers.frequency(of: 0))   // 2  
print(numbers.frequency(of: 1))   // 3
print(numbers.frequency(of: 2))   // 1
print(numbers.frequency(of: 3))   // 1
print(numbers.frequency(of: 4))   // 1
print(numbers.frequency(of: 5))   // 2

By extending Collection it supports Strings as well

"aab".frequency   // ["a": 2, "b": 1]

Answer 2

Create a dictionary, store the number the first time it is found and initialize the key with 1. Increment it otherwise.

let numArray = [1, 2, 2, 2, 5]
var numCount:[Int:Int] = [:]

for item in numArray {
    numCount[item] = (numCount[item] ?? 0) + 1

for (key, value) in numCount {
    println("\(key) occurs \(value) time(s)")
}