How to count occurrences of an element in a Swift array?

With Swift 5, according to your needs, you may choose one of the 7 following Playground sample codes to count the occurrences of hashable items in an array.

#1. Using Array's reduce(into:_:) and Dictionary's subscript(_:default:) subscript

let array = [4, 23, 97, 97, 97, 23]
let dictionary = array.reduce(into: [:]) { counts, number in
    counts[number, default: 0] += 1
}
print(dictionary) // [4: 1, 23: 2, 97: 3]

#2. Using repeatElement(_:count:) function, zip(_:_:) function and Dictionary's init(_:uniquingKeysWith:)initializer

let array = [4, 23, 97, 97, 97, 23]

let repeated = repeatElement(1, count: array.count)
//let repeated = Array(repeating: 1, count: array.count) // also works

let zipSequence = zip(array, repeated)

let dictionary = Dictionary(zipSequence, uniquingKeysWith: { (current, new) in
    return current + new
})
//let dictionary = Dictionary(zipSequence, uniquingKeysWith: +) // also works

print(dictionary) // prints [4: 1, 23: 2, 97: 3]

#3. Using a Dictionary's init(grouping:by:) initializer and mapValues(_:) method

let array = [4, 23, 97, 97, 97, 23]

let dictionary = Dictionary(grouping: array, by: { $0 })

let newDictionary = dictionary.mapValues { (value: [Int]) in
    return value.count
}

print(newDictionary) // prints: [97: 3, 23: 2, 4: 1]

#4. Using a Dictionary's init(grouping:by:) initializer and map(_:) method

let array = [4, 23, 97, 97, 97, 23]

let dictionary = Dictionary(grouping: array, by: { $0 })

let newArray = dictionary.map { (key: Int, value: [Int]) in
    return (key, value.count)
}

print(newArray) // prints: [(4, 1), (23, 2), (97, 3)]

#5. Using a for loop and Dictionary's subscript(_:) subscript

extension Array where Element: Hashable {

    func countForElements() -> [Element: Int] {
        var counts = [Element: Int]()
        for element in self {
            counts[element] = (counts[element] ?? 0) + 1
        }
        return counts
    }

}

let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [4: 1, 23: 2, 97: 3]

#6. Using NSCountedSet and NSEnumerator's map(_:) method (requires Foundation)

import Foundation

extension Array where Element: Hashable {

    func countForElements() -> [(Element, Int)] {
        let countedSet = NSCountedSet(array: self)
        let res = countedSet.objectEnumerator().map { (object: Any) -> (Element, Int) in
            return (object as! Element, countedSet.count(for: object))
        }
        return res
    }

}

let array = [4, 23, 97, 97, 97, 23]
print(array.countForElements()) // prints [(97, 3), (4, 1), (23, 2)]

#7. Using NSCountedSet and AnyIterator (requires Foundation)

import Foundation

extension Array where Element: Hashable {

    func counForElements() -> Array<(Element, Int)> {
        let countedSet = NSCountedSet(array: self)
        var countedSetIterator = countedSet.objectEnumerator().makeIterator()
        let anyIterator = AnyIterator<(Element, Int)> {
            guard let element = countedSetIterator.next() as? Element else { return nil }
            return (element, countedSet.count(for: element))
        }
        return Array<(Element, Int)>(anyIterator)
    }

}

let array = [4, 23, 97, 97, 97, 23]
print(array.counForElements()) // [(97, 3), (4, 1), (23, 2)]

Credits:

  • Swift Idioms
  • generic on Collection, using Dictionary

array.filter{$0 == element}.count

Swift 3 and Swift 2:

You can use a dictionary of type [String: Int] to build up counts for each of the items in your [String]:

let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]

for item in arr {
    counts[item] = (counts[item] ?? 0) + 1
}

print(counts)  // "[BAR: 1, FOOBAR: 1, FOO: 2]"

for (key, value) in counts {
    print("\(key) occurs \(value) time(s)")
}

output:

BAR occurs 1 time(s)
FOOBAR occurs 1 time(s)
FOO occurs 2 time(s)

Swift 4:

Swift 4 introduces (SE-0165) the ability to include a default value with a dictionary lookup, and the resulting value can be mutated with operations such as += and -=, so:

counts[item] = (counts[item] ?? 0) + 1

becomes:

counts[item, default: 0] += 1

That makes it easy to do the counting operation in one concise line using forEach:

let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
var counts: [String: Int] = [:]

arr.forEach { counts[$0, default: 0] += 1 }

print(counts)  // "["FOOBAR": 1, "FOO": 2, "BAR": 1]"

Swift 4: reduce(into:_:)

Swift 4 introduces a new version of reduce that uses an inout variable to accumulate the results. Using that, the creation of the counts truly becomes a single line:

let arr = ["FOO", "FOO", "BAR", "FOOBAR"]
let counts = arr.reduce(into: [:]) { counts, word in counts[word, default: 0] += 1 }

print(counts)  // ["BAR": 1, "FOOBAR": 1, "FOO": 2]

Or using the default parameters:

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

Finally you can make this an extension of Sequence so that it can be called on any Sequence containing Hashable items including Array, ArraySlice, String, and String.SubSequence:

extension Sequence where Element: Hashable {
    var histogram: [Element: Int] {
        return self.reduce(into: [:]) { counts, elem in counts[elem, default: 0] += 1 }
    }
}

This idea was borrowed from this question although I changed it to a computed property. Thanks to @LeoDabus for the suggestion of extending Sequence instead of Array to pick up additional types.

Examples:

print("abacab".histogram)

["a": 3, "b": 2, "c": 1]

print("Hello World!".suffix(6).histogram)

["l": 1, "!": 1, "d": 1, "o": 1, "W": 1, "r": 1]

print([1,2,3,2,1].histogram)

[2: 2, 3: 1, 1: 2]

print([1,2,3,2,1,2,1,3,4,5].prefix(8).histogram)

[1: 3, 2: 3, 3: 2]

print(stride(from: 1, through: 10, by: 2).histogram)

[1: 1, 3: 1, 5: 1, 7: 1, 9: 1]

Tags:

Arrays

Ios

Nsarray

Swift

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