Get Subset of array based on the occurrence of elements

Question

I have an array like:

var arr = [4,1,5,5,3]

I want to fetch subset from the array based on the occurrence of elements in it.

For example:

Elements with frequency 1 is {4,1,3}
Elements with frequency 2 is {5,5}

I followed this StackOverflow question but unable to figure out how to do the above thing.

Is there any way I can do this?

Answer 1

You can use an NSCountedSet to get the count of all elements in arr, then you can build a Dictionary, where the keys will be the number of occurencies for the elements and the values will be Arrays of the elements with key number of occurences. By iterating through Set(arr) rather than simply arr to build the Dictionary, you can make sure that repeating elements are only added once to the Dictionary (so for instance with your original example, 5 wouldn't be added twice as having a frequency of 2).

For the printing, you just need to iterate through the keys of the Dictionary and print the keys along with their corresponding values. I just sorted the keys to make the printing go in ascending order of number of occurences.

let arr = [4,1,5,5,3,2,3,6,2,7,8,2,7,2,8,8,8,7]
let counts = NSCountedSet(array: arr)
var countDict = [Int:[Int]]()
for element in Set(arr) {
    countDict[counts.count(for: element), default: []].append(element)
}
countDict

for freq in countDict.keys.sorted() {
    print("Elements with frequency \(freq) are {\(countDict[freq]!)}")
}

Output:

Elements with frequency 1 are {[4, 6, 1]}
Elements with frequency 2 are {[5, 3]}
Elements with frequency 3 are {[7]}
Elements with frequency 4 are {[2, 8]}

Swift 3 version:

let arr = [4,1,5,5,3,2,3,6,2,7,8,2,7,2,8,8,8,7]
let counts = NSCountedSet(array: arr)
var countDict = [Int:[Int]]()
for element in Set(arr) {
    if countDict[counts.count(for: element)] != nil {
        countDict[counts.count(for: element)]!.append(element)  
    } else {
        countDict[counts.count(for: element)] = [element]
    }
}

for freq in countDict.keys.sorted() {
    print("Elements with frequency \(freq) are {\(countDict[freq]!)}")
}

Answer 2

You just need to get the occurrences of the elements and filter the elements that only occurs once or more than once as shown in this answer:

extension Array where Element: Hashable {
    // Swift 4 or later
    var occurrences: [Element: Int] {
        return reduce(into: [:]) { $0[$1, default: 0] += 1 }
    }
    // // for Swift 3 or earlier
    // var occurrences: [Element: Int] {
    //     var result: [Element: Int] = [:]
    //     forEach{ result[$0] = (result[$0] ?? 0) + 1}
    //     return result
    // }
    func frequencies(where isIncluded: (Int) -> Bool) -> Array {
        return filter{ isIncluded(occurrences[$0] ?? 0) }
    }
}

---

Playground Testing:

let arr = [5, 4, 1, 5, 5, 3, 5, 3]

let frequency1 = arr.frequencies {$0 == 1}       // [4, 1]
let frequency2 = arr.frequencies {$0 == 2}       // [3, 3]
let frequency3orMore = arr.frequencies {$0 >= 3} // [5, 5, 5, 5]

Answer 3

This is it:

func getSubset(of array: [Int], withFrequency frequency: Int) -> [Int]
{
    var counts: [Int: Int] = [:]

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

    let filtered  = counts.filter{ $0.value == frequency}

    return Array(filtered.keys)
}

This is pure Swift (not using good old Next Step classes) and is using ideas from the SO link you supplied.

The counts dictionary contains the frequencies (value) of each of the int-values (key) in your array: [int-value : frequency].