How do you count the amount of unique items in an Array?
Example:
let array:Array<Int> = [1,3,2,4,6,1,3,2]
Count function:
array.count will give 8
but I want to count unique items and this will give 5
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6 Answers
As of Swift 1.2, Swift has a native Set type. Use the Set constructor to create a set from your array, and then the count property will tell you how many unique items you have:
let array = [1,3,2,4,6,1,3,2]
let set = Set(array)
print(set.count) // prints "5"
For Swift 1.1 and earlier:
Turn your array into an NSSet:
let array = [1,3,2,4,6,1,3,2]
let set = NSSet(array: array)
println(set.count) // prints "5"
You can read more about it here.
If you are interested in how many of each item you have, you can use a dictionary to count the items:
var counts = [Int:Int]()
for item in array {
counts[item] = (counts[item] ?? 0) + 1
}
print(counts) // prints "[6: 1, 2: 2, 3: 2, 1: 2, 4: 1]"
print(counts.count) // prints "5"
print("There are \(counts[1] ?? 0) ones.") // prints "There are 2 ones."
print("There are \(counts[7] ?? 0) sevens.") // prints "There are 0 sevens."
Comments
implement the function countDistinct(numbers: [Int]) to return the number of distinct elements in the array. NSSet documentation for Swift https://developer.apple.com/documentation/foundation/nsset
func countDistinct(numbers: [Int]) -> Int {
let array:Array<Int> = numbers
let count = NSSet(array: array).count
return count
}
print(countDistinct(numbers: [20, 10, 10, 30, 20]))
Comments
If you prefer to stick with pure swift, a possible solution consists of:
- sorting the array
- traversing the and count the number of times an element differs from the previous one
Translated in code:
let start: (Int, Int?) = (0, nil)
let count = array.sorted(<).reduce(start) { initial, value in
(initial.0 + (initial.1 == value ? 0 : 1), value)
}
let uniqueElements = count.0
the result is stored in the element 0 of the count tuple.
Explanation: the start tuple is initialized with 0 and nil, and passed as the initial value to the reduce method, called on a sorted copy of the array.
At each iteration, a new tuple is returned, containing the current array element and the current counter, increased by one if the current element is different than the previous one.
Comments
You can also use following generic method for counting unique values inside an array.
func countUniques<T: Comparable>(_ array: Array<T>) -> Int {
let sorted = array.sorted()
let initial: (T?, Int) = (.none, 0)
let reduced = sorted.reduce(initial) {
($1, $0.0 == $1 ? $0.1 : $0.1 + 1)
}
return reduced.1
}
Comments
let numbersInArray = [1, 2, 3, 4, 5, 1, 2, 3]
let uniqueNumbers = Array(Set(numbersInArray))
print(uniqueNumbers.count)
2 Comments
Jeremy Caney
Thank you for contributing to the Stack Overflow community. This may be a correct answer, but it’d be really useful to provide additional explanation of your code so developers can understand your reasoning. This is especially useful for new developers who aren’t as familiar with the syntax or struggling to understand the concepts. Would you kindly edit your answer to include additional details for the benefit of the community?
J. Martin
This is a great answer. Use Set() to ensure that each element only appears once in the collection, then put that result into a new array and simply count it.