How to return first 5 objects of Array in Swift?

Question

In Swift, is there a clever way of using the higher order methods on Array to return the 5 first objects? The obj-c way of doing it was saving an index, and for-loop throug

Answer 1

With Swift 5, according to your needs, you may choose one of the 6 following Playground codes in order to solve your problem.

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#1. Using subscript(_:) subscript

let array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
let arraySlice = array[..<5]
//let arraySlice = array[0..<5] // also works
//let arraySlice = array[0...4] // also works
//let arraySlice = array[...4] // also works
let newArray = Array(arraySlice)
print(newArray) // prints: ["A", "B", "C", "D", "E"]

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#2. Using prefix(_:) method

Complexity: O(1) if the collection conforms to RandomAccessCollection; otherwise, O(k), where k is the number of elements to select from the beginning of the collection.

let array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
let arraySlice = array.prefix(5)
let newArray = Array(arraySlice)
print(newArray) // prints: ["A", "B", "C", "D", "E"]

Apple states for prefix(_:):

If the maximum length exceeds the number of elements in the collection, the result contains all the elements in the collection.

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#3. Using prefix(upTo:) method

Complexity: O(1)

let array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
let arraySlice = array.prefix(upTo: 5)
let newArray = Array(arraySlice)
print(newArray) // prints: ["A", "B", "C", "D", "E"]

Apple states for prefix(upTo:):

Using the prefix(upTo:) method is equivalent to using a partial half-open range as the collection's subscript. The subscript notation is preferred over prefix(upTo:).

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#4. Using prefix(through:) method

let array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
let arraySlice = array.prefix(through: 4)
let newArray = Array(arraySlice)
print(newArray) // prints: ["A", "B", "C", "D", "E"]

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#5. Using removeSubrange(_:) method

Complexity: O(n), where n is the length of the collection.

var array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
array.removeSubrange(5...)
print(array) // prints: ["A", "B", "C", "D", "E"]

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#6. Using dropLast(_:) method

Complexity: O(1) if the collection conforms to RandomAccessCollection; otherwise, O(k), where k is the number of elements to drop.

let array = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L"]
let distance = array.distance(from: 5, to: array.endIndex)
let arraySlice = array.dropLast(distance)
let newArray = Array(arraySlice)
print(newArray) // prints: ["A", "B", "C", "D", "E"]

Answer 2

Swift 4

To get the first N elements of a Swift array you can use prefix(_ maxLength: Int):

Array(largeArray.prefix(5))

Answer 3

Update for swift 4:

[0,1,2,3,4,5].enumerated().compactMap{ $0 < 10000 ? $1 : nil }

For swift 3:

[0,1,2,3,4,5].enumerated().flatMap{ $0 < 10000 ? $1 : nil }

Answer 4

Plain & Simple

extension Array {
    func first(elementCount: Int) -> Array {
          let min = Swift.min(elementCount, count)
          return Array(self[0..<min])
    }
}

Answer 5

let a: [Int] = [0, 0, 1, 1, 2, 2, 3, 3, 4]
let b: [Int] = Array(a.prefix(5))
// result is [0, 0, 1, 1, 2]

Answer 6

Swift 4 with saving array types

extension Array {
    func take(_ elementsCount: Int) -> [Element] {
        let min = Swift.min(elementsCount, count)
        return Array(self[0..<min])
    }
}