sort()Swift 針對合乎Comparable協定的陣列提供排序能力，相關函式宣告如下：
func sorted(isOrderedBefore: (T, T) -> Bool) -> Array<T>
mutating func sort(isOrderedBefore: (T, T) -> Bool)
sorted()回傳排序陣列結果，不影響原陣列內容。sort()回傳排序陣列結果，並將結果存回原陣列。var numbers = [2, 4, 6, 8, 1, 3, 5, 7]
numbers.sorted(<) //return: [1, 2, 3, 4, 5, 6, 7, 8]
numbers //return: [2, 4, 6, 8, 1, 3, 5, 7]
numbers.sort(>)
numbers //return: [8, 7, 6, 5, 4, 3, 2, 1]
只要class與struct繼承並實作Comparable協定，就能透過<或>決定排序結果為升冪或降冪。
下例示範依照各點與原點的距離做升冪排序。
struct Point {
var x: Int
var y: Int
}
extension Point: Comparable {}
func ==(a: Point, b: Point) -> Bool {
return (a.x*a.x + a.y*a.y) == (b.x*b.x + b.y*b.y)
}
func <(a: Point, b: Point) -> Bool {
return (a.x*a.x + a.y*a.y) < (b.x*b.x + b.y*b.y)
}
var p1 = Point(x: 1, y: 5)
var p2 = Point(x: 3, y: 3)
var p3 = Point(x: -4, y: 2)
var points = [p1, p2, p3] //return: [{x 1, y 5}, {x 3, y 3}, {x -4, y 2}]
points.sort(<) //return: [{x 3, y 3}, {x -4, y 2}, {x 1, y 5}]
filter()filter() 宣告如下：
func filter(includeElement: (T) -> Bool) -> Array<T>
includeElement 表示傳入的 function 或 closure，用來判斷陣列個元素是否符合條件。filter() 回傳結果為陣列。以下範例展示透過在 closure 回傳每個元素使否符合判斷式的結果，決定最後 filter() 回傳的元素集合。
var numbers = Array(1...8)
var evenNumbers = numbers.filter { (x) -> Bool in
x % 2 == 0
}
var oddNumbers = numbers.filter { (x) -> Bool in
x % 2 != 0
}
numbers //return: [1, 2, 3, 4, 5, 6, 7, 8]
evenNumbers //return: [2, 4, 6, 8]
oddNumbers //return: [1, 3, 5, 7]
下例為將一、三象限的點過濾出來的範例。
struct Point {
var x: Int
var y: Int
}
var p1 = Point(x: 1, y: 1)
var p2 = Point(x: -1, y: 1)
var p3 = Point(x: -1, y: -1)
var p4 = Point(x: 1, y: -1)
var points = [p1, p2, p3, p4]
var quadrant_1 = points.filter { (p) -> Bool in
return p.x > 0 && p.y > 0
}
var quadrant_3 = points.filter { (p) -> Bool in
return p.x < 0 && p.y < 0
}
points //return: [{x 1, y 1}, {x -1, y 1}, {x -1, y -1}, {x 1, y -1}]
quadrant_1 //return: [{x 1, y 1}]
quadrant_3 //return: [{x -1, y -1}]
map()map() 宣告如下：
func map<U>(transform: (T) -> U) -> Array<U>
transform 表示傳入的 function 或 closure，用來轉換每個陣列的元素。輸入參數與回傳結果型別可以不同。map() 回傳結果為陣列。以下示範如何透過map()將整數陣列轉化成兩倍數值與字串的陣列。
var numbers = Array(1...8)
var doubles = numbers.map { (n) -> Int in
return n*2
}
var strings = numbers.map { (n) -> String in
let digital = [0:"零", 1:"壹", 2:"貳", 3:"参", 4:"肆", 5:"伍", 6:"陸", 7:"柒", 8:"捌", 9:"玖"]
return digital[n] ?? "啥"
}
numbers //return: [1, 2, 3, 4, 5, 6, 7, 8]
doubles //return: [2, 4, 6, 8, 10, 12, 14, 16]
strings //return: ["壹", "貳", "参", "肆", "伍", "陸", "柒", "捌"]
接著示範map()如何將一組二度空間的點對 X 軸與 Y 軸做鏡射。
struct Point {
var x: Int
var y: Int
}
var points = [Point(x: 1, y: 2), Point(x: -2, y: -1)]
var mirror_x = points.map { (p) -> Point in
return Point(x: -p.x, y: p.y)
}
var mirror_y = points.map { (p) -> Point in
return Point(x: p.x, y: -p.y)
}
points //return: [{x 1, y 2}, {x -2, y -1}]
mirror_x //return: [{x -1, y 2}, {x 2, y -1}]
mirror_y //return: [{x 1, y -2}, {x -2, y 1}]
reduce()reduce()宣告如下：
func reduce<U>(initial: U, combine: @noescape (U, T) -> U) -> U
reduce()處理一組陣列，得到一個回傳值作為結果。initial表示combine處理函式的初始值。combine表示傳入的 function 或 closure，用來將上次遞迴的結果U與元素T結合，產生新的結果U將作為下次遞迴函式的輸入值。以下示範透過reduce()將一組整數做總和，total1作法中規中矩，total2精簡語法拿掉 closure 型別宣告。
var numbers = Array(1...10)
var total1 = numbers.reduce(0, combine: { (sum, num) -> Int in //return: 55
return sum + num
})
var total2 = numbers.reduce(0) { $0 + $1 } //return: 55
接著示範reduce()如何得到一組二度空間的點對原點的距離和，total1作法中規中矩，total2精簡語法拿掉 closure 型別宣告。
import UIKit
struct Point {
var x: Double
var y: Double
}
var points = [Point(x: 1, y: 5), Point(x: 3, y: 3), Point(x: -4, y: 2)]
var total1 = points.reduce(0.0, combine: { (total, point) -> Double in //return: 13.81379615571165
return total + sqrt(point.x*point.x + point.y*point.y)
})
var total2 = points.reduce(0.0) { $0 + sqrt($1.x*$1.x + $1.y*$1.y) } //return: 13.81379615571165
sqrt()為平方根運算 (square root)。
sum_1：將filter、map、reduce結果存成變數再傳遞給另一個函式處理。sum_2：透過 function chaining 技巧，將filter、map、reduce結果直接傳遞給另一個函式處理。雖然運算式變長，但依舊維持可讀性。var numbers = Array(1...10)
var odds = numbers.filter { $0 % 2 != 0 } //return: [1, 3, 5, 7, 9]
var triple = odds.map { $0 * 3 } //return: [3, 9, 15, 21, 27]
var sum_1 = triple.reduce(0) { $0 + $1 } //return: 75
var sum_2 = numbers.filter{ $0 % 2 != 0 }.map{ $0 * 3 }.reduce(0){ $0 + $1 } //return: 75