有没有办法在Swift map或reduceSwift中获取数组的索引？我正在寻找类似each_with_indexRuby的东西。

func lunhCheck(number : String) -> Bool
{
    var odd = true;
    return reverse(number).map { String($0).toInt()! }.reduce(0) {
        odd = !odd
        return $0 + (odd ? ($1 == 9 ? 9 : ($1 * 2) % 9) : $1)
    }  % 10 == 0
}

lunhCheck("49927398716")
lunhCheck("49927398717")

我想摆脱上面的odd变量。

您可以使用enumerate将整数计数器和元素配对在一起的元组序列（Array，String等）转换为元组序列。那是：

let numbers = [7, 8, 9, 10]
let indexAndNum: [String] = numbers.enumerate().map { (index, element) in
    return "\(index): \(element)"
}
print(indexAndNum)
// ["0: 7", "1: 8", "2: 9", "3: 10"]

链接到enumerate定义

请注意，这与获取集合的索引不同，它enumerate为您提供了一个整数计数器。这与数组的索引相同，但是在字符串或字典上并不是很有用。要获得每个元素的实际索引，可以使用zip：

let actualIndexAndNum: [String] = zip(numbers.indices, numbers).map { "\($0): \($1)" }
print(actualIndexAndNum)
// ["0: 7", "1: 8", "2: 9", "3: 10"]

当使用带有的枚举序列时reduce，将无法在元组中分离索引和元素，因为方法签名中已经具有累加/当前元组。相反，您需要在闭包的第二个参数上使用.0和：.1reduce

let summedProducts = numbers.enumerate().reduce(0) { (accumulate, current) in
    return accumulate + current.0 * current.1
    //                          ^           ^
    //                        index      element
}
print(summedProducts)   // 56

Swift 3.0及更高版本

由于Swift 3.0的语法完全不同。
另外，您可以使用短语法/内联在字典上映射数组：

let numbers = [7, 8, 9, 10]
let array: [(Int, Int)] = numbers.enumerated().map { ($0, $1) }
//                                                     ^   ^
//                                                   index element

产生：

[(0, 7), (1, 8), (2, 9), (3, 10)]

因为Swift 2.1我写了下一个函数：

extension Array {

 public func mapWithIndex<T> (f: (Int, Element) -> T) -> [T] {     
     return zip((self.startIndex ..< self.endIndex), self).map(f)
   }
 }

然后像这样使用它：

    let numbers = [7, 8, 9, 10]
    let numbersWithIndex: [String] = numbers.mapWithIndex { (index, number) -> String in
        return "\(index): \(number)" 
    }
    print("Numbers: \(numbersWithIndex)")

使用Swift 3，当您有一个符合Sequence协议的对象并且想要将其内部的每个元素与其索引链接时，可以使用enumerated()method。

例如：

let array = [1, 18, 32, 7]
let enumerateSequence = array.enumerated() // type: EnumerateSequence<[Int]>
let newArray = Array(enumerateSequence)
print(newArray) // prints: [(0, 1), (1, 18), (2, 32), (3, 7)]

let reverseRandomAccessCollection = [1, 18, 32, 7].reversed()
let enumerateSequence = reverseRandomAccessCollection.enumerated() // type: EnumerateSequence<ReverseRandomAccessCollection<[Int]>>
let newArray = Array(enumerateSequence)
print(newArray) // prints: [(0, 7), (1, 32), (2, 18), (3, 1)]

let reverseCollection = "8763".characters.reversed()
let enumerateSequence = reverseCollection.enumerated() // type: EnumerateSequence<ReverseCollection<String.CharacterView>>
let newArray = enumerateSequence.map { ($0.0 + 1, String($0.1) + "A") }
print(newArray) // prints: [(1, "3A"), (2, "6A"), (3, "7A"), (4, "8A")]

因此，在最简单的情况下，您可以在Playground中实现Luhn算法，如下所示：

let array = [8, 7, 6, 3]
let reversedArray = array.reversed()
let enumerateSequence = reversedArray.enumerated()

let luhnClosure = { (sum: Int, tuple: (index: Int, value: Int)) -> Int in
    let indexIsOdd = tuple.index % 2 == 1
    guard indexIsOdd else { return sum + tuple.value }
    let newValue = tuple.value == 9 ? 9 : tuple.value * 2 % 9
    return sum + newValue
}

let sum = enumerateSequence.reduce(0, luhnClosure)
let bool = sum % 10 == 0
print(bool) // prints: true

如果从开始String，则可以这样实现：

let characterView = "8763".characters
let mappedArray = characterView.flatMap { Int(String($0)) }
let reversedArray = mappedArray.reversed()
let enumerateSequence = reversedArray.enumerated()

let luhnClosure = { (sum: Int, tuple: (index: Int, value: Int)) -> Int in
    let indexIsOdd = tuple.index % 2 == 1
    guard indexIsOdd else { return sum + tuple.value }
    let newValue = tuple.value == 9 ? 9 : tuple.value * 2 % 9
    return sum + newValue
}

let sum = enumerateSequence.reduce(0, luhnClosure)
let bool = sum % 10 == 0
print(bool) // prints: true

如果需要重复这些操作，则可以将代码重构为扩展名：

extension String {

    func luhnCheck() -> Bool {
        let characterView = self.characters
        let mappedArray = characterView.flatMap { Int(String($0)) }
        let reversedArray = mappedArray.reversed()
        let enumerateSequence = reversedArray.enumerated()

        let luhnClosure = { (sum: Int, tuple: (index: Int, value: Int)) -> Int in
            let indexIsOdd = tuple.index % 2 == 1
            guard indexIsOdd else { return sum + tuple.value }
            let newValue = tuple.value == 9 ? 9 : tuple.value * 2 % 9
            return sum + newValue
        }

        let sum = enumerateSequence.reduce(0, luhnClosure)
        return sum % 10 == 0
    }

}

let string = "8763"
let luhnBool = string.luhnCheck()
print(luhnBool) // prints: true

或者，以一种非常简洁的方式：

extension String {

    func luhnCheck() -> Bool {
        let sum = characters
            .flatMap { Int(String($0)) }
            .reversed()
            .enumerated()
            .reduce(0) {
                let indexIsOdd = $1.0 % 2 == 1
                guard indexIsOdd else { return $0 + $1.1 }
                return $0 + ($1.1 == 9 ? 9 : $1.1 * 2 % 9)
        }
        return sum % 10 == 0
    }

}

let string = "8763"
let luhnBool = string.luhnCheck()
print(luhnBool) // prints: true

除了Nate Cook的示例之外map，您还可以将此行为应用于reduce。

let numbers = [1,2,3,4,5]
let indexedNumbers = reduce(numbers, [:]) { (memo, enumerated) -> [Int: Int] in
    return memo[enumerated.index] = enumerated.element
}
// [0: 1, 1: 2, 2: 3, 3: 4, 4: 5]

请注意，EnumerateSequence传递到闭包中的as enumerated不能以嵌套方式分解，因此，元组的成员必须在闭包内进行分解（即enumerated.index）。

这是使用throw和rethrows的swift 2.1 的有效CollectionType扩展：

extension CollectionType {

    func map<T>(@noescape transform: (Self.Index, Self.Generator.Element) throws -> T) rethrows -> [T] {
        return try zip((self.startIndex ..< self.endIndex), self).map(transform)
    }

}

我知道这不是您要的，但可以解决您的问题。您可以尝试这种快速的2.0 Luhn方法，而无需扩展任何内容：

func luhn(string: String) -> Bool {
    var sum = 0
    for (idx, value) in string.characters.reverse().map( { Int(String($0))! }).enumerate() {
        sum += ((idx % 2 == 1) ? (value == 9 ? 9 : (value * 2) % 9) : value)
    }
    return sum > 0 ? sum % 10 == 0 : false
}