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Is there a way to get the index of the array in map or reduce in Swift? I'm looking for something like each_with_index in Ruby.

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") 

I would like to get rid of the odd variable above.

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5 Answers 5

475

You can use enumerate to convert a sequence (Array, String, etc.) to a sequence of tuples with an integer counter and and element paired together. That is:

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"] 

Link to enumerate definition

Note that this isn't the same as getting the index of the collection—enumerate gives you back an integer counter. This is the same as the index for an array, but on a string or dictionary won't be very useful. To get the actual index along with each element, you can use zip:

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

When using an enumerated sequence with reduce, you won't be able to separate the index and element in a tuple, since you already have the accumulating/current tuple in the method signature. Instead, you'll need to use .0 and .1 on the second parameter to your reduce closure:

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

Swift 3.0 and above

Since Swift 3.0 syntax is quite different.
Also, you can use short-syntax/inline to map array on dictionary:

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

That produces:

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

Don't want to steal the fun of finishing, so if you need it I put the modified Luhn check in a gist instead of in the answer: gist.github.com/natecook1000/1eb756d6b10297006137
In swift 2.0 you need to do: numbers.enumerate().map { (index, element) in ...
@CharlieMartin: You can use .reduce after enumerate() or zip.
In Swift 5 enumerate is now enumerated
it's actually called enumerated()
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11

For Swift 2.1 I wrote next function:

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

And then use it like this:

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

Comments

10

With Swift 3, when you have an object that conforms to Sequence protocol and you to want to link each element inside of it with its index, you can use enumerated() method.

For example:

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")] 

Therefore, in the simplest case, you can implement a Luhn algorithm in a Playground like this:

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 

If you start from a String, you can implement it like this:

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 

If you need to repeat those operations, you can refactor your code into an extension:

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 

Or, in a much concise way:

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 

Comments

2

In addition to Nate Cook's example of map, you can also apply this behavior to 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] 

Note that the EnumerateSequence passed into the closure as enumerated cannot be decomposed in a nested fashion, thus the members of the tuple must be decomposed inside the closure (ie. enumerated.index).

Comments

2

This is a working CollectionType extension for swift 2.1 using throws and rethrows:

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) } } 

I know this is not what you were asking, but solves your issue. You can try this swift 2.0 Luhn method without extending anything:

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 } 

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