Map or abbreviation using an index in Swift

Question

Is there a way to get the index of an array in map or reduce in Swift? I am 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 .

Answer 1

You can use enumerate to convert a sequence ( Array , String , etc.) into a sequence of tuples using an integer counter, and the element together. I.e:

 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 is not the same as getting the collection index - enumerate returns you an integer counter. This is the same as the index for the array, but a string or dictionary will not 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 cannot separate the index and the element in the tuple, since you already have the cumulative / current tuple in the method signature. Instead, you will need to use .0 and .1 for the second parameter to close reduce :

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

Swift 3.0

Since the syntax of Swift 3.0 is completely different.
Alternatively, you can use short-syntax / inline to display an array in a dictionary:

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

It produces:

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

Answer 2

For Swift 2.1 I wrote the following 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)") 

Answer 3

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

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")] 

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Therefore, in the simplest case, you can implement the Moon algorithm on a playground, for example:

 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 

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If you start with 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 

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If you need to repeat these operations, you can reorganize 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 short form:

 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 

Answer 4

In addition to the Nate Cook map example, 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 an EnumerateSequence passed to the closure as enumerated cannot be expanded in a nested manner, so tuple members must be expanded inside the closure (i.e. enumerated.index ).

Answer 5

This is a working CollectionType extension for quick 2.1 using throws and reps:

 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 that this is not what you asked for, but it solves your problem. You can try this quick moon 2.0 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 }