Swift - Convert Array to Dictionary - arrays

Question

I just want convert an array of Player Names into a dictionary Scoreboard, giving everyone an initial score of 0.
Meaning...
var playerNames = ["Harry", "Ron", "Hermione"]
becomes...
var scoreBoard: [String:Int] = [ "Ron":0, "Harry":0, "Hermione":0 ]
This is my first time asking a question, but I’m totally completely stuck on what feels so simple and all how-to's/questions I've found are off in some way. I have tried using reduce in a variety of ways, but always end up short. Thanks in advance!

Answer 1

Here's a quick one liner that I like to use:
let scoreboard = playerNames.reduce(into: [String: Int]()) { $0[$1] = 0 }

Answer 2

reduce is definitely one of the more difficult builtin functions to use correctly, but it is what you want here.
let names = ["Harry", "Ron", "Hermione"]
let scoreboard: [String: Int] = names.reduce(into: [:], { result, next in
result[next] = 0
})
It takes 2 parameters: the initial value (in our case, an empty dictionary [:]), and a closure that updates the result with each element in the array. This closure has 2 parameters, result and next. We want to update result based on the next element. Our closure does this by setting result[next] to 0.

Answer 3

If the player names are known to be all different then you can do
let playerNames = ["Harry", "Ron", "Hermione", "Ron"]
var scoreBoard = Dictionary(uniqueKeysWithValues: zip(playerNames,
repeatElement(0, count: playerNames.count)))
print(scoreBoard) // ["Harry": 0, "Ron": 0, "Hermione": 0]
Here zip is used to create a sequence of player/score pairs, from which the dictionary is created.
Remark: Originally I had used AnySequence { 0 } to generate the zeros. Using repeatElement() instead was suggested by Alexander and has the advantage that the correct required capacity is passed to the dictionary intializer.

Answer 4

You can use reduce(into:) as you suspected. You simply need to declare the initial value as [String:Int]() to be an empty Dictionary of the required type, then simply set the value of all keys in playerNames to 0.
var playerNames = ["Harry", "Ron", "Hermione"]
var scoreBoard = playerNames.reduce(into: [String:Int](), { currentScores,player in
currentScores[player] = 0
})

Answer 5

Using reduce(into:_:):
var playerNames = ["Harry", "Ron", "Hermione"]
let playerScore = playerNames.reduce(into: [:]) { counts, playerNames in
counts[playerNames, default: 0] += 0
}
print(playerScore)
To keep a count of the players names (eg. duplicate names):
counts[myArray, default: 0] += 1
So for example if Ron had two entries before the game started (score > 0) then you would know.
Without using reduce(into:_:) method and as an extension:
var playerNames = ["Harry", "Ron", "Hermione"]
extension Sequence where Self.Iterator.Element: Hashable {
func freq() -> [Self.Iterator.Element: Int] {
return reduce([:]) {
( iter: [Self.Iterator.Element: Int], element) in
var dict = iter
dict[element] = 0
return dict
}
}
}
print(playerNames.freq())
// ["Harry": 0, "Hermione": 0, "Ron": 0]
keep a count (eg. duplicate names):
dict[element, default: -1 ] += 1

Answer 6

Here is how you can do that:
var playerNames = ["Harry", "Ron", "Hermione"]
var dictionary = [String: Int]()
for player in playerNames {
dictionary[player] = 0
}

Answer 7

Here is another way to do it:
// Implementation
extension Dictionary {
static func from(_ array: [Value], key: KeyPath<Value, Key>) -> Dictionary<Key, Value> {
var dict: Dictionary<Key, Value> = [:]
array.forEach { dict[$0[keyPath: key]] = $0}
return dict
}
}
/// Usage
let array: [String] = ["Banana", "Apple"]
Dictionary.from(array, key: \.self)
// or if you have a more complex object
struct Foo {
let id: Int
}
let array2: [Foo] = [Foo(id: 1), Foo(id: 2)]
Dictionary.from(array2, key: \.id)

Answer 8

Based on jmad8 answer
Details
Swift 5.3
Xcode 12.0.1 (12A7300)
Solution
extension Sequence {
func toDictionary<Key: Hashable, Value>(where closure: (Element) -> (Key, Value)) -> [Key: Value] {
reduce(into: [Key: Value]()) { (result, element) in
let components = closure(element)
result[components.0] = components.1
}
}
func toCompactDictionary<Key: Hashable, Value>(where closure: (Element) -> ((Key, Value)?)) -> [Key: Value] {
reduce(into: [Key: Value]()) { (result, element) in
guard let components = closure(element) else { return }
result[components.0] = components.1
}
}
}
Usage
// Sample 1
print(languages.toDictionary { (string) -> (Character, String) in
return (string.first!, string)
})
print(languages.toCompactDictionary { (string) -> (Character, String)? in
guard let character = string.first, character != Character("J") else { return nil }
return (character, string)
})
// Sample 2
print(languages.enumerated().toDictionary { (data) -> (Int, String) in
return (data.offset, data.element)
})
// Shorter version of sample 2
print(languages.enumerated().toDictionary { ($0.offset, $0.element) })
// Sample 3
struct Order {
let id: Int
let desctiption: String
}
let orders = [
Order(id: 0, desctiption: "Apple"),
Order(id: 1, desctiption: "Banana"),
Order(id: 2, desctiption: "watermelon")
]
print(orders.toDictionary { ($0.id, $0) })