Question

我有一组Person个对象：

class Person {
   let name:String
   let position:Int
}

，数组是：

let myArray = [p1,p1,p3]

我想将myArray映射为[position:name]经典解决方案的词典：

var myDictionary =  [Int:String]()

for person in myArray {
    myDictionary[person.position] = person.name
}

使用功能方法map，flatMap ......或其他现代Swift风格，Swift有没有优雅的方式来做到这一点

Answer 1

在Swift 4中，您可以使用into reduce版本struct Person { let name: String let position: Int } let myArray = [Person(name:"h", position: 0), Person(name:"b", position:4), Person(name:"c", position:2)] let myDict = myArray.reduce(into: [Int: String]()) { $0[$1.position] = $1.name } print(myDict)更清晰有效地执行@ Tj3n的方法。它删除了临时字典和返回值，因此它更快更容易阅读。

{{1}}

注意：在Swift 3中不起作用。

Answer 2

好的map不是一个很好的例子，因为它与循环相同，你可以使用reduce来代替每个对象组合并变成单个值：

let myDictionary = myArray.reduce([Int: String]()) { (dict, person) -> [Int: String] in
    var dict = dict
    dict[person.position] = person.name
    return dict
}

//[2: "b", 3: "c", 1: "a"]

Answer 3

由于 Swift 4 ，您可以轻松完成此操作。有two new个初始值设定项可以从一系列元组（键和值对）构建字典。如果密钥保证是唯一的，您可以执行以下操作：

let persons = [Person(name: "Franz", position: 1),
               Person(name: "Heinz", position: 2),
               Person(name: "Hans", position: 3)]

Dictionary(uniqueKeysWithValues: persons.map { ($0.position, $0.name) })

=＆GT; [1: "Franz", 2: "Heinz", 3: "Hans"]

如果任何密钥重复，则会因运行时错误而失败。在这种情况下，您可以使用此版本：

let persons = [Person(name: "Franz", position: 1),
               Person(name: "Heinz", position: 2),
               Person(name: "Hans", position: 1)]

Dictionary(persons.map { ($0.position, $0.name) }) { _, last in last }

=＆GT; [1: "Hans", 2: "Heinz"]

这表现为你的for循环。如果你不想＆＃34;覆盖＆＃34;值并坚持第一个映射，你可以使用这个：

Dictionary(persons.map { ($0.position, $0.name) }) { first, _ in first }

=＆GT; [1: "Franz", 2: "Heinz"]

Swift 4.2 添加了一个third初始值设定项，用于将序列元素分组到字典中。字典键由闭包派生。具有相同键的元素按照与序列中相同的顺序放入数组中。这使您可以获得与上面类似的结果。例如：

Dictionary(grouping: persons, by: { $0.position }).mapValues { $0.last! }

=＆GT; [1: Person(name: "Hans", position: 1), 2: Person(name: "Heinz", position: 2)]

Dictionary(grouping: persons, by: { $0.position }).mapValues { $0.first! }

=＆GT; [1: Person(name: "Franz", position: 1), 2: Person(name: "Heinz", position: 2)]

Answer 4

您可以为Dictionary类型编写自定义初始值设定项，例如来自元组：

extension Dictionary {
    public init(keyValuePairs: [(Key, Value)]) {
        self.init()
        for pair in keyValuePairs {
            self[pair.0] = pair.1
        }
    }
}

然后使用map作为Person的数组：

var myDictionary = Dictionary(keyValuePairs: myArray.map{($0.position, $0.name)})

Answer 5

您可以使用reduce功能。首先，我为Person类

创建了一个指定的初始值设定项

class Person {
  var name:String
  var position:Int

  init(_ n: String,_ p: Int) {
    name = n
    position = p
  }
}

后来，我初始化了一个值数组

let myArray = [Person("Bill",1), 
               Person("Steve", 2), 
               Person("Woz", 3)]

最后，字典变量的结果如下：

let dictionary = myArray.reduce([Int: Person]()){
  (total, person) in
  var totalMutable = total
  totalMutable.updateValue(person, forKey: total.count)
  return totalMutable
}

Answer 6

也许是这样吗？

myArray.forEach({ myDictionary[$0.position] = $0.name })

Answer 7

基于KeyPath的解决方案如何？

extension Array {

    func dictionary<Key, Value>(withKey key: KeyPath<Element, Key>, value: KeyPath<Element, Value>) -> [Key: Value] {
        return reduce(into: [:]) { dictionary, element in
            let key = element[keyPath: key]
            let value = element[keyPath: value]
            dictionary[key] = value
        }
    }
}

这是您将如何使用它：

struct HTTPHeader {

    let field: String, value: String
}

let headers = [
    HTTPHeader(field: "Accept", value: "application/json"),
    HTTPHeader(field: "User-Agent", value: "Safari"),
]

let allHTTPHeaderFields = headers.dictionary(withKey: \.field, value: \.value)

// allHTTPHeaderFields == ["Accept": "application/json", "User-Agent": "Safari"]

Answer 8

这就是我一直在使用的

struct Person {
    let name:String
    let position:Int
}
let persons = [Person(name: "Franz", position: 1),
               Person(name: "Heinz", position: 2),
               Person(name: "Hans", position: 3)]

var peopleByPosition = [Int: Person]()
persons.forEach{peopleByPosition[$0.position] = $0}

如果有办法合并最后两行，那么peopleByPosition可以是let，那就太好了。

我们可以对Array做一个扩展来做到这一点！

extension Array {
    func mapToDict<T>(by block: (Element) -> T ) -> [T: Element] where T: Hashable {
        var map = [T: Element]()
        self.forEach{ map[block($0)] = $0 }
        return map
    }
}

然后我们就可以了

let peopleByPosition = persons.mapToDict(by: {$0.position})

Answer 9

extension Array {
    func mapToDict<T>(by block: (Element) -> T ) -> [T: Element] where T: Hashable {
        var map = [T: Element]()
        self.forEach{ map[block($0)] = $0 }
        return map
    }
}

将对象数组映射到Swift中的Dictionary

9 个答案: