我有两种结构，Header 和Session，它们都符合协议(protocol)TimelineItem。
我有一个 Array 由 TimelineItem 组成，如下所示:
[Header1, SessionA, SessionB, Header2, SessionC, SessionD]

我的需要是将 Session 分组到相关的 Header 下，如下所示:
[ [Header1: [SessionA, SessionB], [Header2: [SessionC, SessionD] ]

我尝试使用filter 方法只检索Header 结构，并使用split 方法检索Session 数组> 数组。这些工作正常，但我无法管理如何协调两者以构建我的最终 [[Header: [Session]]] 对象。

这是我的示例代码:

enum TimelineItemType: String {  
  case Header = "header"   
  case Session = "session"
}

protocol TimelineItem {
  var id: Int { get }
  var type: TimelineItemType { get }
  var startDate: NSDate { get }
}

Header 结构

struct Header: TimelineItem, Decodable, Hashable, Equatable {
  let id: Int
  let type: TimelineItemType = .Header
  let startDate: NSDate
  let text: String

  init?(json: JSON) {
    guard let id: Int = "id" <~~ json,
      let type: TimelineItemType = "type" <~~ json,
      let startDate: NSDate = "startDate" <~~ json,
      let text: String = "text" <~~ json where type == .Header else {
        return nil
    }
    self.id = id
    self.startDate = startDate
    self.text = text
  }

  var hashValue: Int {
    // As id is unique, we can use it for hash purpose
    return id
  }
}

Session 结构

struct Session: TimelineItem, Decodable, Equatable {
  let id: Int
  let type: TimelineItemType = .Session
  let startDate: NSDate
  let name: String
  let syllabus: String
  let speaker: Speaker
  let language: String
  let room: String
  let duration: Int

  init?(json: JSON) {
    guard let id: Int = "id" <~~ json,
      let type: TimelineItemType = "type" <~~ json,
      let startDate: NSDate = Decoder.decodeDateISO8601("startDate")(json),
      let name: String = "name" <~~ json,
      let speaker: Speaker = "speaker" <~~ json,
      let syllabus: String = "syllabus" <~~ json,
      let language: String = "language" <~~ json,
      let room: String = "room" <~~ json,
      let duration: Int = "duration" <~~ json where type == .Session else {
        return nil
    }
    self.id = id
    self.startDate = startDate
    self.name = name
    self.speaker = speaker
    self.syllabus = syllabus
    self.language = language
    self.room = room
    self.duration = duration
  }
}

最后是我尝试拆分数组的代码:

func timelineFromItems(timelineItems: [TimelineItem]) -> [[Header: [Session]]]? {

    let slicedSessions = timelineItems.split { $0 is Header }
    let sessions = Array(slicedSessions)
    let headers = timelineItems.filter { $0.type == .Header }
    var timeline = [[Header: [Session]]]()
    // HOW TO FILL THE TIMELINE ??
}

如何填写时间表？

最佳答案

我通过删除所有不必要的信息来减少这个例子

protocol P {}
struct A: P, Hashable {
    var i:Int
    var hashValue: Int { return i }
}
func ==(lhs: A, rhs: A)->Bool {
    return lhs.i == rhs.i
}
struct B: P {
    var i:Int
}

// your current data
let arr:[P] = [A(i: 1),B(i: 1),B(i: 2), A(i: 2), B(i: 3), B(i: 4), B(i: 5)]

将数据转换为所需格式的函数

func foo(arr: [P])->[[A:[B]]]? {

    var dict:[A:[B]] = [:]
    var arrb:[B] = []
    let arrk:[A] = arr.filter { $0 is A }.map { $0 as! A }
    guard var key = arr[0] as? A else { return nil }

    arr.forEach { (p) in

        if let a = p as? A {
            dict[key] = arrb
            arrb = []
            key = a
        }
        if let b = p as? B {
            arrb.append(b)
        }
    }
    dict[key] = arrb
    var arrr:[[A:[B]]] = []
    arrk.forEach { (a) in
        if let arrb = dict[a] {
            arrr.append([a:arrb])
        }
    }
    return arrr
}

现在生成的数组符合您的要求(我希望 :-))

if let result = foo(arr) {
    print(result) // [[A(i: 1): [B(i: 1), B(i: 2)]], [A(i: 2): [B(i: 3), B(i: 4), B(i: 5)]]]
}

另一个测试数据

let arr:[P] = [A(i: 1),B(i: 1),B(i: 2), A(i: 2), A(i: 3), B(i: 3)]

给你

[[A(i: 1): [B(i: 1), B(i: 2)]], [A(i: 2): []], [A(i: 3): [B(i: 3)]]]

所以，即使 A 后面没有 B，它也能工作

关于ios - Swift:将 [A, B, B, B, A, B, B, B] 数组转换为哈希数组 [ [A: [B, B, B], [A: [B, B, B] ] ]，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35474735/