Question

我有2种结构，Header和Session，都符合协议TimelineItem。
我有一个Array由TimelineItem组成，如下所示：
[Header1, SessionA, SessionB, Header2, SessionC, SessionD]

我需要将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方法仅检索Header结构，并使用split方法检索Session数组的数组。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。

以下是我的示例代码：

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
  }
}

最后我尝试拆分我的数组代码：

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 ??
}

如何填充时间表？

Answer 1

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)]

将您的数据转换为所需格式的功能

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，即使没有B跟随A

，它仍然有效

Answer 2

我没有看到填写时间表的任何障碍，以下是我将如何做到这一点：

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]]]()

    if sessions.count == headers.count { // Check to be sure you have as much header as sessions
        for (index, value) in headers.enumerate() {
            let dictionary = [value: sessions.getElement(index)]
            timeline.append(dictionary)
        }
    }

    return timeline
}

Answer 3

你实际上可以选择以下简单的东西：

func deserialize(input: [TimelineItem]) -> [Header: [Session]] {
    var result = [Header: [Session]]()

    var latestHeader: Header! = nil

    input.forEach() {
        if let header = $0 as? Header {
            latestHeader = header
            result[header] = []
        } else if let session = $0 as? Session {
            result[latestHeader]!.append(session)
        }
    }

    return result
}

<强> UPD

不是那么简单，但仍然简洁（上面生成以[Session]为元素的字典，而@ user3441734指出你需要一个单元素字典数组）：

func deserialize(input: [TimelineItem]) -> [[Header: [Session]]] {
    var result = [[Header: [Session]]]()

    var latestHeader: Header? = nil

    input.forEach() {
        if let header = $0 as? Header {
            latestHeader = header
            result.append([header: []])
        } else if let session = $0 as? Session, let header = latestHeader {
            var headerDict = result.popLast()!
            headerDict[header]!.append(session)
            result.append(headerDict)
        }
    }

    return result
}

Answer 4

感谢大家，我设法满足了我对以下代码的需求：

```

typealias Section = [header: Header, sessions: [TimelineItem]]

private static func groupBySection(timelineItems: [TimelineItem]) -> [Section]? {
    let slicedSessions = timelineItems.split { $0.type == .Header }
    let headers = timelineItems.filter { $0.type == .Header }
    var timeline = [Section]()

    for (index, value) in headers.enumerate() {
      guard index < slicedSessions.count else {
        break
      }

      let sessionIndex = slicedSessions.startIndex.advancedBy(index)
      let arraySessions = Array(slicedSessions[sessionIndex])
      let header = value as! Header
      let section = (header: header, sessions: arraySessions)
      timeline.append(section)
    }
    return timeline.count > 0 ? timeline : nil
  }

```

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]] ]]

4 个答案: