Swift：對字典排序（按鍵或按值）並返回有序數組（按鍵或值）

Question

我有一本字典，需要從中派生一個鍵數組和一個按鍵或值排序的值數組。

我的用例是文件夾列表。 該詞典包含文件夾名稱（鍵）和文件夾中的項目計數（值）。 我想按鍵名（A到Z或Z到A）和計數大小（從大到小或從小到大）進行排序。

在Swift中對鍵進行排序很容易。 但是我求助於迭代以提供排序的值列表。 這似乎不像Swift的做事方式，但是我對Swift中的map / sort / etc的理解不足以讓我看到一種更聰明的做事方式。

誰能解釋一下實現此目標的明智而簡潔的Swift方式？

我當前的代碼是：

let dictionary = ["Alpha" : 24, "Beta" : 47, "Gamma" : 12, "Delta" : 33]

enum FolderOrder : Int {
    case AlphaAtoZ
    case AlphaZtoA
    case SizeLargeToSmall
    case SizeSmallToLarge
}

func folderList(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> [String] {
    switch orderedBy {
    case .AlphaAtoZ:
        return fromDictionary.keys.sort() {$0 < $1}

    case .AlphaZtoA:
        return fromDictionary.keys.sort() {$1 < $0}

    case .SizeSmallToLarge:
        return fromDictionary.keys.sort(){fromDictionary[$0] < fromDictionary [$1]}

    case .SizeLargeToSmall:
        return fromDictionary.keys.sort(){fromDictionary[$1] < fromDictionary [$0]}
    }
}

func folderCounts(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> [Int]
{
    let orderedKeys = folderList(fromDictionary, orderedBy: orderedBy)
    var orderedValues = [Int]()

    for key in orderedKeys {
        orderedValues.append(fromDictionary[key] ?? 0)
    }

    return orderedValues
}

folderList(dictionary, orderedBy: .AlphaAtoZ)
// ["Alpha", "Beta", "Delta", "Gamma"]

folderList(dictionary, orderedBy: .AlphaZtoA)
// ["Gamma", "Delta", "Beta", "Alpha"]

folderList(dictionary, orderedBy: .SizeSmallToLarge)
// ["Gamma", "Alpha", "Delta", "Beta"]

folderList(dictionary, orderedBy: .SizeLargeToSmall)
//["Beta", "Delta", "Alpha", "Gamma"]

folderCounts(dictionary, orderedBy: .AlphaAtoZ)
// [24, 47, 33, 12]

folderCounts(dictionary, orderedBy: .SizeLargeToSmall)
// [47, 33, 24, 12]

更新資料

感謝兩個有用的答案，尤其是@nRewik，我簡化了代碼並提高了對Swift的理解。

修改后的代碼，加上注釋說明了我最初不清楚的內容，因此可能對其他人有幫助：

let dictionary = ["Alpha" : 24, "Beta" : 47, "Gamma" : 12, "Delta" : 33]

enum FolderOrder {
    case AlphaAtoZ
    case AlphaZtoA
    case SizeLargeToSmall
    case SizeSmallToLarge
}

func folderListAndCounts(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> ([String], [Int]) {

    var sortedDictionary : [(String, Int)]

    switch orderedBy {

    // The closure when sort() is applied to a dictionary takes two tuples as parameters
    // where the tuples are of the form (key, value). The first tuple can be accessed as $0.
    // Its key can be accessed as $0.0 and its value as $0.1

    case .AlphaAtoZ:
        sortedDictionary = fromDictionary.sort{ $0.0 < $1.0 } // item(n).key < item(n+1).key
    case .AlphaZtoA:
        sortedDictionary = fromDictionary.sort{ $0.0 > $1.0 } // item(n).key > item(n+1).key
    case .SizeSmallToLarge:
        sortedDictionary = fromDictionary.sort{ $0.1 < $1.1 } // item(n).value < item(n+1).value
    case .SizeLargeToSmall:
        sortedDictionary = fromDictionary.sort{ $0.1 > $1.1 } // item(n).value < item(n+1).value
    }

    // The sorted dictionary has the type: [(String, Int)], i.e. it's an array of tuples.
    // The closure when map is applied to an array of tuples is a tuple. The tuple can be
    // accessed as $0. Its key can be accessed as $0.0 and its value as $0.1

    let sortedKeys = sortedDictionary.map{$0.0}
    let sortedValues = sortedDictionary.map{$0.1}

    // Returns a tuple (arrayOfKeys, arrayOfValues)
    return (sortedKeys, sortedValues)
}

let (keys, counts) = folderListAndCounts(dictionary, orderedBy: .SizeSmallToLarge)

Answer 1

函數式編程風格怎么樣？

typealias DictSorter = ((String,Int),(String,Int)) -> Bool

let alphaAtoZ: DictSorter = { $0.0 < $1.0 }
let alphaZtoA: DictSorter = { $0.0 > $1.0 }
let sizeSmallToLarge: DictSorter = { $0.1 < $1.1 }
let sizeLargeToSmall: DictSorter = { $0.1 > $1.1 }

// selector
let listSelector: (String,Int)->String = { $0.0 }
let countSelector: (String,Int)->Int = { $0.1 }

// Usage
let dict = ["Alpha" : 24, "Beta" : 47, "Gamma" : 12, "Delta" : 33]

let folderListByAlphaAtoZ = dict.sort(alphaAtoZ).map(listSelector)
let folderListByAlphaZtoA = dict.sort(alphaZtoA).map(listSelector)
let folderListBySizeSmallToLarge = dict.sort(sizeSmallToLarge).map(listSelector)
let folderListBySizeLargeToSmall = dict.sort(sizeLargeToSmall).map(listSelector)

let folderCountByAlphaAtoZ = dict.sort(alphaAtoZ).map(countSelector)
let folderCountByAlphaZtoA = dict.sort(alphaZtoA).map(countSelector)
let folderCountBySizeSmallToLarge = dict.sort(sizeSmallToLarge).map(countSelector)
let folderCountBySizeLargeToSmall = dict.sort(sizeLargeToSmall).map(countSelector)

Answer 2

那這個呢？ 這將更改您的代碼以對Dictionary條目進行排序，然后再獲取鍵或值。

let dictionary = ["Alpha" : 24, "Beta" : 47, "Gamma" : 12, "Delta" : 33]

enum FolderOrder : Int {
  case AlphaAtoZ
  case AlphaZtoA
  case SizeLargeToSmall
  case SizeSmallToLarge
}

func entryList(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> [(String, Int)] {
  switch orderedBy {
  case .AlphaAtoZ:
    return fromDictionary.sort { $0.0 < $1.0 }

  case .AlphaZtoA:
    return fromDictionary.sort { $0.0 > $1.0 }

  case .SizeSmallToLarge:
    return fromDictionary.sort { $0.1 < $1.1 }

  case .SizeLargeToSmall:
    return fromDictionary.sort { $0.1 > $1.1 }
  }
}

func folderList(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> [String] {
  return entryList(fromDictionary, orderedBy: orderedBy).map { $0.0 }
}

func folderCounts(fromDictionary: [String: Int], orderedBy : FolderOrder = .AlphaAtoZ) -> [Int] {
  return entryList(fromDictionary, orderedBy: orderedBy).map { $0.1 }
}

folderList(dictionary, orderedBy: .AlphaAtoZ)
// ["Alpha", "Beta", "Delta", "Gamma"]

folderList(dictionary, orderedBy: .AlphaZtoA)
// ["Gamma", "Delta", "Beta", "Alpha"]

folderList(dictionary, orderedBy: .SizeSmallToLarge)
// ["Gamma", "Alpha", "Delta", "Beta"]

folderList(dictionary, orderedBy: .SizeLargeToSmall)
//["Beta", "Delta", "Alpha", "Gamma"]

folderCounts(dictionary, orderedBy: .AlphaAtoZ)
// [24, 47, 33, 12]

folderCounts(dictionary, orderedBy: .SizeLargeToSmall)
// [47, 33, 24, 12]