Question

我有一个字典，我需要从中导出一个键数组和一个在键或值上排序为EITHER的值数组。

我的用例是文件夹列表。字典包含文件夹名称（键）和文件夹中的项目数（值）。我想按键名（A到Z或Z到A）以及计数大小（从大到小或从小到大）排序。

在Swift中对键进行排序很容易。但我正在使用迭代来提供排序的值列表。这似乎不是Swift的做事方式，但我对Swift中map / sort /等的理解对我来说还不够好，看不出更聪明的方法。

任何人都能解释一种聪明而简洁的Swift实现这一目标的方法吗？

我目前的代码是：

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

感谢两个有用的答案，尤其是@nRewik，我简化了我的代码并提高了我对Swift的理解。

修改后的代码，其中的评论拼写出我最初不清楚的内容，因此可能对其他人有所帮助：

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

关于函数式编程风格如何？

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

这个怎么样？这会更改您的代码以对Dictionary条目进行排序，然后再获取键或值。

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]

Swift：排序字典（通过键或值）并返回有序数组（键或值）

更新

2 个答案: