在Swift 3的Dictionary中对数组值进行排序

Question

我有一个.plist文件与根字典和加载字母键如下：

["A": ["AXB", "ACD", "ABC"], ... ]

这是正确的：

["A": ["ABC", "ACD", "AXB"], ... ]

然后，我想排序这个数组的A指数。 所以，我尝试这样做：

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate {

var musicalGroups : NSDictionary = NSDictionary()
var keysOfMusicalGroups : NSMutableArray = NSMutableArray()

override func viewDidLoad() {
    super.viewDidLoad()

    let bundle = Bundle.main
    let path = bundle.path(forResource: "bandas", ofType: "plist")
    musicalGroups = NSDictionary(contentsOfFile: path!)!
    keysOfMusicalGroups = NSMutableArray(array: musicalGroups.allKeys)
    keysOfMusicalGroups.sort(using: NSSelectorFromString("compare:"))

}

我只使用keysOfMusicalGroups.sort代码对Dictionary进行了排序

任何帮助将不胜感激。 提前致谢！

Answer 1

您只需对字典键进行排序，并将每个值（数组）附加到生成的数组2D中：

let dict = ["A": ["Angra", "Aerosmith", "ACDC", "Avantasia"],"B": ["Barao Vermelho", "Bon Jovi", "Bee Gees"],"C": ["Cachorro Loco", "Coldplay", "Creed"] ]

let sortedArray2D = dict.sorted{ $0.key < $1.key }.map { $0.value.sorted() }

print(sortedArray2D)  // "[["ACDC", "Aerosmith", "Angra", "Avantasia"], ["Barao Vermelho", "Bee Gees", "Bon Jovi"], ["Cachorro Loco", "Coldplay", "Creed"]]\n"

您的最终表格视图应如下所示：

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate {
    var musicalGroups: [[String]] = []
    var musicalTitles: [String] = []
    override func viewDidLoad() {
        super.viewDidLoad()
        let dict = NSDictionary(contentsOfFile: Bundle.main.path(forResource: "bandas", ofType: "plist")!) as? [String: [String]] ?? [:]
        musicalTitles = dict.keys.sorted()
        musicalGroups = dict.sorted{ $0.key < $1.key }.map { $0.value.sorted() }
    }
    func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        return musicalGroups[section].count
    }
    func tableView(_ tableView: UITableView, titleForHeaderInSection section: Int) -> String? {
        return musicalTitles[section]
    }
    func numberOfSections(in tableView: UITableView) -> Int {
        return musicalTitles.count
    }
    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = UITableViewCell(style: .default, reuseIdentifier: "exemplo")
        cell.textLabel?.text = musicalGroups[indexPath.section][indexPath.row]
        return cell
    }
}

Answer 2

基于@ Leo-Dabus的更具表现力的答案是：

let dict = ["A": ["Angra", "Aerosmith", "ACDC", "Avantasia"],"B": ["Barao Vermelho", "Bon Jovi", "Bee Gees"],"C": ["Cachorro Loco", "Coldplay", "Creed"] ]

let sortedArray2D: [[String]] = dict.keys.sorted().map{ dict[$0]!.sorted() }

print(sortedArray2D)  // "[["ACDC", "Aerosmith", "Angra", "Avantasia"], ["Barao Vermelho", "Bee Gees", "Bon Jovi"], ["Cachorro Loco", "Coldplay", "Creed"]]\n"

这样做效果更好，因为编译器现在知道sortedArray2D与dict具有相同的大小。

此外，这种方式sortedArray2D是一个常量（ let ），而不再是var （允许进一步的性能增强）。