按键值对字典的Swift数组进行排序

Question

我正在尝试对由字典组成的Swift数组进行排序。我准备了下面的工作例子。目标是按照字典中的"d“元素对整个数组进行排序。我准备了这个工作示例，可以放在Swift项目中：

var myArray = Array<AnyObject>()
var dict = Dictionary<String, AnyObject>()

dict["a"] = "hickory"
dict["b"] = "dickory"
dict["c"] = "dock"
dict["d"] = 5

myArray.append(dict)

dict["a"] = "three"
dict["b"] = "blind"
dict["c"] = "mice"
dict["d"] = 6

myArray.append(dict)

dict["a"] = "larry"
dict["b"] = "moe"
dict["c"] = "curly"
dict["d"] = 2

myArray.append(dict)

println(myArray[0])
println(myArray[1])
println(myArray[2])
}

这将导致日志的以下输出：

{
    a = hickory;
    b = dickory;
    c = dock;
    d = 5;

}
{
    a = three;
    b = blind;
    c = mice;
    d = 6;
}
{
    a = larry;
    b = moe;
    c = curly;
    d = 2;
}

目标是按照"d“元素对数组进行排序，以便将上述输出更改为以下内容(基于”d“的数字顺序：'2，5，6')：

{
    a = larry;
    b = moe;
    c = curly;
    d = 2;
}
{
    a = hickory;
    b = dickory;
    c = dock;
    d = 5;
}
{
    a = three;
    b = blind;
    c = mice;
    d = 6;
}

还有一些其他的问题看起来很相似，但是当你看到它们的时候，很明显它们并没有解决这个问题。谢谢你的帮助。

Answer 1

要声明，如果需要将其保留为AnyObject，则必须显式强制转换：

var myArray = Array<AnyObject>()
var dict = Dictionary<String, AnyObject>()

dict["a"] = ("hickory" as! AnyObject)
dict["b"] = ("dickory" as! AnyObject)
dict["c"] = ("dock" as! AnyObject)
dict["d"] = (6 as! AnyObject)

myArray.append(dict as! AnyObject)

dict["a"] = ("three" as! AnyObject)
dict["b"] = ("blind" as! AnyObject)
dict["c"] = ("mice" as! AnyObject)
dict["d"] = (5 as! AnyObject)


myArray.append(dict as! AnyObject)

dict["a"] = ("larry" as! AnyObject)
dict["b"] = ("moe" as! AnyObject)
dict["c"] = ("curly" as! AnyObject)
dict["d"] = (4 as! AnyObject)

myArray.append(dict as! AnyObject)

无需附加，您就可以这样做：

var myArray: [AnyObject] = [ ([
    "a" : ("hickory" as! AnyObject),
    "b" : ("dickory" as! AnyObject),
    "c" : ("dock" as! AnyObject),
    "d" : (6 as! AnyObject)
  ] as! AnyObject), ([
    "a" : ("three" as! AnyObject),
    "b" : ("blind" as! AnyObject),
    "c" : ("mice" as! AnyObject),
    "d" : (5 as! AnyObject)
  ] as! AnyObject), ([
    "a" : ("larry" as! AnyObject),
    "b" : ("moe" as! AnyObject),
    "c" : ("curly" as! AnyObject),
    "d" : (4 as! AnyObject)
  ] as! AnyObject)
]

给你同样的结果。尽管，如果只需要更改字典中的值对象，则不需要强制转换数组的元素：

var myArray: [Dictionary<String, AnyObject>] = [[
    "a" : ("hickory" as! AnyObject),
    "b" : ("dickory" as! AnyObject),
    "c" : ("dock" as! AnyObject),
    "d" : (6 as! AnyObject)
  ], [
    "a" : ("three" as! AnyObject),
    "b" : ("blind" as! AnyObject),
    "c" : ("mice" as! AnyObject),
    "d" : (5 as! AnyObject)
  ], [
    "a" : ("larry" as! AnyObject),
    "b" : ("moe" as! AnyObject),
    "c" : ("curly" as! AnyObject),
    "d" : (4 as! AnyObject)
  ]
]

然后，要排序，您可以使用sort()闭包，它对数组进行适当的排序。您提供的闭包包含两个参数( $0和$1)，并返回一个Bool。如果$0是在$1之前订购的，则闭包应该返回true，如果后面是false，则返回false。要做到这一点，你必须投很多钱：

//myArray starts as: [
//  ["d": 6, "b": "dickory", "c": "dock", "a": "hickory"],
//  ["d": 5, "b": "blind", "c": "mice", "a": "three"],
//  ["d": 4, "b": "moe", "c": "curly", "a": "larry"]
//]

myArray.sort{
  (($0 as! Dictionary<String, AnyObject>)["d"] as? Int) < (($1 as! Dictionary<String, AnyObject>)["d"] as? Int)
}

//myArray is now: [
//  ["d": 4, "b": "moe", "c": "curly", "a": "larry"],
//  ["d": 5, "b": "blind", "c": "mice", "a": "three"],
//  ["d": 6, "b": "dickory", "c": "dock", "a": "hickory"]
//]

Answer 2

Swift 3&4中字典的排序阵列

let sortedResults = (userArray as NSArray).sortedArray(using: [NSSortDescriptor(key: "name", ascending: true)]) as! [[String:AnyObject]]

Answer 3

编辑/更新：Xcode 11·Swift 5

var array: [[String:Any]] = []
var dict: [String: Any] = [:]

dict["a"] = "hickory"
dict["b"] = "dickory"
dict["c"] = "dock"
dict["d"] = 5

array.append(dict)

dict["a"] = "three"
dict["b"] = "blind"
dict["c"] = "mice"
dict["d"] = 6

array.append(dict)

dict["a"] = "larry"
dict["b"] = "moe"
dict["c"] = "curly"
dict["d"] = 2

array.append(dict)

let sortedArray = array.sorted { $0["d"] as? Int ?? .zero < $1["d"] as? Int ?? .zero }

print(sortedArray)  // "[[b: moe, a: larry, d: 2, c: curly], [b: dickory, a: hickory, d: 5, c: dock], [b: blind, a: three, d: 6, c: mice]]"