按键的值对字典的 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。 我在下面准备了一个工作示例。 目标是按字典中的“d”元素对整个数组进行排序。 我准备了这个工作示例，可以将其放入 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])
}

这会导致以下输出到日志：

{
    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“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;
}

还有一些其他问题看起来很相似，但是当您查看它们时，很明显它们没有解决这个问题。 谢谢您的帮助。

Answer 1

要声明，如果您需要将其保留为 AnyObject，则必须显式转换：

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)
]

这给你同样的结果。 虽然，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，为了排序，你使用 sort() 闭包，它对一个 Array 进行排序。 您提供的闭包采用两个参数（名为 $0 和 $1），并返回一个 Bool。 如果 $0 在 $1 之前排序，则闭包应该返回 true，如果在 $1 之后排序，则返回 false。 要做到这一点，你必须投很多：

//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

在 Swift 3 & 4 中对字典数组进行排序

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]]"

Answer 4

在斯威夫特

let mySortedArray = myArray.sorted(by: {(int1, int2)  -> Bool in
    return ((int1 as! NSDictionary).value(forKey: "d") as! Int) < ((int2 as! NSDictionary).value(forKey: "d") as! Int) // It sorted the values and return to the mySortedArray
  }) 

print(mySortedArray)
    
myArray.removeAllObjects() // Remove all objects and reuse it
    
myArray.addObjects(from: mySortedArray)
    
print(mySortedArray)

按升序排列字典值数组很容易。 它不需要任何循环。

Answer 5

当我们解析数据时，我们可以使用 NSSortDescriptor 进行排序

let dataDict1 = responseDict.valueForKey("Data")
self.customerArray = dataDict1!.valueForKey("Customers") as! NSMutableArray

var tempArray = NSMutableArray()
for  index in self.customerArray {
   tempArray.addObject(index.valueForKey("Customer") as! NSMutableDictionary)
}

let descriptor: NSSortDescriptor =  NSSortDescriptor(key: "name", ascending: true, selector: "caseInsensitiveCompare:")
let sortedResults: NSArray = tempArray.sortedArrayUsingDescriptors([descriptor])
self.customerArray = NSMutableArray(array: sortedResults)