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。目标是通过字典中的“d”元素对整个数组进行排序。我准备了这个可以放入Swift项目的工作示例：

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;
}

目标是通过“d”元素对数组进行排序，以便将上面的输出更改为以下内容（基于数字顺序“d”：'2,5,6'）：< / p>

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

还有其他一些看似相似的问题，但是当你看到它们时，很明显他们没有解决这个问题。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)

如果没有追加，你可以这样做：

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)
]

这会给你相同的结果。虽然，如果只需要更改字典中的值对象，则不需要转换数组的元素：

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，要进行排序，可以使用sort（）闭包，它会对数组进行排序。你提供的闭包有两个参数（名为$ 0和$ 1），并返回一个Bool。如果在$ 1之前订购$ 0，则闭包应返回true，否则返回false。要做到这一点，你必须付出太多的代价：

//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和4中对字典进行排序

import

Answer 3

var myArray: [AnyObject] = []
var dict:[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)

let myArraySorted = myArray.sorted{$1["d"] as? Int > $0["d"] as? Int} as! [[String: AnyObject]]

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

或

var myArray: [[String:AnyObject]] = []
var dict:[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)

let myArraySorted = myArray.sorted{$1["d"] as? Int > $0["d"] as? Int}

println(myArraySorted)  // "[[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

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)

Answer 5

在Swift中

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.addObject(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。

按键的值

5 个答案: