Swift中的数组支持+ =运算符，可将一个数组的内容添加到另一个数组。有没有简单的方法可以做到这一点的字典？

例如：

var dict1 = ["a" : "foo"]
var dict2 = ["b" : "bar"]

var combinedDict = ... (some way of combining dict1 & dict2 without looping)

您可以为定义+=运算符Dictionary，例如

func += <K, V> (left: inout [K:V], right: [K:V]) { 
    for (k, v) in right { 
        left[k] = v
    } 
}

在Swift 4中，应该使用merging(_:uniquingKeysWith:)：

例：

let dictA = ["x" : 1, "y": 2, "z": 3]
let dictB = ["x" : 11, "y": 22, "w": 0]

let resultA = dictA.merging(dictB, uniquingKeysWith: { (first, _) in first })
let resultB = dictA.merging(dictB, uniquingKeysWith: { (_, last) in last })

print(resultA) // ["x": 1, "y": 2, "z": 3, "w": 0]
print(resultB) // ["x": 11, "y": 22, "z": 3, "w": 0]

怎么样

dict2.forEach { (k,v) in dict1[k] = v }

这会将dict2的所有键和值添加到dict1中。

当前，在查看Swift标准库字典参考时，无法轻松地用另一本字典更新字典。

你可以写一个扩展来做

var dict1 = ["a" : "foo"]
var dict2 = ["b" : "bar"]

extension Dictionary {
    mutating func update(other:Dictionary) {
        for (key,value) in other {
            self.updateValue(value, forKey:key)
        }
    }
}

dict1.update(dict2)
// dict1 is now ["a" : "foo", "b" : "bar]

Swift 4提供了merging(_:uniquingKeysWith:)，因此适合您的情况：

let combinedDict = dict1.merging(dict2) { $1 }

速记闭包返回$1，因此当与键冲突时将使用dict2的值。

它没有内置在Swift库中，但是您可以添加运算符重载，例如：

func + <K,V>(left: Dictionary<K,V>, right: Dictionary<K,V>) 
    -> Dictionary<K,V> 
{
    var map = Dictionary<K,V>()
    for (k, v) in left {
        map[k] = v
    }
    for (k, v) in right {
        map[k] = v
    }
    return map
}

这会使+字典的运算符超载，您现在可以使用它为+运算符添加字典，例如：

var dict1 = ["a" : "foo"]
var dict2 = ["b" : "bar"]

var dict3 = dict1 + dict2 // ["a": "foo", "b": "bar"]

斯威夫特3：

extension Dictionary {

    mutating func merge(with dictionary: Dictionary) {
        dictionary.forEach { updateValue($1, forKey: $0) }
    }

    func merged(with dictionary: Dictionary) -> Dictionary {
        var dict = self
        dict.merge(with: dictionary)
        return dict
    }
}

let a = ["a":"b"]
let b = ["1":"2"]
let c = a.merged(with: b)

print(c) //["a": "b", "1": "2"]

雨燕2.0

extension Dictionary {

    mutating func unionInPlace(dictionary: Dictionary) {
        dictionary.forEach { self.updateValue($1, forKey: $0) }
    }

    func union(var dictionary: Dictionary) -> Dictionary {
        dictionary.unionInPlace(self)
        return dictionary
    }
}

一成不变的

我更喜欢将不可变字典与+运算符结合/统一，所以我实现了像这样：

// Swift 2
func + <K,V> (left: Dictionary<K,V>, right: Dictionary<K,V>?) -> Dictionary<K,V> {
    guard let right = right else { return left }
    return left.reduce(right) {
        var new = $0 as [K:V]
        new.updateValue($1.1, forKey: $1.0)
        return new
    }
}

let moreAttributes: [String:AnyObject] = ["Function":"authenticate"]
let attributes: [String:AnyObject] = ["File":"Auth.swift"]

attributes + moreAttributes + nil //["Function": "authenticate", "File": "Auth.swift"]    
attributes + moreAttributes //["Function": "authenticate", "File": "Auth.swift"]
attributes + nil //["File": "Auth.swift"]

可变的

// Swift 2
func += <K,V> (inout left: Dictionary<K,V>, right: Dictionary<K,V>?) {
    guard let right = right else { return }
    right.forEach { key, value in
        left.updateValue(value, forKey: key)
    }
}

let moreAttributes: [String:AnyObject] = ["Function":"authenticate"]
var attributes: [String:AnyObject] = ["File":"Auth.swift"]

attributes += nil //["File": "Auth.swift"]
attributes += moreAttributes //["File": "Auth.swift", "Function": "authenticate"]

现在无需任何字典扩展。Swift（Xcode 9.0+）字典为此提供了功能。在这里看看。以下是有关如何使用它的示例

  var oldDictionary = ["a": 1, "b": 2]
  var newDictionary = ["a": 10000, "b": 10000, "c": 4]

  oldDictionary.merge(newDictionary) { (oldValue, newValue) -> Int in
        // This closure return what value to consider if repeated keys are found
        return newValue 
  }
  print(oldDictionary) // Prints ["b": 10000, "a": 10000, "c": 4]

使用扩展名的更具可读性的变体。

extension Dictionary {    
    func merge(dict: Dictionary<Key,Value>) -> Dictionary<Key,Value> {
        var mutableCopy = self        
        for (key, value) in dict {
            // If both dictionaries have a value for same key, the value of the other dictionary is used.           
            mutableCopy[key] = value 
        }        
        return mutableCopy
    }    
}

你可以试试这个

var dict1 = ["a" : "foo"]
var dict2 = ["b" : "bar"]

var temp = NSMutableDictionary(dictionary: dict1);
temp.addEntriesFromDictionary(dict2)

您也可以使用reduce合并它们。在操场上尝试

let d1 = ["a":"foo","b":"bar"]
let d2 = ["c":"car","d":"door"]

let d3 = d1.reduce(d2) { (var d, p) in
   d[p.0] = p.1
   return d
}

我推荐SwifterSwift库。但是，如果您不想使用整个库及其所有出色的附加功能，则可以利用它们的Dictionary扩展：

迅捷3+

public extension Dictionary {
    public static func +=(lhs: inout [Key: Value], rhs: [Key: Value]) {
        rhs.forEach({ lhs[$0] = $1})
    }
}

您可以遍历键值组合ob想要合并的值，并通过updateValue（forKey :)方法添加它们：

dictionaryTwo.forEach {
    dictionaryOne.updateValue($1, forKey: $0)
}

现在，dictionaryTwo的所有值都添加到dictionaryOne中。

与@farhadf的答案相同，但被Swift 3采纳：

let sourceDict1 = [1: "one", 2: "two"]
let sourceDict2 = [3: "three", 4: "four"]

let result = sourceDict1.reduce(sourceDict2) { (partialResult , pair) in
    var partialResult = partialResult //without this line we could not modify the dictionary
    partialResult[pair.0] = pair.1
    return partialResult
}

Swift 3，字典扩展：

public extension Dictionary {

    public static func +=(lhs: inout Dictionary, rhs: Dictionary) {
        for (k, v) in rhs {
            lhs[k] = v
        }
    }

}

Swift 4的一些更简化的重载：

extension Dictionary {
    static func += (lhs: inout [Key:Value], rhs: [Key:Value]) {
        lhs.merge(rhs){$1}
    }
    static func + (lhs: [Key:Value], rhs: [Key:Value]) -> [Key:Value] {
        return lhs.merging(rhs){$1}
    }
}

您可以添加如下Dictionary扩展名：

extension Dictionary {
    func mergedWith(otherDictionary: [Key: Value]) -> [Key: Value] {
        var mergedDict: [Key: Value] = [:]
        [self, otherDictionary].forEach { dict in
            for (key, value) in dict {
                mergedDict[key] = value
            }
        }
        return mergedDict
    }
}

然后使用是简单的，如下：

var dict1 = ["a" : "foo"]
var dict2 = ["b" : "bar"]

var combinedDict = dict1.mergedWith(dict2)
// => ["a": "foo", "b": "bar"]

如果您希望框架还包含一些更方便的功能，请签出HandySwift。只需将其导入到您的项目中，就可以使用上面的代码，而无需自己添加任何扩展到项目中。

不再需要扩展或任何其他功能。您可以这样写：

firstDictionary.merge(secondDictionary) { (value1, value2) -> AnyObject in
        return object2 // what you want to return if keys same.
    }

您可以使用，

func addAll(from: [String: Any], into: [String: Any]){
    from.forEach {into[$0] = $1}
}

您可以使用bridgeToObjectiveC（）函数使字典成为NSDictionary。

将如下所示：

var dict1 = ["a":"Foo"]
var dict2 = ["b":"Boo"]

var combinedDict = dict1.bridgeToObjectiveC()
var mutiDict1 : NSMutableDictionary! = combinedDict.mutableCopy() as NSMutableDictionary

var combineDict2 = dict2.bridgeToObjectiveC()

var combine = mutiDict1.addEntriesFromDictionary(combineDict2)

然后，您可以将NSDictionary（combine）转换回去或执行任何操作。

import Foundation

let x = ["a":1]
let y = ["b":2]

let out = NSMutableDictionary(dictionary: x)
out.addEntriesFromDictionary(y)

结果是NSMutableDictionary而不是Swift类型的字典，但是使用它的语法是相同的（out["a"] == 1在这种情况下），因此只有在使用期望使用Swift字典的第三方代码时，才有问题需要类型检查。

简短的答案是您实际上必须循环。即使您没有明确输入它，这也是您正在调用的方法（addEntriesFromDictionary：此处）的作用。我建议如果您不清楚为什么会出现这种情况，则应考虑如何合并两个B树的叶子节点。

如果您确实确实需要Swift本机字典类型作为回报，我建议：

let x = ["a":1]
let y = ["b":2]

var out = x
for (k, v) in y {
    out[k] = v
}

这种方法的缺点是字典索引（尽管已完成）可能会在循环中重建几次，因此在实践中，这比NSMutableDictionary方法慢大约10倍。

所有这些响应都很复杂。这是我的Swift 2.2解决方案：

    //get first dictionnary
    let finalDictionnary : NSMutableDictionary = self.getBasicDict()
    //cast second dictionnary as [NSObject : AnyObject]
    let secondDictionnary : [NSObject : AnyObject] = self.getOtherDict() as [NSObject : AnyObject]
    //merge dictionnary into the first one
    finalDictionnary.addEntriesFromDictionary(secondDictionnary) 

我的需求有所不同，我需要合并不完整的嵌套数据集而不会造成麻烦。

merging:
    ["b": [1, 2], "s": Set([5, 6]), "a": 1, "d": ["x": 2]]
with
    ["b": [3, 4], "s": Set([6, 7]), "a": 2, "d": ["y": 4]]
yields:
    ["b": [1, 2, 3, 4], "s": Set([5, 6, 7]), "a": 2, "d": ["y": 4, "x": 2]]

这比我想要的要难。挑战在于从动态类型到静态类型的映射，我使用协议来解决这个问题。

同样值得注意的是，当您使用字典文字语法时，实际上会获得基础类型，而基础类型不会使用协议扩展。我放弃了支持这些工作的努力，因为我无法轻松地验证集合元素的一致性。

import UIKit


private protocol Mergable {
    func mergeWithSame<T>(right: T) -> T?
}



public extension Dictionary {

    /**
    Merge Dictionaries

    - Parameter left: Dictionary to update
    - Parameter right:  Source dictionary with values to be merged

    - Returns: Merged dictionay
    */


    func merge(right:Dictionary) -> Dictionary {
        var merged = self
        for (k, rv) in right {

            // case of existing left value
            if let lv = self[k] {

                if let lv = lv as? Mergable where lv.dynamicType == rv.dynamicType {
                    let m = lv.mergeWithSame(rv)
                    merged[k] = m
                }

                else if lv is Mergable {
                    assert(false, "Expected common type for matching keys!")
                }

                else if !(lv is Mergable), let _ = lv as? NSArray {
                    assert(false, "Dictionary literals use incompatible Foundation Types")
                }

                else if !(lv is Mergable), let _ = lv as? NSDictionary {
                    assert(false, "Dictionary literals use incompatible Foundation Types")
                }

                else {
                    merged[k] = rv
                }
            }

                // case of no existing value
            else {
                merged[k] = rv
            }
        }

        return merged
    }
}




extension Array: Mergable {

    func mergeWithSame<T>(right: T) -> T? {

        if let right = right as? Array {
            return (self + right) as? T
        }

        assert(false)
        return nil
    }
}


extension Dictionary: Mergable {

    func mergeWithSame<T>(right: T) -> T? {

        if let right = right as? Dictionary {
            return self.merge(right) as? T
        }

        assert(false)
        return nil
    }
}


extension Set: Mergable {

    func mergeWithSame<T>(right: T) -> T? {

        if let right = right as? Set {
            return self.union(right) as? T
        }

        assert(false)
        return nil
    }
}



var dsa12 = Dictionary<String, Any>()
dsa12["a"] = 1
dsa12["b"] = [1, 2]
dsa12["s"] = Set([5, 6])
dsa12["d"] = ["c":5, "x": 2]


var dsa34 = Dictionary<String, Any>()
dsa34["a"] = 2
dsa34["b"] = [3, 4]
dsa34["s"] = Set([6, 7])
dsa34["d"] = ["c":-5, "y": 4]


//let dsa2 = ["a": 1, "b":a34]
let mdsa3 = dsa12.merge(dsa34)
print("merging:\n\t\(dsa12)\nwith\n\t\(dsa34) \nyields: \n\t\(mdsa3)")

斯威夫特2.2

func + <K,V>(left: [K : V], right: [K : V]) -> [K : V] {
    var result = [K:V]()

    for (key,value) in left {
        result[key] = value
    }

    for (key,value) in right {
        result[key] = value
    }
    return result
}

我只会使用Dollar库。

https://github.com/ankurp/Dollar/#merge---merge-1

将所有字典合并在一起，后一个字典将覆盖给定键的值

let dict: Dictionary<String, Int> = ["Dog": 1, "Cat": 2]
let dict2: Dictionary<String, Int> = ["Cow": 3]
let dict3: Dictionary<String, Int> = ["Sheep": 4]
$.merge(dict, dict2, dict3)
=> ["Dog": 1, "Cat": 2, "Cow": 3, "Sheep": 4]

这是我写的一个不错的扩展...

extension Dictionary where Value: Any {
    public func mergeOnto(target: [Key: Value]?) -> [Key: Value] {
        guard let target = target else { return self }
        return self.merging(target) { current, _ in current }
    }
}

使用：

var dict1 = ["cat": 5, "dog": 6]
var dict2 = ["dog": 9, "rodent": 10]

dict1 = dict1.mergeOnto(target: dict2)

然后，将dict1修改为

["cat": 5, "dog": 6, "rodent": 10]