Swift 3 - 扩展 Dictionary<String, Any> 数组答案

【问题标题】：Swift 3 - extend an Array of Dictionary<String, Any>Swift 3 - 扩展 Dictionary<String, Any> 数组
【发布时间】：2017-02-27 16:44:57
【问题描述】：

我有这个字典数组：

var dicts = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]

我应该如何扩展Array 以便拥有这个功能：

dicts.values(of: "key1") // result: ["value1", "value3"]

我试试这个：

extension Array where Iterator.Element == [String: Any] { // ERROR!

    func values(of key: String) -> [Any]? {
        var result: [Any]?
        for value in self {
            let val = value as! Dictionary<String, Any>
            for k in val.keys {
                if k == key {
                    if result == nil {
                        result = [Any]()
                    }
                    result?.append(val[k])
                    break
                }
            }
        }
        return result
    }
}

但我总是在extension Array 声明处标记错误：

错误：相同类型的要求使泛型参数“元素”非泛型

我应该怎么做才能修复这个错误？

用for value in self 迭代Array 是否正确？

谢谢

【问题讨论】：

标签： arrays dictionary swift3 extend

【解决方案1】：

extension Sequence where Iterator.Element == [String: Any] {

    func values(of key: String) -> [Any]? {
        var result: [Any] = []
        for value in self {
            let val = value 
            for (k, v) in val {
                if k == key {
                    result.append(v)
                    break
                }
            }
        }
        return result.isEmpty ? nil : result
    }
}

var dicts: [[String: Any]] = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]
dicts.values(of: "key1")   // ["value1", "value3"]

或更短的版本：

extension Sequence where Iterator.Element == [String: Any] {

    func values(of key: String) -> [Any]? {
        let result: [Any] = reduce([]) { (result, dict) -> [Any] in
            var result = result
            result.append(dict.filter{ $0.key == key }.map{ $0.value })
            return result
        }
        return result.isEmpty ? nil : result
    }
}

var dicts: [[String: Any]] = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]
dicts.values(of: "key1")

并且没有减少功能：

extension Sequence where Iterator.Element == [String: Any] {

    func values(of key: String) -> [Any]? {
        var result: [Any] = []
        forEach {
            result.append($0.filter{ $0.key == key }.map{ $0.value })
        }
        return result.isEmpty ? nil : result
    }
}

var dicts: [[String: Any]] = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]
dicts.values(of: "key1")

【讨论】：

【解决方案2】：

使用过滤nil 值的compactMap（Swift 3 中的flatMap）整合解决方案

let dicts : [[String:Any]] = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]

extension Sequence where Iterator.Element == [String:Any] {

  func values(of key: String) -> [Any] {
    return self.compactMap {$0[key]}
  }
}

dicts.values(of:"key1")

【讨论】：

【解决方案3】：

感谢@JMI，您的解决方案非常优雅。

我想为你的作品添加更多细节：

extension Sequence where Iterator.Element == [String: Any] {
    func values(of key: String) -> [Any]? {
        var result = [Any]()
        forEach {
            let tmp = $0.filter{$0.key == key}.map{$0.value}
            if tmp.count > 0 {
                result.append(tmp[0])
            }
        }
        return result.isEmpty ? nil : result
    }
}

var dicts: [[String: Any]] = [["key1": "value1", "key2": "value2"], ["key1": "value3", "key2": "value4"]]

if let result = dicts.values(of: "key1") {
    for element in result {
        print("\(element) -> \(type(of: element))")
    }
}
else {
    print("no result")
}

最好的，

【讨论】：