一个解决方案是使用reduce(into:_:)：

let output = typingAttributes.reduce(into: [String: Any]()) { partialResult, tuple in
    let newKey = //Get new key from tuple.key
    partialResult[newKey] = tuple.value
}

字符串
在您的例子中，由于您使用NSAttributedString.Key作为字典键，并且您想要原始字符串值：

let newKey = tuple.key.rawValue

型
这可以简化为：

let output = typingAttributes.reduce(into: [String: Any]()) { 
    $0[$1.key.rawValue] = $1.value
}

型
带扩展名：

extension Dictionary {
    func mapKeys<T>(_ key: (Key) -> T) -> [T: Value] {
        reduce(into: [T: Value]()) {
            let newKey = key($1.key)
            $0[newKey] = $1.value
        }
    }
}

型
用法：

let output = typingAttributes.mapKeys { $0.rawValue }

型
其他样品用途：

//Convert the keys into their int value (it'd crash if it's not an int)
let testInt = ["1": "a", "2": "b"].mapKeys { Int($0)! }
print(testInt)

//Keep only the first character of the keys
let testPrefix = ["123": "a", "234": "b"].mapKeys { String($0.prefix(1)) }
print(testPrefix)

//Fixing keys into our owns, more "readable" and "Swift friendly for instance
let keysToUpdate = ["lat": "latitude", "long": "longitude", "full_name": "fullName"]
let testFixKeys = ["lat": 0.4, "long": 0.5, "full_name": "Seed", "name": "John", "age": 34].mapKeys { keysToUpdate[$0] ?? $0 }
print(testFixKeys)

//etc.

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