Question

我想将从API获得的值转换为特定格式。

[String:Any] // format received
[Int:[ContentType:Int]] // required format

ContentType是一个枚举

数据示例可能如下所示：

["123":["Tables":"25","Chairs":"14"]] // input
[123:[.Tables:25,.Chairs:14]] // output

我想我需要在地图中包含一张地图才能正常工作，但是我正在努力寻找前进的方向。我可能完全把错误的树种了。我真的不想手动遍历并一次添加每个项目。如果可能，我正在寻找比这更聪明的东西。

enum ContentType: String  {
    case Tables,Chairs
}

let original_values: [String:Any]
    =  ["1234":["Tables":"5","Chairs":"2"]]
let values: [Int:[ContentType:Int]]
    = Dictionary(uniqueKeysWithValues: original_values.map {
        (
            Int($0.key)!,
            (($0.value as? [String:String]).map { // Error on this line - expects 1 argument but two were used
                (
                    ContentType(rawValue: $1.key)!, // $1 is presumably wrong here?
                    Int($1.value)
                )
            }) as? [ContentType:Int]
        )
    })

有人有创意吗？

Answer 1

我想将从API获得的值转换为特定格式。

您可以使您的枚举Decodable

enum ContentType: String, Decodable {
    case tables, chairs
    enum CodingKeys: String, CodingKey {
        case Tables = "Tables"
        case Chairs = "Chairs"
    }
}

然后，您可以解码收到的Data，然后compactMap将其格式化为(Int, [ContentType: Int])。您可以使用设计的初始值设定项将这些元组转换为Dictionary

do {
    let decoded = try JSONDecoder().decode([String: [ContentType: Int]].self, from: data)
    let mapped = Dictionary(uniqueKeysWithValues: decoded.compactMap { (key,value) -> (Int, [ContentType: Int])? in
        if let int = Int(key) {
            return (int, value)
        } else {
            return nil
        }
    })
} catch {
    print(error)
}

Answer 2

在这一行：

(($0.value as? [String:String]).map {

您使用的不是Sequence.map，而是Optional.map。

工作解决方案：

/// First let's map plain types to our types
let resultArray = original_values
    .compactMap { (key, value) -> (Int, [ContentType: Int])? in
        guard let iKey = Int(key), let dValue = value as? [String: String] else { return nil }
        let contentValue = dValue.compactMap { (key, value) -> (ContentType, Int)? in
            guard let cKey = ContentType(rawValue: key), let iValue = Int(value) else { return nil }
            return (cKey, iValue)
        }
        let contentDict = Dictionary(uniqueKeysWithValues: contentValue)
        return (iKey, contentDict)
    }
let result = Dictionary(uniqueKeysWithValues: resultArray)

要提高打印输出，请添加符合CustomStringConvertible的内容：

extension ContentType: CustomStringConvertible {
    var description: String {
        switch self {
        case .Tables:
            return "Tables"
        case .Chairs:
            return "Chairs"
        }
    }
}

Answer 3

这是Swift 5正确的语法

enum ContentType: String  {
    case tables = "Tables"
    case chairs = "Chairs"
}

let originalValues: [String: [String: String]]
    =  ["1234": ["Tables": "5", "Chairs": "2"]]

    let values: [Int: [ContentType: Int]] = Dictionary(uniqueKeysWithValues:
        originalValues.map { arg in
            let (key, innerDict) = arg
            let outMap: [ContentType: Int] = Dictionary(uniqueKeysWithValues:
                innerDict.map { innerArg in
                let (innerKey, innerValue) = innerArg
                return (ContentType.init(rawValue: innerKey)!, Int(innerValue)!)
            }
            )
            return (Int(key)!, outMap)
        }
    )
    print(values)

[1234：[__ lldb_expr_5.ContentType.tables：5，__ lldb_expr_5.ContentType.chairs：2]]

快速字典：地图内的地图

3 个答案: