reduce 函数正在打印一个空字典 [:]

Question

我已在此question 中成功地将我的字典键减少为没有真正 json 模型的伪代码。我在上一个问题中实现的目标是只返回具有匹配值的键。所以输出是一个类似于["WoW": ["@jade", "@kalel"] 的字典。正是我需要的。当然可能还有其他比赛，我也想退回这些比赛。

现在我有了一个合适的 json 模型，reduce 函数打印出一个空字典 [:]。是.reduce(into: [String:[String]]() 中的类型导致了问题吗？

所有数据都在打印，因此 json 和模型结构必须正确。

json

[
{
    "id": "tokenID-tqkif48",
    "name": "@jade",
    "game": "WoW",
    "age": "18"
},
{
    "id": "tokenID-fvkif21",
    "name": "@kalel",
    "game": "WoW",
    "age": "20"
}
]

用户模型

public typealias Users = [UserModel]
public struct UserModel: Codable {

public let name: String
public let game: String
// etc...

enum CodingKeys: String, CodingKey {
    case name
    case game
    // etc...

游乐场

guard let url = Bundle.main.url(forResource: "Users", withExtension: "json") else {
    fatalError()
}
guard let data = try? Data(contentsOf: url) else {
    fatalError()
}

let decoder = JSONDecoder()

do {
    let response = try decoder.decode([UserModel].self, from: data)
    for userModel in response {

        let userDict: [String:String] = [ userModel.name:userModel.game ]

        let reduction = Dictionary(grouping: userDict.keys) { userDict[$0] ?? "" }.reduce(into: [String:[String]](), { (result, element) in
            if element.value.count > 1 {
                result[element.key] = element.value
            }
        })
        // error catch etc
}

Answer 1

您的代码太复杂了。您可以通过 game 对数组进行分组，只需使用

let response = try decoder.decode([UserModel].self, from: data)
let reduction = Dictionary(grouping: response, by: {$0.game}).mapValues{ usermodel in usermodel.map{ $0.name}}

Answer 2

更新我可能弄错了你想得到什么。下面还有一个代码，请检查结果并选择一个你想要的。

如果你想使用reduce(into:updateAccumulatingResult:)，你可以这样写。

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(name: String, game: String)] = response.map {($0.name, $0.game)}

    let reduction = userArray.reduce(into: [String:[String]]()) {result, element in
        if !element.game.isEmpty {
            result[element.name, default: []].append(element.game)
        }
    }
    print(reduction)
} catch {
    print(error)
}

如果您更喜欢 Dictionary 的初始化程序，这可能会起作用：

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(name: String, games: [String])] = response.map {
        ($0.name, $0.game.isEmpty ? [] : [$0.game])
    }

    let reduction = Dictionary(userArray) {old, new in old + new}
    print(reduction)
} catch {
    print(error)
}

两个输出：

["@jade": ["WoW"], "@kalel": ["WoW"]]

无论如何，您将循环、Dictionary(grouping:) 和 reduce(into:) 以及 userDict.keys 组合在一起的方式使事情变得过于复杂。

---

ADDITION当你想得到一个带有键的字典作为游戏时：

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(game: String, name: String)] = response.compactMap {
        $0.game.isEmpty ? nil : ($0.game, $0.name)
    }

    let reduction = userArray.reduce(into: [String:[String]]()) {result, element in
        result[element.game, default: []].append(element.name)
    }
    print(reduction)
} catch {
    print(error)
}

或者：

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(game: String, names: [String])] = response.compactMap {
        $0.game.isEmpty ? nil : ($0.game, [$0.name])
    }

    let reduction = Dictionary(userArray) {old, new in old + new}
    print(reduction)
} catch {
    print(error)
}

输出：

["WoW": ["@jade", "@kalel"]]