reduce 函数正在打印一个空字典 [:]

Question

我在这个中成功地减少了我的字典键作为伪代码，没有真正的 json 模型。我在上一个问题中完成的目标是 return 只有具有匹配值的键 。所以输出是一个看起来像这样的字典 ["WoW": ["@jade", "@kalel"]。正是我所需要的。当然可能还有其他比赛，我也想return那些。

现在我有了一个合适的 json 模型，reduce 函数正在打印一个空字典 [:]。是 .reduce(into: [String:[String]]() 中的类型导致了问题吗？

所有数据都在打印，所以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[[=12=]] ?? "" }.reduce(into: [String:[String]](), { (result, element) in
            if element.value.count > 1 {
                result[element.key] = element.value
            }
        })
        // error catch etc
}

Answer 1

你的代码太复杂了。您可以通过 game 简单地使用

对数组进行分组

let response = try decoder.decode([UserModel].self, from: data)
let reduction = Dictionary(grouping: response, by: {[=10=].game}).mapValues{ usermodel in usermodel.map{ [=10=].name}}

Answer 2

更新我可能误会了你想要得到的东西。下面还有一个代码，请检查结果并选择一个你想要的。

如果想用reduce(into:updateAccumulatingResult:)，可以这样写

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(name: String, game: String)] = response.map {([=10=].name, [=10=].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 的初始值设定项，这可能有效：

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(name: String, games: [String])] = response.map {
        ([=11=].name, [=11=].game.isEmpty ? [] : [[=11=].game])
    }

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

双输出：

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

无论如何，除了 userDict.keys 之外，您组合循环 Dictionary(grouping:) 和 reduce(into:) 的方式使事情变得过于复杂。

ADDITION 当你想获得一个带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 {
        [=13=].game.isEmpty ? nil : ([=13=].game, [=13=].name)
    }

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

或：

do {
    let response = try decoder.decode([UserModel].self, from: data)
    let userArray: [(game: String, names: [String])] = response.compactMap {
        [=14=].game.isEmpty ? nil : ([=14=].game, [[=14=].name])
    }

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

输出：

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

reduce 函数正在打印一个空字典 [:]

reduce function is printing an empty dictionary [:]

reduce

json

dictionary

swift

codable