在 Swift4 中按键对嵌套字典进行排序答案

【问题标题】：Sorting nested dictionary in Swift4 by Key在 Swift4 中按键对嵌套字典进行排序
【发布时间】：2019-08-12 19:41:39
【问题描述】：

目前，我有一个嵌套字典声明为

var userSchedule = [String:[String:String]]()

我得到如下结果：

["2": ["time": "tbd", "date": "1/15", "name": "Chris"], 
 "0": ["time": "6:16 PM", "date": "1/15", "name": "Bob"], 
 "1": ["time": "1:15PM", "date": "1/15", "name": "John"]]

我想对该结果进行排序，以便结果按第一个键排序。

所以：

["0": ["time": "6:16 PM", "date": "1/15", "name": "Bob"], 
 "1": ["time": "1:15PM", "date": "1/15", "name": "John"], 
 "2": ["time": "tbd", "date": "1/15", "name": "Chris"]]

如何在 Swift4 中有效地做到这一点？请帮忙！

【问题讨论】：

Dictionary 是无序的，不能排序。无论如何，您可以从中创建Array
字典是无序集合，不能排序。您需要使用其他一些集合并实现逻辑来转换此字典并对其进行排序。提示：字典数组，其中每个索引表示其对应字典键的数值。
我建议您创建结构而不是使用字典。然后，您可以拥有一个结构数组并对其进行排序
How to sort Dictionary by keys where values are array of objects in Swift 4?的可能重复
Sort Dictionary by keys的可能重复

标签： ios swift sorting dictionary swift4

【解决方案1】：

创建一个结构来保存值并将您的字典映射到该结构的数组

struct Schedule {
    let key: Int?
    let time: String?
    let date: String?
    let name: String?
}

var orderedSchedule = userSchedule.map { Schedule(key: Int($0.key), 
                                                  time: $0.value["time"], 
                                                  date: $0.value["date"], 
                                                  name: $0.value["name"]) }

orderedSchedule.sort { ($0.key ?? 0) < ($1.key ?? 0) }

【讨论】：

【解决方案2】：

Dictionary 无序，无法排序。

但是，在你的情况下，我会摆脱字典，我只会有你的模型数组。

创建代表您的事件的自定义 struct，然后您可以按模型的 id 属性对数组进行排序

struct Event {
    let time, date, name: String
    let id: Int
}

var userSchedule = [Event(time: "tbd", date: "1/15", name: "name", id: 2),
                    Event(time: "6:16 PM", date: "1/15", name: "Chris", id: 0),
                    Event(time: "1:15PM", date: "1/15", name: "John", id: 1)]

userSchedule.sort { $0.id < $1.id }

例如，如果您要从服务器获取字典作为响应，则可以使用 compactMap 将其重新映射到 Event 数组

struct Event {
    let time, date, name: String
    let id: Int
}

var userSchedule = ["2": ["time": "tbd", "date": "1/15", "name": "Chris"],
                    "0": ["time": "6:16 PM", "date": "1/15", "name": "Bob"],
                    "1": ["time": "1:15PM", "date": "1/15", "name": "John"]]

var schedule = userSchedule.compactMap { (event) -> Event? in
    guard let time = event.value["time"], let date = event.value["date"], let name = event.value["name"], let id = Int(event.key) else { return nil }
    return Event(time: time, date: date, name: name, id: id)
}

schedule.sort { $0.id < $1.id }

【讨论】：

【解决方案3】：

你可以这样做：首先将它排序为一个数组，然后用字典元素构建一个数组。很简单，只有两行代码。

var userSchedule = ["2": ["time": "tbd", "date": "1/15", "name": "Chris"],
"0": ["time": "6:16 PM", "date": "1/15", "name": "Bob"],
"1": ["time": "1:15PM", "date": "1/15", "name": "John"]]

let userScheduleArray = userSchedule.sorted{$0.key < $1.key}

print(userScheduleArray) 
//[(key: "0", value: ["time": "6:16 PM", "date": "1/15", "name": "Bob"]), (key: "1", value: ["time": "1:15PM", "date": "1/15", "name": "John"]), (key: "2", value: ["time": "tbd", "date": "1/15", "name": "Chris"])]

let separatedDict = userScheduleArray.map{Dictionary(uniqueKeysWithValues: [$0])}

print(separatedDict)
//[["0": ["time": "6:16 PM", "date": "1/15", "name": "Bob"]], 
// ["1": ["time": "1:15PM", "date": "1/15", "name": "John"]], 
// ["2": ["time": "tbd", "date": "1/15", "name": "Chris"]]]

print(type(of: separatedDict))
//[[String:[String:String]]] i.e. Array<Dictionary<String, Dictionary<String, String>>>

【讨论】：