通过Key对Swift4中的嵌套字典进行排序

Question

目前，我有一个声明为的嵌套字典

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

我得到的结果如下：

["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"]]

我想对这个结果进行排序，以便第一个键按顺序排列结果。

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"]]

我怎样才能在Swift4中有效地做到这一点？请帮忙！

Answer 1

创建一个结构来保存值并将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) }

Answer 2

Dictionary是无序的，你不能对它进行排序。

---

但是，在你的情况下，我会摆脱字典，我会有你的模型的数组。

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 }

---

如果您要将字典作为服务器的响应，您可以使用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 }

Answer 3

你可以这样做：首先将它排序到一个数组，然后用字典元素构建一个数组。这很简单，因为它只有两行代码。

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>>>