我想在Swift中对一个字典进行排序。 我有一个类似于以下的字典：
"A" => Array[]
"Z" => Array[]
"D" => Array[]
等等，我希望它像这样
"A" => Array[]
"D" => Array[]
"Z" => Array[]
我在 Stack Overflow 上尝试了许多解决方案，但没有一个适用于我。我正在使用 XCode6 Beta 5，在它上面有些方法会出现编译错误，有些解决方案会引发异常。所以，如果有人能够发布一个可用的字典排序代码，请帮忙。
在iOS 9 & xcode 7.3，swift 2.2中，“sorted”是不可能的，需要将“sorted”更改为“sort”，就像这样：
let dictionary = ["main course": 10.99, "dessert": 2.99, "salad": 5.99]
let sortedKeysAndValues = Array(dictionary).sort({ $0.0 < $1.0 })
print(sortedKeysAndValues)
//sortedKeysAndValues = ["desert": 2.99, "main course": 10.99, "salad": 5.99]
数组，而不是一个字典。 - user4886069对于Swift 3，以下排序方法按键值返回已排序的字典：
let unsortedDictionary = ["4": "four", "2": "two", "1": "one", "3": "three"]
let sortedDictionary = unsortedDictionary.sorted(by: { $0.0.key < $0.1.key })
print(sortedDictionary)
// ["1": "one", "2": "two", "3": "three", "4": "four"]
Swift 3已按(by:<)排序
let dictionary = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedKeys = Array(dictionary.keys).sorted(by:<) // ["A", "D", "Z"]
对于 Swift 3，以下方法适用于我，而 Swift 2 的语法无法使用：
// menu is a dictionary in this example
var menu = ["main course": 10.99, "dessert": 2.99, "salad": 5.99]
let sortedDict = menu.sorted(by: <)
// without "by:" it does not work in Swift 3
sortedDict是什么？menu又是什么？而且这段代码似乎也不正确。 - random_user_name按键排序的Swift字典
由于Dictionary[关于]是基于哈希函数的，因此无法像我们通常使用的那样使用排序函数（如Array）。作为替代方案，您可以实现一种类似于Java TreeMap/TreeSet的方式。
最简单的方法是使用.sorted(by:)函数。它返回一个已排序的key/value元组的Array，易于处理。
例如：
struct Person {
let name: String
let age: Int
}
struct Section {
let header: String
let rows: [Int]
}
let persons = [
Person(name: "A", age: 1),
Person(name: "B", age: 1),
Person(name: "C", age: 1),
Person(name: "A", age: 2),
Person(name: "B", age: 2),
Person(name: "C", age: 2),
Person(name: "A", age: 3),
Person(name: "B", age: 3),
Person(name: "C", age: 3),
]
//grouped
let personsGroupedByName: [String : [Person]] = Dictionary(grouping: persons, by: { $0.name })
/**
personsGroupedByName
[0] = {
key = "B"
value = 3 values {
[0] = (name = "B", age = 1)
[1] = (name = "B", age = 2)
[2] = (name = "B", age = 3)
}
}
[1] = {
key = "A"
value = 3 values {
[0] = (name = "A", age = 1)
[1] = (name = "A", age = 2)
[2] = (name = "A", age = 3)
}
}
[2] = {
key = "C"
value = 3 values {
[0] = (name = "C", age = 1)
[1] = (name = "C", age = 2)
[2] = (name = "C", age = 3)
}
}
*/
//sort by key
let sortedPersonsGroupedByName: [Dictionary<String, [Person]>.Element] = personsGroupedByName.sorted(by: { $0.0 < $1.0 })
/**
sortedPersonsGroupedByName
[0] = {
key = "A"
value = 3 values {
[0] = (name = "A", age = 1)
[1] = (name = "A", age = 2)
[2] = (name = "A", age = 3)
}
}
[1] = {
key = "B"
value = 3 values {
[0] = (name = "B", age = 1)
[1] = (name = "B", age = 2)
[2] = (name = "B", age = 3)
}
}
[2] = {
key = "C"
value = 3 values {
[0] = (name = "C", age = 1)
[1] = (name = "C", age = 2)
[2] = (name = "C", age = 3)
}
}
*/
//handle
let sections: [Section] = sortedPersonsGroupedByName.compactMap { (key: String, value: [Person]) -> Section in
let rows = value.map { person -> Int in
return person.age
}
return Section(header: key, rows: rows)
}
/**
sections
[0] = {
header = "A"
rows = 3 values {
[0] = 1
[1] = 2
[2] = 3
}
}
[1] = {
header = "B"
rows = 3 values {
[0] = 1
[1] = 2
[2] = 3
}
}
[2] = {
header = "C"
rows = 3 values {
[0] = 1
[1] = 2
[2] = 3
}
}
*/
个人意见，因为所有答案似乎都没有注意到我们有一个字典，并且我们确实想要一个字典:
var menu = ["main course": 10.99, "dessert": 2.99, "salad": 5.99]
let sortedMenu = menu.sorted(by: <) // is an ARRAY
print(type(of: sortedMenu))
let sortedDict = Dictionary( uniqueKeysWithValues: menu.sorted(by: <) )
print(type(of: sortedDict))
以上我按值和键进行了排序：
让我们按照键来排序：
let sorted1 = menu.sorted { (kv1, kv2) in return kv1.key < kv2.key } 或者使用构造函数将其转换为字典。
KeyType替换为您特定字典中键的类型（看起来您正在使用字符串）。 - Sergey Kalinichenko