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I want to sort a dictionary in Swift. I have a dictionary like:
"A" => Array[]
"Z" => Array[]
"D" => Array[]
etc. I want it to be like
"A" => Array[]
"D" => Array[]
"Z" => Array[]
etc.
I have tried many solutions on SO but no one worked for me. I am using XCode6 Beta 5 and on it some are giving compiler error and some solutions are giving exceptions. So anyone who can post the working copy of dictionary sorting.
let dictionary = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedKeys = Array(dictionary.keys).sorted(<) // ["A", "D", "Z"]
EDIT:
The sorted array from the above code contains keys only, while values have to be retrieved from the original dictionary. However, 'Dictionary' is also a 'CollectionType' of (key, value) pairs and we can use the global 'sorted' function to get a sorted array containg both keys and values, like this:
let sortedKeysAndValues = sorted(dictionary) { $0.0 < $1.0 }
println(sortedKeysAndValues) // [(A, [1, 2]), (D, [5, 6]), (Z, [3, 4])]
EDIT2: The monthly changing Swift syntax currently prefers
let sortedKeys = Array(dictionary.keys).sort(<) // ["A", "D", "Z"]
The global sorted is deprecated.
To be clear, you cannot sort Dictionaries. But you can out put an array, which is sortable.
Swift 2.0
Updated version of Ivica M's answer:
let wordDict = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedDict = wordDict.sort { $0.0 < $1.0 }
print("\(sortedDict)") //
Swift 3
wordDict.sorted(by: { $0.0 < $1.0 })
In Swift 5, in order to sort Dictionary by KEYS
let sortedYourArray = YOURDICTIONARY.sorted( by: { $0.0 < $1.0 })
In order to sort Dictionary by VALUES
let sortedYourArray = YOURDICTIONARY.sorted( by: { $0.1 < $1.1 })
If you want to iterate over both the keys and the values in a key sorted order, this form is quite succinct
let d = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
Swift 1,2:
for (k,v) in Array(d).sorted({$0.0 < $1.0}) {
println("\(k):\(v)")
}
Swift 3+:
for (k,v) in Array(d).sorted(by: {$0.0 < $1.0}) {
println("\(k):\(v)")
}
I tried all of the above, in a nutshell all you need is
let sorted = dictionary.sorted { $0.key < $1.key }
let keysArraySorted = Array(sorted.map({ $0.key }))
let valuesArraySorted = Array(sorted.map({ $0.value }))
In swift 4 you can write it smarter:
let d = [ 1 : "hello", 2 : "bye", -1 : "foo" ]
d = [Int : String](uniqueKeysWithValues: d.sorted{ $0.key < $1.key })
Swift 4 & 5
For string keys sorting:
dictionary.keys.sorted(by: {$0.localizedStandardCompare($1) == .orderedAscending})
Example:
var dict : [String : Any] = ["10" : Any, "2" : Any, "20" : Any, "1" : Any]
dictionary.keys.sorted()
["1" : Any, "10" : Any, "2" : Any, "20" : Any]
dictionary.keys.sorted(by: {$0.localizedStandardCompare($1) == .orderedAscending})
["1" : Any, "2" : Any, "10" : Any, "20" : Any]
Swift 5
Input your dictionary that you want to sort alphabetically by keys.
// Sort inputted dictionary with keys alphabetically.
func sortWithKeys(_ dict: [String: Any]) -> [String: Any] {
let sorted = dict.sorted(by: { $0.key < $1.key })
var newDict: [String: Any] = [:]
for sortedDict in sorted {
newDict[sortedDict.key] = sortedDict.value
}
return newDict
}
dict.sorted(by: { $0.key < $1.key }) by it self returns a tuple (value, value) instead of a dictionary [value: value]. Thus, the for loop parses the tuple to return as a dictionary. That way, you put in a dictionary & get a dictionary back.
For Swift 4 the following has worked for me:
let dicNumArray = ["q":[1,2,3,4,5],"a":[2,3,4,5,5],"s":[123,123,132,43,4],"t":[00,88,66,542,321]]
let sortedDic = dicNumArray.sorted { (aDic, bDic) -> Bool in
return aDic.key < bDic.key
}
This is an elegant alternative to sorting the dictionary itself:
As of Swift 4 & 5
let sortedKeys = myDict.keys.sorted()
for key in sortedKeys {
// Ordered iteration over the dictionary
let val = myDict[key]
}
"sorted" in iOS 9 & xcode 7.3, swift 2.2 is impossible, change "sorted" to "sort", like this:
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]
For Swift 3, the following sort returnes sorted dictionary by keys:
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"]
For Swift 3 the following has worked for me and the Swift 2 syntax has not worked:
// 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
Swift 3 is sorted(by:<)
let dictionary = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedKeys = Array(dictionary.keys).sorted(by:<) // ["A", "D", "Z"]
Swift Sort Dictionary by keys
Since Dictionary[About] is based on hash function it is not possible to use a sort function as we are used to(like Array). As an alternative you are able to implement a kind of Java TreeMap/TreeSet
Easiest way is to use .sorted(by:) function. It returns an sorted Array of key/value tupples which is easy to handle
Example:
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
}
}
*/
my two cents, as all answers seem to miss we have a Dict and we do want a Dict:
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))
Above I sorted by value, by Key:
let sorted1 = menu.sorted { (kv1, kv2) in
return kv1.key < kv2.key
}
and/or apply conversion to Dict using constructor.
I want to convert a dictionary to an Array, by showing each [String: [String]] element as a string in the array (but the value may be empty).
So ["A": ["1","2"], "b": [], "c": ["5", "6"]] will give ["b", "A1", "A2", "c5", "c6"].
I want to use map to do this, as the code I have feels unwieldy:
let messages: [String: [String]] = ["A": ["1","2"], "b": [], "c": ["5", "6"]]
var result: [String] = []
for message in messages {
if !message.value.isEmpty {
for value in message.value {
result.append(message.key + value)
}
} else {
result.append(message.key)
}
}
How can I create the solution using map?
A little bit shortened version using map
let messages: [String: [String]] = ["A": ["1","2"], "b": [], "c": ["5", "6"]]
let result = messages.map { dict in
return dict.value.isEmpty ? ["\(dict.key)"] : dict.value.map { "\(dict.key)\($0)" }
}.flatMap{$0}
A possible solution is to use reduce(into:_:), (I prefer this one):
let messages: [String: [String]] = ["A": ["1","2"], "b": [], "c": ["5", "6"]]
let output = messages.reduce(into: [String]()) { result, current in
guard !current.value.isEmpty else { result.append(current.key); return }
let sub = current.value.map { current.key + $0 }
result.append(contentsOf: sub)
}
print("Output: \(output)")
Output:
$> Output: ["c5", "c6", "A1", "A2", "b"]
With a map and a flatMap, I found it less intuitive:
let output2 = messages.map { element -> [String] in
guard !element.value.isEmpty else { return [element.key] }
let sub = element.value.map { element.key + $0 }
return sub
}.flatMap { $0 }
print("Output2: \(output2)")
Which then can be simplified with a direct call to flatMap, but that (map + flatMap) was to show the process, since this one will save you an iteration (you do only one instead of two):
let output3 = messages.flatMap { element -> [String] in
guard !element.value.isEmpty else { return [element.key] }
let sub = element.value.map { element.key + $0 }
return sub
}
print("Output3: \(output3)")
I made the closure explicit enough, but that's up to you if you want to shorten them with ternary if etc.
Given this dictionary as an example:
var d = ["a": ["b": ["c": 30, "d": 20] ] ]
I would like to write a function that could add/update a value into this dictionary.
First example, if I call :
d.set(40, at: ["a", "b", "c"])
The function would progressively go into key a, then b, then c, and transform this dictionary into:
var d = ["a": ["b": ["c": 40, "d": 20] ] ]
Now if I call this instead:
d.set(60, at: ["a", "e", "f"])
I would like to get this dictionary instead:
var d = ["a": ["b": ["c": 40, "d": 20] ], "e": ["f": 60] ]
Could you help me writing this function?
Thank you for your help
Your example only requires this. Is your question actually larger than that? If so, please amend!
extension Dictionary {
subscript<Value>(_ key0: Key, _ key1: Key, _ key2: Key) -> Value?
where Self.Value == [Key: [Key: Value]] {
get { self[key0]?[key1]?[key2] }
set { self[key0, default: [:]][key1, default: [:]][key2] = newValue }
}
}
d["a", "b", "c"] // 30
d["a", "b", "c"] = 40
d["a", "e", "f"] // nil
d["a", "e", "f"] = 60
This seems to do the trick:
Edit Now this covers all the provided uses cases
extension Dictionary where Value == Any {
mutating func set<V>(_ value: V, at keyPath: [Key]) {
var keyPath = keyPath
guard !keyPath.isEmpty else { return }
let topLevel = keyPath.removeFirst()
guard !keyPath.isEmpty else {
self[topLevel] = value
return
}
guard keyPath.count > 1 else {
self = merge(with: [topLevel: [keyPath.last!: value]])
return
}
let leafKey = keyPath.removeLast()
let nested = [leafKey: value].nest(in: keyPath)
guard let child = self[topLevel] as? [Key: Any] else {
self[topLevel] = nested
return
}
self = self.merge(with: [topLevel: child.merge(with: nested)])
}
func merge(with other: [Key: Any] ) -> [Key: Any] {
var result = self
for key in other.keys {
if !keys.contains(key) {
result[key] = other[key]
} else {
switch (self[key], other[key]) {
case let (lhs, rhs) as ([Key: Any], [Key: Any]):
result[key] = lhs.merge(with: rhs)
default:
result[key] = other[key]
}
}
}
return result
}
func nest(in keys: [Key]) -> [Key: Any] {
let initial: [Key: Any] = self as [Key: Any]
let result: [Key: Any] = keys.reversed().reduce(into: initial) { result, key in
result = [key: result]
}
return result
}
}
var d: [String: Any] = ["a": ["b": ["c": 30, "d": 20] ] ] // ["a": ["b": ["c": 30, "d": 20]]]
d.set(40, at: ["a", "b", "c"]) // ["a": ["b": ["c": 40, "d": 20]]]
d.set(60, at: ["a", "e", "f"]) // ["a": ["b": ["c": 40, "d": 20], "e": ["f": 60]]]
d.set(["f": 42], at: ["a", "d"]) // ["a": ["b": ["c": 40, "d": 20], "e": ["f": 60], "d": ["f": 42]]]
I am wanted to make a search on NSMUtableArray with key values to catch the hole object not just an array of values so for example my NSMUtableArray return me
{id : 1, name : x1}
{id : 2, name : x2}
{id : 3, name : x3}
{id : 4, name : y1}
{id : 5, name : y2}
if I am searching of x my code return me
{x1,x2,x3}
or I want it to return
{id : 1, name : x1}
{id : 2, name : x2}
{id : 3, name : x3}
my code is :
let arrayy = lstResults.value(forKey: "name")
let searchTest = NSPredicate(format: "SELF BEGINSWITH [c] %#", searchBar.text!)
let array = (arrayy as! NSArray).filtered(using: searchTest)
filteredTableDataName = array as NSArray
please help I tried a lot of stuff but nothing worked form me.
Instead of extracting and filtering the "name" values you can filter the array directly, using the corresponding key path instead of “SELF” in the predicate:
let array = NSArray(array: [
["id": 1, "name": "x1"], ["id": 2, "name": "x2"], ["id": 3, "name": "y3"]])
let predicate = NSPredicate(format: "name BEGINSWITH [c] %#", "x")
let filtered = array.filtered(using: predicate)
print(filtered)
Result:
[{
id = 1;
name = x1;
}, {
id = 2;
name = x2;
}]
Another option is to cast the NS(Mutable)Array to a Swift array of dictionaries, and use the filter() function:
if let a = array as? [[String: Any]] {
let filtered = a.filter {
($0["name"] as? String)?
.range(of: "x", options: [.caseInsensitive, .anchored]) != nil
}
print(filtered)
// [["id": 1, "name": x1], ["name": x2, "id": 2]]
}
NSMutableArray (or NSArray for that matter) are leftovers from the 'Objective-c' era.
For this purpose there is not real need using it or not using Swift's standard generic Array<Element> => https://developer.apple.com/documentation/swift/array
Might I consider creating a struct instead of using a dictionary:
import Foundation
struct Obj {
let id: Int
let name: String
}
let array = [
Obj(id: 1, name: "x1"),
Obj(id: 2, name: "x2"),
Obj(id: 3, name: "X3"),
Obj(id: 4, name: "y1"),
Obj(id: 5, name: "y2")
]
let filteredArray = array.filter { $0.name.localizedLowercase.hasPrefix("x") }
// filteredArray = [
// Obj(id: 1, name: "x1"),
// Obj(id: 2, name: "x2"),
// Obj(id: 3, name: "X3")
// ]
Else you would have something like this:
import Foundation
let array: [[String: Any]] = [
[ "id": 1, "name": "x1" ],
[ "id": 2, "name": "x2" ],
[ "id": 3, "name": "X3" ],
[ "id": 4, "name": "y1" ],
[ "id": 5, "name": "y2" ]
]
let filteredArray = array.filter { ($0["name"] as? String)?.localizedLowercase.hasPrefix("x") == true }
I want to sort a dictionary in Swift. I have a dictionary like:
"A" => Array[]
"Z" => Array[]
"D" => Array[]
etc. I want it to be like
"A" => Array[]
"D" => Array[]
"Z" => Array[]
etc.
I have tried many solutions on SO but no one worked for me. I am using XCode6 Beta 5 and on it some are giving compiler error and some solutions are giving exceptions. So anyone who can post the working copy of dictionary sorting.
let dictionary = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedKeys = Array(dictionary.keys).sorted(<) // ["A", "D", "Z"]
EDIT:
The sorted array from the above code contains keys only, while values have to be retrieved from the original dictionary. However, 'Dictionary' is also a 'CollectionType' of (key, value) pairs and we can use the global 'sorted' function to get a sorted array containg both keys and values, like this:
let sortedKeysAndValues = sorted(dictionary) { $0.0 < $1.0 }
println(sortedKeysAndValues) // [(A, [1, 2]), (D, [5, 6]), (Z, [3, 4])]
EDIT2: The monthly changing Swift syntax currently prefers
let sortedKeys = Array(dictionary.keys).sort(<) // ["A", "D", "Z"]
The global sorted is deprecated.
To be clear, you cannot sort Dictionaries. But you can out put an array, which is sortable.
Swift 2.0
Updated version of Ivica M's answer:
let wordDict = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedDict = wordDict.sort { $0.0 < $1.0 }
print("\(sortedDict)") //
Swift 3
wordDict.sorted(by: { $0.0 < $1.0 })
In Swift 5, in order to sort Dictionary by KEYS
let sortedYourArray = YOURDICTIONARY.sorted( by: { $0.0 < $1.0 })
In order to sort Dictionary by VALUES
let sortedYourArray = YOURDICTIONARY.sorted( by: { $0.1 < $1.1 })
If you want to iterate over both the keys and the values in a key sorted order, this form is quite succinct
let d = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
Swift 1,2:
for (k,v) in Array(d).sorted({$0.0 < $1.0}) {
println("\(k):\(v)")
}
Swift 3+:
for (k,v) in Array(d).sorted(by: {$0.0 < $1.0}) {
println("\(k):\(v)")
}
I tried all of the above, in a nutshell all you need is
let sorted = dictionary.sorted { $0.key < $1.key }
let keysArraySorted = Array(sorted.map({ $0.key }))
let valuesArraySorted = Array(sorted.map({ $0.value }))
In swift 4 you can write it smarter:
let d = [ 1 : "hello", 2 : "bye", -1 : "foo" ]
d = [Int : String](uniqueKeysWithValues: d.sorted{ $0.key < $1.key })
Swift 4 & 5
For string keys sorting:
dictionary.keys.sorted(by: {$0.localizedStandardCompare($1) == .orderedAscending})
Example:
var dict : [String : Any] = ["10" : Any, "2" : Any, "20" : Any, "1" : Any]
dictionary.keys.sorted()
["1" : Any, "10" : Any, "2" : Any, "20" : Any]
dictionary.keys.sorted(by: {$0.localizedStandardCompare($1) == .orderedAscending})
["1" : Any, "2" : Any, "10" : Any, "20" : Any]
Swift 5
Input your dictionary that you want to sort alphabetically by keys.
// Sort inputted dictionary with keys alphabetically.
func sortWithKeys(_ dict: [String: Any]) -> [String: Any] {
let sorted = dict.sorted(by: { $0.key < $1.key })
var newDict: [String: Any] = [:]
for sortedDict in sorted {
newDict[sortedDict.key] = sortedDict.value
}
return newDict
}
dict.sorted(by: { $0.key < $1.key }) by it self returns a tuple (value, value) instead of a dictionary [value: value]. Thus, the for loop parses the tuple to return as a dictionary. That way, you put in a dictionary & get a dictionary back.
For Swift 4 the following has worked for me:
let dicNumArray = ["q":[1,2,3,4,5],"a":[2,3,4,5,5],"s":[123,123,132,43,4],"t":[00,88,66,542,321]]
let sortedDic = dicNumArray.sorted { (aDic, bDic) -> Bool in
return aDic.key < bDic.key
}
This is an elegant alternative to sorting the dictionary itself:
As of Swift 4 & 5
let sortedKeys = myDict.keys.sorted()
for key in sortedKeys {
// Ordered iteration over the dictionary
let val = myDict[key]
}
"sorted" in iOS 9 & xcode 7.3, swift 2.2 is impossible, change "sorted" to "sort", like this:
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]
For Swift 3, the following sort returnes sorted dictionary by keys:
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"]
For Swift 3 the following has worked for me and the Swift 2 syntax has not worked:
// 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
Swift 3 is sorted(by:<)
let dictionary = [
"A" : [1, 2],
"Z" : [3, 4],
"D" : [5, 6]
]
let sortedKeys = Array(dictionary.keys).sorted(by:<) // ["A", "D", "Z"]
Swift Sort Dictionary by keys
Since Dictionary[About] is based on hash function it is not possible to use a sort function as we are used to(like Array). As an alternative you are able to implement a kind of Java TreeMap/TreeSet
Easiest way is to use .sorted(by:) function. It returns an sorted Array of key/value tupples which is easy to handle
Example:
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
}
}
*/
my two cents, as all answers seem to miss we have a Dict and we do want a Dict:
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))
Above I sorted by value, by Key:
let sorted1 = menu.sorted { (kv1, kv2) in
return kv1.key < kv2.key
}
and/or apply conversion to Dict using constructor.