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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.
I am attempting to remove "" & " " from the back of a string array until the last item contains some text, but my implementation isn't picking up " ".
My implementation so far:
var array = ["A", "B", "", "C", "D", " ", " ", ""]
while true {
if (array.last == " " || array.last == "") {
array.removeLast()
} else {
break
}
}
My desired output is
["A", "B", "", "C", "D"]
, but my current output is
["A", "B", "", "C", "D", " ", " "]
, where the while loop simply breaks after encountering " "
Any advice why is it not picking up the " "?
I don't know why they have drop(while:) and did not implement dropLast(while:). The implementation bellow works on any collection:
extension Collection {
func dropLast(while predicate: (Element) throws -> Bool) rethrows -> SubSequence {
guard let index = try indices.reversed().first(where: { try !predicate(self[$0]) }) else {
return self[startIndex..<startIndex]
}
return self[...index]
}
}
"123".dropLast(while: \.isWholeNumber) // ""
"abc123".dropLast(while: \.isWholeNumber) // "abc"
"123abc".dropLast(while: \.isWholeNumber) // "123abc"
And extending RangeReplaceableCollection we can implement remove(while:) and removeLast(while:) as well:
extension RangeReplaceableCollection {
mutating func remove(while predicate: (Element) throws -> Bool) rethrows {
guard let index = try indices.first(where: { try !predicate(self[$0]) }) else {
removeAll()
return
}
removeSubrange(..<index)
}
mutating func removeLast(while predicate: (Element) throws -> Bool) rethrows {
guard let index = try indices.reversed().first(where: { try !predicate(self[$0]) }) else {
removeAll()
return
}
removeSubrange(self.index(after: index)...)
}
}
var string = "abc123"
string.removeLast(while: \.isWholeNumber)
string // "abc"
var string2 = "abc123"
string2.remove(while: \.isLetter)
string2 // "123"
var array = ["A", "B", "", "C", "D", " ", " ", ""]
array.removeLast { $0 == "" || $0 == " " }
array // ["A", "B", "", "C", "D"]
One way to solve this is to reverse the collection (which is done lazily) and drop the unwanted items until you encounter the wanted ones. Afterwards, reverse the collection again.
let array = ["A", "B", "", "C", "D", " ", " ", ""]
let filtered = array.reversed().drop(while: {
$0.trimmingCharacters(in: .whitespacesAndNewlines).isEmpty
}).reversed() as [String]
print(filtered) // "["A", "B", "", "C", "D"]\n"
Note that the check for " " may fail if it's not a normal space, for example a non-breaking space (Unicode checkpoint U+00A0). This may be the issue you're having in the first place. So trim the string (it removes characters from the start and end only) and check whether the result is an empty string.
Move your condition to while and make sure you're checking on the correct array after the operation.
var array = ["A", "B", "", "C", "D", " ", " ", ""]
while array.last == " " || array.last == "" {
array.removeLast()
}
print(array) // ["A", "B", "", "C", "D"]
Just for fun, lets extend Array with this functionality in a generic way while also externally providing the condition for more flexibility.
Similar to Arrays having a drop(while:), we can make a dropLast(while:) like so:
extension Array {
func dropLast(while handler: (Element)->Bool) -> Array {
var array = self
while let last = array.last, handler(last) {
array.removeLast()
}
return array
}
}
Usage Example:
let array = ["", "A", "B", "", "C", "D", " ", " ", ""]
let modified = array.dropLast { $0.trimmingCharacters(in: .whitespaces).isEmpty }
print(modified) //["", "A", "B", "", "C", "D"]
Bonus:
It can handle other types of arrays too, and since the condition is not baked into the functionality, it's flexible and reusable.
let array = [0, 1, 2, 3, 0, 5, 6, 7, 0, -1, 0, -2]
//Drop (from tail) all numbers less than 1
let modified = array.dropLast(while: { (val) -> Bool in
return val < 1
})
print(modified) //[0, 1, 2, 3, 0, 5, 6, 7]
Basically your solution plays fine. But you can make it more generic for "", " ", "", " ", .... :
import Foundation
var array = ["A", "B", "", "C", "D", " ", " ", ""]
while true {
let shouldRemoveLast = array.last?.trimmingCharacters(in: .whitespacesAndNewlines).isEmpty ?? false
if (shouldRemoveLast) { array.removeLast() } else { break }
}
For completeness, here's a one liner, functional programming style, extension, that removes the last elements that satisfy the given criteria:
extension Collection {
func dropLast(while predicate: (Element) -> Bool) -> SubSequence {
indices.reversed().first { !predicate(self[$0]) }.map(prefix(through:)) ?? prefix(0)
}
}
Same code, but in a more imperative style:
extension Collection {
func dropLast(while predicate: (Element) -> Bool) -> SubSequence {
if let lastValidIndex = indices.reversed().first(where: { !predicate(self[$0]) }) {
return prefix(through: lastValidIndex)
} else {
return prefix(0)
}
}
}
I would like my struct to print its entries in alphabetical order first, then arrange the data in descending order. So the final result would be: "lukes 9", "lukes 4", "smiths 4"
struct MyData {
var company = String()
var score: Int
}
let data = [
MyData(company: "smiths", score: 4 ),
MyData(company: "lukes", score: 4),
MyData(company: "lukes", score: 9)
]
There's 2 ways you can do this. Both would require you to pass your array to sort (Swift 2), now sorted (Swift 3).
A very easy implementation:
struct MyData {
var company = String()
var score: Int
}
let data = [
MyData(company: "smiths", score: 4),
MyData(company: "lukes", score: 4),
MyData(company: "lukes", score: 9)
]
let sortedArray = data.sorted(by: { ($0.company, $1.score) < ($1.company, $0.score) })
You could also make MyData conform to Comparable. This keeps the comparison logic within the MyData type, and you can just run the sorted() function to return a new array:
struct MyData {
var company = String()
var score: Int
}
extension MyData: Equatable {
static func ==(lhs: MyData, rhs: MyData) -> Bool {
return (lhs.company, lhs.score) == (rhs.company, rhs.score)
}
}
extension MyData: Comparable {
static func <(lhs: MyData, rhs: MyData) -> Bool {
return (rhs.company, lhs.score) > (lhs.company, rhs.score)
}
}
let data = [
MyData(company: "smiths", score: 4),
MyData(company: "lukes", score: 4),
MyData(company: "lukes", score: 9)
]
let sortedArray = data.sorted()
These 2 implementations both output your desired result: "lukes 9", "lukes 4", "smiths 4"
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.