How can I sort dictionary [duplicate] - swift

Question

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.

Answer 1

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.

Answer 2

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 })

Answer 3

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 })

Answer 4

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)")
}

Answer 5

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 }))

Answer 6

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 })

Answer 7

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]

Answer 8

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.

Answer 9

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
}

Answer 10

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]
}

Answer 11

"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]

Answer 12

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"]

Answer 13

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

Answer 14

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"]

Answer 15

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
}
}
*/

Answer 16

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.