Sort Dictionary by keys

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

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 4

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 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"]