Sort Dictionary by values in Swift
Try:
let dict = ["a":1, "c":3, "b":2]
extension Dictionary {
func sortedKeys(isOrderedBefore:(Key,Key) -> Bool) -> [Key] {
return Array(self.keys).sort(isOrderedBefore)
}
// Slower because of a lot of lookups, but probably takes less memory (this is equivalent to Pascals answer in an generic extension)
func sortedKeysByValue(isOrderedBefore:(Value, Value) -> Bool) -> [Key] {
return sortedKeys {
isOrderedBefore(self[$0]!, self[$1]!)
}
}
// Faster because of no lookups, may take more memory because of duplicating contents
func keysSortedByValue(isOrderedBefore:(Value, Value) -> Bool) -> [Key] {
return Array(self)
.sort() {
let (_, lv) = $0
let (_, rv) = $1
return isOrderedBefore(lv, rv)
}
.map {
let (k, _) = $0
return k
}
}
}
dict.keysSortedByValue(<)
dict.keysSortedByValue(>)
Updated:
Updated to the new array syntax and sort semantics from beta 3. Note that I'm using sort
and not sorted
to minimize array copying. The code could be made more compact, by looking at the earlier version and replacing sort
with sorted
and fixing the KeyType[]
to be [KeyType]
Updated to Swift 2.2:
Changed types from KeyType
to Key
and ValueType
to Value
. Used new sort
builtin to Array
instead of sort(Array)
Note performance of all of these could be slightly improved by using sortInPlace
instead of sort
You could use something like this perhaps:
var dict = ["cola" : 10, "fanta" : 12, "sprite" : 8]
var myArr = Array(dict.keys)
var sortedKeys = sort(myArr) {
var obj1 = dict[$0] // get ob associated w/ key 1
var obj2 = dict[$1] // get ob associated w/ key 2
return obj1 > obj2
}
myArr // ["fanta", "cola", "sprite"]
Just one line code to sort dictionary by Values in Swift 4, 4.2 and Swift 5:
let sortedByValueDictionary = myDictionary.sorted { $0.1 < $1.1 }