Sort Dictionary by Key Value

You need to sort your dictionary values, not your keys. You can create an array of tuples from your dictionary sorting it by its values as follow:

Xcode 9 • Swift 4 or Xcode 8 • Swift 3

let fruitsDict = ["apple": 5, "pear": 9, "grape": 1]
let fruitsTupleArray = fruitsDict.sorted{ $0.value > $1.value }

fruitsTupleArray // [(.0 "pear", .1 9), (.0 "apple", .1 5), (.0 "grape", .1 1)]

for (fruit,votes) in fruitsTupleArray {
    print(fruit,votes)
}

fruitsTupleArray.first?.key   // "pear"
fruitsTupleArray.first?.value   // 9

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To sort your dictionary using your keys

let fruitsTupleArray = fruitsDict.sorted{ $0.key > $1.key }
fruitsTupleArray  // [(key "pear", value 9), (key "grape", value 1), (key "apple", value 5)]

To sort your dictionary using its keys and localized comparison:

let fruitsTupleArray = fruitsDict.sorted { $0.key.localizedCompare($1.key) == .orderedAscending  }

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edit/update:

We can also extend Sequence protocol and implement a custom sort that takes a predicate and sort using a keypath property as long as it conforms to Comparable:

extension Sequence {
    func sorted<T: Comparable>(_ predicate: (Element) -> T, by areInIncreasingOrder: ((T,T)-> Bool) = (<)) -> [Element] {
        sorted(by: { areInIncreasingOrder(predicate($0), predicate($1)) })
    }
}

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Usage:

let sortedFruitsAscending = fruitsDict.sorted(\.value)
print(sortedFruitsAscending)

let sortedFruitsDescending = fruitsDict.sorted(\.value, by: >)
print(sortedFruitsDescending)

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This will print

[(key: “grape”, value: 1), (key: “apple”, value: 5), (key: “pear”, value: 9)]

[(key: “pear”, value: 9), (key: “apple”, value: 5), (key: “grape”, value: 1)]

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edit/update:

For Xcode 13 or later you can use a new generic structure called KeyPathComparator:

let fruitsTupleArray = fruitsDict.sorted(using: KeyPathComparator(\.value, order: .reverse))