How do I get the key at a specific index from a Dictionary in Swift?

That's because keys returns LazyMapCollection<[Key : Value], Key>, which can't be subscripted with an Int. One way to handle this is to advance the dictionary's startIndex by the integer that you wanted to subscript by, for example:

let intIndex = 1 // where intIndex < myDictionary.count
let index = myDictionary.index(myDictionary.startIndex, offsetBy: intIndex)
myDictionary.keys[index]

Another possible solution would be to initialize an array with keys as input, then you can use integer subscripts on the result:

let firstKey = Array(myDictionary.keys)[0] // or .first

Remember, dictionaries are inherently unordered, so don't expect the key at a given index to always be the same.

Swift 3 : Array() can be useful to do this .

Get Key :

let index = 5 // Int Value
Array(myDict)[index].key

Get Value :

Array(myDict)[index].value

Here is a small extension for accessing keys and values in dictionary by index:

extension Dictionary {
    subscript(i: Int) -> (key: Key, value: Value) {
        return self[index(startIndex, offsetBy: i)]
    }
}

You can iterate over a dictionary and grab an index with for-in and enumerate (like others have said, there is no guarantee it will come out ordered like below)

let dict = ["c": 123, "d": 045, "a": 456]

for (index, entry) in enumerate(dict) {
    println(index)   // 0       1        2
    println(entry)   // (d, 45) (c, 123) (a, 456)
}

If you want to sort first..

var sortedKeysArray = sorted(dict) { $0.0 < $1.0 }
println(sortedKeysArray)   // [(a, 456), (c, 123), (d, 45)]

var sortedValuesArray = sorted(dict) { $0.1 < $1.1 }
println(sortedValuesArray) // [(d, 45), (c, 123), (a, 456)]

then iterate.

for (index, entry) in enumerate(sortedKeysArray) {
    println(index)    // 0   1   2
    println(entry.0)  // a   c   d
    println(entry.1)  // 456 123 45
}

If you want to create an ordered dictionary, you should look into Generics.