Is it possible to have a range as a key in a Swift Dictionary? - swift

Question

For simplification. Lets say i have some unique values -> the numbers from 1 to 10
Now I want 1-5 map to the value "first" and I want 6-10 map to the value "second"
Is there a way I can create or extend a dictionary to work like the following?
let dict: [Range<Int> : String]
The goal is to have the following results:
print(dict[1]) // prints first
print(dict[2]) // prints first
print(dict[3]) // prints first
print(dict[7]) // prints second
print(dict[8]) // prints second
print(dict[9]) // prints second
The way I am currently doing it is to simply have the multiple keys map to the same value. But my dictionary can have sometimes 60k values. So I am wondering if a range can work.
I know I can make the value into a class instead of a struct so that multiple keys can map to the same class object, but I was wondering if simply creating a Dictionary that worked like above was possible?

Answer 1

If you insist on using Dictionary, you have to wait until Swift 3.1 (currently in beta):
extension CountableClosedRange : Hashable {
public var hashValue: Int {
return "\(lowerBound) to \(upperBound)".hashValue
}
}
// This feature is called concrete-type extension and requires Swift 3.1
extension Dictionary where Key == CountableClosedRange<Int> {
subscript(rawValue rawValue: Int) -> Value? {
for k in self.keys {
if k ~= rawValue {
return self[k]
}
}
return nil
}
}
let dict : [CountableClosedRange<Int>: String] = [
1...5: "first",
6...10: "second"
]
print(dict[rawValue: 1])
print(dict[rawValue: 2])
print(dict[rawValue: 3])
print(dict[rawValue: 7])
print(dict[rawValue: 8])
print(dict[rawValue: 9])
However, it's a lot clearer if you implement your own data model:
struct MyRange {
var ranges = [CountableClosedRange<Int>]()
var descriptions = [String]()
mutating func append(range: CountableClosedRange<Int>, description: String) {
// You can check for overlapping range here if you want
self.ranges.append(range)
self.descriptions.append(description)
}
subscript(value: Int) -> String? {
for (i, range) in self.ranges.enumerated() {
if range ~= value {
return descriptions[i]
}
}
return nil
}
}
var range = MyRange()
range.append(range: 1...5, description: "one")
range.append(range: 6...10, description: "second")
print(range[1])
print(range[2])
print(range[6])
print(range[7])
print(range[100])

Answer 2

This is in Swift 3.0, it may not be as nice as Code Different's answer though.
class MyRange: Hashable, Equatable {
public var hashValue: Int {
get {
return (self.range.lowerBound + self.range.upperBound).hashValue
}
}
var range: Range<Int>!
public static func ==(_ lhs: MyRange, _ rhs: MyRange) -> Bool {
return lhs.range == rhs.range
}
init(range: Range<Int>) {
self.range = range
}
}
extension Dictionary where Key: MyRange, Value: ExpressibleByStringLiteral {
internal subscript(index: Int) -> [String] {
return self.filter({$0.key.range.contains(index)}).map({$0.value as! String})
}
}
Now, you can make your dictionary like so:
var dict = Dictionary<MyRange, String>()
dict[MyRange(range: 0..<5)] = "first"
dict[MyRange(range: 5..<10)] = "second"
Getting values works with Integers and Ranges:
print(dict[1]) // ["first"]
print(dict[5]) // ["second"]
print(dict[11]) // []
print(dict[MyRange(range: 0..<5)]) // "first"
print(dict[MyRange(range: 0..<6)]) // nil
The dictionary should look like this:
print(dict)
// [MyRange: "first", MyRange: "second"]