是否可以将范围作为Swift词典中的键？

Question

为简化起见.让我们说我有一些独特的价值 - >来自的数字1 to 10

现在我想要1-5映射到值"first",我想要6-10映射到值"second"

有没有办法可以创建或扩展字典以便像下面一样工作？

let dict: [Range<Int> : String]

目标是获得以下结果:

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

我目前的做法是简单地将多个键映射到相同的值.但我的词典有时可以有60k的值.所以我想知道一个范围是否可行.

我知道我可以将值设置为a class而不是struct使得多个键可以映射到同一个类对象,但我想知道是否只是创建一个像上面那样工作的Dictionary是可能的吗？

Answer 1

如果你坚持使用Dictionary,你必须等到Swift 3.1(目前处于测试阶段):

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])

但是,如果您实现自己的数据模型,它会更清晰:

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])