Swift - `范围内的刻度数 min - max

Question

如何缩放数字以适应指定范围？例如，如果我有：

Input_min = 10.1;
Input_max = 750.0;

Range_min = 0.0;
Range_max = 350.0;

example Input_n = 503.9;

如何使用swift在最小和最大范围内缩放任何数字？

Answer 1

编辑/更新： Xcode 11.5 • Swift 5.2

extension FloatingPoint {
    func converting(from input: ClosedRange<Self>, to output: ClosedRange<Self>) -> Self {
        let x = (output.upperBound - output.lowerBound) * (self - input.lowerBound)
        let y = (input.upperBound - input.lowerBound)
        return x / y + output.lowerBound
    }
}

extension BinaryInteger {
    func converting(from input: ClosedRange<Self>, to output: ClosedRange<Self>) -> Self {
        let x = (output.upperBound - output.lowerBound) * (self - input.lowerBound)
        let y = (input.upperBound - input.lowerBound)
        return x / y + output.lowerBound
    }
}

let integer = 380
let result = integer.converting(from: 10...750, to: 100...350) // 225

let double = 750.0
let result = double.converting(from: 10.1...750.0, to: 0...350) // 350

Answer 2

这是 Swift 3 兼容版本，它也支持泛型（不只为 Doubles 绑定 Rescale）。

struct Rescale<Type : BinaryFloatingPoint> {
    typealias RescaleDomain = (lowerBound: Type, upperBound: Type)

    var fromDomain: RescaleDomain
    var toDomain: RescaleDomain

    init(from: RescaleDomain, to: RescaleDomain) {
        self.fromDomain = from
        self.toDomain = to
    }

    func interpolate(_ x: Type ) -> Type {
        return self.toDomain.lowerBound * (1 - x) + self.toDomain.upperBound * x;
    }

    func uninterpolate(_ x: Type) -> Type {
        let b = (self.fromDomain.upperBound - self.fromDomain.lowerBound) != 0 ? self.fromDomain.upperBound - self.fromDomain.lowerBound : 1 / self.fromDomain.upperBound;
        return (x - self.fromDomain.lowerBound) / b
    }

    func rescale(_ x: Type )  -> Type {
        return interpolate( uninterpolate(x) )
    }
}

示例是：

Rescale(from: (15, 85), to: (0, 100)).rescale(85)