从 CGPoints 数组中获取最大值和最小值

Question

We have an array of CGPoints:

let points = [(1234.0, 1053.0), (1241.0, 1111.0), (1152.0, 1043.0)]

What I'm trying to do is get the CGPoint with the highest x value and the one with the highest y value in the array. I will be using these points to create a CGRect:

extension CGRect {
    init(p1: CGPoint, p2: CGPoint) {
        self.init(x: min(p1.x, p2.x),
                  y: min(p1.y, p2.y),
                  width: abs(p1.x - p2.x),
                  height: abs(p1.y - p2.y))
    }
}

I know there a way to get max and min values in an array by doing something like this:

points.min()
points.max()

but these don't seem to work since its an array of CGPoints. Is it possible to get these values from the array?

Answer 1

You can use max(by:) .

let minXPoint = points.min(by: {$0.x < $1.x}) //(1152.0, 1043.0)
let maxYPoint = points.max(by: {$0.y < $1.y}) //(1241.0, 1111.0)

Answer 2

You can't do that with just a single call to min() / max()

off of the highest and lowest of the y and x coordinates

The point with the minimal x value could be different from the point with the minimal x value. So any method that returns only one of the points (as min() / max() do) isn't sufficient.

You either need two separate calls to each, like:

let minX = points.min(by: { $0.x < $1.x })!.x
let minY = points.min(by: { $0.y < $1.y })!.y
let maxX = points.min(by: { $0.x > $1.x })!.x
let maxY = points.min(by: { $0.y > $1.y })!.y

or you could try doing it all in one swoop with reduce:

let initialAccumulator = (
    minX: CGFloat.max, minY: CGFloat.max, maxX: CGFloat.min, maxY: CGFloat.min)
let (minX, minY, maxX, maxY) = points.reduce(initialAccumulator) { accumulator, point in
    return (
        minX: min(accumulator.minX, point.x),
        minY: min(accumulator.minY, point.y),
        maxX: max(accumulator.maxX, point.x),
        maxY: max(accumulator.maxY, point.y)
    )
}

I would probably do this like so:

extension CGRect {
    static func minimalRect(containing points: [CGPoint]) -> CGRect? {
        if points.isEmpty { return nil }

        let minX = points.min(by: { $0.x < $1.x })!.x
        let minY = points.min(by: { $0.y < $1.y })!.y
        let maxX = points.min(by: { $0.x > $1.x })!.x
        let maxY = points.min(by: { $0.y > $1.y })!.y

        return CGRect(
            x: minX,
            y: minY,
            width: (minX - maxX).magnitude,
            height: (minY - maxY).magnitude
        )
    }
}


let points = [(1234.0, 1053.0), (1241.0, 1111.0), (1152.0, 1043.0)].map(CGPoint.init)
let r = CGRect.minimalRect(containing: points)
print(r ?? .zero) // => (1152.0, 1043.0, 89.0, 68.0)

Answer 3

You can map values to find the min and max in x and y coordinates like below. If you're not sure if points array contains any data, use guard statement to avoid force unwrapping:

let xArray = points.map(\.x)
let yArray = points.map(\.y)
guard let minX = xArray.min(),
      let maxX = xArray.max(),
      let minY = yArray.min(),
      let maxY = yArray.max() else { return }

And from there:

let minPoint = CGPoint(x: minX, y: minY)
let maxPoint = CGPoint(x: minY, y: minY)

then you can modify your extension function because you already know which values are min and max:

extension CGRect {
    init(minPoint: CGPoint, maxPoint: CGPoint) {
        self.init(x: minPoint.x,
                  y: minPoint.y,
                  width: maxPoint.x - minPoint.x,
                  height: maxPoint.y - minPoint.y)
    }
}

As Leo Dabus suggested in the comment below, you can do it all in one go inside failable initializer extension:

extension CGRect {
    init?(points: [CGPoint]) {
        let xArray = points.map(\.x)
        let yArray = points.map(\.y)
        if  let minX = xArray.min(),
            let maxX = xArray.max(),
            let minY = yArray.min(),
            let maxY = yArray.max() {

            self.init(x: minX,
                      y: minY,
                      width: maxX - minX,
                      height: maxY - minY)
        } else {
            return nil
        }
    }
}