查找两个 CGPoints 之间的点
[英]Find points between two CGPoints
问题描述
试图找到起点和终点之间最近的点。
点阵：
let pointsArray = [(10.0, 10.0), (70.0, 10.0), (10.0, 200.0), (70.0, 200.0), (73.0, 10.0), (133.0, 10.0), (73.0, 200.0), (133.0, 200.0), (135.5, 10.0), (195.5, 10.0), (135.5, 200.0), (195.5, 200.0), (198.5, 10.0), (258.5, 10.0), (198.5, 200.0), (258.5, 200.0), (261.5, 10.0), (321.5, 10.0), (261.5, 200.0), (321.5, 200.0), (324.0, 10.0), (384.0, 10.0), (324.0, 200.0), (384.0, 200.0), (387.0, 10.0), (447.0, 10.0), (387.0, 200.0), (447.0, 200.0), (450.0, 10.0), (510.0, 10.0), (450.0, 200.0), (510.0, 200.0), (512.5, 10.0), (572.5, 10.0), (512.5, 200.0), (572.5, 200.0), (575.5, 10.0), (635.5, 10.0), (575.5, 200.0), (635.5, 200.0), (638.5, 10.0), (698.5, 10.0), (638.5, 200.0), (698.5, 200.0), (701.0, 10.0), (761.0, 10.0), (701.0, 200.0), (761.0, 200.0), (764.0, 10.0), (824.0, 10.0), (764.0, 200.0), (824.0, 200.0), (10.0, 390.0), (70.0, 390.0), (73.0, 390.0), (133.0, 390.0), (135.5, 390.0), (195.5, 390.0), (198.5, 390.0), (258.5, 390.0), (261.5, 390.0), (321.5, 390.0), (324.0, 390.0), (384.0, 390.0), (387.0, 390.0), (447.0, 390.0), (450.0, 390.0), (510.0, 390.0), (512.5, 390.0), (572.5, 390.0), (575.5, 390.0), (635.5, 390.0), (638.5, 390.0), (698.5, 390.0), (701.0, 390.0), (761.0, 390.0), (764.0, 390.0), (824.0, 390.0), (10.0, 580.0), (70.0, 580.0), (73.0, 580.0), (133.0, 580.0), (135.5, 580.0), (195.5, 580.0), (198.5, 580.0), (258.5, 580.0)]
let startPoint = CGPoint(x: 80, y: 20)
let endPoint = CGPoint(x: 170, y: 440)
在这里，我试图从现有的点数组中找到起点和终点之间的点。
使用以下从特定点开始的延伸距离，但我无法仅获得 startPoint 和 Endpoint 之间的特定点
extension CGPoint {
func distance(to point: CGPoint) -> CGFloat {
return sqrt(pow(x - point.x, 2) + pow(y - point.y, 2))
}
}
1 个解决方案
解决方案1
1 已采纳 2022-01-25 07:52:12
这绝不是唯一的解决方案，但这是我会采取的一种方法。
1. 从数组中检索有效点
我们只想从位于起点和终点之间的数组中获取有效点。 所以为了形象化它，我假设：
我假设
- 开始 x 和开始 y 将是 <= 结束 x 和结束 y。
- 起点和终点可以是一条直线
- 起点和终点可以是同一个点
- 但 end 不能在 start 的下方或左侧
因此，为了支持这一点，我添加到您的 CGPoint 扩展中以检查该点是否存在于该区域中
extension CGPoint
{
func distance(to point: CGPoint) -> CGFloat
{
return sqrt(pow(x - point.x, 2) + pow(y - point.y, 2))
}
/// Checks if the current point exists in a region. The x and y coordinate of
/// `regionStart` has to be less than or equal to `regionEnd` for a
/// valid check to occur.
/// - Parameters:
/// - regionStart: The bottom left of the region
/// - regionEnd: The top right of the region
/// - Returns: True if the current point falls within the region
func doesExistInRegion(regionStart: CGPoint, regionEnd: CGPoint) -> Bool
{
// Check if we have an invalid region
if regionStart.x > regionEnd.x || regionStart.y > regionEnd.y
{
return false
}
// Check if the current point is outside the region
if x < regionStart.x ||
y < regionStart.y ||
x > regionEnd.x ||
y > regionEnd.y
{
return false
}
// The point is within the region
return true
}
}
然后我使用这样的扩展名仅提取有效点：
let pointsArray = [(10.0, 10.0), (70.0, 10.0), (10.0, 200.0), (70.0, 200.0), (73.0, 10.0), (133.0, 10.0), (73.0, 200.0), (133.0, 200.0), (135.5, 10.0), (195.5, 10.0), (135.5, 200.0), (195.5, 200.0), (198.5, 10.0), (258.5, 10.0), (198.5, 200.0), (258.5, 200.0), (261.5, 10.0), (321.5, 10.0), (261.5, 200.0), (321.5, 200.0), (324.0, 10.0), (384.0, 10.0), (324.0, 200.0), (384.0, 200.0), (387.0, 10.0), (447.0, 10.0), (387.0, 200.0), (447.0, 200.0), (450.0, 10.0), (510.0, 10.0), (450.0, 200.0), (510.0, 200.0), (512.5, 10.0), (572.5, 10.0), (512.5, 200.0), (572.5, 200.0), (575.5, 10.0), (635.5, 10.0), (575.5, 200.0), (635.5, 200.0), (638.5, 10.0), (698.5, 10.0), (638.5, 200.0), (698.5, 200.0), (701.0, 10.0), (761.0, 10.0), (701.0, 200.0), (761.0, 200.0), (764.0, 10.0), (824.0, 10.0), (764.0, 200.0), (824.0, 200.0), (10.0, 390.0), (70.0, 390.0), (73.0, 390.0), (133.0, 390.0), (135.5, 390.0), (195.5, 390.0), (198.5, 390.0), (258.5, 390.0), (261.5, 390.0), (321.5, 390.0), (324.0, 390.0), (384.0, 390.0), (387.0, 390.0), (447.0, 390.0), (450.0, 390.0), (510.0, 390.0), (512.5, 390.0), (572.5, 390.0), (575.5, 390.0), (635.5, 390.0), (638.5, 390.0), (698.5, 390.0), (701.0, 390.0), (761.0, 390.0), (764.0, 390.0), (824.0, 390.0), (10.0, 580.0), (70.0, 580.0), (73.0, 580.0), (133.0, 580.0), (135.5, 580.0), (195.5, 580.0), (198.5, 580.0), (258.5, 580.0)]
let startPoint = CGPoint(x: 80, y: 20)
let endPoint = CGPoint(x: 170, y: 440)
let validPoints = extractValidPoints()
private func extractValidPoints() -> [CGPoint]
{
var validPoints: [CGPoint] = []
for point in pointsArray
{
let coordinate = CGPoint(x: point.0, y: point.1)
if coordinate.doesExistInRegion(regionStart: startPoint, regionEnd: endPoint)
{
validPoints.append(coordinate)
}
}
return validPoints
}
2.找到对之间的最短距离
从上面的数组中，我得到了该区域内的 4 个有效坐标，它们存储在validPoints数组中：
(133.0, 200.0)
(135.5, 200.0)
(133.0, 390.0)
(135.5, 390.0)
现在我们可以遍历这些点来获得距离。 首先，我创建了一个方便的结构来更好地组织事物
struct PointPair: Comparable, Hashable
{
private(set) var startPoint = CGPoint.zero
private(set) var endPoint = CGPoint.zero
private(set) var distance = CGFloat.zero
init(withStartPoint start: CGPoint, andEndPoint end: CGPoint)
{
startPoint = start
endPoint = end
distance = startPoint.distance(to: endPoint)
// Just for convenience
display()
}
func display()
{
print("Distance (\(startPoint.x), \(startPoint.y)) and (\(endPoint.x), \(endPoint.y)): \(distance)")
}
// Needed to implement this so that we conform to Comparable and
// can compare 2 points
static func < (lhs: PointPair, rhs: PointPair) -> Bool
{
return lhs.distance < rhs.distance
}
// Need to implement this to conform to Hashable so we can insert a PointPair
// into dictionaries and data strcutures that work with Hashable types
func hash(into hasher: inout Hasher)
{
hasher.combine(startPoint.x)
hasher.combine(startPoint.y)
hasher.combine(endPoint.x)
hasher.combine(endPoint.y)
}
}
现在我可以遍历validPoints数组并检查这样的对：
if let nearestPoint = retrieveClosestPairUsingSort(fromPoints: validPoints)
{
print("The nearest pair using sort O(n log n) is")
print(nearestPoint.display())
}
private func retrieveClosestPairUsingSort(fromPoints points: [CGPoint]) -> PointPair?
{
var pairs: [PointPair] = []
// Loop through all the points
for index in 0 ..< points.count
{
for secondIndex in index + 1 ..< points.count
{
let pointPair = PointPair(withStartPoint: points[index],
andEndPoint: points[secondIndex])
pairs.append(pointPair)
}
}
return pairs.sorted().first
}
为此的output如下：
Distance (133.0, 200.0) and (135.5, 200.0): 2.5
Distance (133.0, 200.0) and (133.0, 390.0): 190.0
Distance (133.0, 200.0) and (135.5, 390.0): 190.01644665659865
Distance (135.5, 200.0) and (133.0, 390.0): 190.01644665659865
Distance (135.5, 200.0) and (135.5, 390.0): 190.0
Distance (133.0, 390.0) and (135.5, 390.0): 2.5
The nearest pair using sort O(n log n) is
Distance (133.0, 200.0) and (135.5, 200.0): 2.5
3. 更进一步
如果您有大量过滤点，您可以考虑将坐标放入最小堆中以检索 O(n) 中最近的对 - 我在这里有一个实现
if let nearestPoint = retrieveClosestPairUsingHeap(fromPoints: validPoints)
{
print("The nearest pair using heap O(n) is")
print(nearestPoint.display())
}
private func retrieveClosestPairUsingHeap(fromPoints points: [CGPoint]) -> PointPair?
{
// Instantiate a min heap so the root will be the closest pair
var heap = Heap<PointPair>(withProperty: .min)
// Loop through all the points
for index in 0 ..< points.count
{
for secondIndex in index + 1 ..< points.count
{
let pointPair = PointPair(withStartPoint: points[index],
andEndPoint: points[secondIndex])
heap.insert(pointPair)
}
}
return heap.peek()
}
这也给出了相同的 output：
Distance (133.0, 200.0) and (135.5, 200.0): 2.5
Distance (133.0, 200.0) and (133.0, 390.0): 190.0
Distance (133.0, 200.0) and (135.5, 390.0): 190.01644665659865
Distance (135.5, 200.0) and (133.0, 390.0): 190.01644665659865
Distance (135.5, 200.0) and (135.5, 390.0): 190.0
Distance (133.0, 390.0) and (135.5, 390.0): 2.5
The nearest pair using heap O(n) is
Distance (133.0, 390.0) and (135.5, 390.0): 2.5
我创建了一个示例，所有这些代码作为一个简单的控制台应用程序一起工作以进行测试 -你可以从这里获取它。
我希望这回答了你的问题。
获取嵌套在UIScrollView中的UIImageView中两个CGPoints之间的距离
[英]Get the distance between two CGPoints in a UIImageView nested in a UIScrollView
子视图中两个CGPoint之间的线-无法确定嵌套子视图的框架
[英]Line between two CGPoints in a subview - Cannot determine the frame of nested subview
如何使用Swift在两个三角形（3点）之间找到仿射变换
[英]How to find affine transformation between two triangles (3 points) with Swift
如何找到两个坐标之间的道路距离并在相同点之间绘制多段线？
[英]How to find road distance between two Coordinates and make a polyline between same points?
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.