我试图创建一个快速的2D点内多边形算法,用于命中测试(例如多边形.contains(p:点))。对有效技术的建议将不胜感激。
当前回答
我已经做了nirg的c++代码的Python实现:
输入
Bounding_points:组成多边形的节点。 Bounding_box_positions:筛选的候选点。(在我从边界框创建的实现中。 (输入为元组列表,格式为:[(xcord, ycord),…])
返回
多边形内的所有点。
def polygon_ray_casting(self, bounding_points, bounding_box_positions):
# Arrays containing the x- and y-coordinates of the polygon's vertices.
vertx = [point[0] for point in bounding_points]
verty = [point[1] for point in bounding_points]
# Number of vertices in the polygon
nvert = len(bounding_points)
# Points that are inside
points_inside = []
# For every candidate position within the bounding box
for idx, pos in enumerate(bounding_box_positions):
testx, testy = (pos[0], pos[1])
c = 0
for i in range(0, nvert):
j = i - 1 if i != 0 else nvert - 1
if( ((verty[i] > testy ) != (verty[j] > testy)) and
(testx < (vertx[j] - vertx[i]) * (testy - verty[i]) / (verty[j] - verty[i]) + vertx[i]) ):
c += 1
# If odd, that means that we are inside the polygon
if c % 2 == 1:
points_inside.append(pos)
return points_inside
同样,这个想法也是从这里得来的
其他回答
Java版本:
public class Geocode {
private float latitude;
private float longitude;
public Geocode() {
}
public Geocode(float latitude, float longitude) {
this.latitude = latitude;
this.longitude = longitude;
}
public float getLatitude() {
return latitude;
}
public void setLatitude(float latitude) {
this.latitude = latitude;
}
public float getLongitude() {
return longitude;
}
public void setLongitude(float longitude) {
this.longitude = longitude;
}
}
public class GeoPolygon {
private ArrayList<Geocode> points;
public GeoPolygon() {
this.points = new ArrayList<Geocode>();
}
public GeoPolygon(ArrayList<Geocode> points) {
this.points = points;
}
public GeoPolygon add(Geocode geo) {
points.add(geo);
return this;
}
public boolean inside(Geocode geo) {
int i, j;
boolean c = false;
for (i = 0, j = points.size() - 1; i < points.size(); j = i++) {
if (((points.get(i).getLongitude() > geo.getLongitude()) != (points.get(j).getLongitude() > geo.getLongitude())) &&
(geo.getLatitude() < (points.get(j).getLatitude() - points.get(i).getLatitude()) * (geo.getLongitude() - points.get(i).getLongitude()) / (points.get(j).getLongitude() - points.get(i).getLongitude()) + points.get(i).getLatitude()))
c = !c;
}
return c;
}
}
我知道这是旧的,但这里是一个在Cocoa实现的光线投射算法,如果有人感兴趣的话。不确定这是最有效的方法,但它可能会帮助别人。
- (BOOL)shape:(NSBezierPath *)path containsPoint:(NSPoint)point
{
NSBezierPath *currentPath = [path bezierPathByFlatteningPath];
BOOL result;
float aggregateX = 0; //I use these to calculate the centroid of the shape
float aggregateY = 0;
NSPoint firstPoint[1];
[currentPath elementAtIndex:0 associatedPoints:firstPoint];
float olderX = firstPoint[0].x;
float olderY = firstPoint[0].y;
NSPoint interPoint;
int noOfIntersections = 0;
for (int n = 0; n < [currentPath elementCount]; n++) {
NSPoint points[1];
[currentPath elementAtIndex:n associatedPoints:points];
aggregateX += points[0].x;
aggregateY += points[0].y;
}
for (int n = 0; n < [currentPath elementCount]; n++) {
NSPoint points[1];
[currentPath elementAtIndex:n associatedPoints:points];
//line equations in Ax + By = C form
float _A_FOO = (aggregateY/[currentPath elementCount]) - point.y;
float _B_FOO = point.x - (aggregateX/[currentPath elementCount]);
float _C_FOO = (_A_FOO * point.x) + (_B_FOO * point.y);
float _A_BAR = olderY - points[0].y;
float _B_BAR = points[0].x - olderX;
float _C_BAR = (_A_BAR * olderX) + (_B_BAR * olderY);
float det = (_A_FOO * _B_BAR) - (_A_BAR * _B_FOO);
if (det != 0) {
//intersection points with the edges
float xIntersectionPoint = ((_B_BAR * _C_FOO) - (_B_FOO * _C_BAR)) / det;
float yIntersectionPoint = ((_A_FOO * _C_BAR) - (_A_BAR * _C_FOO)) / det;
interPoint = NSMakePoint(xIntersectionPoint, yIntersectionPoint);
if (olderX <= points[0].x) {
//doesn't matter in which direction the ray goes, so I send it right-ward.
if ((interPoint.x >= olderX && interPoint.x <= points[0].x) && (interPoint.x > point.x)) {
noOfIntersections++;
}
} else {
if ((interPoint.x >= points[0].x && interPoint.x <= olderX) && (interPoint.x > point.x)) {
noOfIntersections++;
}
}
}
olderX = points[0].x;
olderY = points[0].y;
}
if (noOfIntersections % 2 == 0) {
result = FALSE;
} else {
result = TRUE;
}
return result;
}
我已经做了nirg的c++代码的Python实现:
输入
Bounding_points:组成多边形的节点。 Bounding_box_positions:筛选的候选点。(在我从边界框创建的实现中。 (输入为元组列表,格式为:[(xcord, ycord),…])
返回
多边形内的所有点。
def polygon_ray_casting(self, bounding_points, bounding_box_positions):
# Arrays containing the x- and y-coordinates of the polygon's vertices.
vertx = [point[0] for point in bounding_points]
verty = [point[1] for point in bounding_points]
# Number of vertices in the polygon
nvert = len(bounding_points)
# Points that are inside
points_inside = []
# For every candidate position within the bounding box
for idx, pos in enumerate(bounding_box_positions):
testx, testy = (pos[0], pos[1])
c = 0
for i in range(0, nvert):
j = i - 1 if i != 0 else nvert - 1
if( ((verty[i] > testy ) != (verty[j] > testy)) and
(testx < (vertx[j] - vertx[i]) * (testy - verty[i]) / (verty[j] - verty[i]) + vertx[i]) ):
c += 1
# If odd, that means that we are inside the polygon
if c % 2 == 1:
points_inside.append(pos)
return points_inside
同样,这个想法也是从这里得来的
我认为这是迄今为止所有答案中最简洁的一个。
例如,假设我们有一个多边形,它带有多边形凹,看起来像这样:
大多边形顶点的二维坐标为
[[139, 483], [227, 792], [482, 849], [523, 670], [352, 330]]
方框顶点的坐标为
[[248, 518], [336, 510], [341, 614], [250, 620]]
空心三角形顶点的坐标为
[[416, 531], [505, 517], [495, 616]]
假设我们想要测试两个点[296,557]和[422,730],如果它们在红色区域内(不包括边缘)。如果我们定位这两个点,它将是这样的:
显然,[296,557]不在读取区域内,而[422,730]在。
我的解决方案是基于圈数算法。下面是我只使用numpy的4行python代码:
def detect(points, *polygons):
import numpy as np
endpoint1 = np.r_[tuple(np.roll(p, 1, 0) for p in polygons)][:, None] - points
endpoint2 = np.r_[polygons][:, None] - points
p1, p2 = np.cross(endpoint1, endpoint2), np.einsum('...i,...i', endpoint1, endpoint2)
return ~((p1.sum(0) < 0) ^ (abs(np.arctan2(p1, p2).sum(0)) > np.pi) | ((p1 == 0) & (p2 <= 0)).any(0))
要测试实现:
points = [[296, 557], [422, 730]]
polygon1 = [[139, 483], [227, 792], [482, 849], [523, 670], [352, 330]]
polygon2 = [[248, 518], [336, 510], [341, 614], [250, 620]]
polygon3 = [[416, 531], [505, 517], [495, 616]]
print(detect(points, polygon1, polygon2, polygon3))
输出:
[False True]
如果你正在寻找一个java脚本库,有一个javascript谷歌maps v3扩展的Polygon类,以检测是否有一个点驻留在它里面。
var polygon = new google.maps.Polygon([], "#000000", 1, 1, "#336699", 0.3);
var isWithinPolygon = polygon.containsLatLng(40, -90);
谷歌扩展Github
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