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

net端口:

    static void Main(string[] args)
    {

        Console.Write("Hola");
        List<double> vertx = new List<double>();
        List<double> verty = new List<double>();

        int i, j, c = 0;

        vertx.Add(1);
        vertx.Add(2);
        vertx.Add(1);
        vertx.Add(4);
        vertx.Add(4);
        vertx.Add(1);

        verty.Add(1);
        verty.Add(2);
        verty.Add(4);
        verty.Add(4);
        verty.Add(1);
        verty.Add(1);

        int nvert = 6;  //Vértices del poligono

        double testx = 2;
        double testy = 5;


        for (i = 0, j = nvert - 1; i < nvert; j = i++)
        {
            if (((verty[i] > testy) != (verty[j] > testy)) &&
             (testx < (vertx[j] - vertx[i]) * (testy - verty[i]) / (verty[j] - verty[i]) + vertx[i]))
                c = 1;
        }
    }

当我还是Michael Stonebraker手下的一名研究员时，我做了一些关于这方面的工作——你知道，就是那位提出了Ingres、PostgreSQL等的教授。

我们意识到最快的方法是首先做一个边界框，因为它非常快。如果它在边界框之外，它就在外面。否则，你就得做更辛苦的工作……

如果你想要一个伟大的算法，看看开源项目PostgreSQL的源代码的地理工作…

我想指出的是，我们从来没有深入了解过左撇子和右撇子(也可以表达为“内”和“外”的问题……

更新

BKB's link provided a good number of reasonable algorithms. I was working on Earth Science problems and therefore needed a solution that works in latitude/longitude, and it has the peculiar problem of handedness - is the area inside the smaller area or the bigger area? The answer is that the "direction" of the verticies matters - it's either left-handed or right handed and in this way you can indicate either area as "inside" any given polygon. As such, my work used solution three enumerated on that page.

此外，我的工作使用单独的函数进行“在线”测试。

.．.因为有人问:我们发现当垂直的数量超过某个数字时，边界盒测试是最好的——如果有必要，在做更长的测试之前做一个非常快速的测试……边界框是通过简单地将最大的x，最小的x，最大的y和最小的y放在一起，组成一个框的四个点来创建的……

另一个提示是:我们在网格空间中进行了所有更复杂的“调光”计算，都是在平面上的正点上进行的，然后重新投影到“真实”的经度/纬度上，从而避免了在经度180线交叉时和处理极地时可能出现的环绕错误。工作好了!

以下是M. Katz基于Nirg方法的答案的JavaScript变体:

function pointIsInPoly(p, polygon) {
    var isInside = false;
    var minX = polygon[0].x, maxX = polygon[0].x;
    var minY = polygon[0].y, maxY = polygon[0].y;
    for (var n = 1; n < polygon.length; n++) {
        var q = polygon[n];
        minX = Math.min(q.x, minX);
        maxX = Math.max(q.x, maxX);
        minY = Math.min(q.y, minY);
        maxY = Math.max(q.y, maxY);
    }

    if (p.x < minX || p.x > maxX || p.y < minY || p.y > maxY) {
        return false;
    }

    var i = 0, j = polygon.length - 1;
    for (i, j; i < polygon.length; j = i++) {
        if ( (polygon[i].y > p.y) != (polygon[j].y > p.y) &&
                p.x < (polygon[j].x - polygon[i].x) * (p.y - polygon[i].y) / (polygon[j].y - polygon[i].y) + polygon[i].x ) {
            isInside = !isInside;
        }
    }

    return isInside;
}

简单的解决方案是将多边形划分为三角形，并按这里解释的那样对三角形进行测试

如果你的多边形是凸多边形，可能有更好的方法。把这个多边形看作是无限条线的集合。每一行将空间一分为二。对于每一个点，很容易判断它是在直线的一边还是另一边。如果一个点在所有直线的同一侧，那么它在多边形内。

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;
}