C# 生成具有边界的随机坐标

我想要一种方法来生成带有边界的随机坐标。大概是这样的：

private Coordinate[] Calculate(Coordinate location1, Coordinate location2, Coordinate location3, Coordinate location4)

public class Coordinate
{
    public double Latitude { set; get; }
    public double Longitude { set; get; }
}

我不知道如何使用指定的无线电生成坐标。

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只需生成从最小到最大的坐标，并排除错误

    private Coordinate[] Calculate(Coordinate location1, Coordinate location2, Coordinate location3,
        Coordinate location4)
    {
        Coordinate[] allCoords = {location1, location2, location3, location4};
        double minLat = allCoords.Min(x => x.Latitude);
        double minLon = allCoords.Min(x => x.Longitude);
        double maxLat = allCoords.Max(x => x.Latitude);
        double maxLon = allCoords.Max(x => x.Longitude);

        Random r = new Random();

        //replase 500 with your number
        Coordinate[] result = new Coordinate[500];
        for (int i = 0; i < result.Length; i++)
        {
            Coordinate point = new Coordinate();
            do
            {
                point.Latitude = r.NextDouble()*(maxLat - minLat) + minLat;
                point.Longitude = r.NextDouble()*(maxLon - minLon) + minLon;
            } while (!IsPointInPolygon(point, allCoords));
            result[i] = point;
        }
        return result;
    }

    //took it from http://codereview.stackexchange.com/a/108903
    //you can use your own one
    private bool IsPointInPolygon(Coordinate point, Coordinate[] polygon)
    {
        int polygonLength = polygon.Length, i = 0;
        bool inside = false;
        // x, y for tested point.
        double pointX = point.Longitude, pointY = point.Latitude;
        // start / end point for the current polygon segment.
        double startX, startY, endX, endY;
        Coordinate endPoint = polygon[polygonLength - 1];
        endX = endPoint.Longitude;
        endY = endPoint.Latitude;
        while (i < polygonLength)
        {
            startX = endX;
            startY = endY;
            endPoint = polygon[i++];
            endX = endPoint.Longitude;
            endY = endPoint.Latitude;
            //
            inside ^= ((endY > pointY) ^ (startY > pointY)) /* ? pointY inside [startY;endY] segment ? */
                      && /* if so, test if it is under the segment */
                      (pointX - endX < (pointY - endY)*(startX - endX)/(startY - endY));
        }
        return inside;
    }

专用坐标[]计算（坐标位置1、坐标位置2、坐标位置3、，
坐标位置（4）
{
坐标[]allCoords={location1，location2，location3，location4}；
double minLat=allCoords.Min（x=>x.纬度）；
double minLon=allCoords.Min（x=>x.Longitude）；
double maxLat=allCoords.Max（x=>x.lation）；
double maxLon=allCoords.Max（x=>x.Longitude）；
随机r=新随机（）；
//用你的号码重播500
坐标[]结果=新坐标[500]；
for（int i=0；ipointY）^（startY>pointY））/*？在[startY；endY]段内有点*/
&&/*如果是，测试它是否在该段下*/
（pointX-endX<（pointY-endY）*（startX-endX）/（startY-endY））；
}
返回内部；
}

只要有一点几何知识，我们就可以使用以下方法：

 private Coordinate Calculate(Coordinate location1, Coordinate location2, Coordinate location3,
        Coordinate location4)
    {
        Random random=new Random(DateTime.Now.Millisecond);
        Coordinate randomCoordinate = new Coordinate()
        {
            Latitude = random.Next((int) Math.Floor(location4.Latitude), (int) Math.Floor(location2.Latitude))
        };
        if (randomCoordinate.Latitude > location1.Latitude)
        {
            double m1 = (location2.Longitude - location1.Longitude)/(location2.Latitude - location1.Latitude);
            double m2 = (location2.Longitude - location3.Longitude)/(location2.Latitude - location3.Latitude);
            double maxLongitude = (randomCoordinate.Latitude - location2.Latitude) *m1;
            double minLongitude = (randomCoordinate.Latitude - location2.Latitude) *m2;
            randomCoordinate.Longitude = random.Next((int) Math.Ceiling(minLongitude), (int) Math.Floor(maxLongitude));
        }
        else
        {
            double m1 = (location4.Longitude - location1.Longitude) / (location4.Latitude - location1.Latitude);
            double m2 = (location4.Longitude - location3.Longitude) / (location4.Latitude - location3.Latitude);
            double maxLongitude = (randomCoordinate.Latitude - location4.Latitude) * m1;
            double minLongitude = (randomCoordinate.Latitude - location4.Latitude) * m2;
            randomCoordinate.Longitude = random.Next((int)Math.Ceiling(minLongitude), (int)Math.Floor(maxLongitude));
        }
        return randomCoordinate;
    }

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这可能有助于您目前所做的工作？我编写了一个方法来生成具有指定半径的坐标。但我想让一些东西和边界一起工作。分享这一点可能是明智的，否则这个问题会变得太广泛，你们无法知道运行时图形的真正形式。因此，您应该使用大量的if。例如，如果位置1不在左侧，而是在下方？它们是顺时针还是逆时针？它可能非常复杂。这只适用于问题的测试用例。如果多边形的形状没有预定义，我建议使用您的方法。然而，对于一个预定义的多边形，使用它并不方便。我想指出的是，在试图在多边形中生成点时，它似乎会无限循环。可能与最小/最大纬度有关？