计算对象是否在一组坐标内？

java coordinates jts

2022-09-01 22:41:26

我有一组X和Y点来构建一个形状，我需要知道一个对象是否在其中，它的计算是什么？

X 和 Y 坐标示例：

522.56055 2389.885
544.96 2386.3406
554.18616 2369.2385
535.21814 2351.396
497.5552 2355.8396

我不太擅长数学:(所以我将不胜感激一些支持，以了解它是如何完成的。

到目前为止，我拥有但似乎不太可靠的示例：

private boolean isInsideShape(Zone verifyZone, Position object)
{
    int corners = verifyZone.getCorners();
    float[] xCoords = verifyZone.getxCoordinates();
    float[] yCoords = verifyZone.getyCoordinates();

    float x = object.getX();
    float y = object.getY();
    float z = object.getZ();

    int i, j = corners - 1;
    boolean inside = false;

    for(i = 0; i < corners; i++)
    {
        if(yCoords[i] < y && yCoords[j] >= y || yCoords[j] < y && yCoords[i] >= y)
            if(xCoords[i] + (y - yCoords[i]) / (yCoords[j] - yCoords[i]) * (xCoords[j] - xCoords[i]) < x)
                inside = !inside;
        j = i;
    }

    return inside;
}

答案 1

你可以从这个开始：http://en.wikipedia.org/wiki/Point_in_polygon

您也可以查看 JTS 拓扑套件。特别是使用此功能。

编辑：这是使用JTS的示例：

import java.util.ArrayList;

import com.vividsolutions.jts.geom.Coordinate;
import com.vividsolutions.jts.geom.GeometryFactory;
import com.vividsolutions.jts.geom.LinearRing;
import com.vividsolutions.jts.geom.Point;
import com.vividsolutions.jts.geom.Polygon;
import com.vividsolutions.jts.geom.impl.CoordinateArraySequence;

public class GeoTest {

  public static void main(final String[] args) {

    final GeometryFactory gf = new GeometryFactory();

    final ArrayList<Coordinate> points = new ArrayList<Coordinate>();
    points.add(new Coordinate(-10, -10));
    points.add(new Coordinate(-10, 10));
    points.add(new Coordinate(10, 10));
    points.add(new Coordinate(10, -10));
    points.add(new Coordinate(-10, -10));
    final Polygon polygon = gf.createPolygon(new LinearRing(new CoordinateArraySequence(points
        .toArray(new Coordinate[points.size()])), gf), null);

    final Coordinate coord = new Coordinate(0, 0);
    final Point point = gf.createPoint(coord);

    System.out.println(point.within(polygon));

  }

}

下面是使用AWT的示例（它更简单，是Java SE的一部分）：

import java.awt.Polygon;

public class JavaTest {

  public static void main(final String[] args) {

    final Polygon polygon = new Polygon();
    polygon.addPoint(-10, -10);
    polygon.addPoint(-10, 10);
    polygon.addPoint(10, 10);
    polygon.addPoint(10, -10);

    System.out.println(polygon.contains(0, 0));

  }

}

答案 2

我一直都是这样做的：

Pick a point you know to be outside the shape.
Make a line between that point and the point you're trying to find whether it's inside the shape or not.
Count the number of sides of the shape the line crosses. 

If the count is odd, the point is inside the shape.
If the count is even, the point is outside the shape.