Question

我有两个GPS坐标连接在一起形成一条线。我还有一个GPS点，它靠近，但从未完全在线上。我的问题是，如何找到沿给定点的最近点？

Answer 1

Game Dev has an answer to this，它在C ++中，但它应该很容易移植。哪个CarlG有kindly done（希望他不介意我重新发布）：

public static Point2D nearestPointOnLine(double ax, double ay, double bx, double by, double px, double py,
        boolean clampToSegment, Point2D dest) {
    // Thanks StackOverflow!
    // https://stackoverflow.com/questions/1459368/snap-point-to-a-line-java
    if (dest == null) {
        dest = new Point2D.Double();
    }

    double apx = px - ax;
    double apy = py - ay;
    double abx = bx - ax;
    double aby = by - ay;

    double ab2 = abx * abx + aby * aby;
    double ap_ab = apx * abx + apy * aby;
    double t = ap_ab / ab2;
    if (clampToSegment) {
        if (t < 0) {
            t = 0;
        } else if (t > 1) {
            t = 1;
        }
    }
    dest.setLocation(ax + abx * t, ay + aby * t);
    return dest;
}

Answer 2

试试这个：

ratio = (((x1-x0)^2+(y1-y0)^2)*((x2-x1)^2 + (y2-y1)^2) - ((x2-x1)(y1-y0) - (x1-x0)(y2-y1))^2)^0.5
        -----------------------------------------------------------------------------------------
                                            ((x2-x1)^2 + (y2-y1)^2)

xc = x1 + (x2-x1)*ratio;
yc = y1 + (y2-y1)*ratio;

Where:
    x1,y1 = point#1 on the line
    x2,y2 = point#2 on the line
    x0,y0 = Another point near the line
    xc,yx = The nearest point of x0,y0 on the line
    ratio = is the ratio of distance of x1,y1 to xc,yc and distance of x1,y1 to x2,y2
    ^2    = square
    ^0.5  = square root

在我们发现从点x0，y0到线（x1，y1 - > x2，y3）的距离之后导出公式。见here

我在这里测试了这段代码（我上面给你的这个特别的代码），但几年前我用过类似的方法，它可以用来试试。

Answer 3

您可以使用JTS。

创建LineSegment（您的行）
创建Coordinate（您想要对齐的点）
使用closestPoint方法

非常简单的代码示例：

// create Line: P1(0,0) - P2(0,10)
LineSegment ls = new LineSegment(0, 0, 0, 10);
// create Point: P3(5,5)
Coordinate c = new Coordinate(5, 5);
// create snapped Point: P4(0,5)
Coordinate snappedPoint = ls.closestPoint(c);

捕捉指向一条线

3 个答案: