在 Android 中查找路径中包含的点

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【中文标题】在 Android 中查找路径中包含的点【英文标题】:Finding Points contained in a Path in Android 【发布时间】:2011-10-26 01:38:02 【问题描述】:

他们决定不在 android 中添加 contains 方法(用于 Path)是否有原因?

我想知道我在路径中有哪些点,并希望它比这里看到的更容易:

How can I tell if a closed path contains a given point?

我创建一个 ArrayList 并将整数添加到数组中会更好吗? (我只在控制语句中检查一次点)即。 if(myPath.contains(x,y)

目前我的选择是:

使用区域 使用 ArrayList 扩展类 您的建议

我只是在寻找最有效的方法来解决这个问题

【问题讨论】:

How can I tell if a closed path contains a given point?的可能重复 【参考方案1】:

不久前我遇到了同样的问题,经过一番搜索,我发现这是最好的解决方案。

Java 有一个 Polygon 类和一个 contains() 方法,这将使事情变得非常简单。不幸的是,Android 不支持java.awt.Polygonclass。但是,我找到了写equivalent class的人。

我认为您无法从 Android Path 类中获取构成路径的各个点,因此您必须以不同的方式存储数据。

该类使用交叉数算法来确定该点是否在给定的点列表内。

/**
 * Minimum Polygon class for Android.
 */
public class Polygon

    // Polygon coodinates.
    private int[] polyY, polyX;

    // Number of sides in the polygon.
    private int polySides;

    /**
     * Default constructor.
     * @param px Polygon y coods.
     * @param py Polygon x coods.
     * @param ps Polygon sides count.
     */
    public Polygon( int[] px, int[] py, int ps )
    
        polyX = px;
        polyY = py;
        polySides = ps;
    

    /**
     * Checks if the Polygon contains a point.
     * @see "http://alienryderflex.com/polygon/"
     * @param x Point horizontal pos.
     * @param y Point vertical pos.
     * @return Point is in Poly flag.
     */
    public boolean contains( int x, int y )
    
        boolean oddTransitions = false;
        for( int i = 0, j = polySides -1; i < polySides; j = i++ )
        
            if( ( polyY[ i ] < y && polyY[ j ] >= y ) || ( polyY[ j ] < y && polyY[ i ] >= y ) )
            
                if( polyX[ i ] + ( y - polyY[ i ] ) / ( polyY[ j ] - polyY[ i ] ) * ( polyX[ j ] - polyX[ i ] ) < x )
                
                    oddTransitions = !oddTransitions;          
                
            
        
        return oddTransitions;
      

【讨论】:

如果不确定多边形的边数怎么办? 从点数可以很容易地推断出边数。一个矩形有 4 个点和 4 个边。一个三角形有 3 个点和 3 个边。由连续路径绘制的多边形具有相同数量的边和点,因此如果您的路径有“n”个点,它也有“n”个边。 太棒了,我很高兴能帮上忙。几周前,这个解决方案也为我省去了很多麻烦。 是的,他们忽略了 contains 方法,这很奇怪。尽管如此,我很高兴你为我节省了大量的上课时间。干杯! 这很有帮助。如果有兴趣,我在这里找到了有关交叉数和绕组数算法的信息(我发现这些信息非常有用):softsurfer.com/Archive/algorithm_0103/algorithm_0103.htm【参考方案2】:

我只想评论@theisenp 的回答:代码有整数数组,如果你查看算法描述网页,它会警告不要使用整数而不是浮点数。

我在上面复制了你的代码,它似乎工作正常,除了一些极端情况,当我制作的线条没有很好地连接到自己时。

通过将所有内容更改为浮点数,我摆脱了这个错误。

【讨论】:

谢谢哥们。我也面临着同样的准确性问题,阅读您的评论使这项工作变得感谢。你得到了 +1 :)【参考方案3】:

尝试了另一个答案,但它对我的案例给出了错误的结果。没有费心去寻找确切的原因,而是从算法中直接翻译了: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

现在代码如下:

/**
 * Minimum Polygon class for Android.
 */
public class Polygon

    // Polygon coodinates.
    private int[] polyY, polyX;

    // Number of sides in the polygon.
    private int polySides;

    /**
     * Default constructor.
     * @param px Polygon y coods.
     * @param py Polygon x coods.
     * @param ps Polygon sides count.
     */
    public Polygon( int[] px, int[] py, int ps )
    
        polyX = px;
        polyY = py;
        polySides = ps;
    

    /**
     * Checks if the Polygon contains a point.
     * @see "http://alienryderflex.com/polygon/"
     * @param x Point horizontal pos.
     * @param y Point vertical pos.
     * @return Point is in Poly flag.
     */
    public boolean contains( int x, int y )
    
        boolean c = false;
        int i, j = 0;
        for (i = 0, j = polySides - 1; i < polySides; j = i++) 
            if (((polyY[i] > y) != (polyY[j] > y))
                && (x < (polyX[j] - polyX[i]) * (y - polyY[i]) / (polyY[j] - polyY[i]) + polyX[i]))
            c = !c;
        
        return c;
      

【讨论】:

谢谢队友。为我省去了很多麻烦。我正在实施@theisenp 答案,但这不符合我的预期。为你 +1。 带有 arc 的路径怎么样?算法使用顶点数组。当路径包含圆弧或更一般地说,路径的圆形部分时,它似乎没有考虑情况。 非常感谢您发布此内容。【参考方案4】:

为了完整起见,我想在这里做几点说明:

从 API 19 开始,路径有一个 intersection operation。您可以在测试点周围创建一个非常小的方形路径,将其与路径相交,然后查看结果是否为空。

您可以将路径转换为区域并执行contains() 操作。然而,区域在整数坐标中工作,我认为它们使用转换后的(像素)坐标,所以你必须使用它。我还怀疑转换过程是计算密集型的。

Hans 发布的边缘交叉算法既好又快,但您必须非常小心某些极端情况,例如光线直接穿过顶点、与水平边缘相交或舍入错误时是个问题,一直都是。

winding number 方法几乎是万无一失的,但涉及大量三角函数并且计算量很大。

This paper by Dan Sunday 提供了一种混合算法,该算法与绕组数一样准确,但在计算上却与光线投射算法一样简单。它的优雅让我大吃一惊。

我的代码

这是我最近用 Java 编写的一些代码,它处理由 弧线组成的路径。 (也是圆圈,但它们本身就是完整的路径,所以这是一种退化的情况。)

package org.efalk.util;

/**
 * Utility: determine if a point is inside a path.
 */
public class PathUtil 
    static final double RAD = (Math.PI/180.);
    static final double DEG = (180./Math.PI);

    protected static final int LINE = 0;
    protected static final int ARC = 1;
    protected static final int CIRCLE = 2;

    /**
     * Used to cache the contents of a path for pick testing.  For a
     * line segment, x0,y0,x1,y1 are the endpoints of the line.  For
     * a circle (ellipse, actually), x0,y0,x1,y1 are the bounding box
     * of the circle (this is how Android and X11 like to represent
     * circles).  For an arc, x0,y0,x1,y1 are the bounding box, a1 is
     * the start angle (degrees CCW from the +X direction) and a1 is
     * the sweep angle (degrees CCW).
     */
    public static class PathElement 
        public int type;
        public float x0,y0,x1,y1;   // Endpoints or bounding box
        public float a0,a1;         // Arcs and circles
    

    /**
     * Determine if the given point is inside the given path.
     */
    public static boolean inside(float x, float y, PathElement[] path) 
        // Based on algorithm by Dan Sunday, but allows for arc segments too.         
        // http://geomalgorithms.com/a03-_inclusion.html
        int wn = 0;
        // loop through all edges of the polygon
        // An upward crossing requires y0 <= y and y1 > y
        // A downward crossing requires y0 > y and y1 <= y
        for (PathElement pe : path) 
            switch (pe.type) 
              case LINE:
                if (pe.x0 < x && pe.x1 < x) // left
                    break;
                if (pe.y0 <= y)            // start y <= P.y
                    if (pe.y1 > y)         // an upward crossing
                        if (isLeft(pe, x, y) > 0) // P left of  edge
                            ++wn;                // have  a valid up intersect
                    
                
                else                               // start y > P.y
                    if (pe.y1 <= y)        // a downward crossing
                        if (isLeft(pe, x, y) < 0) // P right of  edge
                            --wn;                // have  a valid down intersect
                    
                
                break;
              case ARC:
                wn += arcCrossing(pe, x, y);
                break;
              case CIRCLE:
                // This should be the only element in the path, so test it
                // and get out.
                float rx = (pe.x1-pe.x0)/2;
                float ry = (pe.y1-pe.y0)/2;
                float xc = (pe.x1+pe.x0)/2;
                float yc = (pe.y1+pe.y0)/2;
                return (x-xc)*(x-xc)/rx*rx + (y-yc)*(y-yc)/ry*ry <= 1;
            
        
        return wn != 0;
    

    /**
     * Return >0 if p is left of line p0-p1; <0 if to the right; 0 if
     * on the line.
     */
    private static float
    isLeft(float x0, float y0, float x1, float y1, float x, float y)
    
        return (x1 - x0) * (y - y0) - (x - x0) * (y1 - y0);
    

    private static float isLeft(PathElement pe, float x, float y) 
        return isLeft(pe.x0,pe.y0, pe.x1,pe.y1, x,y);
    

    /**
     * Determine if an arc segment crosses the test ray up or down, or not
     * at all.
     * @return winding number increment:
     *      +1 upward crossing
     *       0 no crossing
     *      -1 downward crossing
     */
    private static int arcCrossing(PathElement pe, float x, float y) 
        // Look for trivial reject cases first.
        if (pe.x1 < x || pe.y1 < y || pe.y0 > y) return 0;

        // Find the intersection of the test ray with the arc. This consists
        // of finding the intersection(s) of the line with the ellipse that
        // contains the arc, then determining if the intersection(s)
        // are within the limits of the arc.
        // Since we're mostly concerned with whether or not there *is* an
        // intersection, we have several opportunities to punt.
        // An upward crossing requires y0 <= y and y1 > y
        // A downward crossing requires y0 > y and y1 <= y
        float rx = (pe.x1-pe.x0)/2;
        float ry = (pe.y1-pe.y0)/2;
        float xc = (pe.x1+pe.x0)/2;
        float yc = (pe.y1+pe.y0)/2;
        if (rx == 0 || ry == 0) return 0;
        if (rx < 0) rx = -rx;
        if (ry < 0) ry = -ry;
        // We start by transforming everything so the ellipse is the unit
        // circle; this simplifies the math.
        x -= xc;
        y -= yc;
        if (x > rx || y > ry || y < -ry) return 0;
        x /= rx;
        y /= ry;
        // Now find the points of intersection. This is simplified by the
        // fact that our line is horizontal. Also, by the time we get here,
        // we know there *is* an intersection.
        // The equation for the circle is x²+y² = 1. We have y, so solve
        // for x = ±sqrt(1 - y²)
        double x0 = 1 - y*y;
        if (x0 <= 0) return 0;
        x0 = Math.sqrt(x0);
        // We only care about intersections to the right of x, so
        // that's another opportunity to punt. For a CCW arc, The right
        // intersection is an upward crossing and the left intersection
        // is a downward crossing.  The reverse is true for a CW arc.
        if (x > x0) return 0;
        int wn = arcXing1(x0,y, pe.a0, pe.a1);
        if (x < -x0) wn -= arcXing1(-x0,y, pe.a0, pe.a1);
        return wn;
    

    /**
     * Return the winding number of the point x,y on the unit circle
     * which passes through the arc segment defined by a0,a1.
     */
    private static int arcXing1(double x, float y, float a0, float a1) 
        double a = Math.atan2(y,x) * DEG;
        if (a < 0) a += 360;
        if (a1 > 0)        // CCW
            if (a < a0) a += 360;
            return a0 + a1 > a ? 1 : 0;
         else             // CW
            if (a0 < a) a0 += 360;
            return a0 + a1 <= a ? -1 : 0;
        
    


编辑:根据要求,添加一些使用此功能的示例代码。

import PathUtil;
import PathUtil.PathElement;

/**
 * This class represents a single geographic area defined by a
 * circle or a list of line segments and arcs.
 */
public class Area 
    public float lat0, lon0, lat1, lon1;    // bounds
    Path path = null;
    PathElement[] pathList;

    /**
     * Return true if this point is inside the area bounds. This is
     * used to confirm touch events and may be computationally expensive.
     */
    public boolean pointInBounds(float lat, float lon) 
        if (lat < lat0 || lat > lat1 || lon < lon0 || lon > lon1)
            return false;
        return PathUtil.inside(lon, lat, pathList);
    

    static void loadBounds() 
        int n = number_of_elements_in_input;
        path = new Path();
        pathList = new PathElement[n];
        for (Element element : elements_in_input) 
            PathElement pe = new PathElement();
            pathList[i] = pe;
            pe.type = element.type;
            switch (element.type) 
              case LINE:        // Line segment
                pe.x0 = element.x0;
                pe.y0 = element.y0;
                pe.x1 = element.x1;
                pe.y1 = element.y1;
                // Add to path, not shown here
                break;
              case ARC: // Arc segment
                pe.x0 = element.xmin;     // Bounds of arc ellipse
                pe.y0 = element.ymin;
                pe.x1 = element.xmax;
                pe.y1 = element.ymax;
                pe.a0 = a0; pe.a1 = a1;
                break;
              case CIRCLE: // Circle; hopefully the only entry here
                pe.x0 = element.xmin;    // Bounds of ellipse
                pe.y0 = element.ymin;
                pe.x1 = element.xmax;
                pe.y1 = element.ymax;
                // Add to path, not shown here
                break;
            
        
        path.close();
       

【讨论】:

您能否通过 arc 示例提供如何正确使用您的库的示例,看起来我做错了什么。实际上不确定要为 x0,y0,x1,y0 发送什么。是 RectF obj 的上、下、左、右点吗? (安卓新手) 是的,Android 通过指定椭圆的边界然后指定起点和终点角度来指定弧。我对此有一些注释efalk.org/Docs/Android/graphics_0.html#arcTo 你能举个例子说明如何使用吗? @ShP 你明白了吗?

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