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.Polygon
class。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 和 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?以上是关于在 Android 中查找路径中包含的点的主要内容,如果未能解决你的问题,请参考以下文章
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