有了一个点列表,我如何确定它们是否是顺时针顺序的?
例如:
point[0] = (5,0)
point[1] = (6,4)
point[2] = (4,5)
point[3] = (1,5)
point[4] = (1,0)
会说它是逆时针的(对某些人来说是逆时针的)
有了一个点列表,我如何确定它们是否是顺时针顺序的?
例如:
point[0] = (5,0)
point[1] = (6,4)
point[2] = (4,5)
point[3] = (1,5)
point[4] = (1,0)
会说它是逆时针的(对某些人来说是逆时针的)
当前回答
这是OpenLayers 2的实现函数。有一个顺时针多边形的条件是面积< 0,这是由这个参考确定的。
function IsClockwise(feature)
{
if(feature.geometry == null)
return -1;
var vertices = feature.geometry.getVertices();
var area = 0;
for (var i = 0; i < (vertices.length); i++) {
j = (i + 1) % vertices.length;
area += vertices[i].x * vertices[j].y;
area -= vertices[j].x * vertices[i].y;
// console.log(area);
}
return (area < 0);
}
其他回答
我将提出另一个解决方案,因为它很简单,不需要大量的数学运算,它只是使用了基本的代数。计算多边形的带符号面积。如果是负的,点是顺时针的,如果是正的,点是逆时针的。(这与Beta的解决方案非常相似。)
计算带符号的面积: A = 1/2 * (x1*y2 - x2*y1 + x2*y3 - x3*y2 +…+ xn*y1 - x1*yn)
或者在伪代码中:
signedArea = 0
for each point in points:
x1 = point[0]
y1 = point[1]
if point is last point
x2 = firstPoint[0]
y2 = firstPoint[1]
else
x2 = nextPoint[0]
y2 = nextPoint[1]
end if
signedArea += (x1 * y2 - x2 * y1)
end for
return signedArea / 2
注意,如果你只是检查顺序,你不需要麻烦除以2。
来源:http://mathworld.wolfram.com/PolygonArea.html
我的c# / LINQ解决方案是基于下面@charlesbretana的交叉积建议的。你可以为线圈指定一个参考法线。只要曲线大部分在向上向量所定义的平面内,它就可以工作。
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
namespace SolidworksAddinFramework.Geometry
{
public static class PlanePolygon
{
/// <summary>
/// Assumes that polygon is closed, ie first and last points are the same
/// </summary>
public static bool Orientation
(this IEnumerable<Vector3> polygon, Vector3 up)
{
var sum = polygon
.Buffer(2, 1) // from Interactive Extensions Nuget Pkg
.Where(b => b.Count == 2)
.Aggregate
( Vector3.Zero
, (p, b) => p + Vector3.Cross(b[0], b[1])
/b[0].Length()/b[1].Length());
return Vector3.Dot(up, sum) > 0;
}
}
}
使用单元测试
namespace SolidworksAddinFramework.Spec.Geometry
{
public class PlanePolygonSpec
{
[Fact]
public void OrientationShouldWork()
{
var points = Sequences.LinSpace(0, Math.PI*2, 100)
.Select(t => new Vector3((float) Math.Cos(t), (float) Math.Sin(t), 0))
.ToList();
points.Orientation(Vector3.UnitZ).Should().BeTrue();
points.Reverse();
points.Orientation(Vector3.UnitZ).Should().BeFalse();
}
}
}
为了它的价值,我使用这个mixin来计算谷歌Maps API v3应用程序的缠绕顺序。
该代码利用了多边形区域的副作用:顺时针旋转顺序的顶点产生一个正的区域,而逆时针旋转顺序的相同顶点产生一个负的区域。该代码还使用了谷歌Maps几何库中的一种私有API。我觉得使用它很舒服——使用风险自负。
示例用法:
var myPolygon = new google.maps.Polygon({/*options*/});
var isCW = myPolygon.isPathClockwise();
完整的单元测试示例@ http://jsfiddle.net/stevejansen/bq2ec/
/** Mixin to extend the behavior of the Google Maps JS API Polygon type
* to determine if a polygon path has clockwise of counter-clockwise winding order.
*
* Tested against v3.14 of the GMaps API.
*
* @author stevejansen_github@icloud.com
*
* @license http://opensource.org/licenses/MIT
*
* @version 1.0
*
* @mixin
*
* @param {(number|Array|google.maps.MVCArray)} [path] - an optional polygon path; defaults to the first path of the polygon
* @returns {boolean} true if the path is clockwise; false if the path is counter-clockwise
*/
(function() {
var category = 'google.maps.Polygon.isPathClockwise';
// check that the GMaps API was already loaded
if (null == google || null == google.maps || null == google.maps.Polygon) {
console.error(category, 'Google Maps API not found');
return;
}
if (typeof(google.maps.geometry.spherical.computeArea) !== 'function') {
console.error(category, 'Google Maps geometry library not found');
return;
}
if (typeof(google.maps.geometry.spherical.computeSignedArea) !== 'function') {
console.error(category, 'Google Maps geometry library private function computeSignedArea() is missing; this may break this mixin');
}
function isPathClockwise(path) {
var self = this,
isCounterClockwise;
if (null === path)
throw new Error('Path is optional, but cannot be null');
// default to the first path
if (arguments.length === 0)
path = self.getPath();
// support for passing an index number to a path
if (typeof(path) === 'number')
path = self.getPaths().getAt(path);
if (!path instanceof Array && !path instanceof google.maps.MVCArray)
throw new Error('Path must be an Array or MVCArray');
// negative polygon areas have counter-clockwise paths
isCounterClockwise = (google.maps.geometry.spherical.computeSignedArea(path) < 0);
return (!isCounterClockwise);
}
if (typeof(google.maps.Polygon.prototype.isPathClockwise) !== 'function') {
google.maps.Polygon.prototype.isPathClockwise = isPathClockwise;
}
})();
对于那些不想“重新发明轮子”的人,我认为值得一提的是,这个检查是在一个名为Shapely (github)的漂亮的Python包中实现的(它基于GEOS C/ c++库):
Shapely is a BSD-licensed Python package for manipulation and analysis of planar geometric objects. It is using the widely deployed open-source geometry library GEOS (the engine of PostGIS, and a port of JTS). Shapely wraps GEOS geometries and operations to provide both a feature rich Geometry interface for singular (scalar) geometries and higher-performance NumPy ufuncs for operations using arrays of geometries. Shapely is not primarily focused on data serialization formats or coordinate systems, but can be readily integrated with packages that are.
来源:https://shapely.readthedocs.io/en/stable/
一个给出OP坐标的小例子:
import numpy as np
from shapely.geometry import Polygon
points = np.array([
(5,0),
(6,4),
(4,5),
(1,5),
(1,0)
])
P = Polygon(points)
这是新构造的多边形:
import matplotlib.pyplot as plt
x,y = P.exterior.coords.xy
plt.plot(x,y)
plt.axis('equal')
plt.grid()
plt.show()
你可以直接使用LinearRing的is_ccw属性来检查多边形是CW还是CCW:
type(P.exterior)
>: shapely.geometry.polygon.LinearRing
P.exterior.is_ccw
>: True
如果颠倒:
points = np.flipud(points)
points
>:
array([[1, 0],
[1, 5],
[4, 5],
[6, 4],
[5, 0]])
P1 = Polygon(points)
P1.exterior.is_ccw
>: True
进一步阅读的文档和参考资料:
shaely is_ccw (github): https://github.com/shapely/shapely/blob/eba985c6e0170ecdd90c83592fd0afa7ae793cb8/shapely/predicates.py#L72-L108 Libgeos (github): https://github.com/libgeos/geos GEOS API参考:https://libgeos.org/doxygen/classgeos_1_1algorithm_1_1Orientation.html#a5af93795969b80f97d7997195974d7c8 GEOS实现(github): https://github.com/libgeos/geos/blob/ab0ce6dafdf7f75ec6d234b6c65bb209037dda17/src/algorithm/Orientation.cpp#L43-L133
解决方案R确定方向和反向如果顺时针(发现这是必要的owin对象):
coords <- cbind(x = c(5,6,4,1,1),y = c(0,4,5,5,0))
a <- numeric()
for (i in 1:dim(coords)[1]){
#print(i)
q <- i + 1
if (i == (dim(coords)[1])) q <- 1
out <- ((coords[q,1]) - (coords[i,1])) * ((coords[q,2]) + (coords[i,2]))
a[q] <- out
rm(q,out)
} #end i loop
rm(i)
a <- sum(a) #-ve is anti-clockwise
b <- cbind(x = rev(coords[,1]), y = rev(coords[,2]))
if (a>0) coords <- b #reverses coords if polygon not traced in anti-clockwise direction