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
另一个解决方案是;
const isClockwise = (vertices=[]) => {
const len = vertices.length;
const sum = vertices.map(({x, y}, index) => {
let nextIndex = index + 1;
if (nextIndex === len) nextIndex = 0;
return {
x1: x,
x2: vertices[nextIndex].x,
y1: x,
y2: vertices[nextIndex].x
}
}).map(({ x1, x2, y1, y2}) => ((x2 - x1) * (y1 + y2))).reduce((a, b) => a + b);
if (sum > -1) return true;
if (sum < 0) return false;
}
把所有的顶点作为一个数组;
const vertices = [{x: 5, y: 0}, {x: 6, y: 4}, {x: 4, y: 5}, {x: 1, y: 5}, {x: 1, y: 0}];
isClockwise(vertices);
其他回答
一些建议的方法在非凸多边形(如新月形)的情况下会失败。这里有一个简单的方法,它可以用于非凸多边形(它甚至可以用于自相交的多边形,如数字8,告诉你它是否主要是顺时针)。
对边求和,(x2−x1)(y2 + y1)如果结果是正的,曲线是顺时针的,如果结果是负的,曲线是逆时针的。(结果是封闭面积的两倍,采用+/-惯例。)
point[0] = (5,0) edge[0]: (6-5)(4+0) = 4
point[1] = (6,4) edge[1]: (4-6)(5+4) = -18
point[2] = (4,5) edge[2]: (1-4)(5+5) = -30
point[3] = (1,5) edge[3]: (1-1)(0+5) = 0
point[4] = (1,0) edge[4]: (5-1)(0+0) = 0
---
-44 counter-clockwise
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();
}
}
}
下面是一个基于@Beta答案的算法的简单c#实现。
让我们假设我们有一个Vector类型,它的X和Y属性为double类型。
public bool IsClockwise(IList<Vector> vertices)
{
double sum = 0.0;
for (int i = 0; i < vertices.Count; i++) {
Vector v1 = vertices[i];
Vector v2 = vertices[(i + 1) % vertices.Count];
sum += (v2.X - v1.X) * (v2.Y + v1.Y);
}
return sum > 0.0;
}
%是执行模运算的模运算符或余数运算符,该运算符(根据维基百科)在一个数除以另一个数后求余数。
根据@MichelRouzic评论的优化版本:
double sum = 0.0;
Vector v1 = vertices[vertices.Count - 1]; // or vertices[^1] with
// C# 8.0+ and .NET Core
for (int i = 0; i < vertices.Count; i++) {
Vector v2 = vertices[i];
sum += (v2.X - v1.X) * (v2.Y + v1.Y);
v1 = v2;
}
return sum > 0.0;
这不仅节省了模运算%,还节省了数组索引。
测试(参见与@WDUK的讨论)
public static bool IsClockwise(IList<(double X, double Y)> vertices)
{
double sum = 0.0;
var v1 = vertices[^1];
for (int i = 0; i < vertices.Count; i++) {
var v2 = vertices[i];
sum += (v2.X - v1.X) * (v2.Y + v1.Y);
Console.WriteLine($"(({v2.X,2}) - ({v1.X,2})) * (({v2.Y,2}) + ({v1.Y,2})) = {(v2.X - v1.X) * (v2.Y + v1.Y)}");
v1 = v2;
}
Console.WriteLine(sum);
return sum > 0.0;
}
public static void Test()
{
Console.WriteLine(IsClockwise(new[] { (-5.0, -5.0), (-5.0, 5.0), (5.0, 5.0), (5.0, -5.0) }));
// infinity Symbol
//Console.WriteLine(IsClockwise(new[] { (-5.0, -5.0), (-5.0, 5.0), (5.0, -5.0), (5.0, 5.0) }));
}
Sean的答案的JavaScript实现:
function calcArea(poly) { if(!poly || poly.length < 3) return null; let end = poly.length - 1; let sum = poly[end][0]*poly[0][1] - poly[0][0]*poly[end][1]; for(let i=0; i<end; ++i) { const n=i+1; sum += poly[i][0]*poly[n][1] - poly[n][0]*poly[i][1]; } return sum; } function isClockwise(poly) { return calcArea(poly) > 0; } let poly = [[352,168],[305,208],[312,256],[366,287],[434,248],[416,186]]; console.log(isClockwise(poly)); let poly2 = [[618,186],[650,170],[701,179],[716,207],[708,247],[666,259],[637,246],[615,219]]; console.log(isClockwise(poly2));
我很确定这是对的。这似乎是有效的:-)
这些多边形看起来是这样的,如果你想知道的话:
这是我使用其他答案中的解释的解决方案:
def segments(poly):
"""A sequence of (x,y) numeric coordinates pairs """
return zip(poly, poly[1:] + [poly[0]])
def check_clockwise(poly):
clockwise = False
if (sum(x0*y1 - x1*y0 for ((x0, y0), (x1, y1)) in segments(poly))) < 0:
clockwise = not clockwise
return clockwise
poly = [(2,2),(6,2),(6,6),(2,6)]
check_clockwise(poly)
False
poly = [(2, 6), (6, 6), (6, 2), (2, 2)]
check_clockwise(poly)
True