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
下面是基于这个答案的一个简单的Python 3实现(反过来,它是基于已接受答案中提出的解决方案)
def is_clockwise(points):
# points is your list (or array) of 2d points.
assert len(points) > 0
s = 0.0
for p1, p2 in zip(points, points[1:] + [points[0]]):
s += (p2[0] - p1[0]) * (p2[1] + p1[1])
return s > 0.0
其他回答
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) }));
}
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
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:
Javascript实现的lhf的答案 (再次强调,这只适用于简单的多边形,即不适用于图8)
let polygon = [ {x:5,y:0}, {x:6,y:4}, {x:4,y:5}, {x:1,y:5}, {x:1,y:0} ] document.body.innerHTML += `Polygon ${polygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(polygon)}` let reversePolygon = [] polygon.forEach(point=>reversePolygon.unshift(point)) document.body.innerHTML += `<br/>Polygon ${reversePolygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(reversePolygon)}` function isPolygonClockwise (polygon) { // From http://www.faqs.org/faqs/graphics/algorithms-faq/ "How do I find the orientation of a simple polygon?" // THIS SOMETIMES FAILS if the polygon is a figure 8, or similar shape where it crosses over itself // Take the lowest point (break ties with the right-most). if (polygon.length < 3) { return true // A single point or two points can't be clockwise/counterclockwise } let previousPoint = polygon[0] let lowestPoint = polygon[1] let nextPoint = polygon[2] polygon.forEach((point, index)=>{ if (point.y > lowestPoint.y || (point.y === lowestPoint.y && point.x > lowestPoint.x)) { // larger y values are lower, in svgs // Break ties with furthest right previousPoint = polygon[(index-1) >= (0) ? (index-1) : (polygon.length-1)] lowestPoint = polygon[index] nextPoint = polygon[(index+1) <= (polygon.length-1) ? (index+1) : (0)] } }) // Check the angle between the previous point, that point, and the next point. // If the angle is less than PI radians, the polygon is clockwise let angle = findAngle(previousPoint, lowestPoint, nextPoint) return angle < Math.PI } function findAngle(A,B,C) { var AB = Math.atan2(B.y-A.y, B.x-A.x); var BC = Math.atan2(C.y-B.y, C.x-B.x); if (AB < 0) AB += Math.PI*2 if (BC < 0) BC += Math.PI*2 return BC-AB; }
正如这篇维基百科文章中所解释的曲线方向,给定平面上的3个点p, q和r(即x和y坐标),您可以计算以下行列式的符号
如果行列式为负(即定向(p, q, r) < 0),则多边形是顺时针方向(CW)。如果行列式为正(即定向(p, q, r) > 0),则多边形是逆时针方向(CCW)。如果点p, q和r共线,行列式为零(即定向(p, q, r) == 0)。
在上面的公式中,由于我们使用的是齐次坐标,我们将1放在p, q和r的坐标前面。