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
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
其他回答
以下是基于上述答案的swift 3.0解决方案:
for (i, point) in allPoints.enumerated() {
let nextPoint = i == allPoints.count - 1 ? allPoints[0] : allPoints[i+1]
signedArea += (point.x * nextPoint.y - nextPoint.x * point.y)
}
let clockwise = signedArea < 0
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
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
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; }
找出y最小的顶点(如果有平手,则x最大)。假设顶点是A,列表中的前一个顶点是B,列表中的下一个顶点是c。现在计算AB和AC的叉乘的符号。
引用:
如何确定一个简单多边形的方向?在 常见问题:计算机。图形。算法。 维基百科的曲线定位。