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

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

虽然这些答案是正确的，但它们在数学上的强度比必要的要大。假设地图坐标，其中最北的点是地图上的最高点。找到最北的点，如果两个点相等，它是最北的，然后是最东的(这是lhf在他的答案中使用的点)。在你的观点中，

点[0]= (5,0)

点[1]= (6,4)

点[2]= (4,5)

点[3]= (1,5)

点[4]= (1,0)

If we assume that P2 is the most north then east point either the previous or next point determine clockwise, CW, or CCW. Since the most north point is on the north face, if the P1 (previous) to P2 moves east the direction is CW. In this case, it moves west, so the direction is CCW as the accepted answer says. If the previous point has no horizontal movement, then the same system applies to the next point, P3. If P3 is west of P2, it is, then the movement is CCW. If the P2 to P3 movement is east, it's west in this case, the movement is CW. Assume that nte, P2 in your data, is the most north then east point and the prv is the previous point, P1 in your data, and nxt is the next point, P3 in your data, and [0] is horizontal or east/west where west is less than east, and [1] is vertical.

if (nte[0] >= prv[0] && nxt[0] >= nte[0]) return(CW);
if (nte[0] <= prv[0] && nxt[0] <= nte[0]) return(CCW);
// Okay, it's not easy-peasy, so now, do the math
if (nte[0] * nxt[1] - nte[1] * nxt[0] - prv[0] * (nxt[1] - crt[1]) + prv[1] * (nxt[0] - nte[0]) >= 0) return(CCW); // For quadrant 3 return(CW)
return(CW) // For quadrant 3 return (CCW)

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

在测试了几个不可靠的实现之后，在CW/CCW方向方面提供令人满意结果的算法是由OP在这个线程(shoelace_formula_3)中发布的算法。

与往常一样，正数表示CW方向，而负数表示CCW方向。

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。