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);

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);
}

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:

一个计算上更简单的方法，如果你已经知道多边形内的一个点:

从原始多边形中选择任意线段，按此顺序选择点及其坐标。 加上一个已知的“内部”点，形成一个三角形。 根据以上三点计算CW或CCW。

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