Leaflet中如何检查一个多边形点是否在另一个多边形内部

Question

我从 Leaflet 的 GeoJSON 地图中选择了两组多边形坐标，包括父级和子级坐标：

var parentCoordinates=[
    [
        32.05898221582174,
        -28.31004731142091
    ],
    [
        32.05898221582174,
        -28.308044824292978
    ],
    [
        32.06134255975485,
        -28.308044824292978
    ],
    [
        32.06134255975485,
        -28.31004731142091
    ],
    [
        32.05898221582174,
        -28.31004731142091
    ]
]
var childCoordinates=[
  [
    32.059904895722866,
    -28.30970726909422
  ],
  [
    32.059904895722866,
    -28.308743809931784
  ],
  [
    32.06089194864035,
    -28.308743809931784
  ],
  [
    32.06089194864035,
    -28.30970726909422
  ],
  [
    32.059904895722866,
    -28.30970726909422
  ]
]

如图所示，子元素被绘制在父区域内：

我正在使用射线投射算法来确定点是否位于多边形内，但是我得到的结果是错误的。请告诉我我的错误在哪里，或者有其他方法可以解决。谢谢。

- forgottofly

---

注意：对于多段线位于多边形内部，它可以正常工作。但是对于多边形位于另一个多边形内部（如图所示），它无法正常工作。 - forgottofly

Answer 1

我对Turf有着良好的使用经验。它使用方便，文档完善，示例已经与leaflet一起展示。

针对您的问题，您可以使用turf.within，其中parentCoordinates作为turf.polygon，而childCoordinates则作为turf.point的数组。

var parentPolygon = turf.polygon([parentCoordinates]);

var inside = true;
childCoordinates.forEach(function(coordinates) {
    point = turf.point(coordinates);
    if (!turf.inside(point, parentPolygon)){
      alert("Oh no! "+ coordinates + " isn't in polygon");
      inside = false;
    }
});

alert("Child polygon inside parent polygon ? " + inside);

这是一个Fiddle示例，可以在这里查看。

- Eric Duminil

Answer 2

我使用了您的算法和另一个在这里找到的https://rosettacode.org/wiki/Ray-casting_algorithm，两者都返回正确的值。

也许这个fiddle可以帮助您实现： https://jsfiddle.net/4psL2hoo/1/ 您的算法

// Data
var parentCoordinates=[
    [
        32.05898221582174,
        -28.31004731142091
    ],
    [
        32.05898221582174,
        -28.308044824292978
    ],
    [
        32.06134255975485,
        -28.308044824292978
    ],
    [
        32.06134255975485,
        -28.31004731142091
    ],
    [
        32.05898221582174,
        -28.31004731142091
    ]
]
var childCoordinates=[
  [
    32.059904895722866,
    -28.30970726909422
  ],
  [
    32.059904895722866,
    -28.308743809931784
  ],
  [
    32.06089194864035,
    -28.308743809931784
  ],
  [
    32.06089194864035,
    -28.30970726909422
  ],
  [
    32.059904895722866,
    -28.30970726909422
  ]
]

// Other algo
function test(point, vs) {
    // ray-casting algorithm based on
    // http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

    var x = point[0], y = point[1];

    var inside = false;
    for (var i = 0, j = vs.length - 1; i < vs.length; j = i++) {
        var xi = vs[i][0], yi = vs[i][1];
        var xj = vs[j][0], yj = vs[j][1];

        var intersect = ((yi > y) != (yj > y))
            && (x < (xj - xi) * (y - yi) / (yj - yi) + xi);
        if (intersect) inside = !inside;
    }

    return inside;
};

for (var i = 0; i < childCoordinates.length; i++) {
     var testPoint = childCoordinates[i];
     console.log(JSON.stringify(testPoint) + '\tin parentCoordinate\t' + test(testPoint, parentCoordinates));
}

Rosetta Code算法

//https://rosettacode.org/wiki/Ray-casting_algorithm
function contains(bounds, lat, lng) {
    //https://rosettacode.org/wiki/Ray-casting_algorithm
    var count = 0;
    for (var b = 0; b < bounds.length; b++) {
        var vertex1 = bounds[b];
        var vertex2 = bounds[(b + 1) % bounds.length];
        if (west(vertex1, vertex2, lng, lat))
            ++count;
    }
    return count % 2;

    /**
     * @return {boolean} true if (x,y) is west of the line segment connecting A and B
     */
    function west(A, B, x, y) {
        if (A.y <= B.y) {
            if (y <= A.y || y > B.y ||
                x >= A.x && x >= B.x) {
                return false;
            } else if (x < A.x && x < B.x) {
                return true;
            } else {
                return (y - A.y) / (x - A.x) > (B.y - A.y) / (B.x - A.x);
            }
        } else {
            return west(B, A, x, y);
        }
    }
}

var square = {name: 'square', bounds: [{x: 32.05898221582174, y: -28.31004731142091}, {x: 32.05898221582174, y: -28.308044824292978}, {x: 32.06134255975485, y: -28.308044824292978}, {x: 32.06134255975485, y: -28.31004731142091}]};

var shapes = [square];
var testPoints = [{lng: 32.059904895722866, lat: -28.30970726909422}, {lng: 32.059904895722866, lat: -28.308743809931784}, {lng: 32.06089194864035, lat: -28.308743809931784},
    {lng: 32.06089194864035, lat: -28.30970726909422}];

for (var s = 0; s < shapes.length; s++) {
    var shape = shapes[s];
    for (var tp = 0; tp < testPoints.length; tp++) {
        var testPoint = testPoints[tp];
        console.log(JSON.stringify(testPoint) + '\tin ' + shape.name + '\t' + contains(shape.bounds, testPoint.lat, testPoint.lng));
    }
}

- Fefux

Answer 3

你可以尝试使用Leaflet的API - contains。您可以使用LatLngBounds创建父多边形，然后再创建子多边形。

parentPolygon.contains(childPolygon)

- Alex Parij

---

LatLngBounds创建一个矩形；海报需要一般的多边形。被踩了。 - Ed Staub