問(wèn)題描述

我從傳單 geoJSON 地圖中選擇了兩組多邊形坐標(biāo).父子坐標(biāo)是坐標(biāo)是:

I'm having two sets of polygon coordinates selected from leaflet geoJSON map. The parent and child coordinates are coordinates are:

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
  ]
]

child在父區(qū)域內(nèi)繪制如圖:

The child is drawn inside the parent area as shown in the picture:

使用 Ray Casting 算法 來(lái)確定是否點(diǎn)位于多邊形內(nèi)，我無(wú)法確定，因?yàn)槲业玫降慕Y(jié)果是錯(cuò)誤的.請(qǐng)讓我知道我做錯(cuò)了什么或任何其他方式來(lái)確定解決方案.謝謝

Using Ray Casting algorithm to determine if the point lies inside polygon I 'm not able to determine as the result I'm getting is false. Please let me know where I'm doing wrong or any other way to determine the solution.Thanks

推薦答案

我嘗試了你的算法和另一個(gè)在這里找到的 https://rosettacode.org/wiki/Ray-casting_algorithm 并且都返回正確的值.

I tried with your algorithm and another found here https://rosettacode.org/wiki/Ray-casting_algorithm and both return the right value.

也許這個(gè)小提琴可以幫助你實(shí)現(xiàn):

Maybe this fiddle can help you in the implementation :

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) + '	in parentCoordinate	' + test(testPoint, parentCoordinates));
}

Rosetta 代碼算法

//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) + '	in ' + shape.name + '	' + contains(shape.bounds, testPoint.lat, testPoint.lng));
    }
}

這篇關(guān)于檢查一個(gè)多邊形點(diǎn)是否在傳單中的另一個(gè)內(nèi)部的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！