首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >如何确定一个点是否只存在于一个多边形中,而不是存在于相邻/附近的多边形中?

如何确定一个点是否只存在于一个多边形中,而不是存在于相邻/附近的多边形中?
EN

Stack Overflow用户
提问于 2019-07-12 15:52:47
回答 1查看 133关注 0票数 1

我正在使用这里给出的算法(许多其他人也在使用它)。但在我针对某个特定点的场景中,它显示了多个多边形处的交点。

我在mysql中使用php和导入的geojson。并使用https://assemblysys.com/php-point-in-polygon-algorithm/中的以下代码

代码语言:javascript
复制
<?php
/*
Description: The point-in-polygon algorithm allows you to check if a point is
inside a polygon or outside of it.
Author: Michaël Niessen (2009)
Website: http://AssemblySys.com

If you find this script useful, you can show your
appreciation by getting Michaël a cup of coffee ;)
PayPal: https://www.paypal.me/MichaelNiessen

As long as this notice (including author name and details) is included and
UNALTERED, this code is licensed under the GNU General Public License version 3:
http://www.gnu.org/licenses/gpl.html
*/

class pointLocation {
    var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices?

    function pointLocation() {
    }

    function pointInPolygon($point, $polygon, $pointOnVertex = true) {
        $this->pointOnVertex = $pointOnVertex;

        // Transform string coordinates into arrays with x and y values
        $point = $this->pointStringToCoordinates($point);
        $vertices = array(); 
        foreach ($polygon as $vertex) {
            $vertices[] = $this->pointStringToCoordinates($vertex); 
        }

        // Check if the point sits exactly on a vertex
        if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {
            return "vertex";
        }

        // Check if the point is inside the polygon or on the boundary
        $intersections = 0; 
        $vertices_count = count($vertices);

        for ($i=1; $i < $vertices_count; $i++) {
            $vertex1 = $vertices[$i-1]; 
            $vertex2 = $vertices[$i];
            if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary
                return "boundary";
            }
            if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) { 
                $xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x']; 
                if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)
                    return "boundary";
                }
                if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {
                    $intersections++; 
                }
            } 
        } 
        // If the number of edges we passed through is odd, then it's in the polygon. 
        if ($intersections % 2 != 0) {
            return "inside";
        } else {
            return "outside";
        }
    }

    function pointOnVertex($point, $vertices) {
        foreach($vertices as $vertex) {
            if ($point == $vertex) {
                return true;
            }
        }

    }

    function pointStringToCoordinates($pointString) {
        $coordinates = explode(" ", $pointString);
        return array("x" => $coordinates[0], "y" => $coordinates[1]);
    }

}
?>

上面的代码对于大多数点都工作得很好,但在某些情况下,它会返回给定点存在于多个多边形中(绘制了它们,它们的长度为80 km )。所以我的疑虑是。1-我是否使用right algo来解决我的问题? 2-上面代码中下面几行的用途是什么?

代码语言:javascript
复制
$xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];
EN

回答 1

Stack Overflow用户

发布于 2019-07-16 15:11:50

回答我自己的问题。由于我没有得到任何可靠的解决方案,我不得不想出一个变通办法。我从xinters中减去pointx,将它们放入一个数组中,然后选取最短的一个。这确保了相交的多边形就是我正在寻找的多边形。

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/57002508

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档