判断坐标点是否在坐标围起来的区域内
场景
有个需求需要判断车辆 gps 上传的坐标点是否在所管辖的区域内
代码
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判断一个坐标是否在一个多边形内(由多个坐标围成的)
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基本思想是利用射线法,计算射线与多边形各边的交点,如果是偶数,则点在多边形外,否则
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在多边形内。还会考虑一些特殊情况,如点在多边形顶点上,点在多边形边上等特殊情况。
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@param $point 指定点坐标 ["lon" => "104.24398627", "lat" => "30.29790868"]
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@param $pts 多边形坐标 顺时针方向
[["lon" => 104.271928,"lat" => 30.494732],
["lon" => 104.280496,"lat" => 30.493443]]
function is_point_in_polygon($point, $pts)
{
$N = count($pts);
$boundOrVertex = true; //如果点位于多边形的顶点或边上,也算做点在多边形内,直接返回true
$intersectCount = 0;//cross points count of x
$precision = 2e-10; //浮点类型计算时候与0比较时候的容差
$p1 = 0;//neighbour bound vertices
$p2 = 0;
$p = $point; //测试点
$p1 = $pts[0];//left vertex
for ($i = 1; $i <= $N; ++$i) {//check all rays
if ($p['lon'] == $p1['lon'] && $p['lat'] == $p1['lat']) {
return $boundOrVertex;//p is an vertex
}
$p2 = $pts[$i % $N];//right vertex
if ($p['lat'] < min($p1['lat'], $p2['lat']) || $p['lat'] > max($p1['lat'], $p2['lat'])) {//ray is outside of our interests
$p1 = $p2;
continue;//next ray left point
}if ($p['lat'] > min($p1['lat'], $p2['lat']) && $p['lat'] < max($p1['lat'], $p2['lat'])) {//ray is crossing over by the algorithm (common part of)
if($p['lon'] <= max($p1['lon'], $p2['lon'])){//x is before of ray
if ($p1['lat'] == $p2['lat'] && $p['lon'] >= min($p1['lon'], $p2['lon'])) {//overlies on a horizontal ray
return $boundOrVertex;
}if ($p1['lon'] == $p2['lon']) {//ray is vertical
if ($p1['lon'] == $p['lon']) {//overlies on a vertical ray
return $boundOrVertex;
}else {//before ray
++$intersectCount;
}
}else {//cross point on the left side
$xinters = ($p['lat'] - $p1['lat']) * ($p2['lon'] - $p1['lon']) / ($p2['lat'] - $p1['lat']) + $p1['lon'];//cross point of lon
if (abs($p['lon'] - $xinters) < $precision) {//overlies on a ray
return $boundOrVertex;
}if ($p['lon'] < $xinters) {//before ray
++$intersectCount;
}
}
}
}else {//special case when ray is crossing through the vertex
if ($p['lat'] == $p2['lat'] && $p['lon'] <= $p2['lon']) {//p crossing over p2
$p3 = $pts[($i+1) % $N]; //next vertex
if ($p['lat'] >= min($p1['lat'], $p3['lat']) && $p['lat'] <= max($p1['lat'], $p3['lat'])) { //p.lat lies between p1.lat & p3.lat
++$intersectCount;
}else{
$intersectCount += 2;
}
}
}
$p1 = $p2;//next ray left point
}
if ($intersectCount % 2 == 0) {//偶数在多边形外
return false;
}else { //奇数在多边形内
return true;
}
}