本文介绍了在最接近latlng的多段线中找到一个点的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有一个polyine，我用从google maps方向服务获得的latlng绘制出来的。
现在我想找到最接近给定点的折线上的点。
显而易见的方式（对我来说）是通过循环然后找到它们与给定点之间的距离，但这是无效的，因为折线上的点可能很大。
我会很高兴听到这样做的任何替代方案。
在此先感谢。
解决方案
查看Bill Chadwick的例子：
它正在寻找鼠标线上最近的点。另请注意，这是一个Google Maps API v2示例（但v3的原理是相同的）。
//代码查找经纬度点和经纬度点折线之间的距离（单位为米）
//全部在WGS84中。免费用于任何用途。
//
// Bill Chadwick 2007
//更新为Google Maps API v3，Lawrence Ross 2014
//根据其经度和纬度构建bdccGeo度
函数bdccGeo（lat，lon）
{
var theta =（lon * Math.PI / 180.0）;
var rlat = bdccGeoGeocentricLatitude（lat * Math.PI / 180.0）;
var c = Math.cos（rlat）;
this.x = c * Math.cos（theta）;
this.y = c * Math.sin（theta）;
this.z = Math.sin（rlat）;
}
bdccGeo.prototype = new bdccGeo（）;
//内部帮助函数===================================== ====
//将地理转换为以地心为中心的纬度（弧度）。
函数bdccGeoGeocentricLatitude（geographicLatitude）
{
var flattening = 1.0 / 298.257223563; // WGS84
var f =（1.0 - flattening）*（1.0 - flattening）;
return Math.atan（（Math.tan（geographicLatitude）* f））;
}
//返回由圆弧geo1到geo2和
// geo3到geo4定义的两个很好的
//圆圈的两个对极点。返回一个点作为Geo，使用.antipode获取另一个点
函数bdccGeoGetIntersection（geo1，geo2，geo3，geo4）
{
var geoCross1 = geo1.crossNormalize（geo2）;
var geoCross2 = geo3.crossNormalize（geo4）;
返回geoCross1.crossNormalize（geoCross2）;
}
//从Radians到Meters
函数bdccGeoRadiansToMeters（rad）
{
return rad * 6378137.0; // WGS84赤道半径（以米为单位）
}
//从米到弧度
函数bdccGeoMetersToRadians（m）
{
return m / 6378137.0; // WGS84赤道半径（以米为单位）
}
//属性=========================== ======================
bdccGeo.prototype.getLatitudeRadians = function（）
{
return（bdccGeoGeographicLatitude（Math.atan2（this.z，
Math.sqrt（（this.x * this.x）+（this.y * this.y）））））;
}
bdccGeo.prototype.getLongitudeRadians = function（）
{
return（Math.atan2（this.y，this.x））;
}
bdccGeo.prototype.getLatitude = function（）
{
return this.getLatitudeRadians（）* 180.0 / Math.PI;
}
bdccGeo.prototype.getLongitude = function（）
{
return this.getLongitudeRadians（）* 180.0 / Math.PI;
}
//方法================================== ===============
//数学
bdccGeo.prototype.dot =函数（b）
{
返回（（this.x * bx）+（this.y * by）+（this.z * bz））;
}
//更多数学
bdccGeo.prototype.crossLength =函数（b）
{
var x =（this.y * bz） - （this.z * by）;
var y =（this.z * b.x） - （this.x * b.z）;
var z =（this.x * b.y） - （this.y * b.x）;
return Math.sqrt（（x * x）+（y * y）+（z * z））;
}
//更多数学
bdccGeo.prototype.scale =函数
{
var r = new bdccGeo（0,0） ;
r.x = this.x * s;
r.y = this.y * s;
r.z = this.z * s;
return r;
//更多数学
bdccGeo.prototype.crossNormalize =函数（b）
{
var x =（this.y * bz） - （this.z * by）;
var y =（this.z * b.x） - （this.x * b.z）;
var z =（this.x * b.y） - （this.y * b.x）;
var L = Math.sqrt（（x * x）+（y * y）+（z * z））;
var r = new bdccGeo（0,0）;
r.x = x / L;
r.y = y / L;
r.z = z / L;
return r;
}
//指向这个点的世界另一端
bdccGeo.prototype.antipode = function（）
{
return this。规模（-1.0）;
}
//以弧度为单位的距离，从这点到点v2
bdccGeo。 prototype.distance = function（v2）
{
return Math.atan2（v2.crossLength（this），v2.dot（this））;
}
//以米为单位返回此点与线段geo1-geo2
//的垂直距离的最小值，以及此点到线段的距离在geo1和geo2中结束
bdccGeo.prototype.distanceToLineSegMtrs =函数（geo1，geo2）
{
//单位球体上方原点和法线到地平面的点geo1，geo2
//可以是飞机的任一侧
var p2 = geo1.crossNormalize（geo2）;
//使用GC geo1 / geo2通过p传递给geo1 / geo2的GC法线交点
var ip = bdccGeoGetIntersection（geo1，geo2，this，p2）;
//需要检查ip或其反对象在p1和p2之间
var d = geo1.distance（geo2）;
var d1p = geo1.distance（ip）;
var d2p = geo2.distance（ip）;
//window.status = d +，+ d1p +，+ d2p; （（d> = d1p）&（d> = d2p））
返回bdccGeoRadiansToMeters（this.distance（ip））;
else
{
ip = ip.antipode（）;
d1p = geo1.distance（ip）;
d2p = geo2.distance（ip）; （（d> = d1p）&&（d> = d2p））
返回bdccGeoRadiansToMeters（this.distance（ip））;
}
。
else
return bdccGeoRadiansToMeters（Math.min（geo1.distance（this），geo2.distance（this）））;
}
//与GLatLng指向GPolyline或GPolygon的距离（单位：百万）b
bc函数bdccGeoDistanceToPolyMtrs（poly，point）
{
var d = 999999999 ;
var i;
var p = new bdccGeo（point.lat（），point.lng（））; （$）
for（i = 0; i {
var p1 = poly.getPath（）。getAt（i） ;
var l1 = new bdccGeo（p1.lat（），p1.lng（））;
var p2 = poly.getPath（）。getAt（i + 1）;
var l2 = new bdccGeo（p2.lat（），p2.lng（））;
var dp = p.distanceToLineSegMtrs（l1，l2）;
if（dp d = dp;
}
return d;
}
//获得一个新的GLatLng距离指南针距离GLatLng点的azimuthDegrees
//的距离 - 精确到140米（14米20米处）的200米以内英国
函数bdccGeoPointAtRangeAndBearing（point，distanceMeters，azimuthDegrees）
{
var latr = point.lat（）* Math.PI / 180.0;
var lonr = point.lng（）* Math.PI / 180.0;
var coslat = Math.cos（latr）;
var sinlat = Math.sin（latr）;
var az = azimuthDegrees * Math.PI / 180.0;
var cosaz = Math.cos（az）;
var sinaz = Math.sin（az）;
var dr = distanceMeters / 6378137.0; //使用WGS84的弧度距离Equalsial Radius
var sind = Math.sin（dr）;
var cosd = Math.cos（dr）; $（Math.asin（（sinlat * cosd）+（coslat * sind * cosaz））* 180.0 / Math.PI，
（Math.atan2（ （sind * sinaz），（coslat * cosd） - （sinlat * sind * cosaz））+ lonr）* 180.0 / Math.PI）;
}
i have a polyine which i have drawn with latlngs obtained from google maps directions service.Now i want to find a point on the polyline that is closest to a given point.
The obvious way (to me) is to kind of loop through all the points in the polyline and find the distance between them and the given point, however this is inefficient because the points on the polyline can potentially be large.
I would be glad to hear any alternatives of doing this.Thanks in advance.
解决方案
See Bill Chadwick's example here:
http://www.bdcc.co.uk/Gmaps/BdccGmapBits.htm
above example ported to v3 (code at bottom of this answer)
on his page under:
from that post:
There is a similar, better demo here http://wtp2.appspot.com/cSnapToRouteDemo.html
It is finding the closest point on the line to the mouse. Also note that it is a Google Maps API v2 example (but the principle with v3 would be the same).
// Code to find the distance in metres between a lat/lng point and a polyline of lat/lng points
// All in WGS84. Free for any use.
//
// Bill Chadwick 2007
// updated to Google Maps API v3, Lawrence Ross 2014
// Construct a bdccGeo from its latitude and longitude in degrees
function bdccGeo(lat, lon)
{
var theta = (lon * Math.PI / 180.0);
var rlat = bdccGeoGeocentricLatitude(lat * Math.PI / 180.0);
var c = Math.cos(rlat);
this.x = c * Math.cos(theta);
this.y = c * Math.sin(theta);
this.z = Math.sin(rlat);
}
bdccGeo.prototype = new bdccGeo();
// internal helper functions =========================================
// Convert from geographic to geocentric latitude (radians).
function bdccGeoGeocentricLatitude(geographicLatitude)
{
var flattening = 1.0 / 298.257223563;//WGS84
var f = (1.0 - flattening) * (1.0 - flattening);
return Math.atan((Math.tan(geographicLatitude) * f));
}
// Returns the two antipodal points of intersection of two great
// circles defined by the arcs geo1 to geo2 and
// geo3 to geo4. Returns a point as a Geo, use .antipode to get the other point
function bdccGeoGetIntersection( geo1, geo2, geo3, geo4)
{
var geoCross1 = geo1.crossNormalize(geo2);
var geoCross2 = geo3.crossNormalize(geo4);
return geoCross1.crossNormalize(geoCross2);
}
//from Radians to Meters
function bdccGeoRadiansToMeters(rad)
{
return rad * 6378137.0; // WGS84 Equatorial Radius in Meters
}
//from Meters to Radians
function bdccGeoMetersToRadians(m)
{
return m / 6378137.0; // WGS84 Equatorial Radius in Meters
}
// properties =================================================
bdccGeo.prototype.getLatitudeRadians = function()
{
return (bdccGeoGeographicLatitude(Math.atan2(this.z,
Math.sqrt((this.x * this.x) + (this.y * this.y)))));
}
bdccGeo.prototype.getLongitudeRadians = function()
{
return (Math.atan2(this.y, this.x));
}
bdccGeo.prototype.getLatitude = function()
{
return this.getLatitudeRadians() * 180.0 / Math.PI;
}
bdccGeo.prototype.getLongitude = function()
{
return this.getLongitudeRadians() * 180.0 / Math.PI ;
}
// Methods =================================================
//Maths
bdccGeo.prototype.dot = function( b)
{
return ((this.x * b.x) + (this.y * b.y) + (this.z * b.z));
}
//More Maths
bdccGeo.prototype.crossLength = function( b)
{
var x = (this.y * b.z) - (this.z * b.y);
var y = (this.z * b.x) - (this.x * b.z);
var z = (this.x * b.y) - (this.y * b.x);
return Math.sqrt((x * x) + (y * y) + (z * z));
}
//More Maths
bdccGeo.prototype.scale = function( s)
{
var r = new bdccGeo(0,0);
r.x = this.x * s;
r.y = this.y * s;
r.z = this.z * s;
return r;
}
// More Maths
bdccGeo.prototype.crossNormalize = function( b)
{
var x = (this.y * b.z) - (this.z * b.y);
var y = (this.z * b.x) - (this.x * b.z);
var z = (this.x * b.y) - (this.y * b.x);
var L = Math.sqrt((x * x) + (y * y) + (z * z));
var r = new bdccGeo(0,0);
r.x = x / L;
r.y = y / L;
r.z = z / L;
return r;
}
// point on opposite side of the world to this point
bdccGeo.prototype.antipode = function()
{
return this.scale(-1.0);
}
//distance in radians from this point to point v2
bdccGeo.prototype.distance = function( v2)
{
return Math.atan2(v2.crossLength(this), v2.dot(this));
}
//returns in meters the minimum of the perpendicular distance of this point from the line segment geo1-geo2
//and the distance from this point to the line segment ends in geo1 and geo2
bdccGeo.prototype.distanceToLineSegMtrs = function(geo1, geo2)
{
//point on unit sphere above origin and normal to plane of geo1,geo2
//could be either side of the plane
var p2 = geo1.crossNormalize(geo2);
// intersection of GC normal to geo1/geo2 passing through p with GC geo1/geo2
var ip = bdccGeoGetIntersection(geo1,geo2,this,p2);
//need to check that ip or its antipode is between p1 and p2
var d = geo1.distance(geo2);
var d1p = geo1.distance(ip);
var d2p = geo2.distance(ip);
//window.status = d + ", " + d1p + ", " + d2p;
if ((d >= d1p) && (d >= d2p))
return bdccGeoRadiansToMeters(this.distance(ip));
else
{
ip = ip.antipode();
d1p = geo1.distance(ip);
d2p = geo2.distance(ip);
}
if ((d >= d1p) && (d >= d2p))
return bdccGeoRadiansToMeters(this.distance(ip));
else
return bdccGeoRadiansToMeters(Math.min(geo1.distance(this),geo2.distance(this)));
}
// distance in meters from GLatLng point to GPolyline or GPolygon poly
function bdccGeoDistanceToPolyMtrs(poly, point)
{
var d = 999999999;
var i;
var p = new bdccGeo(point.lat(),point.lng());
for(i=0; i<(poly.getPath().getLength()-1); i++)
{
var p1 = poly.getPath().getAt(i);
var l1 = new bdccGeo(p1.lat(),p1.lng());
var p2 = poly.getPath().getAt(i+1);
var l2 = new bdccGeo(p2.lat(),p2.lng());
var dp = p.distanceToLineSegMtrs(l1,l2);
if(dp < d)
d = dp;
}
return d;
}
// get a new GLatLng distanceMeters away on the compass bearing azimuthDegrees
// from the GLatLng point - accurate to better than 200m in 140km (20m in 14km) in the UK
function bdccGeoPointAtRangeAndBearing (point, distanceMeters, azimuthDegrees)
{
var latr = point.lat() * Math.PI / 180.0;
var lonr = point.lng() * Math.PI / 180.0;
var coslat = Math.cos(latr);
var sinlat = Math.sin(latr);
var az = azimuthDegrees* Math.PI / 180.0;
var cosaz = Math.cos(az);
var sinaz = Math.sin(az);
var dr = distanceMeters / 6378137.0; // distance in radians using WGS84 Equatorial Radius
var sind = Math.sin(dr);
var cosd = Math.cos(dr);
return new google.maps.LatLng(Math.asin((sinlat * cosd) + (coslat * sind * cosaz)) * 180.0 / Math.PI,
(Math.atan2((sind * sinaz), (coslat * cosd) - (sinlat * sind * cosaz)) + lonr) * 180.0 / Math.PI);
}
这篇关于在最接近latlng的多段线中找到一个点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！