Javascript 平滑GPS跟踪路线坐标
我有一些我记录的坐标数据。不幸的是,他们似乎不是很好。他们有时跳过地图。因此,现在我正在寻找一些使路线看起来更真实的平坦或过滤算法 目前,我唯一的过滤器是计算在一秒钟内行驶的最大可能米数(在公共汽车、汽车或步行中),并将其与坐标进行比较,丢弃那些在时间范围内不可能实现的坐标。因此,如果一个人能在一秒钟内走到2.5米,我有两个相互距离10米的coords,它们在两秒钟内被记录下来,我试着找到它们并扔掉。这有点帮助 代码如下:Javascript 平滑GPS跟踪路线坐标,javascript,google-maps,polyline,geography,Javascript,Google Maps,Polyline,Geography,我有一些我记录的坐标数据。不幸的是,他们似乎不是很好。他们有时跳过地图。因此,现在我正在寻找一些使路线看起来更真实的平坦或过滤算法 目前,我唯一的过滤器是计算在一秒钟内行驶的最大可能米数(在公共汽车、汽车或步行中),并将其与坐标进行比较,丢弃那些在时间范围内不可能实现的坐标。因此,如果一个人能在一秒钟内走到2.5米,我有两个相互距离10米的coords,它们在两秒钟内被记录下来,我试着找到它们并扔掉。这有点帮助 代码如下: filters.max_possible_travel = functi
filters.max_possible_travel = function(data) {
//http://en.wikipedia.org/wiki/Preferred_walking_speed
//I switched to 16, as the route was made by driving with a bus...
var maxMetersPerSec = 16,
i, m, last, result = [];
for(i=0;i<data.length;i++) {
m = data[i];
if (last) {
// seconds between current and last coord
var diff = (m.created.getTime() - last.created.getTime()) / 1000;
// the maximum amount of meters a person,bus,car etc can make per sec.
var maxDistance = diff * maxMetersPerSec;
// the actual distance traveled
var traveledDistance = google.maps.geometry.spherical.computeDistanceBetween(last.googLatLng, m.googLatLng);
if (traveledDistance > maxDistance) {
continue;
} else {
result.push(m);
}
}
last = m;
}
return result;
};
filters.max\u可能的行程=功能(数据){
//http://en.wikipedia.org/wiki/Preferred_walking_speed
//我换成了16路,因为路线是由一辆公交车决定的。。。
var MaxMeterPersec=16,
i、 m,last,result=[];
对于(i=0;i最大距离){
继续;
}否则{
结果:推力(m);
}
}
last=m;
}
返回结果;
};
为了让事情变得更简单,我创建了这个fiddle,它已经实现了我的第一个过滤器,还提供了添加新过滤器的功能
我还有一些更进一步的想法:
- 扔掉特定半径内的所有坐标。这最终会消除一些令人不安的协调,如果你只是站在周围几分钟
- 将所有坐标按n秒帧分组,并尝试确定此块中最相关的坐标。不幸的是,我不知道如何:(
/*基于堆栈的Douglas Peucker线简化例程
返回的是一个简化的google.maps.LatLng数组
在Gary J.Robinson博士编写代码之后,
环境系统科学中心,
雷丁大学,雷丁,英国
*/
函数GDouglasPeucker(源,扭结)
/*source[]在google.maps.LatLngs中输入坐标*/
/*扭结以米为单位,超过此深度的扭结保持不变*/
/*扭结深度是三角形abc的高度,其中a-b和b-c是两条连续线段*/
{
变量n_源、n_堆栈、n_目的、开始、结束、i、sig;
var dev_sqr、max dev_sqr、band_sqr;
变量x12、y12、d12、x13、y13、d13、x23、y23、d23;
var F=((数学PI/180.0)*0.5);
var index=new Array();/*要包含在缩减行中的源点索引的数组*/
var sig_start=new Array();/*工作段开始和结束的索引*/
var sig_end=新数组();
/*检查简单的案例*/
如果(震源长度<3)
返回(源);/*一个或两个点*/
/*更复杂的情况。初始化堆栈*/
n_source=source.length;
band_sqr=扭结*360.0/(2.0*Math.PI*6378137.0);/*现在以度为单位*/
band_sqr*=band_sqr;
n_dest=0;
sig_start[0]=0;
sig_end[0]=n_源-1;
n_堆栈=1;
/*虽然堆栈不是空的*/
而(n_堆栈>0){
/*…从堆栈中弹出最上面的条目*/
start=sig_start[n_stack-1];
end=sig_end[n_stack-1];
n_堆栈--;
如果((结束-开始)>1){/*任何中间点?*/
/*…是的,所以找到最不正常的中间点
连接起点和终点的线的任一侧*/
x12=(源[end].lng()-source[start].lng());
y12=(源[end].lat()-source[start].lat());
如果(数学abs(x12)>180.0)
x12=360.0-数学abs(x12);
x12*=Math.cos(F*(source[end].lat()+source[start].lat());/*使用平均lat减少lng*/
d12=(x12*x12)+(y12*y12);
对于(i=start+1,sig=start,max_dev_sqr=-1.0;i180.0)
x13=360.0-数学abs(x13);
x13*=Math.cos(F*(源[i].lat()+源[start].lat());
d13=(x13*x13)+(y13*y13);
x23=(源[i].lng()-源[end].lng());
y23=(源[i].lat()-source[end].lat());
如果(数学绝对值(x23)>180.0)
x23=360.0-数学绝对值(x23);
x23*=Math.cos(F*(源[i].lat()+源[end].lat());
d23=(x23*x23)+(y23*y23);
如果(d13>=(d12+d23))
dev_sqr=d23;
否则如果(d23>=(d12+d13))
dev_sqr=d13;
其他的
dev_sqr=(x13*y12-y13*x12)*(x13*y12-y13*x12)/d12;//求解三角形
如果(开发sqr>最大开发sqr){
sig=i;
/* Stack-based Douglas Peucker line simplification routine
returned is a reduced google.maps.LatLng array
After code by Dr. Gary J. Robinson,
Environmental Systems Science Centre,
University of Reading, Reading, UK
*/
function GDouglasPeucker (source, kink)
/* source[] Input coordinates in google.maps.LatLngs */
/* kink in metres, kinks above this depth kept */
/* kink depth is the height of the triangle abc where a-b and b-c are two consecutive line segments */
{
var n_source, n_stack, n_dest, start, end, i, sig;
var dev_sqr, max_dev_sqr, band_sqr;
var x12, y12, d12, x13, y13, d13, x23, y23, d23;
var F = ((Math.PI / 180.0) * 0.5 );
var index = new Array(); /* aray of indexes of source points to include in the reduced line */
var sig_start = new Array(); /* indices of start & end of working section */
var sig_end = new Array();
/* check for simple cases */
if ( source.length < 3 )
return(source); /* one or two points */
/* more complex case. initialize stack */
n_source = source.length;
band_sqr = kink * 360.0 / (2.0 * Math.PI * 6378137.0); /* Now in degrees */
band_sqr *= band_sqr;
n_dest = 0;
sig_start[0] = 0;
sig_end[0] = n_source-1;
n_stack = 1;
/* while the stack is not empty ... */
while ( n_stack > 0 ){
/* ... pop the top-most entries off the stacks */
start = sig_start[n_stack-1];
end = sig_end[n_stack-1];
n_stack--;
if ( (end - start) > 1 ){ /* any intermediate points ? */
/* ... yes, so find most deviant intermediate point to
either side of line joining start & end points */
x12 = (source[end].lng() - source[start].lng());
y12 = (source[end].lat() - source[start].lat());
if (Math.abs(x12) > 180.0)
x12 = 360.0 - Math.abs(x12);
x12 *= Math.cos(F * (source[end].lat() + source[start].lat()));/* use avg lat to reduce lng */
d12 = (x12*x12) + (y12*y12);
for ( i = start + 1, sig = start, max_dev_sqr = -1.0; i < end; i++ ){
x13 = (source[i].lng() - source[start].lng());
y13 = (source[i].lat() - source[start].lat());
if (Math.abs(x13) > 180.0)
x13 = 360.0 - Math.abs(x13);
x13 *= Math.cos (F * (source[i].lat() + source[start].lat()));
d13 = (x13*x13) + (y13*y13);
x23 = (source[i].lng() - source[end].lng());
y23 = (source[i].lat() - source[end].lat());
if (Math.abs(x23) > 180.0)
x23 = 360.0 - Math.abs(x23);
x23 *= Math.cos(F * (source[i].lat() + source[end].lat()));
d23 = (x23*x23) + (y23*y23);
if ( d13 >= ( d12 + d23 ) )
dev_sqr = d23;
else if ( d23 >= ( d12 + d13 ) )
dev_sqr = d13;
else
dev_sqr = (x13 * y12 - y13 * x12) * (x13 * y12 - y13 * x12) / d12;// solve triangle
if ( dev_sqr > max_dev_sqr ){
sig = i;
max_dev_sqr = dev_sqr;
}
}
if ( max_dev_sqr < band_sqr ){ /* is there a sig. intermediate point ? */
/* ... no, so transfer current start point */
index[n_dest] = start;
n_dest++;
}
else{
/* ... yes, so push two sub-sections on stack for further processing */
n_stack++;
sig_start[n_stack-1] = sig;
sig_end[n_stack-1] = end;
n_stack++;
sig_start[n_stack-1] = start;
sig_end[n_stack-1] = sig;
}
}
else{
/* ... no intermediate points, so transfer current start point */
index[n_dest] = start;
n_dest++;
}
}
/* transfer last point */
index[n_dest] = n_source-1;
n_dest++;
/* make return array */
var r = new Array();
for(var i=0; i < n_dest; i++)
r.push(source[index[i]]);
return r;
}