平滑GPS跟踪的路线坐标

时间:2013-04-20 14:09:05

标签: javascript google-maps polyline geography

我记录了一些坐标数据。不幸的是,他们似乎并不是真的很好。他们有时会跳过地图。所以现在我正在寻找一些平整或过滤算法,使路线看起来更逼真。

目前我唯一的过滤器是计算一秒钟内(在公共汽车或汽车或步行中)行走的最大可能米数,并将它们与坐标进行比较,将这些坐标丢弃,这在时间范围内是不可能的。因此,如果一个人可以在一秒钟内行走2.5米,并且我有两个距离彼此相距10米的坐标,并且它们在两秒钟内被记录下来,我试图找到它们并将它们扔掉。这有点帮助。

这是代码:

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;
};

为了让您更轻松,我创建了这个小提琴,它已经实现了我的第一个过滤器,并且还可以添加新的过滤器。

http://jsfiddle.net/z4hB7/7/

我有一些进一步的想法:

  • 抛出特定半径范围内的所有坐标。如果你只是站了几分钟,这最终会消除一些令人不安的坐标
  • 将所有坐标分组n秒帧并尝试确定此块中最相关的坐标。不幸的是,我不知道如何:(

所以我认为这是一个真正令人讨厌的问题,我希望你能理解我所谈论的一切。我感谢你们的帮助!

编辑:我发现了一些关于线性最小二乘和卡尔曼滤波器的信息。虽然因为我绝对不是数学专家,但我很感激。在这方面我会感激不尽。

编辑2 进展:)我实现了@geocodezip提升给我的DouglasPeucker算法。单独的算法并没有解决所有问题,但是我现在的“max_possible_travel”的组合看起来几乎是完美的。如果我与第二个参数玩一点点就会变得非常有用。请查看新的小提琴,并确保检查过滤器“walkfilter”和“gdouglaspeucker”。 http://jsfiddle.net/z4hB7/8/

1 个答案:

答案 0 :(得分:10)

您可以尝试Douglas Peuker algorithm

  

Ramer-Douglas-Peucker算法是一种用于减少曲线中由一系列点近似的点数的算法。

至少有one implementation for the Google Maps API v3

perl implementation

来自Bill Chadwick's site的Javascript实施代码:

/* 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;

}