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Smoothing GPS tracked route-coordinates

I have some data of coordinates that I recorded. Unfortunatelly they seem to be not realy good. They jump sometimes over the map. So now I´m searching for some flattening or filtering algorithm that would make the route look more realistic.

Currently my only filter is to calculate the max possible meters travelled in a second (in bus or car or walking) and compare them with the coordinates, throwing those away, that are just not possible within a timeframe. So if a person can walk up to 2.5 meters in a second, and I have two coords that are 10 meters away from each other and they were recorded within two seconds, I try to find them and throw them away. This helps a little bit.

This is the code:

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;
};

To make things easier for you, I created this fiddle that already implements my first filter and also gives you the ability to add a new filter.

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

Some futher ideas I have:

  • throw all coords away that are in a specific radius. This eventually would remove some disturbing coords, if you just stand around for a few minutes
  • group all coords by n seconds frames and try to determine the most relevant in this block. Unfortunatelly I dont have any idea how :(

So I think this is a realy interessting issue, I hope you understand everything I was talking about. I thank You guys for any help!

Edit: I found something about Linear least squares and Kalman Filter. I´m into it, though because I´m absolutely not the math expert, I would appreciate any help in this.

EDIT 2 Progress :) I implemented the DouglasPeucker algorithm which @geocodezip promoted to me. The algorithm alone does not fix it all, but the combination of my current "max_possible_travel" it looks almost perfect. If I play a little bit with the second param it will get interessting. Please look at the new fiddle and make sure you check both the filters "walkfilter" and "gdouglaspeucker". http://jsfiddle.net/z4hB7/8/

like image 243
Luke Avatar asked Dec 27 '22 05:12

Luke


1 Answers

You can try the Douglas Peuker algorithm

The Ramer–Douglas–Peucker algorithm is an algorithm for reducing the number of points in a curve that is approximated by a series of points.

There is at least one implementation for the Google Maps API v3

perl implementation

Javascript implementation code from Bill Chadwick's site:

/* 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;

}
like image 101
geocodezip Avatar answered Jan 12 '23 16:01

geocodezip