Skip to main content

Extracting road curves from an image


I have taken several gps readings while driving around town.


gpsPositions=GeoPosition[{{33.657, -84.5197}, {33.6687, -84.4977}, 
{33.692, -84.4907}, {33.7057, -84.4287}, {33.7431, -84.4027}, {33.7285, 
-84.3493}}]

map = GeoGraphics[GeoMarker[gpsPositions], 
  GeoRange ->Entity["City",{"Atlanta","Georgia","UnitedStates"}],ImageSize->Large]

Mathematica graphics



Were are able to extract the main roads by the following functions:


    bgnd = GeoGraphics[ GeoRange ->Entity["City",{"Atlanta","Georgia","UnitedStates"}],ImageSize->Large];
    roads = DeleteSmallComponents@
      Thinning@ImageAdd[
        Dilation[#, DiskMatrix[3]] & /@ 
         DominantColors[bgnd, 7, {"CoverageImage", "Color"}][[5 ;; 6, 1]]];
    ImageCompose[map, ColorNegate@SetAlphaChannel[roads, roads]]

Mathematica graphics


Road Extraction worked fine. Now convert the gpsPositions and map them to the road image.



range = Abs[#1 - #2] & @@@ map[[8, 2]]    
origin = #1 & @@@ map[[8, 2]];
max = Reverse@ImageDimensions[roads];
waypoints = gpsPositions[[1]];
points = Reverse@(Round[(max (#)/range), 1]) & /@ ((# - origin) & /@ 
     waypoints);
Show[ColorNegate@roads, 
 Graphics[{Red, PointSize[Large], Point[points]}]]

Mathematica graphics



I would like to be able to segment the lines between the dots. Then, set up an interpolation function between each dot so we can simulate the placement of the vehicle at any point of the road as needed.


Any thoughts on how this can be achieved?



Answer



I'm not 100% sure what you want: "segmentation" has a well-defined meaning in image processing, and I think that's not what you want. Also, I couldn't reproduce your results (I think DominantColors isn't guaranteed to give the same order or even the same results every time it's run). So this may or may not help...


First, this seems more reproducible than DominantColors:


roadColors = List @@@ {Yellow, Blue};

colorDist = 
  Total[(# - 
        Transpose[

         ImageData[bgnd][[All, All, ;; 3]], {2, 3, 1}])^2] & /@ 
   roadColors;

roads = Thinning[
  MorphologicalBinarize[
   Closing[Image[1 - MapThread[Min, colorDist, 2]], 
    DiskMatrix[3]], {0.43, 0.8}]]

Next, I convert the white pixel locations to a graph:


whitePixels = PixelValuePositions[roads, 1];


nearestFn = Nearest[whitePixels -> Automatic];

edges = MapThread[
   Function[{point, potentialNeighbors}, 
    point <-> # & /@ 
     Select[potentialNeighbors, 
      0 < EuclideanDistance[whitePixels[[#]], whitePixels[[point]]] < 
        2 &]],
   {Range[Length[whitePixels]], nearestFn[#, 10] & /@ whitePixels}];


g = Graph[DeleteDuplicates@Flatten[edges]];

Now, if I choose point locations in the image:


pts = {{133, 97}, {464, 461}};
LocatorPane[Dynamic[pts], ColorNegate@roads]

I can use nearestFn to find the nearest road pixel, and FindShortestPath to find the shortest path between two points:


path = FindShortestPath[g, nearestFn[pts[[1]]][[1]], 
   nearestFn[pts[[2]]][[1]]];


Show[Image@bgnd, 
 Graphics[{Red, Line[whitePixels[[path]]], PointSize[Large], 
   Point[whitePixels[[path[[;; ;; 10]]]]]}]]

enter image description here


We can even make a little Indiana Jones-style animation:


Animate[Show[Image@bgnd, 
  Graphics[{Red, Thick, Line[whitePixels[[path[[;; i ;; 2]]]]]}]], {i,
   1, Length[path], 1}]


(unfortunately, I don't think I can upload videos to Mathematica.SE)


Comments

Post a Comment

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...
Post a Comment
Read more

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...
Post a Comment
Read more