Skip to main content

image processing - How to track objects in video


I have a video that contains object moving in one direction, is it possible to track the objects and get information like object moving speed, object length and coordinates etc.? An example would be I have a video of highway traffic, and I want to get the car size, lane number and speed for each car.




Simplified Problem


Here is a simple version of the problem:


Say I have several rectangles moving in one direction, and I want to know their length and position.


The following code creates 100 frames of moving rectangles:


N0 = 100;
lengthRange = {2, 3, 4, 5};
positionRange = {1, 2, 3, 4, 5};

objectsDimension = RandomChoice[lengthRange, {N0}];(*Different possible length of rectangle*)


objectsPosition = RandomChoice[positionRange, {N0}];(* Track number for the rectangle*)

objectsEndPoins = Transpose /@ Transpose[{Transpose[{objectsPosition - 0.2, objectsPosition + 0.2}], Partition[Prepend[Accumulate[objectsDimension], 0], 2, 1] /. {x_, y_} -> {x + 0.2, y - 0.2}}];
(* coordinates for rectangles*)

timePlot = Graphics[Rectangle @@@ objectsEndPoins, ImageSize -> {200, Automatic}, AspectRatio -> 8];(* the whole time plot*)

frames = Table[Show[timePlot, PlotRange -> {{1 - 0.5, 5 + 0.5}, {n, n + 50}}, AspectRatio -> 1, ImageSize -> {200, Automatic}], {n, 0, 300, 3}];(*show different part of the whole plot, to create video frames*)


frames = Graphics[#[[1]], Sequence @@ Options[#]] & /@ frames;(*workaround of [this][1] problem*)

ListAnimate[frames]

enter image description here


So can we get the coordinates and the length of the rectangles in the video?



What I tried


I first use the FindGeometricTransform find the speed for the moving objects, then stack each frame together to get a whole time plot, then measure the rectangles.


stepPerFrame = Mean@Table[FindGeometricTransform[frames[[i]], frames[[i + 1]], TransformationClass -> "Translation"][[2, 1, 2, 3]], {i, 1, 20}]

  (*calculate the object moving speed per frame. Assume constant speed, average speed for the first 20 frames.*)

blank = Graphics[{}, ImageSize -> ImageDimensions[timePlot]];(*constract a blank plot*)

result = ImagePad[Block[{tmp = blank},
    Do[tmp = 
       ImageCompose[tmp, 
        frames[[i]], {0, (i - 1)*stepPerFrame}, {0, 0}];, {i, 1, 
      Length@frames}];
    tmp], {{0, 0}, {10, 0}}, 

   White];
(*stack each frame together with a offset of object moving speed.*)

imageData = ImageData[Binarize[ColorNegate[result]]];(*binarize the image*)

lines = Reverse /@ 
   Transpose[
    imageData[[All, 
     Range[20, 180, 40]]]];(*outlines at each lane center*)


position = 
  Table[lines[[
      n]] //. {x__, PatternSequence[0, 1], y__} -> {x, 0, "start", 
       y} //. {x__, PatternSequence[1, 0], y__} -> {x, "stop", 0, 
      y}, {n, 1, 5}];(*mark object start and stop in the data *)

objects = 
  Table[(Transpose@{Flatten[Position[position[[n]], "start"]], 
      Flatten[Position[position[[n]], "stop"]]}), {n, 1, 
    5}];(*select the object start and stop position*)


recoveredOjbDim = 
  Flatten[Differences /@ Sort@Flatten[objects, 1]] /. {5 -> 2, 9 -> 3,
     13 -> 4, 17 -> 5};(*calculate the object length*)

recoveredOjbDim == objectsDimension(*compare to the correct answer*)
(*True*)


Questions:




  1. Could you show your methods, I'm sure I can learn a lot.

  2. In the problem, all objects have the same constant speed, what can we do for objects with different speed?




Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more