How to use estimateFlow between two images that are not video frames
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chris crowley
on 7 Aug 2020
Commented: Walter Roberson
on 8 Nov 2022
I am interested in computing optical flow between images that are not part of a video sequence using the built-in MATLAB function estimateFlow. That is, I would like to compute optical flow between two images in my workspace, Img1 and Img2. How can I do this?
Currently, I am combining the images into an array and running a for loop to obtain the optical flow:
ImgStack(:,:,1) = Img1; % combine the images to a sequence
ImgStack(:,:,2) = Img2;
reset(opticFlow); % clear previous optical flow
opticFlow = opticalFlowLKDoG; % At this point, I don't realy care what method is used
for t = 1:2
flow = estimateFlow(opticFlow,ImgStack(:,:,t));
end
This procedure is inelegant and quite cumbersome. I am doing this on a large number of images and I would like a more efficient way to do this. It would be nice if I could just do something like flow = estimateFlow(opticFlow,[Img1,Img2]).
Dose anyone know how to do this?
2 Comments
Image Analyst
on 7 Aug 2020
What do you have to start with? A bunch of arrays called Img1, Img2, Img3, etc.? Or one array ImgStack?
Accepted Answer
Divya Gaddipati
on 11 Aug 2020
You can use opticalFlow function for this purpose.
img1 = im2double(Img1); % the images should be in double
img2 = im2double(Img2);
opflow = opticalFlow(img1, img2);
h = figure;
movegui(h);
hViewPanel = uipanel(h,'Position',[0 0 1 1],'Title','Plot of Optical Flow Vectors');
hPlot = axes(hViewPanel);
imshow(img1)
hold on
plot(opflow,'DecimationFactor',[5 5],'ScaleFactor',60,'Parent',hPlot);
hold off
For more information, refer to the documentation on opticalFlow
3 Comments
Walter Roberson
on 8 Nov 2022
It depends exactly what is in the images. The images just might happen to be sensor readings from a flow meter that is directly estimating velocity at each point, perhaps using sonar or polarization techniques. The image files would have to be TIFF in order to store negative values properly...
So the call here is not definitely wrong... just likely wrong.
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