drawing on an image in loop
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How can I draw on an from loop iterations, without 'imread' and 'imwrite'?
EDIT: Inside the loop, I am doing some processing to determine the pixel coordinates of certain features. Once I have these coordinates, I am using them to draw a square on the image at these locations. I would like the loop to draw the first square on the first iteration, and also draw the second square on the second iteration and so on... until I have 100 squares. Is there a way to update the image like this without having to load and save the image each time?
At the moment, I am using:
for i=1:100
img = imread('img.png');
% !! DO SOME PROCESSING HERE !!
imwrite(output,'img.png');
end
However, the use of 'imread' and 'imwrite' takes a long time because the images are large.
Can the loop be altered to create an updated image, and then use 'imwrite' after the loop to save the result?
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Réponses (2)
Sean de Wolski
le 2 Juil 2012
Modifié(e) : Sean de Wolski
le 2 Juil 2012
You are literally overwriting the same file each time? If so, then you absolutely don't need to save it on each iteration.
img = imread('img.png');
for ii = 1:100;
img = do_stuff_with_image(img); %stuff will be done 100x
end
imwrite(img,'img.png');
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Ryan
le 2 Juil 2012
Modifié(e) : Ryan
le 2 Juil 2012
If you are using imwrite() to save the image file with a square marked on it then I'm assuming you have a matrix with the square on it already, so why not just redifine the 'starting image' at the end of the loop. For example:
I = imread('image.png');
for m = 1:100
% Draw your rectangles and do your processing on I
I = ProcessedImage;
end
imwrite(I,'image.png')
Image Analyst
le 2 Juil 2012
Modifié(e) : Image Analyst
le 2 Juil 2012
Philip: Why do you need to read and write it every iteration? Just read it once, enter the loop and "burn" the box into the image by writing into the image array directly (i.e. don't use plot() or rectangle() because those just place boxes in the overlay), then call cla and imshow(). Finally when the loop exits, call imwrite. This will be much faster and accomplish the same thing.
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