can u make black color region in to some other color say red or green in the given image but dont change the white color?
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here i have attached the image in which i want green or red color instead of black color region for my academic work. please give me accurate answer.
https://mail.google.com/mail/?ui=2&ik=bcbccaefc4&view=att&th=131d66f90017f805&attid=0.1&disp=inline&realattid=f_grfxa06t0&zw
thanks in advance
9 Kommentare
Jan
am 17 Aug. 2011
Please post more details by editing your original question: Does "binary image" mean a LOGICAL array? If you "detect some portion", in which format is this portion stored?
The more details, the more likely is a meaningful answer. Otherwise we waste time with guessing.
@Image Analyst: Thanks. Converting a message twice by non-native speakers with different mother-tongues leads to strange results, usually.
Walter Roberson
am 17 Aug. 2011
http://www.mathworks.com/matlabcentral/answers/7924-where-can-i-upload-images-and-files-for-use-on-matlab-answers
Akzeptierte Antwort
Image Analyst
am 17 Aug. 2011
OK, you say in a comment that you already have the binary image and you want the black parts of it to be red (or green). Something like this should do it:
% Extract the individual red, green, and blue color channels.
redChannel = rgbImage(:, :, 1);
greenChannel = rgbImage(:, :, 2);
blueChannel = rgbImage(:, :, 3);
% Make the black parts pure red.
redChannel(~binaryImage) = 255;
greenChannel (~binaryImage) = 0;
blueChannel (~binaryImage) = 0;
% Make the above lines 0, 255, and 0 if you want green instead.
% Now recombine to form the output image.
rgbImageOut = cat(3, redChannel, greenChannel, blueChannel);
Weitere Antworten (4)
Jan
am 17 Aug. 2011
If you have an UINT8 RGB image and want to replace [0,0,0] pixels:
rgb = uint8(floor(rand(640, 480, 3) * 10));
isBlack = all(rgb == uint8(0), 3);
r = rgb(:, :, 1);
r(isBlack) = 1;
g = rgb(:, :, 2);
g(isBlack) = 0;
b = rgb(:, :, 3);
b(isBlack) = 0;
rgb2 = cat(3, r, g, b);
0 Kommentare
Daniel Shub
am 17 Aug. 2011
While there are faster and better ways ...
x = rand(100, 100, 3);
x(1:10, 1:10, :) = 0;
y = logical([100, 100]);
z = x;
for i = 1:100
for j = 1:100
y(i, j) = all(x(i, j, :) == [0, 0, 0]);
end
end
figure
image(x)
figure
image(z)
1 Kommentar
Jan
am 17 Aug. 2011
"all(x(i, j) == [0,0,0])" is not correct. This checks only the red channel of the 3D image array. I assume you want: "all(x(i, j, :) == [0,0,0])". But then a comparison with a scalar 0 is enough and faster. Vectorization is efficient here: replace the loops by: "all(x==0, 3)".
Image Analyst
am 17 Aug. 2011
See my color segmentation demos: http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862. With the proper method of segmentation you should be able to achieve the color replacement you want, regardless if the color is a single [r,g,b] triplet or a more spread out gamut.
0 Kommentare
Andrei Bobrov
am 17 Aug. 2011
rgb = uint8(rand(5,5,3)<.25)
rgb(rgb~=0) = uint8(randi([1 255],nnz(rgb),1))
bk = im2uint8(all(rgb==0,3))
rgb2 = rgb;
choose one from variants
% red
rgb2(:,:,1) = bk + rgb2(:,:,1)
% green
rgb2(:,:,2) = bk + rgb2(:,:,2)
% blue
rgb2(:,:,3) = bk + rgb2(:,:,3)
OR
rgb = +(rand(5,5,3)<.25)
rgb(rgb~=0) = rand(nnz(rgb),1)
bk = all(rgb==0,3)
rgb2 = rgb
rgb2(:,:,1) = rgb2(:,:,1)+bk
image(rgb2) % red
ADD MORE (about binary image)
d = +(rand(10)<.3) %binary image
d1 = d(:,:,[ 1 1 1])
d1(:,:,1)=ones(size(d)) % red
image(d1)
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