rgb to gray image indeces matching

I have an RGB color image with dimensions: 460x640x3 and a gray image with the same dimensions: 460x640. They are of the same type, that is, uint8. I want to get the intensity information of the gray image using the coordinate values from color image if that coordinate value is not black (intensity value ~= 0). if not the intensity value of gray image should be 0 (black).
If I loop through each values for example like below:
img_length = 460x640;
for i = 1:img_length
if rgbImage(i) ~= 0;
depthImage(i) = i;
else if rgbImage(i) == 0;
depthImage(i) = 0;
i = i+1;
end;
As for color image how can I use only large dimensions as if it is a gray image?

2 Comments

Don't call a variable length, it is a builtin function that you just over-rode.
ok, it is just for reference, it could be any name. I changed it. Thank you, by the way!

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Answers (1)

Image Analyst
Image Analyst on 27 Apr 2017
Edited: Image Analyst on 27 Apr 2017
Try this:
sumImage = rgbImage(:,:,1) + rgbImage(:,:,2) + rgbImage(:,:,3);
mask = sumImage == 0;
grayValues = grayImage(mask); % 1-d list of gray levels where the color image is black.

6 Comments

Was this not what you wanted?
thanks for suggestion, but it didn't give me what I wanted. What does this line do?
mask = sumImage == 0;
Please, clear up some things!
It gives you a "map" or image of all the places where the values are exactly 0 for red, green, and blue, in other words, a map of where true, pure black pixels live.
Thank you. but it didn't help. Is there another way of doing this? When I refer to i in my above code, it should only count first two dimensions like it does on gray images.
I don't understand how it does not do what you want. First it finds all the locations (pixels) where the image is totally black - each of red, green, and blue is exactly zero. Then it gives you the gray levels from the gray scale image at those locations. Your code does nothing like that (nothing like what you asked for) whereas mine does.
Sorry sir, I combined your code with other code and got some side effects. Therefore I thought it didn't work. Thank you!

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on 27 Apr 2017

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