change the color of the pixels that are present in an image

I would like to change the color of pixels in an image.

Specifically, I want to turn white pixels to black and make the number in the lower right-hand corner disappear as in the figure below. Of course, the size of the figure must remain the same.

I am currently using this code:

Im = imread('./Images/Example.jpg');
figure(1);
imshow(Im);

Answer 1

[DGM]

Updated to accomodate new images and fix the bad masking expression. In this example, I'm going to use this composite image:

% this image is a composite of all the images given so far
inpict = imread('loopall.png');

% select objects to keep
% [Hmin Hmax; Smin Smax; Vmin Vmax]
%L = [0.907 0; 0.101 0.429; 0 1]; % for pink only
L = [0.617 0.024; 0.081 0.657; 0 1]; % for pink, purple, blue
hsvpict = rgb2hsv(inpict);
mask = ((hsvpict(:,:,1) >= L(1,1)) | (hsvpict(:,:,1) <= L(1,2))) & ...
    ((hsvpict(:,:,2) >= L(2,1)) & (hsvpict(:,:,2) <= L(2,2))) & ...
    ((hsvpict(:,:,3) >= L(3,1)) & (hsvpict(:,:,3) <= L(3,2)));

% clean up the mask, dilate
mask = bwareaopen(mask,100);
mask = imdilate(mask,strel('disk',3));

% clear areas not covered by the mask
outpict = inpict;
outpict(repmat(~mask,[1 1 3])) = 0;

% display modified image
imshow(outpict)

[Alberto Acri]

I thank you @DGM, that's exactly what I wanted to achieve as a result!

Is it possible to make the code more generic? I have several figures with different colors, like these:

Note:

• The colors of the numbers at the bottom of the image always remain the same (orange, red, light green).
• The colors of the curves, as you can see in the images, can be different (e.g. pink, purple, blue).

[Alberto Acri]

In case, to have a simpler code, it would be useful to change only the lines:

rangeR = [0.907 0; 0.101 0.429; 0 1];

and

mask = (hsvpict(:,:,1) >= rangeR(1,1)) | (hsvpict(:,:,1) >= rangeR(1,2)) & ...
    (hsvpict(:,:,2) >= rangeR(2,1)) & (hsvpict(:,:,2) >= rangeR(2,2)) & ...
    (hsvpict(:,:,3) >= rangeR(3,1)) & (hsvpict(:,:,3) >= rangeR(3,2));

In particular, it would be more useful to make this line generic to isolate objects with any color (example: purple, blue, pink):

rangeR = [0.907 0; 0.101 0.429; 0 1];

p.s. How were the numbers [0.907 0; 0.101 0.429; 0 1] obtained?

And change the values in the "mask" for the three colors (orange, red and blue numbers) . The number colored in "orange" disappeared using these lines:

mask = (hsvpict(:,:,1) >= rangeR(1,1)) | (hsvpict(:,:,1) >= rangeR(1,2)) & ...
    (hsvpict(:,:,2) >= rangeR(2,1)) & (hsvpict(:,:,2) >= rangeR(2,2)) & ...
    (hsvpict(:,:,3) >= rangeR(3,1)) & (hsvpict(:,:,3) >= rangeR(3,2));    

[DGM]

I updated the answer. You were right that my mask combination expression was messed up. Good catch.

As to where the limits came from, I just used the ColorThresholder app to find limits that worked, although similar can be done by displaying the HSV image in imshow() and using the data cursor to inspect the pixel values.

If we're talking about picking the values and generating the selection mask, it's worth noting the apparent inconsistency in the above code. Note that Hmin is larger than Hmax, and that the first major term of the selection expression is a logical OR instead of AND.

L = [0.617 0.024; 0.081 0.657; 0 1]; % for pink, purple, blue
hsvpict = rgb2hsv(inpict);
mask = ((hsvpict(:,:,1) >= L(1,1)) | (hsvpict(:,:,1) <= L(1,2))) & ...
       ((hsvpict(:,:,2) >= L(2,1)) & (hsvpict(:,:,2) <= L(2,2))) & ...
       ((hsvpict(:,:,3) >= L(3,1)) & (hsvpict(:,:,3) <= L(3,2)));

This is because H is cyclic and our selection crosses zero. DeMorgan's theorem could be applied just the same, but I find that it's easier to overlook the inversion that's happening.

L = [0.024 0.617; 0.081 0.657; 0 1]; % for pink, purple, blue
hsvpict = rgb2hsv(inpict);
mask = ~((hsvpict(:,:,1) >= L(1,1)) & (hsvpict(:,:,1) <= L(1,2))) & ...
        ((hsvpict(:,:,2) >= L(2,1)) & (hsvpict(:,:,2) <= L(2,2))) & ...
        ((hsvpict(:,:,3) >= L(3,1)) & (hsvpict(:,:,3) <= L(3,2)));

Either way it's written, mask generation on H will have to be conditionally dependent on whether the selection crosses H = 0.

In the modified example, I only used one selection to include the three pink, purple, blue colors, assuming they may vary within that range. This type of mask generation by selecting limits is essentially selecting a range of colors that lie within a rectangular box in the RGB space. Making a box fit multiple colors may mean that the box has to be fairly large, depending on where those colors are.

Long story short, the more disparate things that need to be included in a single box selection, the harder it tends to be to exclude unwanted colors from the box. Depending on the range of other possible colors that might show up, it may be more effective to select the known colors individually and then combine those masks, since selecting one color only requires a small tightly-fitted box. Likewise, if colors may vary, it may be worth determining whether it's easiest to select the colors that need to be kept or whether it's easier to select the colors that need to be excluded instead.

[Alberto Acri]

Thank you for the explanation. I had not used the ColorThresholder app yet.

I wanted more help to improve the pink pixel capture because for example with the image "ABC.jpg" I happened to lose pixels:

I tried to make some changes to the row values:

L = [0.617 0.024; 0.081 0.657; 0 1]; % for pink, purple, blue

but I was unsuccessful in the task.

[DGM]

That's not an issue with the color-based segmentation part. That's being caused during mask cleanup.

This line

mask = bwareaopen(mask,100);

removes all mask regions with area < 100px. This is there to help eliminate any unattached speckles that might show up near object edges (e.g. due to JPG artifacts, etc).

If you want to preserve small spots, you can either reduce the area parameter or just omit that line. Considering that the spot size is only 7px, there might not be much point trying to filter anything smaller.

Answer 2

[KALYAN ACHARJYA]

It can be done in multiple ways, use morphological operations or see the regionprops function (very useful). Easyest way for specific case-

BW2 = bwareaopen(BW1,200);
figure, imshow(BW2);

Ensure that BW2 is a binary image

Hope it Helps!