How best to change the colors of pixels in an image with known X (column) Y (row) coordinates?

The obvious way is to loop over every XY pair and change the color of the identified pixel, but this is very slow. I can do a lot better using indexing, but it is a clumsy approach, as it does the same thing three times. Is there a better, more intuitive way (even if it's not much faster)?

Image = uint8(repmat(randi(100, [9, 9]), [1, 1, 3]));
Image2 = Image;
Image3 = Image;

Color = uint8([1, 0.25, 0.1] .* 255);

Xs = [2; 2; 3; 3; 3; 4; 5; 6; 7; 7; 7; 8; 8];
Ys = [3; 6; 2; 3; 7; 8; 8; 8; 2; 3; 7; 3; 6];
% Adding lots of duplicates to make this simple problem harder.
Xs = repmat(Xs, [1000000, 1]);
Ys = repmat(Ys, [1000000, 1]);

% Looping is slow.
tic
for ii = 1:numel(Xs)
    Image(Ys(ii), Xs(ii), :) = Color;
end
t1 = toc;

% The best I could come up with.  It's faster, but inelegant.
tic
idx = sub2ind(size(Image2, [1, 2]), Ys, Xs);
Image2(idx)                                 = Color(1);
Image2(idx + prod(size(Image3, [1, 2])))    = Color(2);
Image2(idx + 2.*prod(size(Image3, [1, 2]))) = Color(3);
t2 = toc;

% Is there a better way?
tic
% Do something...
t3 = toc;

montage({Image, Image2, Image3}, 'Size', [1, 3])

disp(['Loop time  = ', num2str(t1, 3), ' seconds.'])
disp(['Index time = ', num2str(t2, 2), ' seconds.'])
disp(['New time   = ', num2str(t3, 2), ' seconds.'])
disp(['Indexing is ~', num2str(round(t1./t2), '%d'), ' times faster than looping.'])

Answer 1

[DGM]

Normally, you'd do region replacement using a mask of some sort.

Image = uint8(repmat(randi(100, [9, 9]), [1, 1, 3]));
Image2 = Image;
Image3 = Image;
Color = uint8([1, 0.25, 0.1] .* 255);

Xs = [2; 2; 3; 3; 3; 4; 5; 6; 7; 7; 7; 8; 8];
Ys = [3; 6; 2; 3; 7; 8; 8; 8; 2; 3; 7; 3; 6];

% Adding lots of duplicates to make this simple problem harder.
Xs = repmat(Xs, [1000000, 1]);
Ys = repmat(Ys, [1000000, 1]);

% Looping is slow.
tic
for ii = 1:numel(Xs)
    Image(Ys(ii), Xs(ii), :) = Color;
end
t1 = toc;

% The best I could come up with.  It's faster, but inelegant.
tic
idx = sub2ind(size(Image2, [1, 2]), Ys, Xs);
Image2(idx)                                 = Color(1);
Image2(idx + prod(size(Image3, [1, 2])))    = Color(2);
Image2(idx + 2.*prod(size(Image3, [1, 2]))) = Color(3);
t2 = toc;

% use a mask
tic
sz = size(Image2, [1, 2]);
idx = sub2ind(sz, Ys, Xs);
mask = false(sz);
mask(idx) = true;

% IPT imoverlay() can work with I/RGB images
% BG must be an image, FG must be an RGB tuple
% mask can only be binarized
% and FG color must be unit-scale float
Image3 = imoverlay(Image3,mask,im2double(Color));

% MIMT replacepixels() can work with I/IA/RGB/RGBA/RGBAAA images
% FG/BG can be images or tuples
% mask can be binarized or smooth, I/RGB
% color can be anything so long as it's correctly-scaled for its class
%Image3 = replacepixels(Color,Image3,mask);
t3 = toc;

montage({Image, Image2, Image3}, 'Size', [1, 3])

disp(['Loop time  = ', num2str(t1, 3), ' seconds.'])

Loop time  = 4.73 seconds.

disp(['Index time = ', num2str(t2, 2), ' seconds.'])

Index time = 0.33 seconds.

disp(['New time   = ', num2str(t3, 2), ' seconds.'])

New time   = 0.29 seconds.

disp(['Indexing is ~', num2str(round(t1./t2), '%d'), ' times faster than looping.'])

Indexing is ~14 times faster than looping.

disp(['Masking is ~', num2str(round(t1./t3), '%d'), ' times faster than looping.'])

Masking is ~16 times faster than looping.

If all you need are binary selections and you're replacing the regions with a single color, then IPT imoverlay() works. If the selected regions are to be replaced by another image, then imoverlay() won't work.

Filling with another image, using only basic tools:

https://www.mathworks.com/matlabcentral/answers/890417-i-want-to-remove-black-color-from-my-image-and-make-the-background-transparent-so-that-i-can-paste-i

Filling with a solid fill, using basic tools or using MIMT replacepixels():

https://www.mathworks.com/matlabcentral/answers/170026-overlay-two-images-of-different-type#answer_1170025