Is there a way to easily plot pixels?

edit

Thanks for the help. Since I want multiple circles, I have to apply the circle function multiple times? Also, doesn't this exceed the size of one pixel? What I mean by this is let's say the circle radius is one, this will technically circle around more than one pixel because the pixel value is just the center, correct?

Other question, let's say I use image.at<uchar>(Point(x, y)) = 0; does uchar work for CV_8U datatype? Sorry, I know that question is really basic. And finally, let's say I want to make the color turquoise instead of black. Do I say: image.at<uchar>(Point(x, y)) = Scalar(R,G,B); where R,G,B are the proper values for whatever color I want?

just as a sidenote, if that operation is after applying a colorMap, it should be: image.at<Vec3b>(y,x) = Scalar(b,g,r);

also, please, Point, not Points

Thanks berak. This might be a dumb question, but why is it (y, x) instead of (x, y)? Is that just the convention when you do <Vec3b>? Or is it because you are using image.at?

EDIT: Nevermind, I looked at the documentation.

Point goes x,y - while cv::Mat is at<type>(row,col); so either:

img.at<type>(y,x); or img.at<type>(Point(x,y));

Yes @TrustMeImAnEngineer, you would need to draw the circle function 4 times. 1 for each point. And @berak was right, since the color map has been applied, you would need to use image.at<Vec3b>(y,x) = Scalar(b,g,r); where b g r are the proper values for the colour turquoise.

In a sense it is a convention, but a historical one. Data storage on a computer is always 1D, so if you look at C-code and you have anarray[y][x], where 'x' is the inner array and 'y' is the outer one. So the convention is to scan in rows like an old TV will do.

There now a huge speed difference between

for(int y=0; y < Ymax; y++) {
    for(int x=0; x < Xmax; x++) {
        int something = array[y][x];
    }
}

and

for(int y=0; y < Ymax; y++) {
    for(int x=0; x < Xmax; x++) {
        int something = array[x][y];
    }
}

where the first example reads the data continous and the second one does not