how can I draw a rectangle on a image?

Hello friends I'm with a problem! I have a vector with x and y coordinates and I need to draw a rectangle over an image.A rectangle with the dimensions 3x3 with the red line from the centroid point how can I do this?

Answer 1

[Walter Roberson]

For any one (x, y) centroid pair:

linelen = 3;

pixbefore = floor((linelen - 1)/2);
pixafter = linelen - 1 - pixbefore;

paintcolor = [255, 0, 0];

for K = 1 : length(paintcolor)
  YourImage(y - pixbefore : y + pixafter, [x - pixbefore, x + pixafter], K) = red(K);  %left and right edge
  YourImage([y - pixbefore + 1, y + pixafter - 1], x - pixbefore : x + pixafter, K) = red(K);  %top and bottom edge
 end

If I was doing a whole bunch of them, I would use a slightly different approach

[Walter Roberson]

Is the question about writing the rectangle into the image array, or is the question about drawing a rectangle on the graphics display on top of an existing image?

If the question is about drawing on the graphics display, see the command rectangle().

If you have the Computer Vision Toolbox, see also http://www.mathworks.com/help/vision/ref/insertshape.html and http://www.mathworks.com/help/vision/ref/vision.shapeinserter-class.html

[Biza Ferreira]

is the question about drawing a rectangle on the graphics display on top of an existing image, I have done all you said and ... It gave me back an error

if true
  % code
Undefined function 'red' for input arguments of type 'double'.
Error in impressoes (line 156)
BSKE(BIFU_XY(:,2) - pixbefore : BIFU_XY(:,2) + pixafter, [BIFU_XY(:,1) - pixbefore, BIFU_XY(:,1) + pixafter], K) = red(K)  %left and right
edge
end

the code:

if true
  % code
BSKE=imread('image.tif')
DIST=DistanciaEuclid(BIFU_XY);
SpuriousMinutae=DIST<D;
[i,j]=find(SpuriousMinutae);
BIFU_XY(i,:)=[];
linelength = 3;
pixbefore = floor((linelength - 1)/2)
pixafter = linelength - 1 - pixbefore
paintcolor = [255, 0, 0];
for K = 1 : length(paintcolor)
BSKE(BIFU_XY(:,2) - pixbefore : BIFU_XY(:,2) + pixafter, [BIFU_XY(:,1) - pixbefore, BIFU_XY(:,1) + pixafter], K) = red(K);  

BSKE([BIFU_XY(:,2) - pixbefore + 1, BIFU_XY(:,2) + pixafter - 1], BIFU_XY(:,1) - pixbefore : BIFU_XY(:,1) + pixafter, K) = red(K);  
end
end

[Image Analyst]

Try paintcolor(K) instead of red(K).

[Walter Roberson]

Right, I renamed a variable as I went.

linelen = 3;

pixbefore = floor((linelen - 1)/2);
pixafter = linelen - 1 - pixbefore;

paintcolor = [255, 0, 0];  %red

for K = 1 : length(paintcolor)
  YourImage(y - pixbefore : y + pixafter, [x - pixbefore, x + pixafter], K) = paintcolor(K);  %left and right edge
  YourImage([y - pixbefore + 1, y + pixafter - 1], x - pixbefore : x + pixafter, K) = paintcolor(K);  %top and bottom edge
 end

[Walter Roberson]

For drawing on the graphics display, you should be using rectangle() unless you have the Computer Vision toolbox.

linelen = 3; %data units now, not pixels
paintcolor = [1 0 0];   %red
rectangle('Position', [x - linelen/2, y - linelen/2, linelen, linelen], 'EdgeColor', paintcolor);

This would be for a rectangle centered around the coordinates. If you want a rectangle whose bottom left is the given coordinates then

linelen = 3; %data units now, not pixels
paintcolor = [1 0 0];   %red
rectangle('Position', [x, y, linelen, linelen], 'EdgeColor', paintcolor);

[Biza Ferreira]

tanks for all

Answer 2

[Dima Lisin]

Hi Biza,

Try using the insertShape function in the Computer Vision System Toolbox.

Answer 3

[Edwards Ramirez]

I had to write it on my own. Here the used function:

function imageReturn = drawRectangle(image, Xmin, Ymin, width, height)
    imageReturn = image;
    intensity = 255;
    X1 = Ymin;
    Y1 = Xmin;
    X2 = Ymin + width;
    Y2 = Xmin + height;
    Xcenter = round((X1+X2)/2);
    Ycenter = round((Y1+Y2)/2);
    
    for i=X1:X2
        imageReturn(i,Y1,1)=intensity;
    end
    for i=X1:X2
        imageReturn(i,Y2,1)=intensity;
    end
    for i=Y1:Y2
        imageReturn(X1,i,1)=intensity;
    end
    for i=Y1:Y2
        imageReturn(X2,i,1)=intensity;
    end
    
    %Comment the following for cycles if you don't want the cross in the centroid
    for i=Xcenter-3:Xcenter+3
        imageReturn(i,Ycenter,1)=intensity;
    end
    for i=Ycenter-3:Ycenter+3
        imageReturn(Xcenter,i,1)=intensity;
    end
end

[Image Analyst]

Well, okay. Your code burns the rectangle into the image (although it's very confusing because you're setting x equal to y and y equal to x instead of just doing it like you should). The poster said he wants to "draw a rectangle over an image", which I take to mean in the graphical overlay over the image, not INTO the image itself. You could put a rectangle into the graphical overlay over the image with the rectangle() function. If you want to burn the rectangle into the image, one could use insertShape like Dima said in his answer (though it requires the Computer Vision System Toolbox).

By the way, your for loops could be vectorized, so

for i=X1:X2
        imageReturn(i,Y1,1)=intensity;
end

would become

imageReturn(Ymin, XMin:(Xmin + width - 1), 1) = intensity;

Notice that I corrected your confusing indexing. It's a very common beginner mistake and that is thinking arrays are indexed as (x,y) instead of (y, x). I know you tried to correct it with your swapping of x and y but I personally think that just made it more confusing. Arrays are indexed as (row, column), so the first index is row, or Y, not X as you had it.

You might want to edit your code and correct that lest anyone every copy it and try to use it and run into problems that confuse them.