How to interpolate the coordinates of a vector on an image using MATLAB?

Question

For instance, if I have got a vector describing a rectangle

xy=[165  88; 
    401  88; 
    401 278; 
    165 278];

on an image.

How can I obtain the following vector

[165  88; % increase X -     hold y
 166  88; 
 167  88;
   ...  ;
 399  88;
 400  88;
 401  88; % hold     x - increase y
 401  89;
 401  90;
 401  91;
   ...  ;
 401 276;
 401 277;
 401 278; % decrease X -     hold y
 400 278;
 399 278;
 398 278;
   ...  ;
 167 278;
 166 278;
 165 278; % hold     x - decrease y
 165 277;
 165 276;
   ...  ;
 165  87];

using a MATLAB inbuilt function or do I need to write it using FOR LOOPS?

The algorithm must work for a generic vector with n-points and xy coordinates.

Answer 1

One solution using IND2SUB:

xy=[165 88; 401 88; 401 278; 165 278];
xmin = min(xy(:,1))-1;
xmax = max(xy(:,1));
ymin = min(xy(:,2))-1;
ymax = max(xy(:,2));

ncol=xmax-xmin;
nrow=ymax-ymin;

[xn yn]=ind2sub([nrow ncol],1:nrow*ncol);
xypairs = [xn'+xmin yn'+ymin];

Answer 2

The quick and dirty way to draw a straights into an off-screen matrix is by evaluating the formula a*X+b*Y=c.

Let h and w be width and height of your buffer:

X = repmat([0:w-1], h, 1)
Y = repmat([0:h-1]', 1, w)

For every pair of points (x1,y1)->(x2,y2) a, b and c are:

a = y2-y1
b = x1-x2
c = x1*y2-x2*y1

Now calculating the straigt:

st = a*X+b*Y-c
st(abs(st)>1) = 1
st = 1 - abs(st)

Matrix st is a w*h matrix containing an anti-aliased straight passing through the points (x1,y1) and (x2,y2). Now let's go from straight to line by masking out the unwanted parts:

[xs] = sort([x1 x2])
st = st .* [zeros(h, xs(1)) ones(h, xs(2)-xs(1)) zeros(h, w-xs(2))]
[ys] = sort([y1 y2])
st = st .* [zeros(ys(1), w) ; ones(ys(2)-ys(1), w) ; zeros(h-ys(2), w)]

We have just manually drawn a single line without any explicit looping. No guarantees about the code's efficiency though :-)

Finally: Add another dimension to every formula above (left as an exercise for the reader).

Answer 3

If you have the image processing toolbox, you can do this by creating an image of the polygon followed by finding the contour:

xy=[165 88; 401 88; 401 278; 165 278];
%# create the image - check the help for impolygon for how to make sure that
%# your line is inside the pixel
img = poly2mask(xy(:,1),xy(:,2),max(xy(:,1))+3,max(xy(:,2))+3);
figure,imshow(img) %# show the image

%# extract the perimeter. Note that you have to inverse x and y, and that I had to
%# add 1 to hit the rectangle - this shows one has to be careful with rectangular 
%# polygons
boundary = bwtraceboundary(logical(img),xy(1,[2,1])+1,'n',8,inf,'clockwise');

%# overlay extracted boundary
hold on, plot(boundary(:,2),boundary(:,1),'.r')

Edited to show how to use bwtraceboundary and to warn of pixel offset with rectangles.