Convert XY coordinates to binary matrix

Hello all,

I have a 1332x2 matrix where X and Y coordinates are given of a glacier surface. When I plot the data with scatter, I just get dots at the given coordinates and it shows the glacier surface by plotting all the dots next to each other. However, I want to create a binary matrix, where 0 = no glacier and 1 = coordinate where the galcier exists. Besides, I want a line to be drawn around the glacier.

How can this be done?

Thanks already!

[Roger Stafford]

You have not given any indication of what determines the points which occur in your matrix. What is it that makes them close to one another on the glacier? Is it somehow related to the topography? It’s too bad you don’t have the third coordinate Z. One could judge by the gradient values.

[Walter Roberson]

All of the x y points are on the surface of the glacier.

Answer 1

[Walter Roberson]

Are the coordinates all positive integers? If they are then one way would be to use full() on sparse() passing in the coordinate vectors. Be careful about whether you want x to correspond to rows or columns.

If they are not positive integers, then hist2 or equivalent might be appropriate.

[Walter Roberson]

To get the line around the points consider using boundary()

Answer 2

[Image Analyst]

Perhaps try scatteredInterpolant() to turn your scattered, randomly located points into a regular 2-D image with a value at every pixel.

[Walter Roberson]

This will not work. The value associated with every specified point is the constant 1. Treat those as scattered points and interpolate anywhere inside or outside the group and you will get the result 1.0 to within numeric accuracy.

Answer 3

[Anibal Carcamo]

I did this for a robot to convert a Distance vs Degree to a matrix of 200x400cm, it may be usefull

 Code:
B=zeros(200,400);
for m=1:1:180
    for l=1:1:200
        for h=1:1:400
            if h==tab(m,5,n)&l==tab(m,4,n)
                    B(l,h)=1;

            end
        end
    end
end

[Walter Roberson]

idx = sub2ind(size(B), squeeze(tab(:,5,:)), squeeze(tab(:,4,:));
B(idx) = 1;