How to find x,y matrix using for sintax

i have been working for my thesis about image segmentation. i have this matrix

a =

     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     1     1     1     0     0
     0     0     0     0     1     1     1     1     1     0
     0     0     0     0     0     1     1     0     0     0
     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     0     0

and then i want to find the first x,y with binary 1 using for sintax this is the one i use but its not working

---------------------------------------------

function [x1,y1,x2,y2,x3,y3,x4,y4]=gedge(x)

 [m,n]=size(x);

 % upper edge xy

 for i=1:m
    for j=1:n
        if(x(i,j)==1)
            x1=i;
            y1=j;
        break
        end
    end
 end

% left edge xy

 for j=1:n
    for i=1:m
        if(x(i,j)==1)
            x2=i;
            y2=j;
        break
        end
    end
 end

% bottom edge xy

 for i=m:-1:1
    for j=n:-1:1
        if(x(i,j)==1)
            x3=i;
            y3=j;
        break
        end
    end
 end

 % right edge xy

 for j=n:-1:1
    for i=m:-1:1
        if(x(i,j)==1)
            x4=i;
            y4=j;
        break
        end
    end
 end

%for the cropping

y=x(x1:x4,y2:y3);

-----------------------------------------------------------------

so i want to erase the 0 binary for my image segmentation. it will be this matrix:

a =

     0     1     1     1     0     
     1     1     1     1     1     
     0     1     1     0     0     
     0     1     0     0     0

please help recorrect my sintax code. Thanks

Answer 1

[random09983492]

Hi,

Here is my implementation. Let me know if there is something you do not understand.

[r,c] = size(a);

% Starts at bottom row and removes rows that have nothing in them.
for i = fliplr(1:r)
    if sum(a(i,:)) == 0
        a(i,:) = [];
    else
        break;
    end
end

% Starts from right-most column and removes columns with nothing in them.
for i = fliplr(1:c)
    if sum(a(:,i)) == 0
        a(:,i) = [];
    else
        break;
    end
end

% Starts at top row, counts number of empty rows until it gets to a
% non-empty row. These cannot be removed immediately or it will screw up
% the for-loop.
top_count = 0;
for i = 1:r
    if sum(a(i,:)) == 0
        top_count = top_count + 1;
    else
        break;
    end
end

% Starts at left-most column, counts number of empty columns until it gets
% to a non-empty column. These cannot be removed immediately or it will
% screw up the for-loop.
left_count = 0;
for i = 1:c
    if sum(a(:,i)) == 0
        left_count = left_count + 1;
    else
        break;
    end
end

% Now left-most empty columns and top empty rows can be safely removed.
if top_count > 0
    a(1:top_count,:) = [];
end
if left_count > 0
    a(:,1:left_count) = [];
end

[rio]

i really-really appreciate Mr Elliot answer, this will help me to solve my thesis. hope you succes in the future as well. one again thanks man.

Answer 2

[Jonathan Sullivan]

It's really simple if you use the functions find and any. See below:

ind1 = find(any(x == 1,2),1);
ind2 = find(any(x == 1,2),1,'last');
ind3 = find(any(x == 1,1),1);
ind4 = find(any(x == 1,1),1,'last');

y = x(ind1:ind2,ind3:ind4);

[rio]

thanks man, i already try your way and its work too. hope you succes in the future as well. very-very help. thank you so much.

Answer 3

[Andrei Bobrov]

out = a(any(a,2),any(a));

[rio]

thanks man, i appreciate it. :)