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How to find x,y matrix using for sintax

Asked by rio on 20 Jun 2013

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

0 Comments

rio

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3 Answers

Answer by Elliot Olsen on 20 Jun 2013
Accepted answer

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

1 Comment

rio on 20 Jun 2013

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.

Elliot Olsen
Answer by Jonathan Sullivan on 20 Jun 2013
Edited by Jonathan Sullivan on 20 Jun 2013

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);

1 Comment

rio on 20 Jun 2013

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.

Jonathan Sullivan
Answer by Andrei Bobrov on 20 Jun 2013
out = a(any(a,2),any(a));

1 Comment

rio on 20 Jun 2013

thanks man, i appreciate it. :)

Andrei Bobrov

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