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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     ```

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Answer by Elliot Olsen on 20 Jun 2013

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.

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.

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. :)