Matrix manipulation syntax help

Hey, I have a 2048x2048 matrix with values of either 0 or 1 which is called "white". I need to find the total number of 3x3 blocks in the matrix where each data point has a value of 1, ie,

1 1 1

1 1 1

1 1 1

Im a first time user and I think that my problem is syntax. My code is as follows,

x=0

for i=2:2047 j=2:2047

if white(i-1,j-1)==1 & white(i-1,j)==1 & white(i-1,j+1)==1 & white(i,j-1)==1 & white(i,j)==1 & white(i,j+1)==1 & white(i+1,j-1)==1 & white(i+1,j)==1 & white(i+1,j+1)==1

x=x+1

end

y=sum(x)

end

Ive been at it for a few hours and any help would be really appreciated. Ill be spending the rest of the afternoon trying to solve this problem....

Thanks for your help!

[Sean de Wolski]

Welcome to MATLAB answers. Your post is well written, you provide: an example, what you've tried, and what you want - everything we typically request.

Answer 1

[Sean de Wolski]

You could do this with your for-loop like this:

n = 0;
for ii = 2:2047
 for jj = 2:2047
  if(all(all(white(ii-1:ii+1,jj-1:jj+1))))
   n = n+1;
  end
 end
end

[Andy]

Derek, also take note of Sean's index variables: ii and jj. If you use i and j as index variables, you overwrite the complex unit i = j = sqrt(-1). Instead, it is common to use ii, jj, ix, and jx as indexing variables for loops, so as to leave i and j for complex computations.

[Derek]

Thanks for your help! I had to make some changes to my original code, but your answer served as a great template for me. I just started a summer job where I will be doing lots of Matlab. This forum is great, thanks again.

Answer 2

[Sean de Wolski]

This will probably frustrate you, but here's a one-liner to do exactly what you want!

n = sum(sum(conv2(double(white),ones(3),'valid')==9));

Algorithm:

• convolve the "white" matrix with a kernel of ones the size you want. Only valid applications apply.
• It will only equal 9 when all of the values in white matching the kernel are == 1 (i.e. (1*1)*9).
• sum it twice to get the total number.
• Enjoy a coffee.

[Matt Tearle]

Of course you actually meant

n = nnz(conv2(double(white),ones(3),'valid')==9);

:)

[Sean de Wolski]

Eww. No. nnz is super slow, avoid it like the plague:

t1 = 0;

t2 = 0;

white = rand(2048)>.4;

for ii = 1:10

tic

n1 = sum(sum(conv2(double(white),ones(3),'valid')==9));

t1 = t1+toc;

tic

n2 = nnz(conv2(double(white),ones(3),'valid')==9);

t2 = t2+toc;

end

disp([t1 t2])

[Matt Fig]

It's a shame too. Because the NNZ solution looks so much cleaner and is easier to read.

[Walter Roberson]

I thought nnz() was the preferred form?

On my system, nnz() is only about 2% slower in the above test. And nnz() would be considerably faster than double-sum on sparse arrays.

[Sean de Wolski]

That's why I keep a handy little one line function around vec

function x = vec(x)

x = reshape(x,numel(x),1);

Then sum(vec(...)) %At least no double sum because those are clumsy

[Sean de Wolski]

Aren't sparse arrays its main (only for my purposes) use?

Answer 3

[Matt Fig]

Another suggestion:

FINDSUBMAT

x2 = length(findsubmat(white,ones(3)));

Also, as far as your code, you were close. This works:

x = 0;

for ii = 2:2047
    for jj = 2:2047
        if white(ii-1,jj-1) & white(ii-1,jj) &...
           white(ii-1,jj+1) & white(ii,jj-1) &...
           white(ii,jj) & white(ii,jj+1) &...
           white(ii+1,jj-1) & white(ii+1,jj) &...
           white(ii+1,jj+1) %#ok
            x = x+1;
        end
    end
end

Answer 4

[Andrei Bobrov]

more

mn = size(white)-2;
C = arrayfun(@(k)cumsum([1:3;ones(k,3)]),mn,'un',0);
C2 = arrayfun(@(s)cumsum(ones(mn),s),[1 2],'un',0);
x = nnz(arrayfun(@(i,j)all(all(white(C{1}(i,:),C{2}(j,:)))),C2{:}));