Counting consecutive repeat values for each row in matrix

I have a matrix for a black and white photo, so 1s and 0s.

I want to horizontally measure the length, in pixels, of each black region in the photo. So the number of 0s.

How can I count the number of consecutive 0's for each row? Bare in mind for each row it is likely there are more than one group of consecutive 0's. I want a count value for each group of consecutive 0s in each row.

For example: I have a matrix

X = [1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0;
     1 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0]

So I want it to tell me:

Row 1: 2 groups of 0's, length of each: 5 and 4 respectively

Row 2: 3 groups of 0's, length of each: 3, 4, and 2 respectively

Any help would be greatly appreciated!

Answer 1

[the cyclist]

First, I would download Jan's RunLength utility from the File Exchange, which was designed to solve exactly this kind of problem.

X = [1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0;
     1 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0];

for nr = 1:size(X,1)
    [b{nr} n{nr}] = RunLength(X(nr,:))
end

Each element of b will be the values of each consecutive "run", and n will have the lengths of those runs. So, in this example, the outputs are

% Row 1 results
[b{1}; n{1}]

ans =

     1     0     1     0
     4     5     3     4
     
     
% Row 2 results
[b{2}; n{2}]

ans =

     1     0     1     0     1     0
     1     3     3     4     3     2

You may need to do another step or two to get the output exactly as you want it, but this could be your first building block.

[William Varcoe]

Amazing, thanks very much!

Answer 2

[David Hill]

X = [1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0;
     1 0 0 0 1 1 1 0 0 0 0 1 1 1 0 0];
for k = 1:size(X,1)
    a=num2str(X(k,:));
    a=a(a~=' ');
    a=regexp(a,'([0]+)','tokenExtents');
    b(k)=max(cellfun(@(y)max(diff(y))+1,a));%b array will have the maximum number of consecutive zeros of each row
end

[b]

David, in your above answer, how to output the positions of b-vector?

For example, if we are counting number of '1' in each row,

b(1)=4

the corrsponding position is

X(1,1:4)

and for

b(2)=3

the corresponding positions are:

X(2,5:7) and X(2,12:14)

How to get these X-values also?

Answer 3

[Guillaume]

It's a job for the image processing toolbox, in particular bwconncomp (with a custom connectivity):

cc = bwconncomp(~yourphoto, [0 0 0; 1 1 1; 0 0 0]);  %only consider connection on a row
rowcount = cell(size(yourphoto, 1), 1);
for idx = 1:cc.NumObjects
    row = mod(cc.PixelIdxList{idx}(1)-1, size(yourphoto, 1)) + 1;  %find which row correspond to the connected component
    rowcount{row} = [rowcount{row}, numel(cc.PixelIdxList{idx}-1)];   %number of pixels in the component
end

edit: fixed lots of typos.

[Daniel M]

This does not work. Is youphoto supposed to be X?

After changing 'yourphoto' to 'youphoto', and 'Pixel' and 'PixelIdx' to 'PixelIdxList', I get the error:

Expected one output from a curly brace or dot indexing expression, but there were 3 results. 

But I'm curious for the solution using image processing. Can you correct your code?

[Guillaume]

Indeed, I shouldn't write answers just before going to bed, there were lots of typos. All fixed now.