function to count number of '1' in each row and column

if i have a (M , N) binary matrix which M is number of rows and N is the number of column , and i want to count Number of 1 in each row and for each column , for example [1 0 1 1 0 1 1 1 1 0 1 0 ] the the output will be in new matrix like this [ 1 2 4 1]

       Matrix = [ 0 1 1 1 0 0 1 0 1 1 
                  1 0 1 1 0 1 0 1 1 0
                  1 1 0 0 1 1 1 1 0 1
                  1 1 1 1 1 0 1 0 0 1
                  0 0 1 0 0 1 1 1 1 0
                  1 1 0 1 1 0 0 1 0 1 ]

Matrix_row =   [3 1 2 0
                1 2 1 2 
                2 4 1 0
                5 1 1 0
                1 4 0 0
                2 2 1 1 ]

 Matrix_col = [0 0 0 0 0 0 0 0 0 0
               0 1 0 2 0 0 0 0 0 1
               3 2 2 1 1 2 1 2 2 2
               1 1 2 1 1 1 3 2 1 1 ]

i need this solution because its a part of graduation project please he me

[Matthew Eicholtz]

I do not understand how Matrix_row and Matrix_col are being computed. Currently, it is not the number of ones in the each row or column (as far as I can tell). Can you provide more details?

[Image Analyst]

Matt, it's basically the lengths of the "runs" of 1's in each direction, then padded with zeros to make it rectangular. Though I'm not sure why Matrix_col has a top row of all zeros - seems like that should have been omitted. So the first row has groupings/runs of 1's of lengths 3, 1, and then 2, each separated by zeros. The last row has 4 separate runs, so the Matrix_row array needs to have (at least) 4 columns. If some row had only , say, 2 runs, you'd have to append two zeros to make that row have the full 4 columns.

By the way, the lengths of the runs can be determined from regionprops() very, very easily:

stats = regionprops(logical(oneRow), 'Area');
allLengths = [stats.Area]; % Would give 3,1,2 for the first row in the example matrix.

By the way, this is a duplicate to http://www.mathworks.com/matlabcentral/answers/275707#answer_215257 where I outlined the solution but didn't give a complete solution because it seemed to be his assignment.

Answer 1

[Azzi Abdelmalek]

M = [ 0 1 1 1 0 0 1 0 1 1
  1 0 1 1 0 1 0 1 1 0
  1 1 0 0 1 1 1 1 0 1
  1 1 1 1 1 0 1 0 0 1
  0 0 1 0 0 1 1 1 1 0
  1 1 0 1 1 0 0 1 0 1 ];
[n,m]=size(M);
b=cell(n,1);
c=cell(1,m);
maxb=1;
maxc=1;
for k=1:n
  a=[0 M(k,:) 0];
  ii1=strfind(a,[0 1]);
  ii2=strfind(a,[1 0]);
  maxb=max(maxb,numel(ii1));
  b{k}=ii2-ii1;
end
for k=1:m
  a=[0 M(:,k)' 0];
  ii1=strfind(a,[0 1]);
  ii2=strfind(a,[1 0]);
  maxc=max(maxc,numel(ii1));
  c{k}=(ii2-ii1)';
end
M_row=cell2mat(cellfun(@(x) [x zeros(1,maxb-numel(x))],b,'un',0))
M_column=cell2mat(cellfun(@(x) [zeros(1,maxc-numel(x));x],c,'un',0))

[Firas Al-Kharabsheh]

Thank You so much

[Firas Al-Kharabsheh]

Can you explain to me how the code working ?