MATLAB: Using 3d matrix inmatlab

3d matrix

A(:,:,1) =

     1     1     0
     0     1     1
     1     1     1

A(:,:,2) =

     1     0     0
     0     0     1
     0     0     1

I used [r,c,v] = ind2sub(size(A),find(A == 1)); to find r,c,v in matrix A that has a value 1. But I want to place a condition that while A == 1 then I(r,c) = v

for example I(1,1) = {1,2} since for v=1,2 A takes a value 1. Please help me out with a code.

Best Answer

 [m,n,p]=size(A);
 s=reshape(1:p,1,1,p);
 A=bsxfun(@times,A,s);

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How can I assign null array by using two indices (`a(:,i,i)`)?

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[m,n,p]=size(a);
a(:,1:n+1:end)=[];
a=reshape(a,m,[],p)

MATLAB: Dividing a column matrix based on the no

A = [0 0 0 1 1 1 1 0 0 0 0 2 2 2 2 0 0 0 0].';

Split sequence (transforming in string)

out = regexp(sprintf('%d',A) ,'(1+0+|2+0+)','match');

Convert to double and store in cell array

c = cellfun(@(x) x.'-'0',out,'un',0)
c{1} % to retrieve the value

Organize eventually in structure with dynamic fields A001, A002, ..., An

numC = numel(c);
s = cell2struct(c, reshape(sprintf('A%03d', 1:numC),4,[]).',numC);
s.A001 % to retrieve the value

EDITED

With a loop

A = [0 0 0 1 1 1 1 0 0 0 0 2 2 2 2 0 0 0 0].';
c = 0;  
s = 0;
% Start condition
if ismembc(A(1,1),1:2) 
   s = 1;
end
for n = 2:numel(A(:,1))-1
    if A(n,1) == 0 && ismembc(A(n+1,1),1:2)
        if s
            c = c+1;
            Out{c} = A(s:n,:);
        end
        s = n+1;
    end
end
% End condition
if s < n
    Out{c+1} = A(s:end,:);
end

Best Answer

Related Solutions

MATLAB: Vectorization with two dimensions at the same time

MATLAB: Dividing a column matrix based on the no

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