Assignment has more non-singleton rhs dimensions than non-singleton subscripts?

Hello, here is my code:

for f = 1:n ind(f,1) = num2cell(find(edges{f,1} == inde_x),2); in_d(f,1) = size(ind{f,:},2); indx = find(in_d ~= 0); end

I have edges matrix 9x1 cell and inde_x = 6

See the edges matrix below [6 9] [1 5] [5 6] [6 9] [6 7] [1 5] [5 9] 7 [1 7]

I'am trying to find values in inde_x = 6 in the edges matrix and ind matrix is formed as follows:

1 [] 2 1 1 [] [] 1 []

As seen above, in the edges matrix is equal to 6 in the row 6 finds the number and writes number of the index. If there is no value equal to 6, the [] sign is written but the number {8,1} does not have a value of 6, but the number 1 occurs and the following error is written:

Error in aaaaaaaaaaaaa (line 125) ind(f,1) = num2cell(find(edges{f,1} == inde_x),2);

How can I solve this problem? Thanks.

[Jan]

The question is not clear. Start with formatting the code using the "{} Code" button, see How to format code in the forum.

The names of the variables ind, inde_x, in_d, indx are really confusing. I recommend to use smarter names.

The show "edge matrix" is not clear. Please use the standard Matlab syntax to define a matrix.

"1 [] 2 1 1 [] [] 1 []" This does not form a matrix. Matrix elements cannot be empty.

"As seen above, in the edges matrix is equal to 6 in the row 6 finds the number and writes number of the index" - I do not understand this sentence. I do not see this above.

"the [] sign is written but the number {8,1}" - this is not clear also. [] is the concatenation operator, not a "sign".

" and the following error is written: ..." - This is only the part of the error message, which tells, where the error occurs. The other parts matter also, which explain what the error is.

[Abdullah Türk]

Thank you for your warnings, Jan.

k = 6;
n = 9;

for f = 1:n
      A(f,1) = num2cell(find(edges{f,1} == k),2);
  end

I have matrix 9x1 named edges.

• edges{1,1} = [6 9]
• edges{2,1} = [1 5]
• edges{3,1} = [5 6]
• edges{4,1} = [6 9]
• edges{5,1} = [6 7]
• edges{6,1} = [1 5]
• edges{7,1} = [5 9]
• edges{8,1} = 7
• edges{9,1} = [1 7]

I want to find subscripts equal to k in the edges matrix and a A matrix has occurred as a result of the loop:

• A{1,1} = 1
• A{2,1} = []
• A{3,1} = 2
• A{4,1} = 1
• A{5,1} = 1
• A{6,1} = []
• A{7,1} = []
• A{8,1} = 1
• A{9,1} = []

and result of the loop:

Assignment has more non-singleton rhs dimensions than non-singleton subscripts

Error in aaaaaaaaaaaaa (line 125)
    ind(f,1) = num2cell(find(edges{f,1} == inde_x),2);

I had to get A {8.1} = [] according to the edges matrix I have, but A {8.1} = 1. Values other than A {8.1} appear to be correct.

Thank you from now.

[Guillaume]

I was going to ask why you'd use a cell array instead of a Nx2 matrix to store edges then saw that one of your edge has only one element. Why? Don't your edge describe a graph? If so all edges should have two elements. If it's a self-loop, the node should be repeated.

Answer 1

[Guillaume]

k = 6;
A = cellfun(@(edge) find(edge == k), edges, 'UniformOutput', false)

The loopy version of that:

k = 6;
%n unused. Never hardcode the number of elements. Ask matlab what it is with size or numel
A = cell(size(edges));
for idx = 1:numel(edges)
   A{idx} = find(edges{idx} == k);
end

Note the difference between A(idx) and A{idx}. Your attempt at using num2cell show that you don't really know how to access cell arrays.

[Abdullah Türk]

Thanks Guillaume. Problem solved. I'm new in Matlab. I know the difference between A(idx) and A{idx} but after seeing your loop, I realized that my error was in the use of num2cell. Thank you again.

Answer 2

[KSSV]

k = 6;
n = 9;
edges{1,1} = [6 9]
edges{2,1} = [1 5]
edges{3,1} = [5 6]
edges{4,1} = [6 9]
edges{5,1} = [6 7]
edges{6,1} = [1 5]
edges{7,1} = [5 9]
edges{8,1} = 7
edges{9,1} = [1 7]
A = cell(n,1) ;

for f = 1:n
    T = num2cell(find(edges{f,1} == k),2) ;
    if ~isempty(T)
        A(f,1) = T;
    else
        A(f,1) = {[]};
    end
end

[Abdullah Türk]

Thank you KSSV.

Unfortunately, the same error occurred.