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Cell Arrays and Indexing?

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James Metz
James Metz on 4 Apr 2020
Edited: James Tursa on 4 Apr 2020
The goal of my code is to take an n-by-n logical array of values and output a cell vector containing the column indices of each true (1) element in the input matrix. The row index of the logical true values is the index of the cell in the cell vector, and the column index is the value(s) within the cell of the corresponding cell in the cell vector.
Here's an example of how it's supposed to work:
function output = input(A)
A =
0 1 0 1
1 1 0 1
1 0 1 0
0 0 0 0
output =
{[2 4], [1 2 4], [1 3], []}
Here's my code:
function opp = hw5_problem5(X)
opp = []; % initialize the matrix to empty
[~, col] = size(X); % setting the size of X to variable so it can change with input
for ii = 1:col
if X(ii) == 1
opp(ii) = opp; {ii}; % add previous cells of opp to new cells as iterations continue
else
opp(ii) = []; % if value of X(ii) == 0 no new values added to opp
end
end
end
I keep getting this error message:
Unable to perform assignment because the indices on the left side are not compatible with the size of the
right side.
Error in hw5_problem5 (line 6)
opp(ii) = opp, {ii}; %add previous cells of opp to new cells as iterations continue
Help please! Thanks :)
  1 Comment
James Tursa
James Tursa on 4 Apr 2020
You need to loop over the rows, not the columns. See below.

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Accepted Answer

James Tursa
James Tursa on 4 Apr 2020
Edited: James Tursa on 4 Apr 2020
This is the reverse of your last assignment. It needs only one loop over the number of rows, and the cell array element for that row contains the column indexes of the 1 elements. You should look into the find( ) function. The inside of your loop needs only one line of code.
  3 Comments
James Tursa
James Tursa on 4 Apr 2020
You are pretty close, but ask yourself what does X(ii) mean? Is it the ii'th row of X? No, it isn't. It is the ii'th element of X using linear indexing. The ii'th row of X is X(ii,:).

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More Answers (1)

Ameer Hamza
Ameer Hamza on 4 Apr 2020
This is an alternative to the looped version
A = [0 1 0 1
1 1 0 1
1 0 1 0
0 0 0 0];
[r, c] = find(A);
out = accumarray(r, c, [], @(x) {x});
if size(out,1) ~= size(A,1)
out{size(out,1)+1:size(A,1)} = [];
end

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