Cell Arrays and Indexing?

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 :)

[James Tursa]

You need to loop over the rows, not the columns. See below.

Answer 1

[James Tursa]

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.

[James Metz]

It's kind of embarassing that I am now needing help with the reverse of my last assignment haha. If you can't tell, I'm not too great at MATLAB, but I am trying. I am familiar with the find() fxn; however, doesn't it out put the indices as if the columns were stacked on top of each other in a vertical list? Does the for loop prevent this? Here's my code with the find() fxn.

function opp = hw5_problem5(X)
opp = [];   
[rows, ~] = size(X); 
for ii = 1:rows
    if X(ii) == 1
        opp = [opp, {find(X(ii) == 1)}];
    else
        opp = [];
    end
end
end

I know you said that I needed just one line of code after the for loop, but won't I need an else statement to ensure that if X(ii) does not equal one, that the function still outputs a cell with an empty matrix to show that there is a row with all false elements?

[James Metz]

Ok, this is what it boils down to I think, but I'm still not sure what is wrong with it. When I input the simple matrix c shown below, that's the output I get.

function opp = hw5_problem5(X)
opp = [];   
[rows, ~] = size(X); 
for ii = 1:rows
        opp = [opp {find(X(ii) == 1)}];
end
end

c = [false true; true false] 
output = {[], [1]}

The second cell is correct, but why isn't the first? Does find(X(ii) == 1) not output the index 2?

[James Tursa]

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,:).

Answer 2

[Ameer Hamza]

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