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copied!<p>Although you can certainly do the math yourself to compute the linear index (as pointed out by <a href="https://stackoverflow.com/questions/919353/5d-array-hash-table/919404#919404">Tal</a>), it would be cleaner and easier to read using the built-in function <a href="http://www.mathworks.com/access/helpdesk/help/techdoc/ref/sub2ind.html" rel="nofollow noreferrer">SUB2IND</a> (as pointed out by <a href="https://stackoverflow.com/questions/919353/5d-array-hash-table/920506#920506">Loren</a>). For your example, you would use it in the following way:</p> <pre><code>index = sub2ind([8 12 3 4 3], idx_r, idx_l, idx_rho, idx_alpha, idx_beta); </code></pre> <p>If all of your indices are 0-based, you would have to add 1 to each of them before passing them to SUB2IND.</p> <p><strong>EDIT:</strong></p> <p>If you would like to see how to correctly compute linear indices by yourself in MATLAB, such that they agree with the results from SUB2IND, here's the code you would use:</p> <pre><code>index = (((idx_beta*4 + idx_alpha)*3 + idx_rho)*12 + idx_l)*8 + idx_r + 1; </code></pre> <p><strong>NOTE</strong> that the indices that need to be used with this equation have to be 0-based, while the indices passed to SUB2IND have to be 1-based. To generalize this equation to an arbitrary number of dimensions <strong>N</strong>:</p> <pre><code>index = (...(idx(N)*sz(N-1) + idx(N-1))*sz(N-2) + ...)*sz(1) + idx(1) + 1; </code></pre> <p>or more succinctly:</p> <pre><code>index = 1 + sum(idx.*cumsum([1 sz(1:(N-1))])); </code></pre> <p>where <strong>idx</strong> is an array of 0-based index values for each dimension and <strong>sz</strong> is an array of the sizes of each dimension.</p>

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