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  1. PO
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    <p>About <code>y=~isnan(x)</code>, you are right.</p> <p>The line with <code>x(k,y(k,:))</code> will give the not-Nans in the <code>k</code>-th row of <code>x</code>. So it seems that <code>z</code> is gathering the not-Nans values of <code>x</code> (in a weird way). Note that <code>y(k,:)</code> acts as a logical index for the columns, where <code>true</code> means "include that column" and <code>false</code> means "do not include".</p> <p>As for your last question: <code>[z{:,:}]</code> is in this case equivalent to <code>[z{:}]</code>, because <code>z</code> has one dimension only, and it will horizontally concatenate the contents of the cell array <code>z</code>. For example, with <code>z{1} = [1; 2]; z{2} = [3 4; 5 6];</code> it will give <code>[1 3 4; 2 5 6]</code>. Thus <code>m</code> will be the <em>common</em> number of rows in the matrices that make up <code>z</code>, and <code>n</code> will be the sum of the numbers of columns (in my example <code>m</code> will be 2 and <code>n</code> will be 3). If there is no such common number of rows it will give an error. For example, if <code>z{1} = [1 2]; z{2} = [3 4; 5 6];</code> then <code>[z{:}]</code> or <code>[z{:,:}]</code> give an error.</p> <p>So the final result <code>d</code> is just a row vector which contains the not-Nans from <code>x</code> ordered by increasing row and then increasing column. That could have been obtained more easily as</p> <pre><code>xt = x.'; d = xt(~isnan(xt(:))).'; </code></pre> <p>which is more compact, more Matlab-like, and probably faster.</p>
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    1. POmatlab, cell arrays, logical array indexing, and array type conversion
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    1. POmatlab, cell arrays, logical array indexing, and array type conversion
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    1. COAh, I made an error typing the line with x(k, (k:)). It should be x(k, z(k:)). I corrected that in the code. The horizontal concatenation, together with indexing by a logical array (in the line I corrected) makes sense as somehow selecting and concatenating non-NaN numbers, but I still am not sure what size/dimension arrays z will hold (that is, what size/dimension arrays are created by indexing by a logical array). Also I think you reversed the meaning of n & m; note how the line is [m,n] = size(d), not [n,m] = size(d).
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    2. COYou're right about m and n. And with your correction the code makes more sense. Please see updated answer.
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    3. COThe code obtains the result (`d`) in a weird way. It's not the standard way to do things in Matlab. Loops can be easily avoided here, the cell array is definitely not needed, and why work by rows? May I ask where you got it from? Some kind of automatically generated or translated code?
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