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copied!<p>First of all, you can skip the matrix <code>m</code>. It is a less-than-optimal way to go about the problem in this case. Also, get rid of the zeros at the end of <code>x_t</code>. They are not necessary. If you need help trimming the zeros off the end, try this:</p> <pre><code>x_t = x_t(1:find(x_t~=0,1,'last')); </code></pre> <p>Now, for the matching. You can find the closest match in <code>x_t</code> to the elements of <code>x</code> with the following:</p> <pre><code>B = bsxfun(@minus, x_t(:), x(:).'); %' [idx,~] = find(bsxfun(@eq, abs(B), min(abs(B)))); y = x_t(idx); </code></pre> <p>Now the approximations of <code>x</code> are found in <code>y</code>, with corresponding index found in <code>idx</code>.</p> <h1>Explanation</h1> <hr> <pre><code>bsxfun(@minus, x, y); </code></pre> <p>This will take a column vector <code>x</code> and subtract it by the each or the elements of the row vector <code>y</code>. For example,</p> <pre><code>>> x = [3 6 9 12].'; y = [1 2 3]; >> ans = bsxfun(@minus, x, y); ans = 2 1 0 5 4 3 8 7 6 11 10 9 </code></pre> <p>Or in other words, what is happening is:</p> <p>ans = </p> <pre><code> (3-1) (3-2) (3-3) (6-1) (6-2) (6-3) (9-1) (9-2) (9-3) (12-1) (12-2) (12-3) </code></pre> <hr> <pre><code>bsxfun(@eq, A, min(A)); </code></pre> <p>This finds the locations in the columns of the matrix <code>A</code> where the column minimum resides. It returns a logical matrix (ones and zeros). For example,</p> <pre><code>>> A = [ 1 2 3 4; 3 3 6 5; 2 2 0 7; 5 1 8 5; 4 5 2 6]; >> ans = bsxfun(@eq, A, min(A)) ans = 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 </code></pre> <hr> <pre><code>[idx,~] = find(A); </code></pre> <p>The <code>find</code> command searches for all of the non-zero elements of the matrix <code>A</code>. Normally you would call it as <code>[row, col] = find(A);</code> to get back the rows and columns, but since I am only interested in the row, I use <code>~</code> to tell Matlab that I don't want to keep the column value.</p> <hr> <pre><code>y = x(idx); </code></pre> <p>This creates a new vector <code>y</code> which contains the elements of <code>x</code> as referenced by <code>idx</code>. Since I already found the indices that minimize the error, I use them to pluck out the <code>x</code> values at those indices.</p> <hr> <p><strong>Putting it all together:</strong></p> <p>At a high level, the code does the following:</p> <p><li> The matrix <code>B</code> compares every value of <code>x</code> (columns) with every value of <code>x_t</code> (rows) <li> For every column (<code>x</code>) of <code>B</code>, find the row (<code>x_t</code>) which minimizes the error. <li> Knowing the rows we can extract the <code>x_t</code> values directly using the row indices.</p> <p>Hope this helps!</p> <h1>Update</h1> <p>I realized that if a value of <code>x</code> was exactly the same distance between two values of <code>x_t</code> then you would end up with a vector that was too long, because multiple answers will be selected. Below is the fix to the problem.</p> <pre><code>B = bsxfun(@minus, x_t(:), x(:).'); %' [idx,col] = find(bsxfun(@eq, abs(B), min(abs(B)))); idx(1 < sum(triu(bsxfun(@eq, col(:), col(:).')))) = []; %' y = x_t(idx); </code></pre> <p>Basically what the added line does is throw away any index values that do not correspond to the first match in a given column.</p>

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