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<h1>4.7. Implementing an efficient rolling average algorithm with stride tricks</h1>
</header>
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<p><a href="/"><img src="https://raw.githubusercontent.com/ipython-books/cookbook-2nd/master/cover-cookbook-2nd.png" align="left" alt="IPython Cookbook, Second Edition" height="130" style="margin-right: 20px; margin-bottom: 10px;" /></a> <em>This is one of the 100+ free recipes of the <a href="/">IPython Cookbook, Second Edition</a>, by <a href="http://cyrille.rossant.net">Cyrille Rossant</a>, a guide to numerical computing and data science in the Jupyter Notebook. The ebook and printed book are available for purchase at <a href="https://www.packtpub.com/big-data-and-business-intelligence/ipython-interactive-computing-and-visualization-cookbook-second-e">Packt Publishing</a>.</em></p>
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<p>Stride tricks can be useful for local computations on arrays, when the computed value at a given position depends on the neighboring values. Examples include dynamical systems, digital filters, and cellular automata.</p>
<p>In this recipe, we will implement an efficient <strong>rolling average</strong> algorithm (a particular type of convolution-based linear filter) with NumPy stride tricks. A rolling average of a 1D vector contains, at each position, the average of the elements around this position in the original vector. Roughly speaking, this process filters out the noisy components of a signal so as to keep only the slower components.</p>
<h2>How to do it...</h2>
<p>The idea is to start from a 1D vector, and make a <em>virtual</em> 2D array where each line is a shifted version of the previous line. When using stride tricks, this process is very efficient as it does not involve any copy.</p>
<p><strong>1. </strong> Let's generate a 1D vector:</p>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">numpy</span> <span class="kn">as</span> <span class="nn">np</span>
<span class="kn">from</span> <span class="nn">numpy.lib.stride_tricks</span> <span class="kn">import</span> <span class="n">as_strided</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">aid</span><span class="p">(</span><span class="n">x</span><span class="p">):</span>
<span class="c1"># This function returns the memory</span>
<span class="c1"># block address of an array.</span>
<span class="k">return</span> <span class="n">x</span><span class="o">.</span><span class="n">__array_interface__</span><span class="p">[</span><span class="s1">'data'</span><span class="p">][</span><span class="mi">0</span><span class="p">]</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="n">n</span> <span class="o">=</span> <span class="mi">5</span>
<span class="n">k</span> <span class="o">=</span> <span class="mi">2</span>
<span class="n">a</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">n</span><span class="p">)</span>
<span class="n">ax</span> <span class="o">=</span> <span class="n">aid</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>
</pre></div>
<p><strong>2. </strong> Let's change the strides of <code>a</code> to add shifted rows:</p>
<div class="highlight"><pre><span></span><span class="n">as_strided</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="p">(</span><span class="n">k</span><span class="p">,</span> <span class="n">n</span><span class="p">),</span> <span class="p">(</span><span class="mi">8</span><span class="p">,</span> <span class="mi">8</span><span class="p">))</span>
</pre></div>
<div class="highlight"><pre><span></span>array([[ 1e+000, 2e+000, 3e+000, 4e+000, 5e+000],
[ 2e+000, 3e+000, 4e+000, 5e+000, 9e-321]])
</pre></div>
<p>The last value indicates an out-of-bounds problem: stride tricks can be dangerous as memory access is not checked. Here, we should take edge effects into account by limiting the shape of the array.</p>
<p><strong>3. </strong> Now, let's implement the computation of the rolling average. The first version (standard method) involves explicit array copies, whereas the second version uses stride tricks:</p>
<div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">shift1</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
<span class="k">return</span> <span class="n">np</span><span class="o">.</span><span class="n">vstack</span><span class="p">([</span><span class="n">x</span><span class="p">[</span><span class="n">i</span><span class="p">:</span><span class="n">n</span> <span class="o">-</span> <span class="n">k</span> <span class="o">+</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">]</span>
<span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">k</span><span class="p">)])</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">shift2</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
<span class="k">return</span> <span class="n">as_strided</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="p">(</span><span class="n">k</span><span class="p">,</span> <span class="n">n</span> <span class="o">-</span> <span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">),</span>
<span class="p">(</span><span class="n">x</span><span class="o">.</span><span class="n">itemsize</span><span class="p">,</span> <span class="n">x</span><span class="o">.</span><span class="n">itemsize</span><span class="p">))</span>
</pre></div>
<p><strong>4. </strong> These two functions return the same result, except that the array returned by the second function refers to the original data buffer:</p>
<div class="highlight"><pre><span></span><span class="n">b</span> <span class="o">=</span> <span class="n">shift1</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="n">b</span>
</pre></div>
<div class="highlight"><pre><span></span>array([[ 1., 2., 3., 4.],
[ 2., 3., 4., 5.]])
</pre></div>
<div class="highlight"><pre><span></span><span class="n">aid</span><span class="p">(</span><span class="n">b</span><span class="p">)</span> <span class="o">==</span> <span class="n">ax</span>
</pre></div>
<div class="highlight"><pre><span></span>False
</pre></div>
<p>And now with the second function:</p>
<div class="highlight"><pre><span></span><span class="n">c</span> <span class="o">=</span> <span class="n">shift2</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="n">c</span>
</pre></div>
<div class="highlight"><pre><span></span>array([[ 1., 2., 3., 4.],
[ 2., 3., 4., 5.]])
</pre></div>
<div class="highlight"><pre><span></span><span class="n">aid</span><span class="p">(</span><span class="n">c</span><span class="p">)</span> <span class="o">==</span> <span class="n">ax</span>
</pre></div>
<div class="highlight"><pre><span></span>True
</pre></div>
<p><strong>5. </strong> Let's generate a signal:</p>
<div class="highlight"><pre><span></span><span class="n">n</span><span class="p">,</span> <span class="n">k</span> <span class="o">=</span> <span class="mi">1000</span><span class="p">,</span> <span class="mi">10</span>
<span class="n">t</span> <span class="o">=</span> <span class="n">np</span><span class="o">.</span><span class="n">linspace</span><span class="p">(</span><span class="mf">0.</span><span class="p">,</span> <span class="mf">1.</span><span class="p">,</span> <span class="n">n</span><span class="p">)</span>
<span class="n">x</span> <span class="o">=</span> <span class="n">t</span> <span class="o">+</span> <span class="o">.</span><span class="mi">1</span> <span class="o">*</span> <span class="n">np</span><span class="o">.</span><span class="n">random</span><span class="o">.</span><span class="n">randn</span><span class="p">(</span><span class="n">n</span><span class="p">)</span>
</pre></div>
<p><strong>6. </strong> We compute the signal rolling average by creating the shifted version of the signal, and averaging along the vertical dimension:</p>
<div class="highlight"><pre><span></span><span class="n">y</span> <span class="o">=</span> <span class="n">shift2</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<span class="n">x_avg</span> <span class="o">=</span> <span class="n">y</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
</pre></div>
<p><strong>7. </strong> Let's plot these arrays:</p>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">matplotlib.pyplot</span> <span class="kn">as</span> <span class="nn">plt</span>
<span class="o">%</span><span class="n">matplotlib</span> <span class="n">inline</span>
</pre></div>
<div class="highlight"><pre><span></span><span class="n">fig</span><span class="p">,</span> <span class="n">ax</span> <span class="o">=</span> <span class="n">plt</span><span class="o">.</span><span class="n">subplots</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="n">figsize</span><span class="o">=</span><span class="p">(</span><span class="mi">8</span><span class="p">,</span> <span class="mi">4</span><span class="p">))</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x</span><span class="p">[:</span><span class="o">-</span><span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">],</span> <span class="s1">'-k'</span><span class="p">,</span> <span class="n">lw</span><span class="o">=</span><span class="mi">1</span><span class="p">,</span> <span class="n">alpha</span><span class="o">=.</span><span class="mi">5</span><span class="p">)</span>
<span class="n">ax</span><span class="o">.</span><span class="n">plot</span><span class="p">(</span><span class="n">x_avg</span><span class="p">,</span> <span class="s1">'-k'</span><span class="p">,</span> <span class="n">lw</span><span class="o">=</span><span class="mi">2</span><span class="p">)</span>
</pre></div>
<p><img alt="<matplotlib.figure.Figure at 0x7f3f49a015f8>" src="https://ipython-books.github.io/pages/chapter04_optimization/07_rolling_average_files/07_rolling_average_29_1.png" /></p>
<p><strong>8. </strong> Let's evaluate the time taken by the first method:</p>
<div class="highlight"><pre><span></span><span class="o">%</span><span class="n">timeit</span> <span class="n">shift1</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>15.4 µs ± 302 ns per loop (mean ± std. dev. of 7 runs,
100000 loops each)
</pre></div>
<div class="highlight"><pre><span></span><span class="o">%%</span><span class="n">timeit</span> <span class="n">y</span> <span class="o">=</span> <span class="n">shift1</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<span class="n">z</span> <span class="o">=</span> <span class="n">y</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>10.3 µs ± 123 ns per loop (mean ± std. dev. of 7 runs,
100000 loops each)
</pre></div>
<p>Here, most of the total time is spent in the array copy (the <code>shift1()</code> function).</p>
<p><strong>9. </strong> Let's benchmark the second method:</p>
<div class="highlight"><pre><span></span><span class="o">%</span><span class="n">timeit</span> <span class="n">shift2</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>4.77 µs ± 70.3 ns per loop (mean ± std. dev. of 7 runs,
100000 loops each)
</pre></div>
<div class="highlight"><pre><span></span><span class="o">%%</span><span class="n">timeit</span> <span class="n">y</span> <span class="o">=</span> <span class="n">shift2</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<span class="n">z</span> <span class="o">=</span> <span class="n">y</span><span class="o">.</span><span class="n">mean</span><span class="p">(</span><span class="n">axis</span><span class="o">=</span><span class="mi">0</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>9 µs ± 179 ns per loop (mean ± std. dev. of 7 runs,
100000 loops each)
</pre></div>
<p>This time, thanks to the stride tricks, most of the time is instead spent in the computation of the average.</p>
<h2>See also</h2>
<ul>
<li>Using stride tricks with NumPy</li>
</ul>
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