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    <p>Here is a general strategy for solving this kind of problem.</p> <p>First, write a small script, with the loop written explicitly in two different functions, and a test at the end making sure that the two functions are exactly the same:</p> <pre><code>import numpy as np from numpy import newaxis def explicit(a): n = a.shape[0] m = np.zeros_like(a) for k in range(n): for i in range(n): for j in range(n): m[k,i] += a[i,j] - a[i,k] - a[k,j] + a[k,k] return m def implicit(a): n = a.shape[0] m = np.zeros_like(a) for k in range(n): for i in range(n): for j in range(n): m[k,i] += a[i,j] - a[i,k] - a[k,j] + a[k,k] return m a = np.random.randn(10,10) assert np.allclose(explicit(a), implicit(a), atol=1e-10, rtol=0.) </code></pre> <p>Then, vectorize the function step by step by editing <code>implicit</code>, running the script at each step to make sure that they continue staying the same:</p> <p>Step 1</p> <pre><code>def implicit(a): n = a.shape[0] m = np.zeros_like(a) for k in range(n): for i in range(n): m[k,i] = (a[i,:] - a[k,:]).sum() - n*a[i,k] + n*a[k,k] return m </code></pre> <p>Step 2</p> <pre><code>def implicit(a): n = a.shape[0] m = np.zeros_like(a) m = - n*a.T + n*np.diag(a)[:,newaxis] for k in range(n): for i in range(n): m[k,i] += (a[i,:] - a[k,:]).sum() return m </code></pre> <p>Step 3</p> <pre><code>def implicit(a): n = a.shape[0] m = np.zeros_like(a) m = - n*a.T + n*np.diag(a)[:,newaxis] m += (a.T[newaxis,...] - a[...,newaxis]).sum(1) return m </code></pre> <p>Et voila'! No loops in the last one. To vectorize that kind of equations, <a href="http://www.scipy.org/EricsBroadcastingDoc" rel="noreferrer">broadcasting</a> is the way to go!</p> <p>Warning: make sure that <code>explicit</code> is the equation that you wanted to vectorize. I wasn't sure if the terms that do not depend on <code>j</code> should also be summed over.</p>
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    1. USMark Dickinson
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    1. COReally nice answer, and great general advice!
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      1. USJoe Kington
    2. COThanks a lot. Great advice for the future :)
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      1. USyassin
 

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