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copied!<p>Regarding <strong>efficiency</strong>: The method you show has an <a href="http://en.wikipedia.org/wiki/Big_O_notation" rel="nofollow">asymptotic runtime complexity</a> of <code>O(m*n)</code> where <code>m</code> and <code>n</code> are the number of elements in <code>reference</code> and <code>done</code>, i.e. if you double the size of both lists, the algorithm will run 4 times longer (times a fixed constant that is uniteresting to theoretical computer scientists). If <code>m</code> and <code>n</code> are very large, you will probably want to choose a faster algorithm, e.g sort the two lists first using the <code>.sort()</code> (runtime complexity: <code>O(n * log(n))</code>) and then go through the lists just once (runtime complexity: <code>O(n)</code>). That algorithm has a worst-case runtime complexity of <code>O(n * log(n))</code>, which is already a big improvement. However, you trade readability and simplicity of the code for efficiency, so I would only advise you to do this if absolutely necessary.</p> <p>Regarding <strong>coding style</strong>: You do not <code>.close()</code> the file handles which you should. Instead of opening and closing the file handle, you could use the <a href="http://docs.python.org/reference/compound_stmts.html#the-with-statement" rel="nofollow"><code>with</code> language construct</a> of python. Also, if you like the functional style, you could replace the for loop by a list expression:</p> <pre><code>for i in reference: if i not in done: print i, </code></pre> <p>then becomes:</p> <pre><code>items = [i.strip() for i in reference if i not in done] print ' '.join(items) </code></pre> <p>However, this way you will not see any progress while the list is being composed.</p> <p>As joaquin already mentions, you can loop over <code>f</code> directly instead of <code>f.readlines()</code> as file handles support the <a href="http://www.python.org/dev/peps/pep-0234/" rel="nofollow">iterator protocol</a>.</p>

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