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copied!<p>Just add items to the dictionary without checking for their existence. I added 100,000 items to a dictionary using 3 different methods and timed it with the timeit module.</p> <ol> <li><code>if k not in d: d[k] = v</code> </li> <li><code>d.setdefault(k, v)</code> </li> <li><code>d[k] = v</code></li> </ol> <p>Option 3 was the fastest, but not by much.</p> <p>[ <em>Actually, I also tried <code>if k not in d.keys(): d[k] = v</code>, but that was slower by a factor of 300 (each iteration built a list of keys and performed a linear search). It made my tests so slow that I left it out here.</em> ]</p> <p>Here's my code:</p> <pre><code>import timeit setup = """ import random random.seed(0) item_count = 100000 # divide key range by 5 to ensure lots of duplicates items = [(random.randint(0, item_count/5), 0) for i in xrange(item_count)] """ in_dict = """ d = {} for k, v in items: if k not in d: d[k] = v """ set_default = """ d = {} for k, v in items: d.setdefault(k, v) """ straight_add = """ d = {} for k, v in items: d[k] = v """ print 'in_dict ', timeit.Timer(in_dict, setup).timeit(1000) print 'set_default ', timeit.Timer(set_default, setup).timeit(1000) print 'straight_add ', timeit.Timer(straight_add, setup).timeit(1000) </code></pre> <p>And the results:</p> <pre><code>in_dict 13.090878085 set_default 21.1309413091 straight_add 11.4781760635 </code></pre> <p><strong>Note:</strong> This is all pretty pointless. We get many questions daily about what's the fastest way to do x or y in Python. In most cases, it is clear that the question was being asked before any performance issues were encountered. My advice? Focus on writing the clearest program you can write and if it's too slow, profile it and optimize where needed. In my experience, I almost never get to to profile and optimize step. From the description of the problem, it seems as if dictionary storage will not be the major bottle-neck in your program. </p>

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