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    <blockquote> <p>If I know the number of keys and exactly what are those keys, is there any way in python to make a dict (or a hashtable) work more efficiently? I vaguely remember that if you know the keys, you can design the hash function smartly (perfect hash?) and allocate the space beforehand.</p> </blockquote> <p>Python doesn't expose a pre-sizing option to speed-up the "growth phase" of a dictionary, nor does it provide any direct controls over "placement" in the dictionary.</p> <p>That said, if the keys are always known in advance, you can store them in a <a href="http://docs.python.org/2.7/library/functions.html#func-set" rel="noreferrer"><em>set</em></a> and build your dictionaries from the set using <a href="http://docs.python.org/2.7/library/stdtypes.html#dict.fromkeys" rel="noreferrer"><em>dict.fromkeys()</em></a>. That classmethod is <a href="http://hg.python.org/cpython/file/09811ecd5df1/Objects/dictobject.c#l1378" rel="noreferrer">optimized to pre-size the dictionary based on the set size</a> and it can populate the dictionary without any new calls to __hash__():</p> <pre><code>&gt;&gt;&gt; keys = {'red', 'green', 'blue', 'yellow', 'orange', 'pink', 'black'} &gt;&gt;&gt; d = dict.fromkeys(keys) # dict is pre-sized to 32 empty slots </code></pre> <p>If reducing collisions is your goal, you can run experiments on the insertion order in the dictionary to minimize pile-ups. (Take a look at <a href="http://www.drdobbs.com/cpp/tables-within-tables-c-and-brents-techni/199900579" rel="noreferrer">Brent's variation on Algorithm D</a> in Knuth's TAOCP to get an idea of how this is done). </p> <p>By instrumenting a pure Python model for dictionaries (such as <a href="http://code.activestate.com/recipes/578375-proof-of-concept-for-a-more-space-efficient-faster/" rel="noreferrer">this one</a>), it is possible to count the weighted-average number of probes for an alternative insertion order. For example, inserting <code>dict.fromkeys([11100, 22200, 44400, 33300])</code> averages 1.75 probes per lookup. That beats the 2.25 average probes per lookup for <code>dict.fromkeys([33300, 22200, 11100, 44400])</code>.</p> <p>Another "trick" is to increase spareness in a fully populated dictionary by fooling it into <a href="http://hg.python.org/cpython/file/09811ecd5df1/Objects/dictobject.c#l1548" rel="noreferrer">increasing its size without adding new key</a>s:</p> <pre><code> d = dict.fromkeys(['red', 'green', 'blue', 'yellow', 'orange']) d.update(dict(d)) # This makes room for additional keys # and makes the set collision-free. </code></pre> <p>Lastly, you can introduce your own custom __hash__() for your keys with the goal of eliminating all collisions (perhaps using a perfect hash generator such as <a href="http://www.gnu.org/software/gperf/" rel="noreferrer"><em>gperf</em></a>).</p>
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