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copied!<p>Note that your results will depend on the precise Python implementation. cpython (and pypy) automatically resize your list and <em>overprovision</em> space for future appends and thereby speed up the <code>append</code> furthermore.</p> <p>Internally, lists are just chunks of memory with a constant size (on the <em>heap</em>). Sometimes you're lucky and can just increase the size of the chunk, but in many cases, an object will already be there. For example, assume you allocated a chunk of size 4 for a list <code>[a,b,c,d]</code>, and some other piece of code allocated a chunk of size 6 for a dictionary:</p> <pre><code>Memory 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |a b c d| | dictionary | </code></pre> <p>Assume your list has 4 elements, and another one is added. Now, you can simply resize the list to size 5:</p> <pre><code>Memory 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |a b c d e| dictionary | </code></pre> <p>However, what do you do if you need another element now?</p> <p>Well, the only thing you can do is acquire a new space and copy the contents of the list.</p> <pre><code>Memory 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | dictionary |a b c d e f | </code></pre> <p>Note that if you acquire space in bulk (the aforementioned overprovisioning), you'll only need to resize (and potentially copy) the list every now and then.</p> <p>In contrast, when you insert at position 0, you always need to copy your list. Let's insert <code>x</code>:</p> <pre><code>Memory 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 orig |a b c d| |dictionary| after |x a b c d|dictionary| </code></pre> <p>Although there was enough space to append x at the end, we had to move (not even copy, which may be less expensive in memory) all the other values.</p>

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