Most efficient way to "pre-grow" a list?

exarkun at exarkun at
Sun Nov 8 02:32:30 CET 2009

On 01:18 am, pavlovevidence at wrote:
>On Nov 7, 5:05 pm, sturlamolden <sturlamol... at> wrote:
>>On 7 Nov, 03:46, gil_johnson <gil_john... at> wrote:>
>> > I don't have the code with me, but for huge arrays, I have used
>> > something like:
>> > >>> arr[0] = initializer
>> > >>> for i in range N:
>> > >>>      arr.extend(arr)
>> > This doubles the array every time through the loop, and you can add
>> > the powers of 2 to get the desired result.
>> > Gil
>>You should really use append instead of extend. The above code is O
>>(N**2), with append it becomes O(N) on average.
>I think the top one is O(N log N), and I'm suspicious that it's even
>possible to grow a list in less than O(N log N) time without knowing
>the final size in advance.  Citation?  Futhermore why would it matter
>to use extend instead of append; I'd assume extend uses the same
>growth algorithm.  (Although in this case since the extend doubles the
>size of the list it most likely reallocates after each call.)
>[None]*N is linear time and is better than growing the list using
>append or extend.

The wikipedia page for 
conveniently uses exactly this example to explain the concept of 
amortized costs.


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