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

Carl Banks pavlovevidence at gmail.com
Sun Nov 8 02:18:38 CET 2009


On Nov 7, 5:05 pm, sturlamolden <sturlamol... at yahoo.no> wrote:
> On 7 Nov, 03:46, gil_johnson <gil_john... at earthlink.net> 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.


Carl Banks



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