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

kj no.email at please.post
Sun Nov 8 01:23:18 CET 2009


In <mailman.44.1257626420.2873.python-list at python.org> Luis Alberto Zarrabeitia Gomez <kyrie at uh.cu> writes:


>Quoting Andre Engels <andreengels at gmail.com>:

>> On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
>> <kyrie at uh.cu> wrote:
>> >
>> > Ok, he has a dict.
>> >
>> > Now what? He needs a non-sparse array.
>> 
>> Let d be your dict.
>> 
>> Call the zeroeth place in your array d[0], the first d[1], the 10000th
>> d[100000].

>Following that reasoning, we could get rid of lists and arrays altogether.

>Here's why that wouldn't work:

>for x,y in zip(d,other):
>    ... do something ...

>Yes, we could also ignore zip and just use range/xrange to iterate for the
>indices... 

>Lists and dictionaries have different semantics. One thing is to argue that you
>shouldn't be thinking on pre-growing a list for performance reasons before being
>sure that it is a bottleneck, and a very different one is to argue that because
>one operation (__setitem__) is the same with both structures, we should not use
>lists for what may need, depending on the problem, list semantics. 

>¿Have you ever tried to read list/matrix that you know it is not sparse, but you
>don't know the size, and it may not be in order? A "grow-able" array would just
>be the right thing to use - currently I have to settle with either hacking
>together my own grow-able array, or preloading the data into a dict, growing a
>list with the [0]*size trick, and updating that list. Not hard, not worthy of a
>PEP, but certainly not so easy to dismiss.

Thanks.  Well said.

Saludos,

kynn



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