On Wed, 11 Nov 2009 08:13:54 -0800, Francis Carr wrote:
> Hmmm. I am trying some algorithms, and timeit reports that a
> list.extend variant is fastest... WTH?! Really this seems like it must
> be a "bug" in implementing the "[None]*x" idiom.
> The straightforward "pythonic" solution is better
Hmmm. I am trying some algorithms, and timeit reports that a
list.extend variant is fastest... WTH?! Really this seems like it
must be a "bug" in implementing the "[None]*x" idiom.
As expected, appending one-at-a-time is slowest (by an order of
magnitude):
% python -m timeit -s "N=100" \
On Nov 9, 10:56 pm, "Gabriel Genellina" wrote:
>
> [much cutting]
>
> def method3a():
> newArray = array.array('I', [INITIAL_VALUE]) * SIZE
> assert len(newArray)==SIZE
> assert newArray[SIZE-1]==INITIAL_VALUE
>
> [more cutting]
>
> So arrays are faster than lists, and in both cases one_it
En Sun, 08 Nov 2009 10:08:35 -0300, gil_johnson
escribió:
On Nov 6, 8:46 pm, gil_johnson wrote:
The problem I was solving was this: I wanted an array of 32-bit
integers to be used as a bit array, and I wanted it initialized with
all bits set, that is, each member of the array had to be set to
On 11/7/2009 5:18 PM Carl Banks said...
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?
Quoting Tim --
Python uses mildly exponential over-allocation, sufficient so
In Luis Alberto Zarrabeitia
Gomez writes:
>¿Have you ever tried to read list/matrix that you know it is not sparse, but
>you
jdon'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
On Nov 6, 8:46 pm, gil_johnson 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 ge
On Nov 6, 8:46 pm, gil_johnson wrote:
> >>> 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
To all who asked about my previous post,
I'm sorry I pos
kj writes:
> The best I can come up with is this:
>
> arr = [None] * 100
>
> Is this the most efficient way to achieve this result?
>
Depends on what you take as given. You can do it with ctypes more
efficiently, but you can also shoot yourself in the foot.
Another possibility is to use a n
On 01:18 am, pavlovevide...@gmail.com wrote:
On Nov 7, 5:05�pm, sturlamolden wrote:
On 7 Nov, 03:46, gil_johnson 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 d
On Nov 7, 5:05 pm, sturlamolden wrote:
> On 7 Nov, 03:46, gil_johnson 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 thr
On 6 Nov, 22:03, kj wrote:
> As I said, this is considered an optimization, at least in Perl,
> because it lets the interpreter allocate all the required memory
> in one fell swoop, instead of having to reallocate it repeatedly
> as the array grows.
Python does not need to reallocate repeatedly
On 7 Nov, 03:46, gil_johnson 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
On 6 Nov, 13:12, kj wrote:
>
> The best I can come up with is this:
>
> arr = [None] * 100
>
> Is this the most efficient way to achieve this result?
Yes, but why would you want to? Appending to a Python list has
amortized O(1) complexity. I am not sure about Perl, but in MATLAB
arrays are p
In Luis Alberto Zarrabeitia
Gomez writes:
>Quoting Andre Engels :
>> On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
>> 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
On Nov 6, 3:12 pm, kj wrote:
> The best I can come up with is this:
>
> arr = [None] * 100
>
> Is this the most efficient way to achieve this result?
It is the most efficient SAFE way to achieve this result.
In fact, there IS the more efficient way, but it's dangerous, unsafe,
unpythonic and
Steven D'Aprano wrote:
On Fri, 06 Nov 2009 18:46:33 -0800, gil_johnson 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
po
Quoting Andre Engels :
> On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
> 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 1th
> d[10].
Following
On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
wrote:
>
> Quoting Bruno Desthuilliers :
>
>> > Another situation where one may want to do this is if one needs to
>> > initialize a non-sparse array in a non-sequential order,
>>
>> Then use a dict.
>
> Ok, he has a dict.
>
> Now what
Quoting Bruno Desthuilliers :
> > Another situation where one may want to do this is if one needs to
> > initialize a non-sparse array in a non-sequential order,
>
> Then use a dict.
Ok, he has a dict.
Now what? He needs a non-sparse array.
--
Luis Zarrabeitia
Facultad de Matemática y Comput
kj a écrit :
>
> As I said, this is considered an optimization, at least in Perl,
> because it lets the interpreter allocate all the required memory
> in one fell swoop, instead of having to reallocate it repeatedly
> as the array grows.
IIRC, CPython has it's own way to optimize list growth.
>
On Fri, 06 Nov 2009 18:46:33 -0800, gil_johnson 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
> po
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