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 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 yet:
%
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_item*N
En Sun, 08 Nov 2009 10:08:35 -0300, gil_johnson
gil_john...@earthlink.net escribió:
On Nov 6, 8:46 pm, gil_johnson gil_john...@earthlink.net 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,
On Nov 6, 8:46 pm, gil_johnson gil_john...@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
In mailman.44.1257626420.2873.python-l...@python.org Luis Alberto Zarrabeitia
Gomez ky...@uh.cu 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 -
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
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
powers of 2 to get
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.
(Of
Quoting Bruno Desthuilliers bdesth.quelquech...@free.quelquepart.fr:
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
On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
ky...@uh.cu wrote:
Quoting Bruno Desthuilliers bdesth.quelquech...@free.quelquepart.fr:
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
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
On Nov 6, 3:12 pm, kj no.em...@please.post 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,
On Nov 7, 5:05 pm, sturlamolden sturlamol...@yahoo.no wrote:
On 7 Nov, 03:46, gil_johnson gil_john...@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
On 01:18 am, pavlovevide...@gmail.com wrote:
On Nov 7, 5:05�pm, sturlamolden sturlamol...@yahoo.no wrote:
On 7 Nov, 03:46, gil_johnson gil_john...@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:
kj no.em...@please.post 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
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