Login | Register For Free | Help
Search for: (Advanced)

Mailing List Archive: Python: Python

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

 

 

First page Previous page 1 2 Next page Last page  View All Python python RSS feed   Index | Next | Previous | View Threaded


no.email at please

Nov 6, 2009, 4:12 AM

Post #1 of 32 (913 views)
Permalink
Most efficient way to "pre-grow" a list?

In Perl one can assign a value to any element of an array, even to
ones corresponding to indices greater or equal than the length of
the array:

my @arr;
$arr[999] = 42;

perl grows the array as needed to accommodate this assignment. In
fact one common optimization in Perl is to "pre-grow" the array to
its final size, rather than having perl grow it piecemeal as required
by assignments like the one above:

my @arr;
$#arr = 999_999;

After assigning to $#arr (the last index of @arr) as shown above,
@arr has length 1,000,000, and all its elements are initialized to
undef.

In Python the most literal translation of the first code snippet
above triggers an IndexError exception:

>>> arr = list()
>>> arr[999] = 42
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list assignment index out of range

In fact, one would need to pre-grow the list sufficiently to be
able to make an assignment like this one. I.e. one needs the
equivalent of the second Perl snippet above.

The best I can come up with is this:

arr = [None] * 1000000

Is this the most efficient way to achieve this result?

TIA!

kynn

--
http://mail.python.org/mailman/listinfo/python-list


http://phr.cx at NOSPAM

Nov 6, 2009, 4:34 AM

Post #2 of 32 (885 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

kj <no.email [at] please> writes:
> >>> arr[999] = 42
> ...
> The best I can come up with is this:
> arr = [None] * 1000000
> Is this the most efficient way to achieve this result?

If you're talking about an array of ints, use the array module.
You might also look at numpy.
--
http://mail.python.org/mailman/listinfo/python-list


joncle at googlemail

Nov 6, 2009, 6:11 AM

Post #3 of 32 (884 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Nov 6, 12:12 pm, kj <no.em...@please.post> wrote:

[snip]

> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one.  I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?

That's a good as it gets I think.

If sparsely populated I might be tempted to use a dict (or maybe
defaultdict):

d = {999: 42, 10673: 123}
for idx in xrange(1000000): # Treat it as though it's a list of
1,000,000 items...
print 'index %d has a value of %d' % (idx, d.get(idx, None))

Efficiency completely untested.

Jon.
--
http://mail.python.org/mailman/listinfo/python-list


brian.curtin at gmail

Nov 6, 2009, 7:04 AM

Post #4 of 32 (884 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Fri, Nov 6, 2009 at 06:12, kj <no.email [at] please> wrote:

>
> In Perl one can assign a value to any element of an array, even to
> ones corresponding to indices greater or equal than the length of
> the array:
>
> my @arr;
> $arr[999] = 42;
>
> perl grows the array as needed to accommodate this assignment. In
> fact one common optimization in Perl is to "pre-grow" the array to
> its final size, rather than having perl grow it piecemeal as required
> by assignments like the one above:
>
> my @arr;
> $#arr = 999_999;
>
> After assigning to $#arr (the last index of @arr) as shown above,
> @arr has length 1,000,000, and all its elements are initialized to
> undef.
>
> In Python the most literal translation of the first code snippet
> above triggers an IndexError exception:
>
> >>> arr = list()
> >>> arr[999] = 42
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> IndexError: list assignment index out of range
>
> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one. I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?
>
> TIA!
>
> kynn
>

Is there a reason you need to pre-allocate the list other than the fact that
you do it that way in Perl?


andreengels at gmail

Nov 6, 2009, 7:12 AM

Post #5 of 32 (885 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Fri, Nov 6, 2009 at 1:12 PM, kj <no.email [at] please> wrote:
>
> In Perl one can assign a value to any element of an array, even to
> ones corresponding to indices greater or equal than the length of
> the array:
>
>  my @arr;
>  $arr[999] = 42;
>
> perl grows the array as needed to accommodate this assignment.  In
> fact one common optimization in Perl is to "pre-grow" the array to
> its final size, rather than having perl grow it piecemeal as required
> by assignments like the one above:
>
>  my @arr;
>  $#arr = 999_999;
>
> After assigning to $#arr (the last index of @arr) as shown above,
> @arr has length 1,000,000, and all its elements are initialized to
> undef.
>
> In Python the most literal translation of the first code snippet
> above triggers an IndexError exception:
>
>>>> arr = list()
>>>> arr[999] = 42
> Traceback (most recent call last):
>  File "<stdin>", line 1, in <module>
> IndexError: list assignment index out of range
>
> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one.  I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?

It depends - what do you want to do with it? My first hunch would be
to use a dictionary instead of a list, then the whole problem
disappears. If there is a reason you don't want to do that, what is
it?


--
André Engels, andreengels [at] gmail
--
http://mail.python.org/mailman/listinfo/python-list


python at rcn

Nov 6, 2009, 9:39 AM

Post #6 of 32 (883 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

[kj]
> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one.  I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?

Yes.


Raymond

--
http://mail.python.org/mailman/listinfo/python-list


emily at nospam

Nov 6, 2009, 10:11 AM

Post #7 of 32 (878 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

"Andre Engels" <andreengels [at] gmail> wrote in message
news:mailman.2696.1257520404.2807.python-list [at] python
>On Fri, Nov 6, 2009 at 1:12 PM, kj <no.email [at] please> wrote:
[snip]
>> arr = [None] * 1000000
>>
>> Is this the most efficient way to achieve this result?
>
> It depends - what do you want to do with it? My first hunch would be
> to use a dictionary instead of a list, then the whole problem
> disappears. If there is a reason you don't want to do that, what is
> it?
>
> --
> André Engels, andreengels [at] gmail

I second this. It might seem a sensible thing to do in perl, but I can't
imagine what you would actually want to do it for! Seems like an odd thing
to want to do!


no.email at please

Nov 6, 2009, 1:03 PM

Post #8 of 32 (875 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

In <hd1os1$aqs$1 [at] cam-news1> "Emily Rodgers" <emily [at] nospam> writes:


>"Andre Engels" <andreengels [at] gmail> wrote in message
>news:mailman.2696.1257520404.2807.python-list [at] python
>>On Fri, Nov 6, 2009 at 1:12 PM, kj <no.email [at] please> wrote:
>[snip]
>>> arr = [None] * 1000000
>>>
>>> Is this the most efficient way to achieve this result?
>>
>> It depends - what do you want to do with it? My first hunch would be
>> to use a dictionary instead of a list, then the whole problem
>> disappears. If there is a reason you don't want to do that, what is
>> it?

>I second this. It might seem a sensible thing to do in perl, but I can't
>imagine what you would actually want to do it for! Seems like an odd thing
>to want to do!

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. (Of course, like with all optimizations,
whether it's worth the bother is another question.)

Another situation where one may want to do this is if one needs to
initialize a non-sparse array in a non-sequential order, e.g. if
that's the way the data is initially received by the code. Of
course, there are many ways to skin such a cat; pre-allocating the
space and using direct list indexing is just one of them. I happen
to think it is a particularly straighforward one, but I realize that
others (you, Andre, etc.) may not agree.

kynn
--
http://mail.python.org/mailman/listinfo/python-list


gil_johnson at earthlink

Nov 6, 2009, 6:46 PM

Post #9 of 32 (867 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Nov 6, 6:12 am, kj <no.em...@please.post> wrote:
> In Perl one can assign a value to any element of an array, even to
> ones corresponding to indices greater or equal than the length of
> the array:
>
>   my @arr;
>   $arr[999] = 42;
>
> perl grows the array as needed to accommodate this assignment.  In
> fact one common optimization in Perl is to "pre-grow" the array to
> its final size, rather than having perl grow it piecemeal as required
> by assignments like the one above:
>
>   my @arr;
>   $#arr = 999_999;
>
> After assigning to $#arr (the last index of @arr) as shown above,
> @arr has length 1,000,000, and all its elements are initialized to
> undef.
>
> In Python the most literal translation of the first code snippet
> above triggers an IndexError exception:
>
> >>> arr = list()
> >>> arr[999] = 42
>
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
> IndexError: list assignment index out of range
>
> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one.  I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?
>
> TIA!
>
> kynn

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
--
http://mail.python.org/mailman/listinfo/python-list


rt8396 at gmail

Nov 6, 2009, 7:06 PM

Post #10 of 32 (870 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Nov 6, 6:12 am, kj <no.em...@please.post> wrote:
> In Perl one can assign a value to any element of an array, even to
> ones corresponding to indices greater or equal than the length of
> the array:
>
>   my @arr;
>   $arr[999] = 42;
>
> perl grows the array as needed to accommodate this assignment.  In
> fact one common optimization in Perl is to "pre-grow" the array to
> its final size, rather than having perl grow it piecemeal as required
> by assignments like the one above:
>
>   my @arr;
>   $#arr = 999_999;
>
> After assigning to $#arr (the last index of @arr) as shown above,
> @arr has length 1,000,000, and all its elements are initialized to
> undef.
>
> In Python the most literal translation of the first code snippet
> above triggers an IndexError exception:
>
> >>> arr = list()
> >>> arr[999] = 42
>
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
> IndexError: list assignment index out of range
>
> In fact, one would need to pre-grow the list sufficiently to be
> able to make an assignment like this one.  I.e. one needs the
> equivalent of the second Perl snippet above.
>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> Is this the most efficient way to achieve this result?
>
> TIA!
>
> kynn

You mean sum'in like dis?

class PerlishList(list):
'''Hand holding list object for even the most demanding Perl
hacker'''
def __init__(self, dim=0):
list.__init__(self)
if dim:
self.__setitem__(dim, None)

def __setitem__(self, idx, v):
lenkeys = len(self)
sup = super(PerlishList, self)
if idx > lenkeys:
for idx in range(lenkeys, idx):
sup.append(None)
sup.__setitem__(idx, v)

def __getitem__(self, idx):
return self[idx]

l = PerlishList(3)
l.append('a')
l.append('b')
print l
l[10] = 10
print l

;-)
--
http://mail.python.org/mailman/listinfo/python-list


steve at REMOVE-THIS-cybersource

Nov 7, 2009, 7:07 AM

Post #11 of 32 (857 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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 the desired result. Gil

Why is it better to grow the list piecemeal instead of just allocating a
list the size you want in one go?

arr = [x]*size_wanted



--
Steven
--
http://mail.python.org/mailman/listinfo/python-list


bdesth.quelquechose at free

Nov 7, 2009, 9:55 AM

Post #12 of 32 (857 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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 course, like with all optimizations,
> whether it's worth the bother is another question.)

My very humble opinion is that unless you spot a bottleneck (that is,
you have real performance issues AND the profiler identified list growth
as the culprit), the answer is a clear and obvious NO.

> 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.

--
http://mail.python.org/mailman/listinfo/python-list


kyrie at uh

Nov 7, 2009, 11:25 AM

Post #13 of 32 (858 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

Quoting Bruno Desthuilliers <bdesth.quelquechose [at] free>:

> > 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 Computación, UH
http://profesores.matcom.uh.cu/~kyrie

--
Participe en Universidad 2010, del 8 al 12 de febrero de 2010
La Habana, Cuba
http://www.universidad2010.cu

--
http://mail.python.org/mailman/listinfo/python-list


andreengels at gmail

Nov 7, 2009, 11:33 AM

Post #14 of 32 (857 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
<kyrie [at] uh> wrote:
>
> Quoting Bruno Desthuilliers <bdesth.quelquechose [at] free>:
>
>> > 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.

Let d be your dict.

Call the zeroeth place in your array d[0], the first d[1], the 10000th
d[100000].


--
André Engels, andreengels [at] gmail
--
http://mail.python.org/mailman/listinfo/python-list


kyrie at uh

Nov 7, 2009, 12:40 PM

Post #15 of 32 (856 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

Quoting Andre Engels <andreengels [at] gmail>:

> On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
> <kyrie [at] uh> 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.

--
Luis Zarrabeitia
Facultad de Matemática y Computación, UH
http://profesores.matcom.uh.cu/~kyrie

--
Participe en Universidad 2010, del 8 al 12 de febrero de 2010
La Habana, Cuba
http://www.universidad2010.cu

--
http://mail.python.org/mailman/listinfo/python-list


tjreedy at udel

Nov 7, 2009, 2:19 PM

Post #16 of 32 (855 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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
>> powers of 2 to get the desired result. Gil
>
> Why is it better to grow the list piecemeal instead of just allocating a
> list the size you want in one go?

It isn't.

> arr = [x]*size_wanted

Is what I would do.

--
http://mail.python.org/mailman/listinfo/python-list


ivan.illarionov at gmail

Nov 7, 2009, 2:24 PM

Post #17 of 32 (855 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Nov 6, 3:12 pm, kj <no.em...@please.post> wrote:
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> 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 plain evil:
>>> import ctypes
>>> ctypes.pythonapi.PyList_New.restype = ctypes.py_object
>>> ctypes.pythonapi.PyList_New(100)


-- Ivan
--
http://mail.python.org/mailman/listinfo/python-list


no.email at please

Nov 7, 2009, 4:23 PM

Post #18 of 32 (853 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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


>Quoting Andre Engels <andreengels [at] gmail>:

>> On Sat, Nov 7, 2009 at 8:25 PM, Luis Alberto Zarrabeitia Gomez
>> <kyrie [at] uh> 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


sturlamolden at yahoo

Nov 7, 2009, 5:03 PM

Post #19 of 32 (853 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On 6 Nov, 13:12, kj <no.em...@please.post> wrote:

>
> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> 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 preallocated because resize has complexity O(n), instead of
amortized O(1). You don't need to worry about that in Python. Python
lists are resized with empty slots at the end, in proportion to the
size of the list. On average, this has the same complexity as pre-
allocation.












--
http://mail.python.org/mailman/listinfo/python-list


sturlamolden at yahoo

Nov 7, 2009, 5:05 PM

Post #20 of 32 (853 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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 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.









--
http://mail.python.org/mailman/listinfo/python-list


sturlamolden at yahoo

Nov 7, 2009, 5:13 PM

Post #21 of 32 (854 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On 6 Nov, 22:03, kj <no.em...@please.post> 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 as a list grows. That is
why it's called a 'list' and not an array.

There will be empty slots at the end of a list you can append to,
without reallocating. When the list is resized, it is reallocated with
even more of these. Thus the need to reallocate becomes more and more
rare as the list grows, and on average the complexity of appending is
just O(1).




--
http://mail.python.org/mailman/listinfo/python-list


pavlovevidence at gmail

Nov 7, 2009, 5:18 PM

Post #22 of 32 (853 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

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 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
--
http://mail.python.org/mailman/listinfo/python-list


exarkun at twistedmatrix

Nov 7, 2009, 5:32 PM

Post #23 of 32 (852 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On 01:18 am, pavlovevidence [at] gmail 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:
>> > >>>      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 http://en.wikipedia.org/wiki/Amortized_analysis
conveniently uses exactly this example to explain the concept of
amortized costs.

Jean-Paul


paul.nospam at rudin

Nov 7, 2009, 11:07 PM

Post #24 of 32 (848 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

kj <no.email [at] please> writes:

> The best I can come up with is this:
>
> arr = [None] * 1000000
>
> 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 numpy array.

--
http://mail.python.org/mailman/listinfo/python-list


gil_johnson at earthlink

Nov 8, 2009, 4:45 AM

Post #25 of 32 (845 views)
Permalink
Re: Most efficient way to "pre-grow" a list? [In reply to]

On Nov 6, 8:46 pm, gil_johnson <gil_john...@earthlink.net> 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 posted such a cryptic note before, I should have waited
until I
had the code available.

I am still new to Python, and I couldn't find a way to create an array
of
size N, with each member initialized to a given value. If there is
one,
please let me know.

I think Paul Rubin and Paul Rudin are right, if they really are 2
different
people, an array is a better solution if you're dealing with integers.
They
both mention numpy, which I know nothing about, or the array module.

kj, what *are* you going to do with this list/array?
As others have pointed out, there are differences between lists,
arrays, and
dictionaries.

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 4294967295. Of course,
you
could set your initializer to 0, or any other 32-bit number.

Originally I found that the doubling method I wrote about before was a
LOT
faster than appending the elements one at a time, and tonight I tried
the
"list = initializer * N" method. Running the code below, the doubling
method is still fastest, at least on my system.

Of course, as long as you avoid the 'one at a time' method, we're
talking
about fractions of a second, even for arrays that I think are huge,
like
the 536,870,912 byte beastie below.

[code]

# Written in Python 3.x

import array
import time

#* * * * * * * * * * * * * * * * * * * * * *
# Doubling method, run time = 0.413938045502

t0 = time.time()

newArray = array.array('I') # 32-bit unsigned
integers

newArray.append(4294967295)

for i in range(27): # 2**27 integers, 2**29
bytes
newArray.extend(newArray)

print(time.time() - t0)

print(newArray[134217727]) # confirm array is fully
initialized

#* * * * * * * * * * * * * * * * * * * * * *
# One at a time, run time = 28.5479729176

t0 = time.time()

newArray2 = array.array('I')

for i in range(134217728): # the same size as
above
newArray2.append(4294967295)

print(time.time() - t0)

print(newArray2[134217727]) # confirm array

#* * * * * * * * * * * * * * * * * * * * * *
# List with "*", run time = 1.06160402298

t0 = time.time()

newList = [4294967295] * 134217728

print(time.time() - t0)

print(newList[134217727]) # confirm list

[/code]

If, instead of 134,217,728 integers, I want something different, like
100,000,000, the method I use is:

[code]

#* * * * * * * * * * * * * * * * * * * * * *
# Not a power of 2, run time = 0.752086162567

t0 = time.time()

newArray = array.array('I')

tempArray = array.array('I')

tempArray.append(4294967295)

size = 100000000

while size: # chew through
'size' until it's gone
if (size & 1): # if this bit of
'size' is 1
newArray.extend(tempArray) # add a copy of the temp array
size >>= 1 # chew off one bit
tempArray.extend(tempArray) # double the size of the temp
array

print(time.time() - t0)

print(newArray[99999999])

#* * * * * * * * * * * * * * * * * * * * * *
# # Not a power of 2, list with "*", run time = 1.19271993637

t0 = time.time()

newList = [4294967295] * 100000000

print(time.time() - t0)

print(newList[99999999])

[/code]

I think it is interesting that the shorter list takes longer than the
one
that is a power of 2 in length. I think this may say that the "list =
initializer * N" method uses something similar to the doubling method.

Also, tempArray (above) gets reallocated frequently, and demonstrates
that reallocation is not a big problem.

Finally, I just looked into calling C functions, and found
PyMem_Malloc,
PyMem_Realloc, PyMem_Free, etc. in the Memory Management section of
the
Python/C API Reference Manual. This gives you uninitialized memory,
and
should be really fast, but it's 6:45 AM here, and I don't have the
energy
to try it.

Gil
--
http://mail.python.org/mailman/listinfo/python-list

First page Previous page 1 2 Next page Last page  View All Python python RSS feed   Index | Next | Previous | View Threaded
 
 


Interested in having your list archived? Contact Gossamer Threads
 
  Web Applications & Managed Hosting Powered by Gossamer Threads Inc.