Am 03.05.2012 15:45, schrieb Robert Kern:
> On Thu, May 3, 2012 at 2:24 PM, Robert Elsner wrote:
>> Hello Everybody,
>>
>> is there any news on the status of np.bincount with respect to "big"
>> numbers? It seems I have just been bitten by #225. Is there an
Hello Everybody,
is there any news on the status of np.bincount with respect to "big"
numbers? It seems I have just been bitten by #225. Is there an efficient
way around? I found the np.histogram function painfully slow.
Below a simple script, that demonstrates bincount failing with a memory
erro
Hey Everybody,
I noticed that the c-api docs (2.0.dev-72ab385) lack a clear statement
what the preferred entry point into the c-api is (from a users point of
view). Normally I would expect a sentence or two stating that the api
entry point is arrayobject.h (or whatever).
Instead the docs ponder ab
5:10 PM, Robert Elsner wrote:
>
>> Boiled it down a bit more to only include code that actually takes time.
>> First time around I found the other variant more instructive because it
>> shows the discrepancy between the DCT and the loop but might be
>> confusing. Thus h
Yes I did. Slicing and Cython do not mix too well. Using an explicit
loop fixes the problem. In case anybody is interested the code is attached.
Thanks for your help
Robert
On 25.07.2011 12:30, Robert Elsner wrote:
> Thanks for the hint. I thought about Cython myself but I was unable to
&g
> On Sun, 24 Jul 2011 18:10:14 -0500, Robert Elsner
> wrote:
>
>> Boiled it down a bit more to only include code that actually takes time.
>> First time around I found the other variant more instructive because it
>> shows the discrepancy between the DCT and the loop b
first derivative from the coefficients of the
Chebyshev polynomials.
Cheers
Robert
On 25.07.2011 00:43, Robert Elsner wrote:
> Hey Everybody,
>
> I am approximating the derivative of nonperiodic functions on [-1,1]
> using Chebyshev polynomials. The implementation is straightforward
Hey Everybody,
I am approximating the derivative of nonperiodic functions on [-1,1]
using Chebyshev polynomials. The implementation is straightforward and
works well but is painfully slow. The routine wastes most of its time on
a trivial operation (see comment in the code)
Unfortunately the spectr
solution you pointed out. Thanks
On 29.06.2011 16:38, Skipper Seabold wrote:
> On Wed, Jun 29, 2011 at 10:32 AM, Robert Elsner wrote:
>> Hello everyone,
>>
>> I would like to solve the following problem (preferably without
>> reshaping / flipping the array a).
>>
>&g
Oh and I forgot to mention: I want to specify the axis so that it is
possible to multiply x along an arbitrary axis of a (given that the
lengths match).
On 29.06.2011 16:32, Robert Elsner wrote:
> Hello everyone,
>
> I would like to solve the following problem (preferably without
&g
your help
Robert Elsner
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I suspect the author meant that instead (or a simple minus in front of
the delta). Just posting because I wondered the same this morning after
looking at it (after the first misunderstanding).
It looks much better to me than the cumsum approach with its hidden test
for true using .astype(numpy.int)
Well happens. The problem description was not 100% clear thus I still
think your line did solve the problem. A simple misunderstanding. So
what do I learn from it?: Always look at the code, not the
description :D
Am Mittwoch, den 09.06.2010, 10:19 +0200 schrieb Francesc Alted:
> A Wednesday 09 Jun
> Given a certain value delta, I would like to get a subset of x, named
> y,
> where (y[i+1] - y[i]) >= delta
So in fact the problem is to find y such that
(y[i(k)+n] - y[i(k)]) >= delta
for n <= len(x) - 1 - i
and i(0) = 0, i(k+1) = i(k) + n
? Well a loop or list comparison seems like a good
Hah beat you to it one minute ;)
Am Mittwoch, den 09.06.2010, 10:08 +0200 schrieb Francesc Alted:
> A Wednesday 09 June 2010 10:00:50 V. Armando Solé escrigué:
> > Well, this seems to be quite close to what I need
> >
> > y = numpy.cumsum((x[1:]-x[:-1])/delta).astype(numpy.int)
> > i1 = numpy.non
There might be an easier way to accomplish that
y = x[(x[1:]-x[:-1]) >= delta]
cheers
Am Mittwoch, den 09.06.2010, 10:00 +0200 schrieb "V. Armando Solé":
> Well, this seems to be quite close to what I need
>
> y = numpy.cumsum((x[1:]-x[:-1])/delta).astype(numpy.int)
> i1 = numpy.nonzero(y[1:]
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