On Wed, Dec 24, 2014 at 10:29 AM, wrote:
>
>
> On Wed, Dec 24, 2014 at 10:30 AM, Julian Taylor <
> jtaylor.deb...@googlemail.com> wrote:
>
>> On 24.12.2014 16:25, Neal Becker wrote:
>> > What would be the most efficient way to compute:
>> >
>> > c[j] = \sum_i (a[i] * b[i,j])
>> >
>> > where a[i]
On Wed, Dec 24, 2014 at 10:30 AM, Julian Taylor <
jtaylor.deb...@googlemail.com> wrote:
> On 24.12.2014 16:25, Neal Becker wrote:
> > What would be the most efficient way to compute:
> >
> > c[j] = \sum_i (a[i] * b[i,j])
> >
> > where a[i] is a 1-d vector, b[i,j] is a 2-d array?
> >
> > This seems
At 06:47 PM 12/23/2014, you wrote:
The performance of fftpack depends very strongly on the array size
-- sizes that are powers of two are good, but also powers of three,
five and seven, or numbers whose only prime factors are from
(2,3,5,7). For problems that can use padding, rounding up the si
On Wed, Dec 24, 2014 at 7:56 AM, Sturla Molden
wrote:
> On 24/12/14 14:34, Sturla Molden wrote:
> > I would rather have SciPy implement this with the overlap-and-add method
> > rather than padding the FFT. Overlap-and-add is more memory efficient
> > for large n:
>
> (eh, the list should be)
>
>
Nathaniel Smith wrote:
> On Wed, Dec 24, 2014 at 3:25 PM, Neal Becker wrote:
>> What would be the most efficient way to compute:
>>
>> c[j] = \sum_i (a[i] * b[i,j])
>>
>> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
>
> I think this formula is just np.dot(a, b). Did you mean c = \sum_j
> \
On Wed, Dec 24, 2014 at 3:25 PM, Neal Becker wrote:
> What would be the most efficient way to compute:
>
> c[j] = \sum_i (a[i] * b[i,j])
>
> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
I think this formula is just np.dot(a, b). Did you mean c = \sum_j
\sum_i (a[i] * b[i, j])?
> This seems
On 24.12.2014 16:25, Neal Becker wrote:
> What would be the most efficient way to compute:
>
> c[j] = \sum_i (a[i] * b[i,j])
>
> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
>
> This seems to be one way:
>
> import numpy as np
> a = np.arange (3)
> b = np.arange (12).reshape (3,4)
> c = n
What would be the most efficient way to compute:
c[j] = \sum_i (a[i] * b[i,j])
where a[i] is a 1-d vector, b[i,j] is a 2-d array?
This seems to be one way:
import numpy as np
a = np.arange (3)
b = np.arange (12).reshape (3,4)
c = np.dot (a, b).sum()
but np.dot returns a vector, which then need
On 24/12/14 14:34, Sturla Molden wrote:
> I would rather have SciPy implement this with the overlap-and-add method
> rather than padding the FFT. Overlap-and-add is more memory efficient
> for large n:
(eh, the list should be)
- Overlap-and-add is more memory efficient for large n.
- It scales
On 24/12/14 04:33, Robert McGibbon wrote:
> Alex Griffing pointed out on github that this feature was recently added
> to scipy in https://github.com/scipy/scipy/pull/3144. Sweet!
I would rather have SciPy implement this with the overlap-and-add method
rather than padding the FFT. Overlap-and-ad
On 24/12/14 13:23, Julian Taylor wrote:
> hm this is a brute force search, probably fast enough but slower than
> scipy's code (if it also were cython)
That was what I tought as well when I wrote it. But it turned out that
regular numbers are so close and abundant that was damn fast, even in
Py
On 24.12.2014 13:07, Sturla Molden wrote:
> On 24/12/14 04:33, Robert McGibbon wrote:
>> Alex Griffing pointed out on github that this feature was recently added
>> to scipy in https://github.com/scipy/scipy/pull/3144. Sweet!
>
>
> I use different padsize search than the one in SciPy. It would be
On 24/12/14 13:07, Sturla Molden wrote:
> v
>
> cdef intp_t checksize(intp_t n):
> while not (n % 5): n /= 5
> while not (n % 3): n /= 3
> while not (n % 2): n /= 2
> return (1 if n == 1 else 0)
>
> def _next_regular(target):
> cdef intp_t n = target
> while not
On 24/12/14 04:33, Robert McGibbon wrote:
> Alex Griffing pointed out on github that this feature was recently added
> to scipy in https://github.com/scipy/scipy/pull/3144. Sweet!
I use different padsize search than the one in SciPy. It would be
interesting to see which is faster.
from numpy c
I still have the plan to add this function as public api to numpy's fft
helper functions, though I didn't get to it yet.
Its a relative simple task if someone wants to contribute.
On 24.12.2014 04:33, Robert McGibbon wrote:
> Alex Griffing pointed out on github that this feature was recently added
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