On 08/16/2011 04:22 AM, Timo Kluck wrote:
> 2011/8/1 Timo Kluck:
>> I just submitted a patch at
>> http://projects.scipy.org/numpy/ticket/1920 . It implements Eric's
>> suggestion. Please review, I'll be happy to adapt it to any of your
>> feedback.
>>
> I submitted a minor patch a while ago. It ha
2011/8/1 Timo Kluck :
> I just submitted a patch at
> http://projects.scipy.org/numpy/ticket/1920 . It implements Eric's
> suggestion. Please review, I'll be happy to adapt it to any of your
> feedback.
>
I submitted a minor patch a while ago. It hasn't been reviewed yet,
but I don't know whether t
2011/8/1 Timo Kluck
> 2011/7/30 Eric Firing
>> Maybe the thing to do is to pre-calculate if len(xp) <= len(x), or some
>> such guess as to which method would be more efficient.
>>
> What you're suggesting is reasonable. The cutoff at len(xp) <= len(x) can
> distinguish between the 'refinement' c
2011/7/30 Eric Firing
> On 07/29/2011 11:18 AM, Timo Kluck wrote:
>
> The current implementation of numpy.interp(x,xp,fp) comes down to: first
> > calculating all the slopes of the linear interpolant (these are
> > len(xp)-1), then use a binary search to find where x is in xp (running
> > time lo
On 07/29/2011 11:18 AM, Timo Kluck wrote:
> Dear numpy developers,
>
> The current implementation of numpy.interp(x,xp,fp) comes down to: first
> calculating all the slopes of the linear interpolant (these are
> len(xp)-1), then use a binary search to find where x is in xp (running
> time log(len(x
Dear numpy developers,
The current implementation of numpy.interp(x,xp,fp) comes down to: first
calculating all the slopes of the linear interpolant (these are len(xp)-1),
then use a binary search to find where x is in xp (running time
log(len(xp)). So we obtain a running time of
O( len(xp) + len
