josef.p...@gmail.com wrote:
> Is it really possible to get the same as np.sum(a*a, axis) with
> tensordot if a.ndim=2 ?
> Any way I try the "something_else", I get extra terms as in np.dot(a.T, a)
Just to answer this question, np.dot(a,a) is equivalent to
np.tensordot(a,a, axis=(0,0))
but the l
2009/10/20 :
> On Tue, Oct 20, 2009 at 3:09 PM, Anne Archibald
> wrote:
>> 2009/10/20 :
>>> On Sun, Oct 18, 2009 at 6:06 AM, Gary Ruben wrote:
Hi Gaël,
If you've got a 1D array/vector called "a", I think the normal idiom is
np.dot(a,a)
For the more general cas
On Tue, Oct 20, 2009 at 3:09 PM, Anne Archibald
wrote:
> 2009/10/20 :
>> On Sun, Oct 18, 2009 at 6:06 AM, Gary Ruben wrote:
>>> Hi Gaël,
>>>
>>> If you've got a 1D array/vector called "a", I think the normal idiom is
>>>
>>> np.dot(a,a)
>>>
>>> For the more general case, I think
>>> np.tensordot
2009/10/20 :
> On Sun, Oct 18, 2009 at 6:06 AM, Gary Ruben wrote:
>> Hi Gaël,
>>
>> If you've got a 1D array/vector called "a", I think the normal idiom is
>>
>> np.dot(a,a)
>>
>> For the more general case, I think
>> np.tensordot(a, a, axes=something_else)
>> should do it, where you should be ab
On Sun, Oct 18, 2009 at 6:06 AM, Gary Ruben wrote:
> Hi Gaël,
>
> If you've got a 1D array/vector called "a", I think the normal idiom is
>
> np.dot(a,a)
>
> For the more general case, I think
> np.tensordot(a, a, axes=something_else)
> should do it, where you should be able to figure out somethin
On Sun, Oct 18, 2009 at 11:37 AM, wrote:
> On Sun, Oct 18, 2009 at 12:06 PM, Skipper Seabold
> wrote:
> > On Sun, Oct 18, 2009 at 8:09 AM, Gael Varoquaux
> > wrote:
> >> On Sun, Oct 18, 2009 at 09:06:15PM +1100, Gary Ruben wrote:
> >>> Hi Gaël,
> >>
> >>> If you've got a 1D array/vector called
Skipper Seabold skrev:
> I'm curious about this as I use ss, which is just np.sum(a*a, axis),
> in statsmodels and didn't much think about it.
>
> Do the number of loops matter in the timings and is dot always faster
> even without the blas dot?
>
The thing is that a*a returns a temporary array
On Sun, Oct 18, 2009 at 12:06 PM, Skipper Seabold wrote:
> On Sun, Oct 18, 2009 at 8:09 AM, Gael Varoquaux
> wrote:
>> On Sun, Oct 18, 2009 at 09:06:15PM +1100, Gary Ruben wrote:
>>> Hi Gaël,
>>
>>> If you've got a 1D array/vector called "a", I think the normal idiom is
>>
>>> np.dot(a,a)
>>
>>>
On Sun, Oct 18, 2009 at 8:09 AM, Gael Varoquaux
wrote:
> On Sun, Oct 18, 2009 at 09:06:15PM +1100, Gary Ruben wrote:
>> Hi Gaël,
>
>> If you've got a 1D array/vector called "a", I think the normal idiom is
>
>> np.dot(a,a)
>
>> For the more general case, I think
>> np.tensordot(a, a, axes=somethin
On Sun, Oct 18, 2009 at 09:06:15PM +1100, Gary Ruben wrote:
> Hi Gaël,
> If you've got a 1D array/vector called "a", I think the normal idiom is
> np.dot(a,a)
> For the more general case, I think
> np.tensordot(a, a, axes=something_else)
> should do it, where you should be able to figure out som
Hi Gaël,
If you've got a 1D array/vector called "a", I think the normal idiom is
np.dot(a,a)
For the more general case, I think
np.tensordot(a, a, axes=something_else)
should do it, where you should be able to figure out something_else for
your particular case.
Gary R.
Gael Varoquaux wrote:
>
On Sat, Oct 17, 2009 at 07:27:55PM -0400, josef.p...@gmail.com wrote:
> >> > Why aren't you using logaddexp ufunc from numpy?
> >> Maybe because it is difficult to find, it doesn't have its own docs entry.
Speaking of which...
I thought that there was a readily-written, optimized function (or uf
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