On Thu, Mar 13, 2014 at 1:03 AM, Alan G Isaac wrote:
> On 3/12/2014 6:04 PM, Nathaniel Smith wrote:
>>https://github.com/numpy/numpy/pull/4351
>
> The Semantics section still begins with 0d, then 2d, then 1d, then nd.
> Given the context of the proposal, the order should be:
>
> 2d (the core n
On 3/12/2014 6:04 PM, Nathaniel Smith wrote:
>https://github.com/numpy/numpy/pull/4351
The Semantics section still begins with 0d, then 2d, then 1d, then nd.
Given the context of the proposal, the order should be:
2d (the core need expressed in the proposal)
nd (which generalizes via broadca
Hi all,
The proposal to add an infix operator to Python for matrix
multiplication is nearly ready for its debut on python-ideas; so if
you want to look it over first, just want to check out where it's
gone, then now's a good time:
https://github.com/numpy/numpy/pull/4351
The basic idea here is
On Fri, Jun 05, 2009 at 06:02:09PM -0400, Alan G Isaac wrote:
> I think something close to this would be possible:
> add dot as an array method.
> A .dot(B) .dot(C)
> is not as pretty as
> A * B * C
> but it is much better than
> np.dot(np.dot(A,B),C)
> In fact it is so much bett
On Fri, Jun 5, 2009 at 5:19 PM, Christopher Barker
wrote:
> Robert Kern wrote:
> x = np.array([1,2,3])
> timeit x.sum()
>>> 10 loops, best of 3: 3.01 µs per loop
> from numpy import sum
> timeit sum(x)
>>> 10 loops, best of 3: 4.84 µs per loop
>
> that is a VERY short array
Robert Kern wrote:
x = np.array([1,2,3])
timeit x.sum()
>> 10 loops, best of 3: 3.01 µs per loop
from numpy import sum
timeit sum(x)
>> 10 loops, best of 3: 4.84 µs per loop
that is a VERY short array, so one extra function call overhead could
make the difference. Is i
On Fri, Jun 5, 2009 at 17:54, Keith Goodman wrote:
> On Fri, Jun 5, 2009 at 3:02 PM, Alan G Isaac wrote:
>> I think something close to this would be possible:
>> add dot as an array method.
>> A .dot(B) .dot(C)
>> is not as pretty as
>> A * B * C
>> but it is much better than
>>
On Fri, Jun 5, 2009 at 3:02 PM, Alan G Isaac wrote:
> I think something close to this would be possible:
> add dot as an array method.
> A .dot(B) .dot(C)
> is not as pretty as
> A * B * C
> but it is much better than
> np.dot(np.dot(A,B),C)
I've noticed that x.sum() is faste
On Fri, Jun 5, 2009 at 4:02 PM, Alan G Isaac wrote:
> On 6/5/2009 5:41 PM Chris Colbert apparently wrote:
> > well, it sounded like a good idea.
>
>
> I think something close to this would be possible:
> add dot as an array method.
>A .dot(B) .dot(C)
> is not as pretty as
>A * B *
> I think something close to this would be possible:
> add dot as an array method.
> A .dot(B) .dot(C)
> is not as pretty as
> A * B * C
> but it is much better than
> np.dot(np.dot(A,B),C)
That is much better.
Matthew
___
Numpy-dis
On 6/5/2009 5:41 PM Chris Colbert apparently wrote:
> well, it sounded like a good idea.
I think something close to this would be possible:
add dot as an array method.
A .dot(B) .dot(C)
is not as pretty as
A * B * C
but it is much better than
np.dot(np.dot(A,B),C)
In fact
>> idx=N.ravel(x)!=0
>> A[:,idx]*x[idx]
On Fri, 14 Dec 2007, dmitrey apparently wrote:
> I expect being vector of length n A and x change
> every iter. A is not sparse.
Still, whenever x has a lot of zeros,
this should have a substantial payoff.
That seems the only exploitable information
yo
Thank you for the tip, however, I expect being vector of length n
A and x change every iter. A is not sparse.
Alan G Isaac wrote:
> On Fri, 14 Dec 2007, dmitrey apparently wrote:
>
>> I guess it doesn't matter, but typical n are 1...1000.
>> However, I need to call the operation hundreds or t
Hi,
How sparse are A, x and Ax?
You probably will not find any time or operation efficiency unless A
and x are relatively sparse. With the small size of n, any efficiency
will typically be small in any case.
As you indicate that this has to be repeated multiple times, perhaps
you should look at y
On Fri, 14 Dec 2007, dmitrey apparently wrote:
> I guess it doesn't matter, but typical n are 1...1000.
> However, I need to call the operation hundreds or thousands times (while
> running NLP solver ralg, so 4..5 * nIter times).
> Number of zeros can be 0...n-1
Do both A and x change every it
I guess it doesn't matter, but typical n are 1...1000.
However, I need to call the operation hundreds or thousands times (while
running NLP solver ralg, so 4..5 * nIter times).
Number of zeros can be 0...n-1
Charles R Harris wrote:
> How big is n?
>
> On Dec 14, 2007 1:24 AM, dmitrey <[EMAIL PROT
How big is n?
On Dec 14, 2007 1:24 AM, dmitrey <[EMAIL PROTECTED]> wrote:
> Hi all,
> I have to get Ax, A is n x n matrix, x is vector of length n.
> Some coords of x are zeros, so I want to economy time/cputime somehow
> w/o connecting sparse module from scipy.
> What's the easiest way to do so
Hi all,
I have to get Ax, A is n x n matrix, x is vector of length n.
Some coords of x are zeros, so I want to economy time/cputime somehow
w/o connecting sparse module from scipy.
What's the easiest way to do so?
Thx, D.
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