>> 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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