[Numpy-discussion] Trick for fast
Fri Feb 3 12:58:43 CST 2012
I am unclear on what you want to say, but all I am doing in the code is
getting inertia tensor for a bunch of particle masses.
So the diagonals are not actually zeros but would have z^2 + y^2 ..
The reason which I said 3secs could be misunderstood .. This code is called
many times over a loop and the bulk time is taken in computing this
After code change, the entire loop finishes off in 3 ses.
Thanks for alertness,
On Fri, Feb 3, 2012 at 12:47 PM, <email@example.com> wrote:
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> 1. Re: Trick for fast (santhu kumar)
> 2. Re: Trick for fast (firstname.lastname@example.org)
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> Date: Fri, 3 Feb 2012 12:29:26 -0600
> From: santhu kumar <email@example.com>
> Subject: Re: [Numpy-discussion] Trick for fast
> To: firstname.lastname@example.org
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> On Fri, Feb 3, 2012 at 1:29 PM, santhu kumar <email@example.com> wrote:
> > Hello all,
> > Thanks for lovely solutions. I have sat on it for some time and wrote it
> > myself :
> > n =x.shape
> > ea = np.array([1,0,0,0,1,0,0,0,1])
> > inert = ((np.tile(ea,(n,1))*((x*x).sum(axis=1)[:,np.newaxis]) -
> > inert.shape = 3,3
> > Does the trick and reduces the time from over 45 secs to 3 secs.
> > I do want to try einsum but my numpy is little old and it does not have
> > Thanks Sebastian (it was tricky to understand your code for me) and Josef
> > (clean).
> Isn't the entire substraction of the first term just to set the
> diagonal of the result to zero.
> It looks to me now just like the weighted dot product and setting the
> diagonal to zero. That shouldn't take 3 secs unless you actual
> dimensions are huge.
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