[Numpy-discussion] Trick for fast
Fri Feb 3 13:33:25 CST 2012
On Fri, Feb 3, 2012 at 1:58 PM, santhu kumar <firstname.lastname@example.org> wrote:
> Hi Josef,
> 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 inertial
> After code change, the entire loop finishes off in 3 ses.
ok, I still had a python sum instead of the numpy sum in there (I
really really like namespaces :)
array([ 1764., 1849., 1936.])
this might match better.
print (mass * x**2).sum() * np.eye(3) - np.dot(x.T, mass*x)
res = - np.dot(x.T, mass*x)
res[np.arange(3), np.arange(3)] += (mass * x**2).sum()
> Thanks for alertness,
> On Fri, Feb 3, 2012 at 12:47 PM, <email@example.com> wrote:
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>> Today's Topics:
>> 1. Re: Trick for fast (santhu kumar)
>> 2. Re: Trick for fast (firstname.lastname@example.org)
>> Message: 1
>> 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
>> Content-Type: text/plain; charset="iso-8859-1"
>> 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]) -
>> > np.hstack([x*x[:,0][:,np.newaxis],x*x[:,1][:,np.newaxis],x*x[:,2][:,np.newaxis]]))*mass[:,np.newaxis]).sum(axis=0)
>> > 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
>> > it.
>> > 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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