[Numpy-discussion] Help with the tensordot function?
Tue Sep 7 15:14:11 CDT 2010
I was confused an unclear in my original question. I subsequently posted a
followup, titled "Simplified question on tensordot' where I explained myself
a lot better, and got some really good help. So, thank you very much for
looking into this issue, but I believe that this has been resolved.
On Tue, Sep 7, 2010 at 1:23 PM, Ondrej Certik <email@example.com> wrote:
> Hi Rick!
> On Fri, Sep 3, 2010 at 4:02 AM, Rick Muller <firstname.lastname@example.org> wrote:
> > Can someone help me replace a slow expression with a faster one based on
> > tensordot? I've read the documentation and I'm still confused.
> > I have two matrices b and d. b is n x m and d is m x m. I want to replace
> > the expression
> >>>> bdb = zeros(n,'d')
> >>>> for i in xrange(n):
> >>>> bdb[i,:] = dot(b[i,:],dot(d,b[i,:])
> I am first trying to reproduce this --- the above is missing one ")"
> and also dot() seems to produce a number, but you are assigning it to
> bdb[i,:], also you declare bdb as a 1D array. So I tried this:
> > with something that doesn't have the for loop and thus is a bit faster.
> > The first step is
> >>>> bd = dot(b,d)
> > However, following this with
> >>>> bdb = dot(bd,b.T)
> > doesn't work, since it yields a n x n matrix instead of an n x 1 vector.
> > Reading the notes on tensordot makes me think it's the function to use,
> > I'm having trouble grokking the axes argument. Can anyone help?
> In the above gist, I did the following:
> bd = dot(b, d)
> bdb = diag(dot(bd, b.T))
> print bdb
> which printed the same as:
> for i in xrange(n):
> bdb[i] = dot(b[i,:], dot(d, b[i, :]))
> print bdb
> but I think that this is not what you want, is it? I think you want to
> do something like:
> b * d * b.T
> but since b is a (n, m) matrix, the result is a matrix, not a vector,
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