[Numpy-discussion] Quikest way to create a symetric (diagonal???) matrix ?

[Hoyt Koepke]

If the rest of the matrix is already zeros and memory wasn't a
problem, you could just use

A_sym = A + A.T - diag(diag(A))

If memory was an issue, I'd suggest weave.inline (if that's a viable
option) or pyrex to do the loop, which would be about as fast as you
could get.

--Hoyt



On Wed, Mar 26, 2008 at 7:22 AM, Alexandre Fayolle
<alexandre.fayolle@logilab.fr> wrote:
> On Wed, Mar 26, 2008 at 09:48:02AM -0400, Pierre GM wrote:
>  > All,
>  > What's the quickest way to create a diagonal matrix ? I already have the
>  > elements above the main diagonal. Of course, I could use loops:
>  > >>>m=5
>  > >>>z = numpy.arange(m*m).reshape(m,m)
>  > >>>for k in range(m):
>  > >>>    for j in range(k+1,m):
>  > >>>        z[j,k] = z[k,j]
>  > But I was looking for something more efficient.
>
>  From your code, you certainly meant "symetric" and not diagonal.
>
>  Maybe you can speed up things a bit by assigning slices:
>
>  >>> for k in range(m):
>  ...     z[k:, k] = z[k, k:]
>
>
>
>  --
>  Alexandre Fayolle                              LOGILAB, Paris (France)
>  Formations Python, Zope, Plone, Debian:  http://www.logilab.fr/formations
>  Développement logiciel sur mesure:       http://www.logilab.fr/services
>  Informatique scientifique:               http://www.logilab.fr/science
>
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