[Numpy-discussion] expm

Charles R Harris charlesr.harris@gmail....
Fri Jul 20 14:45:48 CDT 2007


On 7/20/07, Kevin Jacobs <jacobs@bioinformed.com> <bioinformed@gmail.com>
wrote:
>
> On 7/20/07, Anne Archibald <peridot.faceted@gmail.com> wrote:
> >
> > On 20/07/07, Nils Wagner <nwagner@iam.uni-stuttgart.de> wrote:
> > > lorenzo bolla wrote:
> > > > hi all.
> > > > is there a function in numpy to compute the exp of a matrix, similar
> >
> > > > to expm in matlab?
> > > > for example:
> > > > expm([[0,0],[0,0]]) = eye(2)
> > > Numpy doesn't provide expm but scipy does.
> > > >>> from scipy.linalg import expm, expm2, expm3
> >
> > Just as a warning, numpy does provide expm1, but it does something
> > different (exp(x)-1, computed directly).
> >
>
> On a separate note, I'm working to provide faster and more accurate
> versions of sqrtm and expm.  The current versions do not take full advantage
> of LAPACK.  Here are some preliminary benchmarks:
>
> Ill-conditioned
> ----------------
> linalg.sqrtm   : error=9.37e-27, 573.38 usec/pass
> sqrtm_svd      : error=2.16e-28, 142.38 usec/pass
> sqrtm_eig      : error=4.79e-27, 270.38 usec/pass
> sqrtm_symmetric: error= 1.04e-27, 239.30 usec/pass
> sqrtm_symmetric2: error=2.73e-27, 190.03 usec/pass
>
> Well-conditioned
> ----------------
> linalg.sqrtm   : error=1.83e-29, 478.67 usec/pass
> sqrtm_svd      : error=8.11e-30, 130.57 usec/pass
> sqrtm_eig      : error=4.50e-30, 255.56 usec/pass
> sqrtm_symmetric: error=2.78e-30, 237.61 usec/pass
> sqrtm_symmetric2: error=3.35e-30, 167.27 usec/pass
>
> Large
> ----------------
> linalg.sqrtm   : error=5.95e-25 , 8450081.68 usec/pass
> sqrtm_svd      : error=1.64e-24, 151206.61 usec/pass
> sqrtm_eig      : error=6.31e-24, 549837.40 usec/pass
> sqrtm_symmetric: error=8.55e-25, 177422.29 usec/pass
>
> where:
>
> def sqrtm_svd(x):
>   u,s,vt = linalg.svd(x)
>   return dot(u,transpose((s**0.5)*transpose(vt)))
>
> def sqrtm_eig(x):
>   d,e = linalg.eig(x)
>   d = (d**0.5).astype(float)
>   return dot(e,transpose(d*e))
>
> def sqrtm_symmetric(x,cond=1e-7):
>   d,e = linalg.eigh(x)
>   d[d<cond] = 0
>   return dot(e,transpose((d**0.5)*e)).astype(float)
>
> def sqrtm_symmetric2(x):
>   # Not as robust due to initial Cholesky step
>   l=linalg.cholesky(x,lower=1)
>   u,s,vt = linalg.svd(l)
>   return dot(u,transpose(s*u))
>
> with SciPy linked against ACML.


I expect using sqrt(x) will be faster than x**.5.

Chuck
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