dolfin team mailing list archive

[Johan Hake <johan.hake@xxxxxxxxx>]

On Monday 25 January 2010 03:10:46 Garth N. Wells wrote:
> Johan Hake wrote:
> > On Sunday 24 January 2010 16:39:14 Garth N. Wells wrote:
> >> Johan Hake wrote:
> >>> On Sunday 24 January 2010 16:12:37 Garth N. Wells wrote:
> >>>> Johan Hake wrote:
> >>>>> On Sunday 24 January 2010 00:03:41 Garth N. Wells wrote:
> >>>>>> Johan Hake wrote:
> >>>>>>> On Saturday 23 January 2010 14:55:08 Garth N. Wells wrote:
> >>>>>>>> Johan Hake wrote:
> >>>>>>>>> On Saturday 23 January 2010 08:42:14 Garth N. Wells wrote:
> >>>>>>>>>> Is it correct that behind the scenes that
> >>>>>>>>>>
> >>>>>>>>>>   U0 = Function(V)
> >>>>>>>>>>   U  = Function(V)
> >>>>>>>>>>   U0.vector()[:] = U.vector()[:]
> >>>>>>>>>>
> >>>>>>>>>> involves a GenericVector::get(..) call and a
> >>>>>>>>>> GenericVector::set(..) call? If so, it isn't ideal since it
> >>>>>>>>>> introduces unnecessary new/delete operations and unnecessary
> >>>>>>>>>> copying of data.
> >>>>>>>>>
> >>>>>>>>> None of GenericVector::get(..) or GenericVector::set(..) are
> >>>>>>>>> invoked, see __getslice__ and __setslice__ in la_post.i.
> >>>>>>>>>
> >>>>>>>>>    U0.vector()[:]
> >>>>>>>>>
> >>>>>>>>> involves
> >>>>>>>>>
> >>>>>>>>>   GenericVector::operator =(..)
> >>>>>>>>>
> >>>>>>>>> and
> >>>>>>>>>
> >>>>>>>>>   U.vector()[:]
> >>>>>>>>>
> >>>>>>>>> involves
> >>>>>>>>>
> >>>>>>>>>   GenericVector::copy()
> >>>>>>>>>
> >>>>>>>>> However the latter is unnecessary as you instead can do:
> >>>>>>>>>
> >>>>>>>>>   U0.vector()[:] = U.vector()
> >>>>>>>>>
> >>>>>>>>> invoking the assignment operator of U0's vector with U's vector.
> >>>>>>>>
> >>>>>>>> What happens if I do
> >>>>>>>>
> >>>>>>>>   x = U.vector()[:]
> >>>>>>>
> >>>>>>> It just triggers the copy method of GenericVector, which is the
> >>>>>>> same behavior as for other itterable Python types.
> >>>>>>>
> >>>>>>>> ? Is x a numpy array?
> >>>>>>>
> >>>>>>> No you need to call array() to accomplish that.
> >>>>>>
> >>>>>> OK. What I'm trying to do is
> >>>>>>
> >>>>>>     # Get vectors
> >>>>>>     u_vec  = u.vector()[:]
> >>>>>>     u0_vec = u0.vector()[:]
> >>>>>>     v0_vec = v0.vector()[:]
> >>>>>>     a0_vec = a0.vector()[:]
> >>>>>
> >>>>> You should not need to make a copy of the vectors here.
> >>>>
> >>>> How can I avoid it?
> >>>
> >>> a_vec and v_vec are new vectors. None of the four vectors below get
> >>> modified by the a_vec and v_vec expressions so no need of copying, and
> >>> the v0 and a0 assignment should work with GenericVectors too.
> >>
> >> Do you mean that just
> >>
> >>   a_vec = 1.0/(2.0*beta)*((u - u0 - v0*dt)/(0.5*dt*dt) \
> >>            - (1.0-2.0*beta)*a0 )
> >>
> >> where u and u0 are GenericVectors should work?
> >
> > Have you tried?
> 
> Can I somehow get a 'reference' to the vector so I don't have to use
> u.vector()[:] in the expressions?

That is what u.vector() gives you.

Johan

> Garth
> 
> > As long as the rest (besides a0, which I assume also is a GenericVector)
> > are scalars everything should just work. The Python LA interface (at
> > least for GenericVector) should work more or less as the NumPy interface
> > which I think is nice :)
> >
> > We cannot take 1./v, where v is a GenericVector.
> >
> >>> Do you get any error messages?
> >>
> >> No errors. What I have now seems to work fine.
> >
> > Ok, and that is because what you do obviously works for NumPy arrays.
> >
> > Johan
> >
> >> Garth
>