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 > _______________________________________________ Mailing list: https://launchpad.net/~dolfin Post to : [email protected] Unsubscribe : https://launchpad.net/~dolfin More help : https://help.launchpad.net/ListHelp
