On Monday 25 January 2010 10:01:18 Garth N. Wells wrote: > Johan Hake wrote: > > 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. > > I figured it out eventually. There's nothing quite like a '&' symbol . . > . . .
Good! In Python every name in a namespace is a reference(pointer) to an instance. Johan > Garth > > > 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
