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 . . . . . 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
