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Anders Logg wrote:
On Mon, Oct 19, 2009 at 12:21:36PM +0100, Garth N. Wells wrote:Anders Logg wrote:On Sat, Oct 17, 2009 at 05:41:54PM +0100, Garth N. Wells wrote:Anders Logg wrote:On Sat, Oct 17, 2009 at 01:54:57PM +0100, Garth N. Wells wrote:Johan Hake wrote:On Saturday 17 October 2009 12:32:54 Garth N. Wells wrote:Will uh = U[1] return a deep or shallow copy of the sub-Function?To avoid confusion with the ufl interface we have limited the interface for SubFunctions in PyDOLFIN to split. split returns a shallow copy by default. pass True to split and it will return a deepcopy. In your case it would be: uh = U.split()[0] and uh = U.split(True)[0] operator[] is used when you define forms. We have not yet managed to merge the two operations into one.OK. Do we still have the function 'sub'?Where is the operator [] define for sub-Functions?The operator you are using is the ufl one, which is defined in ufl.exproperators.py.OK, but it's not clear to me then what's happing with this extract of my code: problem = VariationalProblem(a, L, bcs) Uh = problem.solve() u = Expression("epx(x[0])", V = Vexact) uh = Uh[1] M = (uh-u)*(uh-u)*dx error = sqrt(assemble(M, mesh=mesh))This should work fine since uh will be a component of the coefficient Uh in the UFL form. Same as when you write say inner(v, u)*dx or anything else that accesses components. So in summary [] can be used for any purpose in forms. It can also be used for plotting, but extracting the vector etc will not work.It works as expected, but what's going on behind the scenes? Who is creating the dof map and is the vector being copied?There's nothing out of the ordinary going on behind the scenes.There must be something going on - either a reference to or a copy of the underlying vector, and a dof map. Where are these coming from?Why must there be a reference? Are we discussing different things here?
This is what I'm doing/want to know: # Compute Uh Uh = my_pde.solve() # Get sub-function. uh is both a UFL and a cpp Function. What's # happening on the cpp side here? uh = Uh[1] Garth
-- AndersGarthWhen you write uh = Uh[1] it's no different from doing a = uh[1]*v*dx or doing gradu = grad(u) a = dot(grad(v), gradu)*dx gradu is not a new Function, it's a UFL expression involving the grad operator and the Function u. gradu can be used in forms and plotted (since the plot will project it to a Function), but you can't do gradu.vector()._______________________________________________ DOLFIN-dev mailing list DOLFIN-dev@xxxxxxxxxx http://www.fenics.org/mailman/listinfo/dolfin-dev ------------------------------------------------------------------------ _______________________________________________ DOLFIN-dev mailing list DOLFIN-dev@xxxxxxxxxx http://www.fenics.org/mailman/listinfo/dolfin-dev
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