dolfin team mailing list archive

["Garth N. Wells" <gnw20@xxxxxxxxx>]

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