On 16/10/2006 21:52 Jonathan Guyer wrote the following:
On Oct 15, 2006, at 9:47 PM, hzhatlboro wrote:
Hi, could you suggest a small piece of code to define such cell
variables in FiPy? Suppose that we have N cell variables phi_i, i=1,
2, ..., N, and we want to define another set of cell variables s_i in
such a way that:
s_i at a nodal point of the grid is defined as s_i=1, if either phi_i
at the nodal point or phi_i at any one of the four nearest
neighboring nodal points is nonzero, and otherwise as s_i=0. Thanks
for your time! Regards, hzhatlboro
It's probably better in the long run to focus on what you're trying to
achieve mathematically, rather than trying to manipulate things on a
cell-by-cell basis.
Good comments and thank you! I am rushing to get one something working
at the moment...
FiPy's just not designed to work that way. Nonetheless,
s_i = (phi_i != 0) | (phi_i.getGrad().getMag() != 0)
should accomplish what you've asked.
Thanks for this!
Realize that FiPy is not designed with the assumption of a Cartesian
grid. There may not be exactly four nearest neighbors. One cell might
have three and another might have eighty-two.
Could you explain this a little bit further? In particular, the case
that one cell has "eighty-two" neighbors. Are your referring to
unstructured mesh?
Two more question, not related to this message, though. On page 170 of
"Programmer's reference" v1.1, you presented one example on how to
override the values of particular cells by using "mesh.getCells()", i.e.:-
>>>v1.setValue(1, cells=mesh.getCells()[2:])
Could you give some examples in the cases that, for example, I only want
to replace the value of cell No. 2, or the values of cell No. 3, 4, and 5?
>>> x=mesh.getCellCenters()[...,0]
By issuing the above command, I should be able to get the x coordinates
of the centers of the cells. The question is: does the sequence of x
follow the sequence of the cell numbers, or are x[0], x[1], x[2], ...,
the coordinates of the centers of cell No. 0, 1, 2, ...?
Thanks for your support!
Regards,
hzhatlboro
