to., 07.10.2010 kl. 17.46 -0600, skrev Nicholas Kinar:
> Hello,
>
> Working again with the FDTD re-arrangement of terms, I am wondering if
> there is any way to run collect() on indices of an IndexedBase object
> using the current master sympy git repository. Rather than having (n+1)
> express
Hello,
Working again with the FDTD re-arrangement of terms, I am wondering if
there is any way to run collect() on indices of an IndexedBase object
using the current master sympy git repository. Rather than having (n+1)
expressed as a power, is it possible to run collect() in the manner
shown
#begin modified sample code
from sympy import *
from sympy.tensor import Indexed, Idx, IndexedBase
from sympy.matrices import *
p = IndexedBase('p')
var('deltax deltay delta deltat A B')
i,j,n = symbols('i j n', integer = True)
def generate_fdtd(expr, nsolve):
w = Wild('w')
expr0 = ex
Hello,
Looking back at a previous thread
(http://groups.google.com/group/sympy/browse_thread/thread/a406176eeb28eb04),
an interesting procedure was discussed to place p[i,j,n+1] terms on the
LHS of the expression. (Thanks again, Aaron!) As shown in another
sample code below, the procedure u
Your problem comes from using atoms(IndexedBase). Here is what that gives:
In [24]: expr.atoms(IndexedBase)
Out[24]: set(p)
But this doesn't have the i, j, n args that you want. I think what you really
want to do is use atoms(Indexed) (Øyvind, correct me if I am wrong here). That
gives:
In
Your problem comes from using atoms(IndexedBase). Here is what that gives:
In [24]: expr.atoms(IndexedBase)
Out[24]: set(p)
But this doesn't have the i, j, n args that you want. I think what you really
want to do is use atoms(Indexed) (Øyvind, correct me if I am wrong here). That
gives:
Yes, this is exactly what I wanted, Aaron; many thanks for pointing me
in the right direction. As shown in the final example code below,
I've done as you've suggested and used atoms(Indexed). I've also
taken a poet's license with the code and called expand() to be able to
distribute through
> Here's a short comment: using Idx(n+1) seems to work a little better for me,
> since I've noticed that if I use Idx(n) with a very complicated expression,
> there is a possibility of terms with (n+1) indices ending up on the same side
> as the (n) indices.
That sounds like a bug. Can you giv
Here's a short comment: using Idx(n+1) seems to work a little better for me,
since I've noticed that if I use Idx(n) with a very complicated expression,
there is a possibility of terms with (n+1) indices ending up on the same side
as the (n) indices.
That sounds like a bug. Can you gi
Here's a short comment: using Idx(n+1) seems to work a little better
for me, since I've noticed that if I use Idx(n) with a very
complicated expression, there is a possibility of terms with (n+1)
indices ending up on the same side as the (n) indices.
That sounds like a bug. Can you give a spe
I think what you are getting is expected behavior. Remember that when you
switch n and n + 1, you are switching the side of the equation that they will
end up on. So in the first one, you have all terms with n on the right-hand
side, and in the second one, you have all the terms with n + 1 on
I think what you are getting is expected behavior. Remember that when you
switch n and n + 1, you are switching the side of the equation that they will
end up on. So in the first one, you have all terms with n on the right-hand
side, and in the second one, you have all the terms with n + 1
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