On Thu, Dec 04, 2008 at 04:08:15AM -0800, Scott wrote:
>
> How can I translate a formula output by sympy into one that is scipy
> friendly?
Here is my attempt with the attached file:
In [1]: run test.py
In [2]: F0??
[...]
def F0(dof,d_dof,U,dt):
args = tuple(dof) + tuple(d_dof) + (U, dt)
return f0(*args)
In [3]: F0([1,2,3,4], [5,6,7,8], 9, 10)
Out[3]: -281.64999999999998
In [4]: %timeit F0([1,2,3,4], [5,6,7,8], 9, 10)
100000 loops, best of 3: 11.2 µs per loop
I hope that helps.
By,
Friedrich
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to [email protected]
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/sympy?hl=en
-~----------~----~----~----~------~----~------~--~---
from sympy import var, lambdify
var("""h1 theta1 v_h1 v_theta1 delta_h
delta_theta delta_v_h delta_v_theta U dt""".split())
term = -delta_v_h/2 - delta_v_theta/20 + dt*(
-2*h1/25 - delta_h/25 -
3*U*(v_theta1 + delta_v_theta/2)/50 -
U*(v_h1 + delta_v_h/2)/20 - U**2*(theta1 + delta_theta/2)/20 )
args = (h1, theta1, v_h1, v_theta1, delta_h,
delta_theta, delta_v_h, delta_v_theta, U, dt)
f0 = lambdify(args, term)
def F0(dof,d_dof,U,dt):
args = tuple(dof) + tuple(d_dof) + (U, dt)
return f0(*args)
