Not to hijack this post, but why does subs(phi(z,t).diff(z),...) work
when phi(z,t).diff(t,2) does not contain any derivatives of phi with
respect to z? It seems like this substitution is saying that
phi(z,t).diff(z) = phi(z,t)
I'm new to sympy, so I apologize if this is a stupid question.
Alex
On 05/19/2015 12:16 PM, Jonathan Lindgren wrote:
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have
some very strange behaviour with subs. The following code
from sympy.abc import phi
import sympy as sp
z=sp.Symbol('z')
t=sp.Symbol('t')
sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
gives me the output
2
∂
───(b(z, t))
2
∂t
but I would expect the output
2
∂
───(φ(z, t))
2
∂t
This was working perfectly in my previous version of sympy.
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