Well, that is exactly the problem, and what I think is a bug....it should
not work like that..
Den tisdag 19 maj 2015 kl. 22:00:46 UTC+2 skrev Alexander Lindsay:
>
> 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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