Not that I can think of. Good catch on the minus sign. And use the `as_Add`
flag unless you are positive that the expression you are working with is
always an Add. Does this work:
sp.Eq(*[-1**i*a for i, a in enumerate(reversed((ht * _lhs - ht *
_rhs).expand().as_independent(rho1, as_Add=True)))]
On Saturday, May 6, 2017 at 3:24:24 PM UTC-5, Jonathan Essen wrote:
> Thank you for the reply, but I noticed that the new right hand side is off
> by an overall minus sign! I fixed it using:
>
> tmp = (ht * _lhs - ht * _rhs).expand().as_independent(rho1)
> sp.Eq(tmp[1], rhs=-tmp[0])
>
> Is there was a cleaner way to do this?
>
>
>
> On Thursday, May 4, 2017 at 8:21:44 AM UTC-7, Chris Smith wrote:
>>
>> Given Eq(L, R) where L and R are the left and right hand sides of the
>> equation, try: `Eq(*list(reversed((L - R).expand().as_indepedent(foo,
>> as_Add=True))))` where foo is the variable of interest to get the new
>> equation.
>>
>> On Tuesday, April 25, 2017 at 2:35:10 PM UTC-5, Jonathan Essen wrote:
>>>
>>> I am attempting to use sympy to generate an implicit scheme for the heat
>>> equation. With the Crank-Nicolson time discretization I have equation 1
>>> (attached).
>>>
>>> Is there any way to isolate all occurrences of $\rho^{n+1}$ (along with
>>> its derivatves) to the lefthand side, as in equation 2 (attached).
>>>
>>> I would be very interested to know! Apparently it has something to do
>>> with the collect function.
>>>
>>> Best,
>>> Jonathan
>>>
>>
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