I like that. I am glad to write a patch if this is the desired behavior of
simplify. Does there need to be a check for whether or not something is a
fraction first? Is checking for denom == 1 the best way to do that?
In [146]: a,b = x2_tf_ds.as_numer_denom()
In [147]: a
Out[147]: Gc*k
In [148]: b
Out[148]: Gc*k + k*m1*s**2 + k*m2*s**2 + m1*m2*s**4
In [149]: b.as_numer_denom()
Out[149]: (Gc*k + k*m1*s**2 + k*m2*s**2 + m1*m2*s**4, 1)
Ryan
On Mon, Jun 8, 2009 at 12:45 PM, Aaron S. Meurer <asmeu...@gmail.com> wrote:
>
> You might look into could_extract_minus_sign:
> >>> sympify('-Gc*k/(-Gc*k - k*m1*s**2 - k*m2*s**2 -
> m1*m2*s**4)').could_extract_minus_sign()
> True
> >>> sympify('Gc*k/(-Gc*k - k*m1*s**2 - k*m2*s**2 -
> m1*m2*s**4)').could_extract_minus_sign()
> False
>
> Perhaps simplify() should do
>
> if expr.could_extract_minus_sign():
> n, d = expr.as_numer_denom()
> expr = sympify(-1)*n/(sympify(-1)*d)
>
> or some similar.
>
>
> Aaron Meurer
> On Jun 8, 2009, at 11:36 AM, Ryan Krauss wrote:
>
> > So, I have yet another simplification question. What is the correct
> > way to factor a minus one out of this:
> >
> > In [108]: type(x2_tf_ds)
> > Out[108]: <class 'sympy.core.mul.Mul'>
> >
> > In [109]: x2_tf_ds
> > Out[109]: -Gc*k/(-Gc*k - k*m1*s**2 - k*m2*s**2 - m1*m2*s**4)
> >
> > This didn't work:
> > In [110]: sympy.simplify(x2_tf_ds)
> > Out[110]: -Gc*k/(-Gc*k - k*m1*s**2 - k*m2*s**2 - m1*m2*s**4)
> >
> > This works, and is fine with me. But I wonder if there is a better
> > way:
> > In [111]: a,b = x2_tf_ds.as_numer_denom()
> >
> > In [112]: -a/-b
> > Out[112]: Gc*k/(Gc*k + k*m1*s**2 + k*m2*s**2 + m1*m2*s**4)
> >
> >
> > Thanks,
> >
> > Ryan
> >
> > >
>
>
> >
>
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