Cool. I can do this in the next couple of days (also need to learn to use git-rebase and clean up my branches so I can see the forest).
On Mon, Jun 8, 2009 at 1:53 PM, Ondrej Certik <ond...@certik.cz> wrote: > > On Mon, Jun 8, 2009 at 12:36 PM, Aaron S. Meurer<asmeu...@gmail.com> > wrote: > > I wouldn't bother checking. If the expression is not a fraction, yes the > > denom term will be 1, but the code I suggested will do nothing to it. > >>>> (-x).could_extract_minus_sign() > > True > >>>> n,d = (-x).as_numer_denom() > >>>> n > > -x > >>>> d > > 1 > >>>> -n/-d > > -x > > One thing you might want to consider is that as_numer_denom() seems to > pull > > together Adds with a common denominator > >>>> (x-y).as_numer_denom() > > (x - y, 1) > >>>> (x-y/(1-x)).as_numer_denom() > > (-y + x⋅(1 - x), 1 - x) > >>>> (x-y/(1-x)).could_extract_minus_sign() > > True > >>>> n, d = (x-y/(1-x)).as_numer_denom() > >>>> -n/-d > > -(y - x*(1 - x))/(1 - x) > >>>> print simplify((x-y/(1-x))) > > (x - y - x**2)/(1 - x) > >>>> ((x - y - x**2)/(1 - x)).could_extract_minus_sign() > > False > > I would recommend putting the code at the end of simplify. As you can > see, > > the above looks much nicer if you run it through simplify first. > > > Yes -- consider maybe writing a new function that will just put a > fraction into the canonical form (using the approach above) and then > call it at the end of simplify. > > Ondrej > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sympy@googlegroups.com To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sympy?hl=en -~----------~----~----~----~------~----~------~--~---
