By the way, the method I was using turned out to give problems. Since
.doit() returned the sympy log, I ended up having two different version of
log in later expressions (depending on whether or not .doit() had been
called), leading to strange results.
I found a better solution, simply setting _should_evalf to false.
from sympy import log as sympy_log
class log(sympy_log):
@classmethod
def _should_evalf(cls, arg):
return -1
On Saturday, September 6, 2014 3:46:41 PM UTC-5, Aaron Meurer wrote:
> Another option would be to name your class something different and
> alias it to log (and also change the printers to print "log"). This
> way is probably simpler, though, as you only have to override the one
> evalf method rather than several printer methods.
>
> Aaron Meurer
>
> On Sat, Sep 6, 2014 at 3:09 PM, Duane Nykamp <[email protected]
> <javascript:>> wrote:
> > Perfect. The following solution seems to work for me. Thanks!
> >
> > class log(Function):
> >
> > def _eval_evalf(self,prec):
> > pass
> >
> > def doit(self, **hints):
> > from sympy import log as sympy_log
> > return sympy_log(self.args[0]).doit(**hints)
> >
> >
> >
> >
> >
> > On Saturday, September 6, 2014 2:20:12 PM UTC-5, Aaron Meurer wrote:
> >>
> >> SymPy automatically defines evaluation based on mpmath function names.
> >> I think you can get around it by defining an empty _eval_evalf(self,
> >> prec) function.
> >>
> >> Aaron Meurer
> >>
> >> On Sat, Sep 6, 2014 at 1:50 PM, Duane Nykamp <[email protected]>
> wrote:
> >> > I really like the evaluate=False flag on parse_expr. But, since
> >> > log(0.1)
> >> > automatically evaluates to a float, I'm trying to create an
> unevaluated
> >> > version of log. However, I'm getting strange results where log is
> >> > implicitly defined.
> >> >
> >> > As shown below, once I define a function log that does nothing, sympy
> >> > somehow implicitly defines it to be the usual log function.
> >> >
> >> > Am I missing how this works or is this aberrant behavior?
> >> >
> >> > My eventual plan is to have log.doit() return the original sympy
> >> > function.
> >> >
> >> > If I call the function something else, like llog, then I don't run
> into
> >> > the
> >> > behavior. But, I'm not sure how to make printers display my llog
> >> > function
> >> > as log.
> >> >
> >> > Thanks,
> >> > Duane
> >> >
> >> >
> >> >
> >> > In [1]: from sympy import Function
> >> >
> >> > In [2]: log
> >> >
> >> >
> ---------------------------------------------------------------------------
> >> > NameError Traceback (most recent call
> >> > last)
> >> > <ipython-input-2-9a115ebc25a0> in <module>()
> >> > ----> 1 log
> >> >
> >> > NameError: name 'log' is not defined
> >> >
> >> > In [3]: class log(Function):
> >> > ...: pass
> >> > ...:
> >> >
> >> > In [4]: log
> >> > Out[4]: __main__.log
> >> >
> >> > In [5]: log(0.1)
> >> > Out[5]: -2.30258509299405
> >> >
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