python - Different results when passing function / method to equation solver -


this question follow-up of previous question pass class method fsolve.

@jim's answer in question difference between function object name , function call clarified confusion , solved problem. however, when tried similar things in sympy:

from sympy.solvers import solve sympy import symbol  class demo():     def __init__(self, var):         self.i = var     def func(self):         return self.i ** 2 - 4  x = symbol('x') def func(v):     return v ** 2 - 4 new = demo(x) solve(new.func(), x) # works fine, function call solve(func(x), x) # works fine, function call

why have different results? (in scipy need pass function name solver while in sympy need pass function call.) because different implementation of 2 libraries? in above example, if substitute function call function name, exception raised:

file "<ipython-input-26-3554c1f86646>", line 13, in <module> solve(new.func, x)  file "anaconda3\lib\site-packages\sympy\solvers\solvers.py", line 817, in solve f, symbols = (_sympified_list(w) w in [f, symbols])  file "anaconda3\lib\site-packages\sympy\solvers\solvers.py", line 817, in <genexpr> f, symbols = (_sympified_list(w) w in [f, symbols])  file "anaconda3\lib\site-packages\sympy\solvers\solvers.py", line 808, in _sympified_list return list(map(sympify, w if iterable(w) else [w]))  file "anaconda3\lib\site-packages\sympy\core\sympify.py", line 324, in sympify raise sympifyerror('could not parse %r' % a, exc)  sympifyerror: sympify of expression 'could not parse '<bound method demo.func of <__main__.demo object @ 0x0000018a37da6518>>'' failed, because of exception being raised: syntaxerror: invalid syntax (<string>, line 1)

when solver provided parameter (func(x), x), equivalent provide (x ** 2 - 4, x), first part of expression containing declared symbol x.

however when parameter list (func, x), variable in function definition, v, undefined , not match provided second argument, x. causes exception raised.

in scipy case, returned expression of function definition has 1 variable, var, solved, , can work properly.


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