I have a variable which may or may not be a sympy class. I want to convert it to a float but I’m having trouble doing this in a general manner:
$ python Python 2.7.3 (default, Dec 18 2014, 19:10:20) [GCC 4.6.3] on linux2 >>> import sympy >>> x = sympy.sqrt(2) >>> type(x) <class 'sympy.core.power.Pow'> >>> y = 1 + x >>> type(y) <class 'sympy.core.add.Add'> >>> z = 3 * x >>> type(z) <class 'sympy.core.mul.Mul'> >>> if isinstance(z, sympy.core): ... z = z.evalf(50) # 50 dp ... Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: isinstance() arg 2 must be a class, type, or tuple of classes and types
I will be testing
z to convert into a float. note that I can't just run
z without checking because they might be integers or floats already and this would raise an exception.
sympy.sympify() unfortunately does not convert to float. if it did then that would be the ideal solution to my problem:
>>> sympy.sympify(z) 3*sqrt(2) >>> type(sympy.sympify(z)) <class 'sympy.core.mul.Mul'>
All sympy objects inherit from
sympy.Basic. To evaluate an sympy expression numerically, just use
In : import sympy In : x = sympy.sqrt(2) In : x Out: sqrt(2) In : x.n() Out: 1.41421356237310 In : if (isinstance(x, sympy.Basic)): ...: print(x.n()) ...: 1.41421356237310
heh, at the end of the day, it turns out you can just convert a sympy object to float using python's native float conversion function! don't know why i didn't try this straight away...
>>> import sympy >>> x = sympy.sqrt(2) >>> float(x) 1.4142135623730951
i will use this since it works even if
x is a python int or already a python float