Sorry for the newbie question but I have not found the answer elsewhere so far.
Say we have a symbol and we multiply it by the float number one:
import sympy
x = sympy.Symbol('x')
x = 1.0 * x
All the expressions that contain this x will also contain factor one whatever I do:
from pprint import pprint
pprint(x)
pprint(sympy.simplify(x))
Output:
1.0⋅x
1.0⋅x
In the example above, I want to output x instead of 1.0⋅x. Is it possible to simplify the expression in the way that it solves all multiplications by a float number one? Should I transform this float number to another datatype?
Looks like it's due to the floating-point vs integer 1 - don't use 1.0 if you can get away with 1, perhaps?
You could check if it's identically 1.0 with if f.as_integer_ratio() == (1, 1), and if so don't prepend the unnecessary '1.0':
>>> xf = 1.0 * x
>>> xi = 1 * x
>>> xf
1.0*x
>>> xi
x
>>> pprint(xf)
1.0*x
>>> pprint(xi)
x
It might be due to floating-point inexactness (I don't know sympy much, can't say more).
EDIT: the runtime of float.as_integer_ratio() is tiny, in response to your question:
>>> python -m timeit -n 1000000 '(1.0).as_integer_ratio() == (1,1)'
1000000 loops, best of 5: 172 nsec per loop
>>> python -m timeit -n 1000000 '(1.1).as_integer_ratio() == (1,1)'
1000000 loops, best of 5: 327 nsec per loop
>>> python -m timeit -n 1000000 '(1.01).as_integer_ratio() == (1,1)'
1000000 loops, best of 5: 331 nsec per loop
>>> python -m timeit -n 1000000 '(0.999999999999).as_integer_ratio() == (1,1)'
1000000 loops, best of 5: 329 nsec per loop