What is the best way to compare floats for almost-equality in Python?

Python 3.5 adds the math.isclose and cmath.isclose functions as described in PEP 485.

If you're using an earlier version of Python, the equivalent function is given in the documentation.

def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):    return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)

rel_tol is a relative tolerance, it is multiplied by the greater of the magnitudes of the two arguments; as the values get larger, so does the allowed difference between them while still considering them equal.

abs_tol is an absolute tolerance that is applied as-is in all cases. If the difference is less than either of those tolerances, the values are considered equal.

Is something as simple as the following not good enough?

return abs(f1 - f2) <= allowed_error

I would agree that Gareth's answer is probably most appropriate as a lightweight function/solution.

But I thought it would be helpful to note that if you are using NumPy or are considering it, there is a packaged function for this.

numpy.isclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)

A little disclaimer though: installing NumPy can be a non-trivial experience depending on your platform.