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

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.

Is something as simple as the following not good enough?

return abs(f1 - f2) <= allowed_error

Use Python's decimal module, which provides the Decimal class.

From the comments:

It is worth noting that if you're doing math-heavy work and you don't absolutely need the precision from decimal, this can really bog things down. Floats are way, way faster to deal with, but imprecise. Decimals are extremely precise but slow.

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.