What is the best way to compare floats for almost
It's well known that comparing floats for equality is a little fiddly due to rounding and precision issues.
For example: https://randomascii.wordpress.com/2012/02/25/comparing-floating-point-numbers-2012-edition/
What is the recommended way to deal with this in Python?
Surely there is a standard library function for this somewhere?
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.
是否像以下那样简单不够好？
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.
链接地址: http://www.djcxy.com/p/4712.html下一篇: 几乎可以比较花车的最佳方式是什么？