Wednesday, 9 May 2018

floating point - General way of comparing numerics in Python




I have been looking around to find a general way of comparing two numerics in Python. In particular, I want to figure out whether they are the same or not.



The numeric types in Python are:



int, long, float & complex


For example, I can compare 2 integers (a type of numeric) by simply saying:




a == b


For floats, we have to be more careful due to rounding precision, but I can compare them within some tolerance.



Question



We get 2 general numerics a and b: How do we compare them? I was thinking of casting both to complex (which would then have a 0 imaginary part if the type is, say, int) and compare in that domain?



This question is more general than simply comparing floats directly. Certainly, it is related to this problem, but it is not the same.



Answer



In Python 3.5 (and in Numpy) you can use isclose



Read the PEP 485 that describes it, Python 3.5 math library listing and numpy.isclose for more. The numpy version works in all versions of Python that numpy is supported.



Examples:



>>> from math import isclose
>>> isclose(1,1.00000000001)
True

>>> isclose(1,1.00001)
False


The relative and absolute tolerance can be changed.



Relative tolerance can be thought of as +- a percentage between the two values:



>>> isclose(100,98.9, rel_tol=0.02)
True

>>> isclose(100,97.1, rel_tol=0.02)
False


The absolute tolerance is a absolute value between the two values. It is the same as the test of abs(a-b)<=tolerance



All numeric types of Python are support with the Python 3.5 version. (Use the cmath version for complex)



I think longer term, this is your better bet for numerics. For older Python, just import the source. There is a version on Github.




Or, (forgoing error checking and inf and NaN support) you can just use:



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

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