73 lines
3.5 KiB
Plaintext
73 lines
3.5 KiB
Plaintext
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This rule raises an issue when direct and indirect equality/inequality checks are made on floats.
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== Why is this an issue?
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Floating point math is imprecise because of the challenges of storing such values in a binary representation.
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In base 10, the fraction `1/3` is represented as `0.333...` which, for a given number of significant digit, will never exactly be `1/3`. The same problem happens when trying to represent `1/10` in base 2, with leads to the infinitely repeating fraction `0.0001100110011...`. This makes floating point representations inherently imprecise.
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Even worse, floating point math is not associative; push a ``++float++`` through a series of simple mathematical operations and the answer will be different based on the order of those operation because of the rounding that takes place at each step.
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Even simple floating point assignments are not simple, as can be vizualized using the `format` function to check for significant digits:
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[source,python]
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----
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>>> format(0.1, ".17g")
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'0.10000000000000001'
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----
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This can also be vizualized as a fraction using the `as_integer_ratio` method:
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[source,python]
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----
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>>> my_float = 0.1
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>>> numerator, denominator = my_float.as_integer_ratio()
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>>> f"{numerator} / {denominator}"
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'3602879701896397 / 36028797018963968'
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----
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Therefore, the use of the equality (``++==++``) and inequality (``++!=++``) operators on ``++float++`` values is almost always erroneous.
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== How to fix it
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Whenever attempting to compare float values, it is important to consider the inherent imprecision of floating-point arithmetic.
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One common solution to this problem is to use a tolerance value (also called epsilon) to define an acceptable range of difference between two floats. A tolerance value may be relative (based on the magnitude of the numbers being compared) or absolute. Note that comparing a value to 0 is a special case: as it has no magnitude, it is important to use an absolute tolerance value.
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The `math.isclose` function allows to compare floats with a relative and absolute tolerance. One should however be careful when comparing values to 0, as by default, the absolute tolerance of `math.isclose` is `0.0` (this case is covered by rule S6727)
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. Depending on the library you're using, equivalent functions exist, with possibly different default tolerances (e.g `numpy.isclose` or `torch.isclose` which are respectively designed to work with `numpy` arrays and `pytorch` tensors).
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If precise decimal arithmetic is needed, another option is to use the `Decimal` class of the `decimal` module, which allows for exact decimal arithmetic.
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=== Code examples
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==== Noncompliant code example
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[source,python,diff-id=1,diff-type=noncompliant]
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----
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def foo(a, b):
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return a == b - 0.1
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----
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==== Compliant solution
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[source,python,diff-id=1,diff-type=compliant]
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----
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import math
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def foo(a, b):
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return math.isclose(a, b - 0.1, rel_tol=1e-09, abs_tol=1e-09)
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----
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== Resources
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=== Documentation
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* Python Documentation - https://docs.python.org/3/tutorial/floatingpoint.html#floating-point-arithmetic-issues-and-limitations[Floating Point Arithmetic: Issues and Limitations]
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* Python Documentation - https://docs.python.org/3/library/decimal.html#module-decimal[Decimal fixed point and floating point arithmetic]
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* NumPy Documentation - https://numpy.org/doc/stable/reference/generated/numpy.isclose.html[numpy.isclose]
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* PyTorch Documentation - https://pytorch.org/docs/stable/generated/torch.isclose.html[torch.isclose]
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=== Related rules
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* S6727 - The `abs_tol` parameter should be provided when using `math.isclose` to compare values to `0`
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