Improve docs of math.isclose(0, x) when x = 0

[guyj-s1]

Documentation

In the documentation of math.isclose it states

When comparing x to 0.0, isclose(x, 0) is computed as abs(x) <= rel_tol * abs(x), which is False for any x and rel_tol less than 1.0.

The statement isn't correct since x = 0 will give True

Linked PRs

• gh-131094: Refine math.isclose docs #131138
• gh-131094: Refine math.isclose docs #131139
• [3.13] gh-131094: Refine math.isclose docs (GH-131139) #131392
• [3.12] gh-131094: Refine math.isclose docs (GH-131139) #131393

[hoodmane]

Seems the correction is to add a single word:

When comparing x to 0.0, isclose(x, 0) is computed as abs(x) <= rel_tol * abs(x), which is False for any nonzero x and rel_tol less than 1.0.

Would you mind making a PR?

[eendebakpt]

When making the PR also update the formatting (the parameters should be emphasized). I think the suggested change to the doc is an improvement, although with floating point arithmetic there are some corner cases:

import math

x = 1e-320
rel_tol = 1e-9
abs_tol = math.nextafter(1, 0)
print(math.isclose(x, 0, rel_tol=rel_tol, abs_tol=abs_tol))

print(abs(x) <= rel_tol * abs(x))
print(abs(x) <= abs(rel_tol * x))

results in

True
False
False

[plashchynski]

@eendebakpt it's true because the difference is less than abs_tol:

import math

x = 1e-320
rel_tol = 1e-9
abs_tol = math.nextafter(1, 0)

diff = abs(0 - x);
print(diff <= abs_tol) # returns True

I think this statement in the documentation holds only for abs_tol being zero (which is the default).

[eendebakpt]

The relevant code is:

cpython/Modules/mathmodule.c

Lines 3202 to 3211 in ad90c5f

	/* now do the regular computation
	this is essentially the "weak" test from the Boost library
	*/
	diff = fabs(b - a);
	return (((diff <= fabs(rel_tol * b)) ||
	(diff <= fabs(rel_tol * a))) ||
	(diff <= abs_tol));
	}

@plashchynski I agree the statement is valid if f abs_tol is zero. The way the documentation is written suggest the statement is true in general though (if b=0).

[plashchynski]

@eendebakpt imho this statements sounds more like a reason to set abs_tol to a value greater than zero:

cpython/Doc/library/math.rst

Lines 351 to 355 in ad90c5f

	*abs_tol* is the absolute tolerance; it defaults to ``0.0`` and it must be
	nonnegative. When comparing ``x`` to ``0.0``, ``isclose(x, 0)`` is computed
	as ``abs(x) <= rel_tol * abs(x)``, which is ``False`` for any ``x`` and
	rel_tol less than ``1.0``. So add an appropriate positive abs_tol argument
	to the call.

[plashchynski]

Btw, if a = math.nextafter(0, 1) then math.isclose(a, 0) will always be true for any rel_tol > 0:

import math
a = math.nextafter(0, 1)
print(math.isclose(a, 0, rel_tol=0.9)) # True

because:

print(math.nextafter(0, 1) * 0.9 == math.nextafter(0, 1)) # True

which is contrary to this statement and to pure mathematics as well.

Is there a policy for such corner cases? Should they be considered and handled?

[skirpichev]

@eendebakpt, I think you misread documentation. It looks like abs_tol disappear in equation of the paragraph, which deals with comparing x to 0, but it's because abs(x) <= rel_tol * abs(x) is valid only in case of abs_tol=0.0 (isclose(x, 0) means the default value of abs_val). Last sentence suggests what to do in this case: "So add an appropriate positive abs_tol argument to the call."

For nonzero abs_tol you must use the generic equation, mentioned right at the beginning of docs: abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol).

So, in your example:

>>> import math
... 
... x = 1e-320
... rel_tol = 1e-9
... abs_tol = math.nextafter(1, 0)
... 
>>> math.isclose(x, 0, rel_tol=rel_tol, abs_tol=abs_tol)
True
>>> abs(x) <= max(rel_tol * abs(x), abs_tol)
True

@hoodmane fix looks good for me. Though, I'm not sure we need such pedancy in docs.

[picnixz]

Thank you for the report and the fix!