What is the best way to compare floats for almostequality in Python?
Question:
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/comparingfloatingpointnumbers2012edition/
What is the recommended way to deal with this in Python?
Is a standard library function for this somewhere?
Answers:
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 mathheavy 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.
I’m not aware of anything in the Python standard library (or elsewhere) that implements Dawson’s AlmostEqual2sComplement
function. If that’s the sort of behaviour you want, you’ll have to implement it yourself. (In which case, rather than using Dawson’s clever bitwise hacks you’d probably do better to use more conventional tests of the form if abs(ab) <= eps1*(abs(a)+abs(b)) + eps2
or similar. To get Dawsonlike behaviour you might say something like if abs(ab) <= eps*max(EPS,abs(a),abs(b))
for some small fixed EPS
; this isn’t exactly the same as Dawson, but it’s similar in spirit.
The common wisdom that floatingpoint numbers cannot be compared for equality is inaccurate. Floatingpoint numbers are no different from integers: If you evaluate “a == b”, you will get true if they are identical numbers and false otherwise (with the understanding that two NaNs are of course not identical numbers).
The actual problem is this: If I have done some calculations and am not sure the two numbers I have to compare are exactly correct, then what? This problem is the same for floatingpoint as it is for integers. If you evaluate the integer expression “7/3*3”, it will not compare equal to “7*3/3”.
So suppose we asked “How do I compare integers for equality?” in such a situation. There is no single answer; what you should do depends on the specific situation, notably what sort of errors you have and what you want to achieve.
Here are some possible choices.
If you want to get a “true” result if the mathematically exact numbers would be equal, then you might try to use the properties of the calculations you perform to prove that you get the same errors in the two numbers. If that is feasible, and you compare two numbers that result from expressions that would give equal numbers if computed exactly, then you will get “true” from the comparison. Another approach is that you might analyze the properties of the calculations and prove that the error never exceeds a certain amount, perhaps an absolute amount or an amount relative to one of the inputs or one of the outputs. In that case, you can ask whether the two calculated numbers differ by at most that amount, and return “true” if they are within the interval. If you cannot prove an error bound, you might guess and hope for the best. One way of guessing is to evaluate many random samples and see what sort of distribution you get in the results.
Of course, since we only set the requirement that you get “true” if the mathematically exact results are equal, we left open the possibility that you get “true” even if they are unequal. (In fact, we can satisfy the requirement by always returning “true”. This makes the calculation simple but is generally undesirable, so I will discuss improving the situation below.)
If you want to get a “false” result if the mathematically exact numbers would be unequal, you need to prove that your evaluation of the numbers yields different numbers if the mathematically exact numbers would be unequal. This may be impossible for practical purposes in many common situations. So let us consider an alternative.
A useful requirement might be that we get a “false” result if the mathematically exact numbers differ by more than a certain amount. For example, perhaps we are going to calculate where a ball thrown in a computer game traveled, and we want to know whether it struck a bat. In this case, we certainly want to get “true” if the ball strikes the bat, and we want to get “false” if the ball is far from the bat, and we can accept an incorrect “true” answer if the ball in a mathematically exact simulation missed the bat but is within a millimeter of hitting the bat. In that case, we need to prove (or guess/estimate) that our calculation of the ball’s position and the bat’s position have a combined error of at most one millimeter (for all positions of interest). This would allow us to always return “false” if the ball and bat are more than a millimeter apart, to return “true” if they touch, and to return “true” if they are close enough to be acceptable.
So, how you decide what to return when comparing floatingpoint numbers depends very much on your specific situation.
As to how you go about proving error bounds for calculations, that can be a complicated subject. Any floatingpoint implementation using the IEEE 754 standard in roundtonearest mode returns the floatingpoint number nearest to the exact result for any basic operation (notably multiplication, division, addition, subtraction, square root). (In case of tie, round so the low bit is even.) (Be particularly careful about square root and division; your language implementation might use methods that do not conform to IEEE 754 for those.) Because of this requirement, we know the error in a single result is at most 1/2 of the value of the least significant bit. (If it were more, the rounding would have gone to a different number that is within 1/2 the value.)
Going on from there gets substantially more complicated; the next step is performing an operation where one of the inputs already has some error. For simple expressions, these errors can be followed through the calculations to reach a bound on the final error. In practice, this is only done in a few situations, such as working on a highquality mathematics library. And, of course, you need precise control over exactly which operations are performed. Highlevel languages often give the compiler a lot of slack, so you might not know in which order operations are performed.
There is much more that could be (and is) written about this topic, but I have to stop there. In summary, the answer is: There is no library routine for this comparison because there is no single solution that fits most needs that is worth putting into a library routine. (If comparing with a relative or absolute error interval suffices for you, you can do it simply without a library routine.)
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=1e05, atol=1e08, equal_nan=False)
A little disclaimer though: installing NumPy can be a nontrivial experience depending on your platform.
I found the following comparison helpful:
str(f1) == str(f2)
For some of the cases where you can affect the source number representation, you can represent them as fractions instead of floats, using integer numerator and denominator. That way you can have exact comparisons.
See Fraction from fractions module for details.
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=1e09, abs_tol=0.0):
return abs(ab) <= 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 asis in all cases. If the difference is less than either of those tolerances, the values are considered equal.
If you want to use it in testing/TDD context, I’d say this is a standard way:
from nose.tools import assert_almost_equals
assert_almost_equals(x, y, places=7) #default is 7
math.isclose() has been added to Python 3.5 for that (source code). Here is a port of it to Python 2. It’s difference from oneliner of Mark Ransom is that it can handle “inf” and “inf” properly.
def isclose(a, b, rel_tol=1e09, abs_tol=0.0):
'''
Python 2 implementation of Python 3.5 math.isclose()
https://hg.python.org/cpython/file/tip/Modules/mathmodule.c#l1993
'''
# sanity check on the inputs
if rel_tol < 0 or abs_tol < 0:
raise ValueError("tolerances must be nonnegative")
# short circuit exact equality  needed to catch two infinities of
# the same sign. And perhaps speeds things up a bit sometimes.
if a == b:
return True
# This catches the case of two infinities of opposite sign, or
# one infinity and one finite number. Two infinities of opposite
# sign would otherwise have an infinite relative tolerance.
# Two infinities of the same sign are caught by the equality check
# above.
if math.isinf(a) or math.isinf(b):
return False
# now do the regular computation
# this is essentially the "weak" test from the Boost library
diff = math.fabs(b  a)
result = (((diff <= math.fabs(rel_tol * b)) or
(diff <= math.fabs(rel_tol * a))) or
(diff <= abs_tol))
return result
This maybe is a bit ugly hack, but it works pretty well when you don’t need more than the default float precision (about 11 decimals).
The round_to function uses the format method from the builtin str class to round up the float to a string that represents the float with the number of decimals needed, and then applies the eval builtin function to the rounded float string to get back to the float numeric type.
The is_close function just applies a simple conditional to the rounded up float.
def round_to(float_num, prec):
return eval("'{:." + str(int(prec)) + "f}'.format(" + str(float_num) + ")")
def is_close(float_a, float_b, prec):
if round_to(float_a, prec) == round_to(float_b, prec):
return True
return False
>>>a = 10.0
10.0
>>>b = 10.0001
10.0001
>>>print is_close(a, b, prec=3)
True
>>>print is_close(a, b, prec=4)
False
Update:
As suggested by @stepehjfox, a cleaner way to build a rount_to function avoiding “eval” is using nested formatting:
def round_to(float_num, prec):
return '{:.{precision}f}'.format(float_num, precision=prec)
Following the same idea, the code can be even simpler using the great new fstrings (Python 3.6+):
def round_to(float_num, prec):
return f'{float_num:.{prec}f}'
So, we could even wrap it up all in one simple and clean ‘is_close’ function:
def is_close(a, b, prec):
return f'{a:.{prec}f}' == f'{b:.{prec}f}'
I liked @Sesquipedal ‘s suggestion but with modification (a special use case when both values are 0 returns False). In my case I was on Python 2.7 and just used a simple function:
if f1 ==0 and f2 == 0:
return True
else:
return abs(f1f2) < tol*max(abs(f1),abs(f2))
Useful for the case where you want to make sure 2 numbers are the same ‘up to precision’, no need to specify the tolerance:

Find minimum precision of the 2 numbers

Round both of them to minimum precision and compare
def isclose(a,b):
astr=str(a)
aprec=len(astr.split('.')[1]) if '.' in astr else 0
bstr=str(b)
bprec=len(bstr.split('.')[1]) if '.' in bstr else 0
prec=min(aprec,bprec)
return round(a,prec)==round(b,prec)
As written, only works for numbers without the ‘e’ in their string representation ( meaning 0.9999999999995e4 < number <= 0.9999999999995e11 )
Example:
>>> isclose(10.0,10.049)
True
>>> isclose(10.0,10.05)
False
To compare up to a given decimal without atol/rtol
:
def almost_equal(a, b, decimal=6):
return '{0:.{1}f}'.format(a, decimal) == '{0:.{1}f}'.format(b, decimal)
print(almost_equal(0.0, 0.0001, decimal=5)) # False
print(almost_equal(0.0, 0.0001, decimal=4)) # True
In terms of absolute error, you can just check
if abs(a  b) <= error:
print("Almost equal")
Some information of why float act weird in Python
https://youtu.be/v4HhvoNLILk?t=1129
You can also use math.isclose for relative errors
Use ==
is a simple good way, if you don’t care about tolerance precisely.
# Python 3.8.5
>>> 1.0000000000001 == 1
False
>>> 1.00000000000001 == 1
True
But watch out for 0
:
>>> 0 == 0.00000000000000000000000000000000000000000001
False
The 0
is always the zero.
Use math.isclose
if you want to control the tolerance.
The default a == b
is equivalent to math.isclose(a, b, rel_tol=1e16, abs_tol=0)
.
If you still want to use ==
with a selfdefined tolerance:
>>> class MyFloat(float):
def __eq__(self, another):
return math.isclose(self, another, rel_tol=0, abs_tol=0.001)
>>> a == MyFloat(0)
>>> a
0.0
>>> a == 0.001
True
So far, I didn’t find anywhere to config it globally for float
. Besides, mock
is also not working for float.__eq__
.
If you want to do it in a testing or TDD context using the pytest
package, here’s how:
import pytest
PRECISION = 1e3
def assert_almost_equal():
obtained_value = 99.99
expected_value = 100.00
assert obtained_value == pytest.approx(expected_value, PRECISION)
Pretty lame but it works anyway
def isEqual(f1, f2):
f1 = float(f1)
f2 = float(f2)
ans1 = f1f2
ans2 = f2f1
return (ans1 == ans2)
Usage
f1 = 4.000444
f2 = 4.0
print(isEqual(f1, f2))
print(isEqual(f2, f2))
You can also round off for approximation equality
import math
def isEqual(f1, f2):
f1 = math.ceil(f1)
f2 = math.ceil(f2)
ans1 = f1f2
ans2 = f2f1
return (ans1 == ans2)
what i don’t know if its efficient for your application!