# floating point – What is the best way to compare floats for almost-equality in Python?

## The Question :

360 people think this question is useful

It’s well known that comparing floats for equality is a little fiddly due to rounding and precision issues.

What is the recommended way to deal with this in Python?

Surely there is a standard library function for this somewhere?

• @tolomea: Since it depends on your application and your data and your problem domain — and it’s only one line of code — why would there be a “standard library function”?
• @S.Lott: all, any, max, min are each basically one-liners, and they aren’t just provided in a library, they’re builtin functions. So the BDFL’s reasons aren’t that. The one line of code that most people write is pretty unsophisticated and often doesn’t work, which is a strong reason to provide something better. Of course any module providing other strategies would have to also provide caveats describing when they’re appropriate, and more importantly when they aren’t. Numeric analysis is hard, it’s no great disgrace that language designers usually don’t attempt tools to help with it.
• @Steve Jessop. Those collection-oriented functions don’t have the application, data and problem domain dependencies that float-point does. So the “one-liner” clearly isn’t as important as the real reasons. Numeric analysis is hard, and can’t be a first-class part of a general-purpose language library.
• @S.Lott: I’d probably agree if the standard Python distribution didn’t come with multiple modules for XML interfaces. Clearly the fact that different applications need to do something differently is no bar at all to putting modules in the base set to do it one way or another. Certainly there are tricks for comparing floats that get re-used a lot, the most basic being a specified number of ulps. So I only partially agree – the problem is that numeric analysis is hard. Python could in principle provide tools to make it somewhat easier, some of the time. I guess nobody has volunteered.
• Also, “it boils down to one hard-to-design line of code” – if it’s still a one-liner once you’re doing it properly, I think your monitor is wider than mine ;-). Anyway, I think the whole area is quite specialized, in the sense that most programmers (including me) very rarely use it. Combined with being hard, it’s not going to hit the top of the “most wanted” list for core libraries in most languages.

354 people think this answer is useful

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.

78 people think this answer is useful

Is something as simple as the following not good enough?

return abs(f1 - f2) <= allowed_error



51 people think this answer is useful

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.

15 people think this answer is useful

Use Python’s decimal module, which provides the Decimal class.

It is worth noting that if you’re doing math-heavy 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.

12 people think this answer is useful

The common wisdom that floating-point numbers cannot be compared for equality is inaccurate. Floating-point 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 floating-point 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 floating-point 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 floating-point implementation using the IEEE 754 standard in round-to-nearest mode returns the floating-point 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 high-quality mathematics library. And, of course, you need precise control over exactly which operations are performed. High-level 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.)

11 people think this answer is useful

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(a-b) <= eps1*(abs(a)+abs(b)) + eps2 or similar. To get Dawson-like behaviour you might say something like if abs(a-b) <= 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.

8 people think this answer is useful

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 one-liner of Mark Ransom is that it can handle “inf” and “-inf” properly.

def isclose(a, b, rel_tol=1e-09, 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 non-negative")

# 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



6 people think this answer is useful

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



2 people think this answer is useful

I found the following comparison helpful:

str(f1) == str(f2)



2 people think this answer is useful

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 0.9999999999995e-4 < number <= 0.9999999999995e11 )

Example:

>>> isclose(10.0,10.049)
True
>>> isclose(10.0,10.05)
False



1 people think this answer is useful

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.

1 people think this answer is useful

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(f1-f2) < tol*max(abs(f1),abs(f2))



0 people think this answer is useful

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



0 people think this answer is useful

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 built-in 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 built-in 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 f-strings (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}'



if abs(a - b) <= error: