| .. testsetup:: | |
| import math | |
| from fractions import Fraction | |
| .. _tut-fp-issues: | |
| ************************************************** | |
| Floating-Point Arithmetic: Issues and Limitations | |
| ************************************************** | |
| Floating-point numbers are represented in computer hardware as base 2 (binary) | |
| fractions. For example, the **decimal** fraction ``0.625`` | |
| has value 6/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.101`` | |
| has value 1/2 + 0/4 + 1/8. These two fractions have identical values, the only | |
| real difference being that the first is written in base 10 fractional notation, | |
| and the second in base 2. | |
| Unfortunately, most decimal fractions cannot be represented exactly as binary | |
| fractions. A consequence is that, in general, the decimal floating-point | |
| numbers you enter are only approximated by the binary floating-point numbers | |
| actually stored in the machine. | |
| The problem is easier to understand at first in base 10. Consider the fraction | |
| 1/3. You can approximate that as a base 10 fraction:: | |
| 0.3 | |
| or, better, :: | |
| 0.33 | |
| or, better, :: | |
| 0.333 | |
| and so on. No matter how many digits you're willing to write down, the result | |
| will never be exactly 1/3, but will be an increasingly better approximation of | |
| 1/3. | |
| In the same way, no matter how many base 2 digits you're willing to use, the | |
| decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base | |
| 2, 1/10 is the infinitely repeating fraction :: | |
| 0.0001100110011001100110011001100110011001100110011... | |
| Stop at any finite number of bits, and you get an approximation. On most | |
| machines today, floats are approximated using a binary fraction with | |
| the numerator using the first 53 bits starting with the most significant bit and | |
| with the denominator as a power of two. In the case of 1/10, the binary fraction | |
| is ``3602879701896397 / 2 ** 55`` which is close to but not exactly | |
| equal to the true value of 1/10. | |
| Many users are not aware of the approximation because of the way values are | |
| displayed. Python only prints a decimal approximation to the true decimal | |
| value of the binary approximation stored by the machine. On most machines, if | |
| Python were to print the true decimal value of the binary approximation stored | |
| for 0.1, it would have to display:: | |
| >>> 0.1 | |
| 0.1000000000000000055511151231257827021181583404541015625 | |
| That is more digits than most people find useful, so Python keeps the number | |
| of digits manageable by displaying a rounded value instead: | |
| .. doctest:: | |
| >>> 1 / 10 | |
| 0.1 | |
| Just remember, even though the printed result looks like the exact value | |
| of 1/10, the actual stored value is the nearest representable binary fraction. | |
| Interestingly, there are many different decimal numbers that share the same | |
| nearest approximate binary fraction. For example, the numbers ``0.1`` and | |
| ``0.10000000000000001`` and | |
| ``0.1000000000000000055511151231257827021181583404541015625`` are all | |
| approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal | |
| values share the same approximation, any one of them could be displayed | |
| while still preserving the invariant ``eval(repr(x)) == x``. | |
| Historically, the Python prompt and built-in :func:`repr` function would choose | |
| the one with 17 significant digits, ``0.10000000000000001``. Starting with | |
| Python 3.1, Python (on most systems) is now able to choose the shortest of | |
| these and simply display ``0.1``. | |
| Note that this is in the very nature of binary floating point: this is not a bug | |
| in Python, and it is not a bug in your code either. You'll see the same kind of | |
| thing in all languages that support your hardware's floating-point arithmetic | |
| (although some languages may not *display* the difference by default, or in all | |
| output modes). | |
| For more pleasant output, you may wish to use string formatting to produce a | |
| limited number of significant digits: | |
| .. doctest:: | |
| >>> format(math.pi, '.12g') # give 12 significant digits | |
| '3.14159265359' | |
| >>> format(math.pi, '.2f') # give 2 digits after the point | |
| '3.14' | |
| >>> repr(math.pi) | |
| '3.141592653589793' | |
| It's important to realize that this is, in a real sense, an illusion: you're | |
| simply rounding the *display* of the true machine value. | |
| One illusion may beget another. For example, since 0.1 is not exactly 1/10, | |
| summing three values of 0.1 may not yield exactly 0.3, either: | |
| .. doctest:: | |
| >>> 0.1 + 0.1 + 0.1 == 0.3 | |
| False | |
| Also, since the 0.1 cannot get any closer to the exact value of 1/10 and | |
| 0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with | |
| :func:`round` function cannot help: | |
| .. doctest:: | |
| >>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1) | |
| False | |
| Though the numbers cannot be made closer to their intended exact values, | |
| the :func:`math.isclose` function can be useful for comparing inexact values: | |
| .. doctest:: | |
| >>> math.isclose(0.1 + 0.1 + 0.1, 0.3) | |
| True | |
| Alternatively, the :func:`round` function can be used to compare rough | |
| approximations: | |
| .. doctest:: | |
| >>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2) | |
| True | |
| Binary floating-point arithmetic holds many surprises like this. The problem | |
| with "0.1" is explained in precise detail below, in the "Representation Error" | |
| section. See `Examples of Floating Point Problems | |
| <https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/>`_ for | |
| a pleasant summary of how binary floating point works and the kinds of | |
| problems commonly encountered in practice. Also see | |
| `The Perils of Floating Point <http://www.indowsway.com/floatingpoint.htm>`_ | |
| for a more complete account of other common surprises. | |
| As that says near the end, "there are no easy answers." Still, don't be unduly | |
| wary of floating point! The errors in Python float operations are inherited | |
| from the floating-point hardware, and on most machines are on the order of no | |
| more than 1 part in 2\*\*53 per operation. That's more than adequate for most | |
| tasks, but you do need to keep in mind that it's not decimal arithmetic and | |
| that every float operation can suffer a new rounding error. | |
| While pathological cases do exist, for most casual use of floating-point | |
| arithmetic you'll see the result you expect in the end if you simply round the | |
| display of your final results to the number of decimal digits you expect. | |
| :func:`str` usually suffices, and for finer control see the :meth:`str.format` | |
| method's format specifiers in :ref:`formatstrings`. | |
| For use cases which require exact decimal representation, try using the | |
| :mod:`decimal` module which implements decimal arithmetic suitable for | |
| accounting applications and high-precision applications. | |
| Another form of exact arithmetic is supported by the :mod:`fractions` module | |
| which implements arithmetic based on rational numbers (so the numbers like | |
| 1/3 can be represented exactly). | |
| If you are a heavy user of floating-point operations you should take a look | |
| at the NumPy package and many other packages for mathematical and | |
| statistical operations supplied by the SciPy project. See <https://scipy.org>. | |
| Python provides tools that may help on those rare occasions when you really | |
| *do* want to know the exact value of a float. The | |
| :meth:`float.as_integer_ratio` method expresses the value of a float as a | |
| fraction: | |
| .. doctest:: | |
| >>> x = 3.14159 | |
| >>> x.as_integer_ratio() | |
| (3537115888337719, 1125899906842624) | |
| Since the ratio is exact, it can be used to losslessly recreate the | |
| original value: | |
| .. doctest:: | |
| >>> x == 3537115888337719 / 1125899906842624 | |
| True | |
| The :meth:`float.hex` method expresses a float in hexadecimal (base | |
| 16), again giving the exact value stored by your computer: | |
| .. doctest:: | |
| >>> x.hex() | |
| '0x1.921f9f01b866ep+1' | |
| This precise hexadecimal representation can be used to reconstruct | |
| the float value exactly: | |
| .. doctest:: | |
| >>> x == float.fromhex('0x1.921f9f01b866ep+1') | |
| True | |
| Since the representation is exact, it is useful for reliably porting values | |
| across different versions of Python (platform independence) and exchanging | |
| data with other languages that support the same format (such as Java and C99). | |
| Another helpful tool is the :func:`sum` function which helps mitigate | |
| loss-of-precision during summation. It uses extended precision for | |
| intermediate rounding steps as values are added onto a running total. | |
| That can make a difference in overall accuracy so that the errors do not | |
| accumulate to the point where they affect the final total: | |
| .. doctest:: | |
| >>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0 | |
| False | |
| >>> sum([0.1] * 10) == 1.0 | |
| True | |
| The :func:`math.fsum` goes further and tracks all of the "lost digits" | |
| as values are added onto a running total so that the result has only a | |
| single rounding. This is slower than :func:`sum` but will be more | |
| accurate in uncommon cases where large magnitude inputs mostly cancel | |
| each other out leaving a final sum near zero: | |
| .. doctest:: | |
| >>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16, | |
| ... -143401161400469.7, 266262841.31058735, -0.003244936839808227] | |
| >>> float(sum(map(Fraction, arr))) # Exact summation with single rounding | |
| 8.042173697819788e-13 | |
| >>> math.fsum(arr) # Single rounding | |
| 8.042173697819788e-13 | |
| >>> sum(arr) # Multiple roundings in extended precision | |
| 8.042178034628478e-13 | |
| >>> total = 0.0 | |
| >>> for x in arr: | |
| ... total += x # Multiple roundings in standard precision | |
| ... | |
| >>> total # Straight addition has no correct digits! | |
| -0.0051575902860057365 | |
| .. _tut-fp-error: | |
| Representation Error | |
| ==================== | |
| This section explains the "0.1" example in detail, and shows how you can perform | |
| an exact analysis of cases like this yourself. Basic familiarity with binary | |
| floating-point representation is assumed. | |
| :dfn:`Representation error` refers to the fact that some (most, actually) | |
| decimal fractions cannot be represented exactly as binary (base 2) fractions. | |
| This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many | |
| others) often won't display the exact decimal number you expect. | |
| Why is that? 1/10 is not exactly representable as a binary fraction. Since at | |
| least 2000, almost all machines use IEEE 754 binary floating-point arithmetic, | |
| and almost all platforms map Python floats to IEEE 754 binary64 "double | |
| precision" values. IEEE 754 binary64 values contain 53 bits of precision, so | |
| on input the computer strives to convert 0.1 to the closest fraction it can of | |
| the form *J*/2**\ *N* where *J* is an integer containing exactly 53 bits. | |
| Rewriting | |
| :: | |
| 1 / 10 ~= J / (2**N) | |
| as :: | |
| J ~= 2**N / 10 | |
| and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``), | |
| the best value for *N* is 56: | |
| .. doctest:: | |
| >>> 2**52 <= 2**56 // 10 < 2**53 | |
| True | |
| That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The | |
| best possible value for *J* is then that quotient rounded: | |
| .. doctest:: | |
| >>> q, r = divmod(2**56, 10) | |
| >>> r | |
| 6 | |
| Since the remainder is more than half of 10, the best approximation is obtained | |
| by rounding up: | |
| .. doctest:: | |
| >>> q+1 | |
| 7205759403792794 | |
| Therefore the best possible approximation to 1/10 in IEEE 754 double precision | |
| is:: | |
| 7205759403792794 / 2 ** 56 | |
| Dividing both the numerator and denominator by two reduces the fraction to:: | |
| 3602879701896397 / 2 ** 55 | |
| Note that since we rounded up, this is actually a little bit larger than 1/10; | |
| if we had not rounded up, the quotient would have been a little bit smaller than | |
| 1/10. But in no case can it be *exactly* 1/10! | |
| So the computer never "sees" 1/10: what it sees is the exact fraction given | |
| above, the best IEEE 754 double approximation it can get: | |
| .. doctest:: | |
| >>> 0.1 * 2 ** 55 | |
| 3602879701896397.0 | |
| If we multiply that fraction by 10\*\*55, we can see the value out to | |
| 55 decimal digits: | |
| .. doctest:: | |
| >>> 3602879701896397 * 10 ** 55 // 2 ** 55 | |
| 1000000000000000055511151231257827021181583404541015625 | |
| meaning that the exact number stored in the computer is equal to | |
| the decimal value 0.1000000000000000055511151231257827021181583404541015625. | |
| Instead of displaying the full decimal value, many languages (including | |
| older versions of Python), round the result to 17 significant digits: | |
| .. doctest:: | |
| >>> format(0.1, '.17f') | |
| '0.10000000000000001' | |
| The :mod:`fractions` and :mod:`decimal` modules make these calculations | |
| easy: | |
| .. doctest:: | |
| >>> from decimal import Decimal | |
| >>> from fractions import Fraction | |
| >>> Fraction.from_float(0.1) | |
| Fraction(3602879701896397, 36028797018963968) | |
| >>> (0.1).as_integer_ratio() | |
| (3602879701896397, 36028797018963968) | |
| >>> Decimal.from_float(0.1) | |
| Decimal('0.1000000000000000055511151231257827021181583404541015625') | |
| >>> format(Decimal.from_float(0.1), '.17') | |
| '0.10000000000000001' |