1.. testsetup:: 2 3 import math 4 5.. _tut-fp-issues: 6 7************************************************** 8Floating Point Arithmetic: Issues and Limitations 9************************************************** 10 11.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net> 12 13 14Floating-point numbers are represented in computer hardware as base 2 (binary) 15fractions. For example, the **decimal** fraction ``0.125`` 16has value 1/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.001`` 17has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only 18real difference being that the first is written in base 10 fractional notation, 19and the second in base 2. 20 21Unfortunately, most decimal fractions cannot be represented exactly as binary 22fractions. A consequence is that, in general, the decimal floating-point 23numbers you enter are only approximated by the binary floating-point numbers 24actually stored in the machine. 25 26The problem is easier to understand at first in base 10. Consider the fraction 271/3. You can approximate that as a base 10 fraction:: 28 29 0.3 30 31or, better, :: 32 33 0.33 34 35or, better, :: 36 37 0.333 38 39and so on. No matter how many digits you're willing to write down, the result 40will never be exactly 1/3, but will be an increasingly better approximation of 411/3. 42 43In the same way, no matter how many base 2 digits you're willing to use, the 44decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base 452, 1/10 is the infinitely repeating fraction :: 46 47 0.0001100110011001100110011001100110011001100110011... 48 49Stop at any finite number of bits, and you get an approximation. On most 50machines today, floats are approximated using a binary fraction with 51the numerator using the first 53 bits starting with the most significant bit and 52with the denominator as a power of two. In the case of 1/10, the binary fraction 53is ``3602879701896397 / 2 ** 55`` which is close to but not exactly 54equal to the true value of 1/10. 55 56Many users are not aware of the approximation because of the way values are 57displayed. Python only prints a decimal approximation to the true decimal 58value of the binary approximation stored by the machine. On most machines, if 59Python were to print the true decimal value of the binary approximation stored 60for 0.1, it would have to display :: 61 62 >>> 0.1 63 0.1000000000000000055511151231257827021181583404541015625 64 65That is more digits than most people find useful, so Python keeps the number 66of digits manageable by displaying a rounded value instead :: 67 68 >>> 1 / 10 69 0.1 70 71Just remember, even though the printed result looks like the exact value 72of 1/10, the actual stored value is the nearest representable binary fraction. 73 74Interestingly, there are many different decimal numbers that share the same 75nearest approximate binary fraction. For example, the numbers ``0.1`` and 76``0.10000000000000001`` and 77``0.1000000000000000055511151231257827021181583404541015625`` are all 78approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal 79values share the same approximation, any one of them could be displayed 80while still preserving the invariant ``eval(repr(x)) == x``. 81 82Historically, the Python prompt and built-in :func:`repr` function would choose 83the one with 17 significant digits, ``0.10000000000000001``. Starting with 84Python 3.1, Python (on most systems) is now able to choose the shortest of 85these and simply display ``0.1``. 86 87Note that this is in the very nature of binary floating-point: this is not a bug 88in Python, and it is not a bug in your code either. You'll see the same kind of 89thing in all languages that support your hardware's floating-point arithmetic 90(although some languages may not *display* the difference by default, or in all 91output modes). 92 93For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits:: 94 95 >>> format(math.pi, '.12g') # give 12 significant digits 96 '3.14159265359' 97 98 >>> format(math.pi, '.2f') # give 2 digits after the point 99 '3.14' 100 101 >>> repr(math.pi) 102 '3.141592653589793' 103 104 105It's important to realize that this is, in a real sense, an illusion: you're 106simply rounding the *display* of the true machine value. 107 108One illusion may beget another. For example, since 0.1 is not exactly 1/10, 109summing three values of 0.1 may not yield exactly 0.3, either:: 110 111 >>> .1 + .1 + .1 == .3 112 False 113 114Also, since the 0.1 cannot get any closer to the exact value of 1/10 and 1150.3 cannot get any closer to the exact value of 3/10, then pre-rounding with 116:func:`round` function cannot help:: 117 118 >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1) 119 False 120 121Though the numbers cannot be made closer to their intended exact values, 122the :func:`round` function can be useful for post-rounding so that results 123with inexact values become comparable to one another:: 124 125 >>> round(.1 + .1 + .1, 10) == round(.3, 10) 126 True 127 128Binary floating-point arithmetic holds many surprises like this. The problem 129with "0.1" is explained in precise detail below, in the "Representation Error" 130section. See `Examples of Floating Point Problems 131<https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/>`_ for 132a pleasant summary of how binary floating-point works and the kinds of 133problems commonly encountered in practice. Also see 134`The Perils of Floating Point <https://www.lahey.com/float.htm>`_ 135for a more complete account of other common surprises. 136 137As that says near the end, "there are no easy answers." Still, don't be unduly 138wary of floating-point! The errors in Python float operations are inherited 139from the floating-point hardware, and on most machines are on the order of no 140more than 1 part in 2\*\*53 per operation. That's more than adequate for most 141tasks, but you do need to keep in mind that it's not decimal arithmetic and 142that every float operation can suffer a new rounding error. 143 144While pathological cases do exist, for most casual use of floating-point 145arithmetic you'll see the result you expect in the end if you simply round the 146display of your final results to the number of decimal digits you expect. 147:func:`str` usually suffices, and for finer control see the :meth:`str.format` 148method's format specifiers in :ref:`formatstrings`. 149 150For use cases which require exact decimal representation, try using the 151:mod:`decimal` module which implements decimal arithmetic suitable for 152accounting applications and high-precision applications. 153 154Another form of exact arithmetic is supported by the :mod:`fractions` module 155which implements arithmetic based on rational numbers (so the numbers like 1561/3 can be represented exactly). 157 158If you are a heavy user of floating-point operations you should take a look 159at the NumPy package and many other packages for mathematical and 160statistical operations supplied by the SciPy project. See <https://scipy.org>. 161 162Python provides tools that may help on those rare occasions when you really 163*do* want to know the exact value of a float. The 164:meth:`float.as_integer_ratio` method expresses the value of a float as a 165fraction:: 166 167 >>> x = 3.14159 168 >>> x.as_integer_ratio() 169 (3537115888337719, 1125899906842624) 170 171Since the ratio is exact, it can be used to losslessly recreate the 172original value:: 173 174 >>> x == 3537115888337719 / 1125899906842624 175 True 176 177The :meth:`float.hex` method expresses a float in hexadecimal (base 17816), again giving the exact value stored by your computer:: 179 180 >>> x.hex() 181 '0x1.921f9f01b866ep+1' 182 183This precise hexadecimal representation can be used to reconstruct 184the float value exactly:: 185 186 >>> x == float.fromhex('0x1.921f9f01b866ep+1') 187 True 188 189Since the representation is exact, it is useful for reliably porting values 190across different versions of Python (platform independence) and exchanging 191data with other languages that support the same format (such as Java and C99). 192 193Another helpful tool is the :func:`math.fsum` function which helps mitigate 194loss-of-precision during summation. It tracks "lost digits" as values are 195added onto a running total. That can make a difference in overall accuracy 196so that the errors do not accumulate to the point where they affect the 197final total: 198 199 >>> sum([0.1] * 10) == 1.0 200 False 201 >>> math.fsum([0.1] * 10) == 1.0 202 True 203 204.. _tut-fp-error: 205 206Representation Error 207==================== 208 209This section explains the "0.1" example in detail, and shows how you can perform 210an exact analysis of cases like this yourself. Basic familiarity with binary 211floating-point representation is assumed. 212 213:dfn:`Representation error` refers to the fact that some (most, actually) 214decimal fractions cannot be represented exactly as binary (base 2) fractions. 215This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many 216others) often won't display the exact decimal number you expect. 217 218Why is that? 1/10 is not exactly representable as a binary fraction. Since at 219least 2000, almost all machines use IEEE 754 binary floating-point arithmetic, 220and almost all platforms map Python floats to IEEE 754 binary64 "double 221precision" values. IEEE 754 binary64 values contain 53 bits of precision, so 222on input the computer strives to convert 0.1 to the closest fraction it can of 223the form *J*/2**\ *N* where *J* is an integer containing exactly 53 bits. 224Rewriting 225:: 226 227 1 / 10 ~= J / (2**N) 228 229as :: 230 231 J ~= 2**N / 10 232 233and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``), 234the best value for *N* is 56:: 235 236 >>> 2**52 <= 2**56 // 10 < 2**53 237 True 238 239That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The 240best possible value for *J* is then that quotient rounded:: 241 242 >>> q, r = divmod(2**56, 10) 243 >>> r 244 6 245 246Since the remainder is more than half of 10, the best approximation is obtained 247by rounding up:: 248 249 >>> q+1 250 7205759403792794 251 252Therefore the best possible approximation to 1/10 in IEEE 754 double precision 253is:: 254 255 7205759403792794 / 2 ** 56 256 257Dividing both the numerator and denominator by two reduces the fraction to:: 258 259 3602879701896397 / 2 ** 55 260 261Note that since we rounded up, this is actually a little bit larger than 1/10; 262if we had not rounded up, the quotient would have been a little bit smaller than 2631/10. But in no case can it be *exactly* 1/10! 264 265So the computer never "sees" 1/10: what it sees is the exact fraction given 266above, the best IEEE 754 double approximation it can get: 267 268 >>> 0.1 * 2 ** 55 269 3602879701896397.0 270 271If we multiply that fraction by 10\*\*55, we can see the value out to 27255 decimal digits:: 273 274 >>> 3602879701896397 * 10 ** 55 // 2 ** 55 275 1000000000000000055511151231257827021181583404541015625 276 277meaning that the exact number stored in the computer is equal to 278the decimal value 0.1000000000000000055511151231257827021181583404541015625. 279Instead of displaying the full decimal value, many languages (including 280older versions of Python), round the result to 17 significant digits:: 281 282 >>> format(0.1, '.17f') 283 '0.10000000000000001' 284 285The :mod:`fractions` and :mod:`decimal` modules make these calculations 286easy:: 287 288 >>> from decimal import Decimal 289 >>> from fractions import Fraction 290 291 >>> Fraction.from_float(0.1) 292 Fraction(3602879701896397, 36028797018963968) 293 294 >>> (0.1).as_integer_ratio() 295 (3602879701896397, 36028797018963968) 296 297 >>> Decimal.from_float(0.1) 298 Decimal('0.1000000000000000055511151231257827021181583404541015625') 299 300 >>> format(Decimal.from_float(0.1), '.17') 301 '0.10000000000000001' 302