1.. testsetup:: 2 3 import math 4 from fractions import Fraction 5 6.. _tut-fp-issues: 7 8************************************************** 9Floating-Point Arithmetic: Issues and Limitations 10************************************************** 11 12.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net> 13.. sectionauthor:: Raymond Hettinger <python at rcn dot com> 14 15 16Floating-point numbers are represented in computer hardware as base 2 (binary) 17fractions. For example, the **decimal** fraction ``0.625`` 18has value 6/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.101`` 19has value 1/2 + 0/4 + 1/8. These two fractions have identical values, the only 20real difference being that the first is written in base 10 fractional notation, 21and the second in base 2. 22 23Unfortunately, most decimal fractions cannot be represented exactly as binary 24fractions. A consequence is that, in general, the decimal floating-point 25numbers you enter are only approximated by the binary floating-point numbers 26actually stored in the machine. 27 28The problem is easier to understand at first in base 10. Consider the fraction 291/3. You can approximate that as a base 10 fraction:: 30 31 0.3 32 33or, better, :: 34 35 0.33 36 37or, better, :: 38 39 0.333 40 41and so on. No matter how many digits you're willing to write down, the result 42will never be exactly 1/3, but will be an increasingly better approximation of 431/3. 44 45In the same way, no matter how many base 2 digits you're willing to use, the 46decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base 472, 1/10 is the infinitely repeating fraction :: 48 49 0.0001100110011001100110011001100110011001100110011... 50 51Stop at any finite number of bits, and you get an approximation. On most 52machines today, floats are approximated using a binary fraction with 53the numerator using the first 53 bits starting with the most significant bit and 54with the denominator as a power of two. In the case of 1/10, the binary fraction 55is ``3602879701896397 / 2 ** 55`` which is close to but not exactly 56equal to the true value of 1/10. 57 58Many users are not aware of the approximation because of the way values are 59displayed. Python only prints a decimal approximation to the true decimal 60value of the binary approximation stored by the machine. On most machines, if 61Python were to print the true decimal value of the binary approximation stored 62for 0.1, it would have to display:: 63 64 >>> 0.1 65 0.1000000000000000055511151231257827021181583404541015625 66 67That is more digits than most people find useful, so Python keeps the number 68of digits manageable by displaying a rounded value instead: 69 70.. doctest:: 71 72 >>> 1 / 10 73 0.1 74 75Just remember, even though the printed result looks like the exact value 76of 1/10, the actual stored value is the nearest representable binary fraction. 77 78Interestingly, there are many different decimal numbers that share the same 79nearest approximate binary fraction. For example, the numbers ``0.1`` and 80``0.10000000000000001`` and 81``0.1000000000000000055511151231257827021181583404541015625`` are all 82approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal 83values share the same approximation, any one of them could be displayed 84while still preserving the invariant ``eval(repr(x)) == x``. 85 86Historically, the Python prompt and built-in :func:`repr` function would choose 87the one with 17 significant digits, ``0.10000000000000001``. Starting with 88Python 3.1, Python (on most systems) is now able to choose the shortest of 89these and simply display ``0.1``. 90 91Note that this is in the very nature of binary floating point: this is not a bug 92in Python, and it is not a bug in your code either. You'll see the same kind of 93thing in all languages that support your hardware's floating-point arithmetic 94(although some languages may not *display* the difference by default, or in all 95output modes). 96 97For more pleasant output, you may wish to use string formatting to produce a 98limited number of significant digits: 99 100.. doctest:: 101 102 >>> format(math.pi, '.12g') # give 12 significant digits 103 '3.14159265359' 104 105 >>> format(math.pi, '.2f') # give 2 digits after the point 106 '3.14' 107 108 >>> repr(math.pi) 109 '3.141592653589793' 110 111It's important to realize that this is, in a real sense, an illusion: you're 112simply rounding the *display* of the true machine value. 113 114One illusion may beget another. For example, since 0.1 is not exactly 1/10, 115summing three values of 0.1 may not yield exactly 0.3, either: 116 117.. doctest:: 118 119 >>> 0.1 + 0.1 + 0.1 == 0.3 120 False 121 122Also, since the 0.1 cannot get any closer to the exact value of 1/10 and 1230.3 cannot get any closer to the exact value of 3/10, then pre-rounding with 124:func:`round` function cannot help: 125 126.. doctest:: 127 128 >>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1) 129 False 130 131Though the numbers cannot be made closer to their intended exact values, 132the :func:`math.isclose` function can be useful for comparing inexact values: 133 134.. doctest:: 135 136 >>> math.isclose(0.1 + 0.1 + 0.1, 0.3) 137 True 138 139Alternatively, the :func:`round` function can be used to compare rough 140approximations: 141 142.. doctest:: 143 144 >>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2) 145 True 146 147Binary floating-point arithmetic holds many surprises like this. The problem 148with "0.1" is explained in precise detail below, in the "Representation Error" 149section. See `Examples of Floating Point Problems 150<https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/>`_ for 151a pleasant summary of how binary floating point works and the kinds of 152problems commonly encountered in practice. Also see 153`The Perils of Floating Point <http://www.indowsway.com/floatingpoint.htm>`_ 154for a more complete account of other common surprises. 155 156As that says near the end, "there are no easy answers." Still, don't be unduly 157wary of floating point! The errors in Python float operations are inherited 158from the floating-point hardware, and on most machines are on the order of no 159more than 1 part in 2\*\*53 per operation. That's more than adequate for most 160tasks, but you do need to keep in mind that it's not decimal arithmetic and 161that every float operation can suffer a new rounding error. 162 163While pathological cases do exist, for most casual use of floating-point 164arithmetic you'll see the result you expect in the end if you simply round the 165display of your final results to the number of decimal digits you expect. 166:func:`str` usually suffices, and for finer control see the :meth:`str.format` 167method's format specifiers in :ref:`formatstrings`. 168 169For use cases which require exact decimal representation, try using the 170:mod:`decimal` module which implements decimal arithmetic suitable for 171accounting applications and high-precision applications. 172 173Another form of exact arithmetic is supported by the :mod:`fractions` module 174which implements arithmetic based on rational numbers (so the numbers like 1751/3 can be represented exactly). 176 177If you are a heavy user of floating-point operations you should take a look 178at the NumPy package and many other packages for mathematical and 179statistical operations supplied by the SciPy project. See <https://scipy.org>. 180 181Python provides tools that may help on those rare occasions when you really 182*do* want to know the exact value of a float. The 183:meth:`float.as_integer_ratio` method expresses the value of a float as a 184fraction: 185 186.. doctest:: 187 188 >>> x = 3.14159 189 >>> x.as_integer_ratio() 190 (3537115888337719, 1125899906842624) 191 192Since the ratio is exact, it can be used to losslessly recreate the 193original value: 194 195.. doctest:: 196 197 >>> x == 3537115888337719 / 1125899906842624 198 True 199 200The :meth:`float.hex` method expresses a float in hexadecimal (base 20116), again giving the exact value stored by your computer: 202 203.. doctest:: 204 205 >>> x.hex() 206 '0x1.921f9f01b866ep+1' 207 208This precise hexadecimal representation can be used to reconstruct 209the float value exactly: 210 211.. doctest:: 212 213 >>> x == float.fromhex('0x1.921f9f01b866ep+1') 214 True 215 216Since the representation is exact, it is useful for reliably porting values 217across different versions of Python (platform independence) and exchanging 218data with other languages that support the same format (such as Java and C99). 219 220Another helpful tool is the :func:`sum` function which helps mitigate 221loss-of-precision during summation. It uses extended precision for 222intermediate rounding steps as values are added onto a running total. 223That can make a difference in overall accuracy so that the errors do not 224accumulate to the point where they affect the final total: 225 226.. doctest:: 227 228 >>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0 229 False 230 >>> sum([0.1] * 10) == 1.0 231 True 232 233The :func:`math.fsum` goes further and tracks all of the "lost digits" 234as values are added onto a running total so that the result has only a 235single rounding. This is slower than :func:`sum` but will be more 236accurate in uncommon cases where large magnitude inputs mostly cancel 237each other out leaving a final sum near zero: 238 239.. doctest:: 240 241 >>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16, 242 ... -143401161400469.7, 266262841.31058735, -0.003244936839808227] 243 >>> float(sum(map(Fraction, arr))) # Exact summation with single rounding 244 8.042173697819788e-13 245 >>> math.fsum(arr) # Single rounding 246 8.042173697819788e-13 247 >>> sum(arr) # Multiple roundings in extended precision 248 8.042178034628478e-13 249 >>> total = 0.0 250 >>> for x in arr: 251 ... total += x # Multiple roundings in standard precision 252 ... 253 >>> total # Straight addition has no correct digits! 254 -0.0051575902860057365 255 256 257.. _tut-fp-error: 258 259Representation Error 260==================== 261 262This section explains the "0.1" example in detail, and shows how you can perform 263an exact analysis of cases like this yourself. Basic familiarity with binary 264floating-point representation is assumed. 265 266:dfn:`Representation error` refers to the fact that some (most, actually) 267decimal fractions cannot be represented exactly as binary (base 2) fractions. 268This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many 269others) often won't display the exact decimal number you expect. 270 271Why is that? 1/10 is not exactly representable as a binary fraction. Since at 272least 2000, almost all machines use IEEE 754 binary floating-point arithmetic, 273and almost all platforms map Python floats to IEEE 754 binary64 "double 274precision" values. IEEE 754 binary64 values contain 53 bits of precision, so 275on input the computer strives to convert 0.1 to the closest fraction it can of 276the form *J*/2**\ *N* where *J* is an integer containing exactly 53 bits. 277Rewriting 278:: 279 280 1 / 10 ~= J / (2**N) 281 282as :: 283 284 J ~= 2**N / 10 285 286and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``), 287the best value for *N* is 56: 288 289.. doctest:: 290 291 >>> 2**52 <= 2**56 // 10 < 2**53 292 True 293 294That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The 295best possible value for *J* is then that quotient rounded: 296 297.. doctest:: 298 299 >>> q, r = divmod(2**56, 10) 300 >>> r 301 6 302 303Since the remainder is more than half of 10, the best approximation is obtained 304by rounding up: 305 306.. doctest:: 307 308 309 310 >>> q+1 311 7205759403792794 312 313Therefore the best possible approximation to 1/10 in IEEE 754 double precision 314is:: 315 316 7205759403792794 / 2 ** 56 317 318Dividing both the numerator and denominator by two reduces the fraction to:: 319 320 3602879701896397 / 2 ** 55 321 322Note that since we rounded up, this is actually a little bit larger than 1/10; 323if we had not rounded up, the quotient would have been a little bit smaller than 3241/10. But in no case can it be *exactly* 1/10! 325 326So the computer never "sees" 1/10: what it sees is the exact fraction given 327above, the best IEEE 754 double approximation it can get: 328 329.. doctest:: 330 331 >>> 0.1 * 2 ** 55 332 3602879701896397.0 333 334If we multiply that fraction by 10\*\*55, we can see the value out to 33555 decimal digits: 336 337.. doctest:: 338 339 >>> 3602879701896397 * 10 ** 55 // 2 ** 55 340 1000000000000000055511151231257827021181583404541015625 341 342meaning that the exact number stored in the computer is equal to 343the decimal value 0.1000000000000000055511151231257827021181583404541015625. 344Instead of displaying the full decimal value, many languages (including 345older versions of Python), round the result to 17 significant digits: 346 347.. doctest:: 348 349 >>> format(0.1, '.17f') 350 '0.10000000000000001' 351 352The :mod:`fractions` and :mod:`decimal` modules make these calculations 353easy: 354 355.. doctest:: 356 357 >>> from decimal import Decimal 358 >>> from fractions import Fraction 359 360 >>> Fraction.from_float(0.1) 361 Fraction(3602879701896397, 36028797018963968) 362 363 >>> (0.1).as_integer_ratio() 364 (3602879701896397, 36028797018963968) 365 366 >>> Decimal.from_float(0.1) 367 Decimal('0.1000000000000000055511151231257827021181583404541015625') 368 369 >>> format(Decimal.from_float(0.1), '.17') 370 '0.10000000000000001' 371