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Georg Brandl116aa622007-08-15 14:28:22 +00001.. _tut-fp-issues:
2
3**************************************************
4Floating Point Arithmetic: Issues and Limitations
5**************************************************
6
7.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
8
9
10Floating-point numbers are represented in computer hardware as base 2 (binary)
11fractions. For example, the decimal fraction ::
12
13 0.125
14
15has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
16
17 0.001
18
19has value 0/2 + 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
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000051Stop at any finite number of bits, and you get an approximation. On most
52machines today, floats are approximated using a binary fraction with
Raymond Hettinger1d180682009-06-28 23:21:38 +000053the numerator using the first 53 bits starting with the most significant bit and
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000054with 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.
Georg Brandl116aa622007-08-15 14:28:22 +000057
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000058Many 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 ::
Georg Brandl116aa622007-08-15 14:28:22 +000063
64 >>> 0.1
65 0.1000000000000000055511151231257827021181583404541015625
66
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000067That is more digits than most people find useful, so Python keeps the number
68of digits manageable by displaying a rounded value instead ::
Georg Brandl116aa622007-08-15 14:28:22 +000069
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000070 >>> 1 / 10
71 0.1
Georg Brandl116aa622007-08-15 14:28:22 +000072
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000073Just remember, even though the printed result looks like the exact value
74of 1/10, the actual stored value is the nearest representable binary fraction.
75
76Interestingly, there are many different decimal numbers that share the same
77nearest approximate binary fraction. For example, the numbers ``0.1`` and
78``0.10000000000000001`` and
79``0.1000000000000000055511151231257827021181583404541015625`` are all
80approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal
Georg Brandlb58f46f2009-04-24 19:06:29 +000081values share the same approximation, any one of them could be displayed
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000082while still preserving the invariant ``eval(repr(x)) == x``.
83
Mark Dickinson6a74d342010-07-29 13:56:56 +000084Historically, the Python prompt and built-in :func:`repr` function would choose
Raymond Hettingereafaf4c2009-06-28 22:30:13 +000085the one with 17 significant digits, ``0.10000000000000001``. Starting with
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000086Python 3.1, Python (on most systems) is now able to choose the shortest of
87these and simply display ``0.1``.
Georg Brandl116aa622007-08-15 14:28:22 +000088
89Note that this is in the very nature of binary floating-point: this is not a bug
90in Python, and it is not a bug in your code either. You'll see the same kind of
91thing in all languages that support your hardware's floating-point arithmetic
92(although some languages may not *display* the difference by default, or in all
93output modes).
94
Ezio Melottie130a522011-10-19 10:58:56 +030095For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits::
Georg Brandl116aa622007-08-15 14:28:22 +000096
Mark Dickinson388122d2010-08-04 20:56:28 +000097 >>> format(math.pi, '.12g') # give 12 significant digits
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000098 '3.14159265359'
Georg Brandl116aa622007-08-15 14:28:22 +000099
Mark Dickinson388122d2010-08-04 20:56:28 +0000100 >>> format(math.pi, '.2f') # give 2 digits after the point
101 '3.14'
102
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000103 >>> repr(math.pi)
104 '3.141592653589793'
105
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000106
107It's important to realize that this is, in a real sense, an illusion: you're
108simply rounding the *display* of the true machine value.
Georg Brandl116aa622007-08-15 14:28:22 +0000109
Raymond Hettingerf0320c72009-04-26 20:10:50 +0000110One illusion may beget another. For example, since 0.1 is not exactly 1/10,
Raymond Hettinger4af36292009-04-26 21:37:46 +0000111summing three values of 0.1 may not yield exactly 0.3, either::
Georg Brandl116aa622007-08-15 14:28:22 +0000112
Raymond Hettinger4af36292009-04-26 21:37:46 +0000113 >>> .1 + .1 + .1 == .3
114 False
115
116Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
1170.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
118:func:`round` function cannot help::
119
120 >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
121 False
122
123Though the numbers cannot be made closer to their intended exact values,
124the :func:`round` function can be useful for post-rounding so that results
Raymond Hettingereafaf4c2009-06-28 22:30:13 +0000125with inexact values become comparable to one another::
Raymond Hettinger4af36292009-04-26 21:37:46 +0000126
Raymond Hettingereafaf4c2009-06-28 22:30:13 +0000127 >>> round(.1 + .1 + .1, 10) == round(.3, 10)
Raymond Hettinger4af36292009-04-26 21:37:46 +0000128 True
Georg Brandl116aa622007-08-15 14:28:22 +0000129
130Binary floating-point arithmetic holds many surprises like this. The problem
131with "0.1" is explained in precise detail below, in the "Representation Error"
132section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
133for a more complete account of other common surprises.
134
135As that says near the end, "there are no easy answers." Still, don't be unduly
136wary of floating-point! The errors in Python float operations are inherited
137from the floating-point hardware, and on most machines are on the order of no
138more than 1 part in 2\*\*53 per operation. That's more than adequate for most
Raymond Hettingereafaf4c2009-06-28 22:30:13 +0000139tasks, but you do need to keep in mind that it's not decimal arithmetic and
Georg Brandl116aa622007-08-15 14:28:22 +0000140that every float operation can suffer a new rounding error.
141
142While pathological cases do exist, for most casual use of floating-point
143arithmetic you'll see the result you expect in the end if you simply round the
144display of your final results to the number of decimal digits you expect.
Benjamin Petersone6f00632008-05-26 01:03:56 +0000145:func:`str` usually suffices, and for finer control see the :meth:`str.format`
146method's format specifiers in :ref:`formatstrings`.
Georg Brandl116aa622007-08-15 14:28:22 +0000147
Raymond Hettingereba99df2008-10-05 17:57:52 +0000148For use cases which require exact decimal representation, try using the
149:mod:`decimal` module which implements decimal arithmetic suitable for
150accounting applications and high-precision applications.
151
152Another form of exact arithmetic is supported by the :mod:`fractions` module
153which implements arithmetic based on rational numbers (so the numbers like
1541/3 can be represented exactly).
155
Guido van Rossum0616b792007-08-31 03:25:11 +0000156If you are a heavy user of floating point operations you should take a look
157at the Numerical Python package and many other packages for mathematical and
158statistical operations supplied by the SciPy project. See <http://scipy.org>.
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000159
160Python provides tools that may help on those rare occasions when you really
161*do* want to know the exact value of a float. The
162:meth:`float.as_integer_ratio` method expresses the value of a float as a
163fraction::
164
165 >>> x = 3.14159
166 >>> x.as_integer_ratio()
Raymond Hettingereafaf4c2009-06-28 22:30:13 +0000167 (3537115888337719, 1125899906842624)
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000168
169Since the ratio is exact, it can be used to losslessly recreate the
170original value::
171
172 >>> x == 3537115888337719 / 1125899906842624
173 True
174
175The :meth:`float.hex` method expresses a float in hexadecimal (base
17616), again giving the exact value stored by your computer::
177
178 >>> x.hex()
179 '0x1.921f9f01b866ep+1'
180
181This precise hexadecimal representation can be used to reconstruct
182the float value exactly::
183
184 >>> x == float.fromhex('0x1.921f9f01b866ep+1')
185 True
186
187Since the representation is exact, it is useful for reliably porting values
188across different versions of Python (platform independence) and exchanging
189data with other languages that support the same format (such as Java and C99).
190
Raymond Hettinger5afb5c62009-04-26 22:01:46 +0000191Another helpful tool is the :func:`math.fsum` function which helps mitigate
192loss-of-precision during summation. It tracks "lost digits" as values are
193added onto a running total. That can make a difference in overall accuracy
194so that the errors do not accumulate to the point where they affect the
195final total:
196
197 >>> sum([0.1] * 10) == 1.0
198 False
199 >>> math.fsum([0.1] * 10) == 1.0
200 True
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000201
Georg Brandl116aa622007-08-15 14:28:22 +0000202.. _tut-fp-error:
203
204Representation Error
205====================
206
207This section explains the "0.1" example in detail, and shows how you can perform
208an exact analysis of cases like this yourself. Basic familiarity with binary
209floating-point representation is assumed.
210
211:dfn:`Representation error` refers to the fact that some (most, actually)
212decimal fractions cannot be represented exactly as binary (base 2) fractions.
213This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000214others) often won't display the exact decimal number you expect.
Georg Brandl116aa622007-08-15 14:28:22 +0000215
216Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
217machines today (November 2000) use IEEE-754 floating point arithmetic, and
218almost all platforms map Python floats to IEEE-754 "double precision". 754
219doubles contain 53 bits of precision, so on input the computer strives to
Benjamin Peterson5c6d7872009-02-06 02:40:07 +0000220convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
Georg Brandl116aa622007-08-15 14:28:22 +0000221an integer containing exactly 53 bits. Rewriting ::
222
223 1 / 10 ~= J / (2**N)
224
225as ::
226
227 J ~= 2**N / 10
228
229and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
230the best value for *N* is 56::
231
Raymond Hettinger1d180682009-06-28 23:21:38 +0000232 >>> 2**52 <= 2**56 // 10 < 2**53
233 True
Georg Brandl116aa622007-08-15 14:28:22 +0000234
235That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
236best possible value for *J* is then that quotient rounded::
237
238 >>> q, r = divmod(2**56, 10)
239 >>> r
Georg Brandlbae1b942008-08-10 12:16:45 +0000240 6
Georg Brandl116aa622007-08-15 14:28:22 +0000241
242Since the remainder is more than half of 10, the best approximation is obtained
243by rounding up::
244
245 >>> q+1
Georg Brandlbae1b942008-08-10 12:16:45 +0000246 7205759403792794
Georg Brandl116aa622007-08-15 14:28:22 +0000247
Raymond Hettinger1d180682009-06-28 23:21:38 +0000248Therefore the best possible approximation to 1/10 in 754 double precision is::
Georg Brandl116aa622007-08-15 14:28:22 +0000249
Raymond Hettinger1d180682009-06-28 23:21:38 +0000250 7205759403792794 / 2 ** 56
Georg Brandl116aa622007-08-15 14:28:22 +0000251
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000252Dividing both the numerator and denominator by two reduces the fraction to::
253
Raymond Hettinger1d180682009-06-28 23:21:38 +0000254 3602879701896397 / 2 ** 55
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000255
Georg Brandl116aa622007-08-15 14:28:22 +0000256Note that since we rounded up, this is actually a little bit larger than 1/10;
257if we had not rounded up, the quotient would have been a little bit smaller than
2581/10. But in no case can it be *exactly* 1/10!
259
260So the computer never "sees" 1/10: what it sees is the exact fraction given
261above, the best 754 double approximation it can get::
262
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000263 >>> 0.1 * 2 ** 55
264 3602879701896397.0
Georg Brandl116aa622007-08-15 14:28:22 +0000265
Raymond Hettinger1d180682009-06-28 23:21:38 +0000266If we multiply that fraction by 10\*\*55, we can see the value out to
26755 decimal digits::
Georg Brandl116aa622007-08-15 14:28:22 +0000268
Raymond Hettinger1d180682009-06-28 23:21:38 +0000269 >>> 3602879701896397 * 10 ** 55 // 2 ** 55
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000270 1000000000000000055511151231257827021181583404541015625
Georg Brandl116aa622007-08-15 14:28:22 +0000271
Raymond Hettinger1d180682009-06-28 23:21:38 +0000272meaning that the exact number stored in the computer is equal to
273the decimal value 0.1000000000000000055511151231257827021181583404541015625.
274Instead of displaying the full decimal value, many languages (including
275older versions of Python), round the result to 17 significant digits::
276
277 >>> format(0.1, '.17f')
278 '0.10000000000000001'
Georg Brandl116aa622007-08-15 14:28:22 +0000279
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000280The :mod:`fractions` and :mod:`decimal` modules make these calculations
281easy::
Georg Brandl116aa622007-08-15 14:28:22 +0000282
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000283 >>> from decimal import Decimal
284 >>> from fractions import Fraction
Raymond Hettinger1d180682009-06-28 23:21:38 +0000285
286 >>> Fraction.from_float(0.1)
287 Fraction(3602879701896397, 36028797018963968)
288
289 >>> (0.1).as_integer_ratio()
290 (3602879701896397, 36028797018963968)
291
292 >>> Decimal.from_float(0.1)
293 Decimal('0.1000000000000000055511151231257827021181583404541015625')
294
295 >>> format(Decimal.from_float(0.1), '.17')
296 '0.10000000000000001'