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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 Hettingerb8aff6a2009-06-29 01:50:51 +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
84Historically, the Python prompt and built-in :func:`repr` function would chose
Georg Brandl7d1e8802009-08-13 08:47:18 +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
Mark Dickinson934896d2009-02-21 20:59:32 +000095Python's built-in :func:`str` function produces only 12 significant digits, and
Georg Brandl116aa622007-08-15 14:28:22 +000096you may wish to use that instead. It's unusual for ``eval(str(x))`` to
97reproduce *x*, but the output may be more pleasant to look at::
98
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +000099 >>> str(math.pi)
100 '3.14159265359'
Georg Brandl116aa622007-08-15 14:28:22 +0000101
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000102 >>> repr(math.pi)
103 '3.141592653589793'
104
105 >>> format(math.pi, '.2f')
106 '3.14'
107
108It's important to realize that this is, in a real sense, an illusion: you're
109simply rounding the *display* of the true machine value.
Georg Brandl116aa622007-08-15 14:28:22 +0000110
Raymond Hettingerf0320c72009-04-26 20:10:50 +0000111One illusion may beget another. For example, since 0.1 is not exactly 1/10,
Raymond Hettinger4af36292009-04-26 21:37:46 +0000112summing three values of 0.1 may not yield exactly 0.3, either::
Georg Brandl116aa622007-08-15 14:28:22 +0000113
Raymond Hettinger4af36292009-04-26 21:37:46 +0000114 >>> .1 + .1 + .1 == .3
115 False
116
117Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
1180.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
119:func:`round` function cannot help::
120
121 >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
122 False
123
124Though the numbers cannot be made closer to their intended exact values,
125the :func:`round` function can be useful for post-rounding so that results
Georg Brandl7d1e8802009-08-13 08:47:18 +0000126with inexact values become comparable to one another::
Raymond Hettinger4af36292009-04-26 21:37:46 +0000127
Georg Brandl7d1e8802009-08-13 08:47:18 +0000128 >>> round(.1 + .1 + .1, 10) == round(.3, 10)
Raymond Hettinger4af36292009-04-26 21:37:46 +0000129 True
Georg Brandl116aa622007-08-15 14:28:22 +0000130
131Binary floating-point arithmetic holds many surprises like this. The problem
132with "0.1" is explained in precise detail below, in the "Representation Error"
133section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
134for a more complete account of other common surprises.
135
136As that says near the end, "there are no easy answers." Still, don't be unduly
137wary of floating-point! The errors in Python float operations are inherited
138from the floating-point hardware, and on most machines are on the order of no
139more than 1 part in 2\*\*53 per operation. That's more than adequate for most
Georg Brandl7d1e8802009-08-13 08:47:18 +0000140tasks, but you do need to keep in mind that it's not decimal arithmetic and
Georg Brandl116aa622007-08-15 14:28:22 +0000141that every float operation can suffer a new rounding error.
142
143While pathological cases do exist, for most casual use of floating-point
144arithmetic you'll see the result you expect in the end if you simply round the
145display of your final results to the number of decimal digits you expect.
Benjamin Petersone6f00632008-05-26 01:03:56 +0000146:func:`str` usually suffices, and for finer control see the :meth:`str.format`
147method's format specifiers in :ref:`formatstrings`.
Georg Brandl116aa622007-08-15 14:28:22 +0000148
Raymond Hettingereba99df2008-10-05 17:57:52 +0000149For use cases which require exact decimal representation, try using the
150:mod:`decimal` module which implements decimal arithmetic suitable for
151accounting applications and high-precision applications.
152
153Another form of exact arithmetic is supported by the :mod:`fractions` module
154which implements arithmetic based on rational numbers (so the numbers like
1551/3 can be represented exactly).
156
Guido van Rossum0616b792007-08-31 03:25:11 +0000157If you are a heavy user of floating point operations you should take a look
158at the Numerical Python package and many other packages for mathematical and
159statistical operations supplied by the SciPy project. See <http://scipy.org>.
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000160
161Python provides tools that may help on those rare occasions when you really
162*do* want to know the exact value of a float. The
163:meth:`float.as_integer_ratio` method expresses the value of a float as a
164fraction::
165
166 >>> x = 3.14159
167 >>> x.as_integer_ratio()
Georg Brandl7d1e8802009-08-13 08:47:18 +0000168 (3537115888337719, 1125899906842624)
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000169
170Since the ratio is exact, it can be used to losslessly recreate the
171original value::
172
173 >>> x == 3537115888337719 / 1125899906842624
174 True
175
176The :meth:`float.hex` method expresses a float in hexadecimal (base
17716), again giving the exact value stored by your computer::
178
179 >>> x.hex()
180 '0x1.921f9f01b866ep+1'
181
182This precise hexadecimal representation can be used to reconstruct
183the float value exactly::
184
185 >>> x == float.fromhex('0x1.921f9f01b866ep+1')
186 True
187
188Since the representation is exact, it is useful for reliably porting values
189across different versions of Python (platform independence) and exchanging
190data with other languages that support the same format (such as Java and C99).
191
Raymond Hettinger5afb5c62009-04-26 22:01:46 +0000192Another helpful tool is the :func:`math.fsum` function which helps mitigate
193loss-of-precision during summation. It tracks "lost digits" as values are
194added onto a running total. That can make a difference in overall accuracy
195so that the errors do not accumulate to the point where they affect the
196final total:
197
198 >>> sum([0.1] * 10) == 1.0
199 False
200 >>> math.fsum([0.1] * 10) == 1.0
201 True
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000202
Georg Brandl116aa622007-08-15 14:28:22 +0000203.. _tut-fp-error:
204
205Representation Error
206====================
207
208This section explains the "0.1" example in detail, and shows how you can perform
209an exact analysis of cases like this yourself. Basic familiarity with binary
210floating-point representation is assumed.
211
212:dfn:`Representation error` refers to the fact that some (most, actually)
213decimal fractions cannot be represented exactly as binary (base 2) fractions.
214This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000215others) often won't display the exact decimal number you expect.
Georg Brandl116aa622007-08-15 14:28:22 +0000216
217Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
218machines today (November 2000) use IEEE-754 floating point arithmetic, and
219almost all platforms map Python floats to IEEE-754 "double precision". 754
220doubles contain 53 bits of precision, so on input the computer strives to
Benjamin Peterson5c6d7872009-02-06 02:40:07 +0000221convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
Georg Brandl116aa622007-08-15 14:28:22 +0000222an integer containing exactly 53 bits. Rewriting ::
223
224 1 / 10 ~= J / (2**N)
225
226as ::
227
228 J ~= 2**N / 10
229
230and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
231the best value for *N* is 56::
232
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000233 >>> 2**52 <= 2**56 // 10 < 2**53
234 True
Georg Brandl116aa622007-08-15 14:28:22 +0000235
236That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
237best possible value for *J* is then that quotient rounded::
238
239 >>> q, r = divmod(2**56, 10)
240 >>> r
Georg Brandlbae1b942008-08-10 12:16:45 +0000241 6
Georg Brandl116aa622007-08-15 14:28:22 +0000242
243Since the remainder is more than half of 10, the best approximation is obtained
244by rounding up::
245
246 >>> q+1
Georg Brandlbae1b942008-08-10 12:16:45 +0000247 7205759403792794
Georg Brandl116aa622007-08-15 14:28:22 +0000248
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000249Therefore the best possible approximation to 1/10 in 754 double precision is::
Georg Brandl116aa622007-08-15 14:28:22 +0000250
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000251 7205759403792794 / 2 ** 56
Georg Brandl116aa622007-08-15 14:28:22 +0000252
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000253Dividing both the numerator and denominator by two reduces the fraction to::
254
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000255 3602879701896397 / 2 ** 55
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000256
Georg Brandl116aa622007-08-15 14:28:22 +0000257Note that since we rounded up, this is actually a little bit larger than 1/10;
258if we had not rounded up, the quotient would have been a little bit smaller than
2591/10. But in no case can it be *exactly* 1/10!
260
261So the computer never "sees" 1/10: what it sees is the exact fraction given
262above, the best 754 double approximation it can get::
263
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000264 >>> 0.1 * 2 ** 55
265 3602879701896397.0
Georg Brandl116aa622007-08-15 14:28:22 +0000266
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000267If we multiply that fraction by 10\*\*55, we can see the value out to
26855 decimal digits::
Georg Brandl116aa622007-08-15 14:28:22 +0000269
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000270 >>> 3602879701896397 * 10 ** 55 // 2 ** 55
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000271 1000000000000000055511151231257827021181583404541015625
Georg Brandl116aa622007-08-15 14:28:22 +0000272
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000273meaning that the exact number stored in the computer is equal to
274the decimal value 0.1000000000000000055511151231257827021181583404541015625.
275Instead of displaying the full decimal value, many languages (including
276older versions of Python), round the result to 17 significant digits::
277
278 >>> format(0.1, '.17f')
279 '0.10000000000000001'
Georg Brandl116aa622007-08-15 14:28:22 +0000280
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000281The :mod:`fractions` and :mod:`decimal` modules make these calculations
282easy::
Georg Brandl116aa622007-08-15 14:28:22 +0000283
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000284 >>> from decimal import Decimal
285 >>> from fractions import Fraction
Raymond Hettingerb8aff6a2009-06-29 01:50:51 +0000286
287 >>> Fraction.from_float(0.1)
288 Fraction(3602879701896397, 36028797018963968)
289
290 >>> (0.1).as_integer_ratio()
291 (3602879701896397, 36028797018963968)
292
293 >>> Decimal.from_float(0.1)
294 Decimal('0.1000000000000000055511151231257827021181583404541015625')
295
296 >>> format(Decimal.from_float(0.1), '.17')
297 '0.10000000000000001'