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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
53the numerator using the first 53 bits following 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.
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
81values share the same approximation, any one of them could be displayed and
82while still preserving the invariant ``eval(repr(x)) == x``.
83
84Historically, the Python prompt and built-in :func:`repr` function would chose
85the one with 17 significant digits, ``0.10000000000000001``, Starting with
86Python 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
111Other surprises follow from this one. For example, after seeing ::
112
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000113 >>> format(0.1, '.17g')
114 '0.10000000000000001'
Georg Brandl116aa622007-08-15 14:28:22 +0000115
116you may be tempted to use the :func:`round` function to chop it back to the
117single digit you expect. But that makes no difference::
118
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000119 >>> format(round(0.1, 1), '.17g')
120 '0.10000000000000001'
Georg Brandl116aa622007-08-15 14:28:22 +0000121
122The problem is that the binary floating-point value stored for "0.1" was already
123the best possible binary approximation to 1/10, so trying to round it again
124can't make it better: it was already as good as it gets.
125
126Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
1270.1 may not yield exactly 1.0, either::
128
129 >>> sum = 0.0
130 >>> for i in range(10):
131 ... sum += 0.1
132 ...
133 >>> sum
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000134 0.9999999999999999
Georg Brandl116aa622007-08-15 14:28:22 +0000135
136Binary floating-point arithmetic holds many surprises like this. The problem
137with "0.1" is explained in precise detail below, in the "Representation Error"
138section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
139for a more complete account of other common surprises.
140
141As that says near the end, "there are no easy answers." Still, don't be unduly
142wary of floating-point! The errors in Python float operations are inherited
143from the floating-point hardware, and on most machines are on the order of no
144more than 1 part in 2\*\*53 per operation. That's more than adequate for most
145tasks, but you do need to keep in mind that it's not decimal arithmetic, and
146that every float operation can suffer a new rounding error.
147
148While pathological cases do exist, for most casual use of floating-point
149arithmetic you'll see the result you expect in the end if you simply round the
150display of your final results to the number of decimal digits you expect.
Benjamin Petersone6f00632008-05-26 01:03:56 +0000151:func:`str` usually suffices, and for finer control see the :meth:`str.format`
152method's format specifiers in :ref:`formatstrings`.
Georg Brandl116aa622007-08-15 14:28:22 +0000153
Raymond Hettingereba99df2008-10-05 17:57:52 +0000154For use cases which require exact decimal representation, try using the
155:mod:`decimal` module which implements decimal arithmetic suitable for
156accounting applications and high-precision applications.
157
158Another form of exact arithmetic is supported by the :mod:`fractions` module
159which implements arithmetic based on rational numbers (so the numbers like
1601/3 can be represented exactly).
161
Guido van Rossum0616b792007-08-31 03:25:11 +0000162If you are a heavy user of floating point operations you should take a look
163at the Numerical Python package and many other packages for mathematical and
164statistical operations supplied by the SciPy project. See <http://scipy.org>.
Raymond Hettinger9fce0ba2008-10-05 16:46:29 +0000165
166Python provides tools that may help on those rare occasions when you really
167*do* want to know the exact value of a float. The
168:meth:`float.as_integer_ratio` method expresses the value of a float as a
169fraction::
170
171 >>> x = 3.14159
172 >>> x.as_integer_ratio()
173 (3537115888337719L, 1125899906842624L)
174
175Since the ratio is exact, it can be used to losslessly recreate the
176original value::
177
178 >>> x == 3537115888337719 / 1125899906842624
179 True
180
181The :meth:`float.hex` method expresses a float in hexadecimal (base
18216), again giving the exact value stored by your computer::
183
184 >>> x.hex()
185 '0x1.921f9f01b866ep+1'
186
187This precise hexadecimal representation can be used to reconstruct
188the float value exactly::
189
190 >>> x == float.fromhex('0x1.921f9f01b866ep+1')
191 True
192
193Since the representation is exact, it is useful for reliably porting values
194across different versions of Python (platform independence) and exchanging
195data with other languages that support the same format (such as Java and C99).
196
197
Georg Brandl116aa622007-08-15 14:28:22 +0000198.. _tut-fp-error:
199
200Representation Error
201====================
202
203This section explains the "0.1" example in detail, and shows how you can perform
204an exact analysis of cases like this yourself. Basic familiarity with binary
205floating-point representation is assumed.
206
207:dfn:`Representation error` refers to the fact that some (most, actually)
208decimal fractions cannot be represented exactly as binary (base 2) fractions.
209This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000210others) often won't display the exact decimal number you expect.
Georg Brandl116aa622007-08-15 14:28:22 +0000211
212Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
213machines today (November 2000) use IEEE-754 floating point arithmetic, and
214almost all platforms map Python floats to IEEE-754 "double precision". 754
215doubles contain 53 bits of precision, so on input the computer strives to
Benjamin Peterson5c6d7872009-02-06 02:40:07 +0000216convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
Georg Brandl116aa622007-08-15 14:28:22 +0000217an integer containing exactly 53 bits. Rewriting ::
218
219 1 / 10 ~= J / (2**N)
220
221as ::
222
223 J ~= 2**N / 10
224
225and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
226the best value for *N* is 56::
227
228 >>> 2**52
Georg Brandlbae1b942008-08-10 12:16:45 +0000229 4503599627370496
Georg Brandl116aa622007-08-15 14:28:22 +0000230 >>> 2**53
Georg Brandlbae1b942008-08-10 12:16:45 +0000231 9007199254740992
Georg Brandl116aa622007-08-15 14:28:22 +0000232 >>> 2**56/10
Georg Brandlbae1b942008-08-10 12:16:45 +0000233 7205759403792794.0
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
248Therefore the best possible approximation to 1/10 in 754 double precision is
249that over 2\*\*56, or ::
250
251 7205759403792794 / 72057594037927936
252
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000253Dividing both the numerator and denominator by two reduces the fraction to::
254
255 3602879701896397 / 36028797018963968
256
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 Hettinger8bd1d4f2009-04-24 03:09:06 +0000267If we multiply that fraction by 10\*\*60, we can see the value of out to
26860 decimal digits::
Georg Brandl116aa622007-08-15 14:28:22 +0000269
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000270 >>> 3602879701896397 * 10 ** 60 // 2 ** 55
271 1000000000000000055511151231257827021181583404541015625
Georg Brandl116aa622007-08-15 14:28:22 +0000272
273meaning that the exact number stored in the computer is approximately equal to
274the decimal value 0.100000000000000005551115123125. Rounding that to 17
275significant digits gives the 0.10000000000000001 that Python displays (well,
276will display on any 754-conforming platform that does best-possible input and
277output conversions in its C library --- yours may not!).
278
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000279The :mod:`fractions` and :mod:`decimal` modules make these calculations
280easy::
Georg Brandl116aa622007-08-15 14:28:22 +0000281
Raymond Hettinger8bd1d4f2009-04-24 03:09:06 +0000282 >>> from decimal import Decimal
283 >>> from fractions import Fraction
284 >>> print(Fraction.from_float(0.1))
285 3602879701896397/36028797018963968
286 >>> print(Decimal.from_float(0.1))
287 0.1000000000000000055511151231257827021181583404541015625