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Georg Brandl8ec7f652007-08-15 14:28:01 +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
Mark Dickinson33e59352010-08-04 21:44:47 +000051Stop at any finite number of bits, and you get an approximation.
52
53On a typical machine running Python, there are 53 bits of precision available
54for a Python float, so the value stored internally when you enter the decimal
55number ``0.1`` is the binary fraction ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000056
Mark Dickinsond5d32562010-07-30 12:58:44 +000057 0.00011001100110011001100110011001100110011001100110011010
Georg Brandl8ec7f652007-08-15 14:28:01 +000058
Mark Dickinsond5d32562010-07-30 12:58:44 +000059which is close to, but not exactly equal to, 1/10.
60
61It's easy to forget that the stored value is an approximation to the original
62decimal fraction, because of the way that floats are displayed at the
63interpreter prompt. Python only prints a decimal approximation to the true
64decimal value of the binary approximation stored by the machine. If Python
65were to print the true decimal value of the binary approximation stored for
660.1, it would have to display ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000067
68 >>> 0.1
69 0.1000000000000000055511151231257827021181583404541015625
70
Mark Dickinsond5d32562010-07-30 12:58:44 +000071That is more digits than most people find useful, so Python keeps the number
72of digits manageable by displaying a rounded value instead ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000073
Mark Dickinsond5d32562010-07-30 12:58:44 +000074 >>> 0.1
75 0.1
Georg Brandl8ec7f652007-08-15 14:28:01 +000076
Mark Dickinsond5d32562010-07-30 12:58:44 +000077It's important to realize that this is, in a real sense, an illusion: the value
78in the machine is not exactly 1/10, you're simply rounding the *display* of the
79true machine value. This fact becomes apparent as soon as you try to do
80arithmetic with these values ::
81
82 >>> 0.1 + 0.2
83 0.30000000000000004
Georg Brandl8ec7f652007-08-15 14:28:01 +000084
Mark Dickinson33e59352010-08-04 21:44:47 +000085Note that this is in the very nature of binary floating-point: this is not a
86bug in Python, and it is not a bug in your code either. You'll see the same
87kind of thing in all languages that support your hardware's floating-point
88arithmetic (although some languages may not *display* the difference by
89default, or in all output modes).
Georg Brandl8ec7f652007-08-15 14:28:01 +000090
Mark Dickinson33e59352010-08-04 21:44:47 +000091Other surprises follow from this one. For example, if you try to round the
92value 2.675 to two decimal places, you get this ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000093
Mark Dickinsond5d32562010-07-30 12:58:44 +000094 >>> round(2.675, 2)
95 2.67
Georg Brandl8ec7f652007-08-15 14:28:01 +000096
Mark Dickinsond5d32562010-07-30 12:58:44 +000097The documentation for the built-in :func:`round` function says that it rounds
98to the nearest value, rounding ties away from zero. Since the decimal fraction
992.675 is exactly halfway between 2.67 and 2.68, you might expect the result
100here to be (a binary approximation to) 2.68. It's not, because when the
Mark Dickinson33e59352010-08-04 21:44:47 +0000101decimal string ``2.675`` is converted to a binary floating-point number, it's
Mark Dickinsond5d32562010-07-30 12:58:44 +0000102again replaced with a binary approximation, whose exact value is ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000103
Mark Dickinsond5d32562010-07-30 12:58:44 +0000104 2.67499999999999982236431605997495353221893310546875
Georg Brandl8ec7f652007-08-15 14:28:01 +0000105
Mark Dickinsond5d32562010-07-30 12:58:44 +0000106Since this approximation is slightly closer to 2.67 than to 2.68, it's rounded
107down.
Georg Brandl8ec7f652007-08-15 14:28:01 +0000108
Mark Dickinsond5d32562010-07-30 12:58:44 +0000109If you're in a situation where you care which way your decimal halfway-cases
110are rounded, you should consider using the :mod:`decimal` module.
111Incidentally, the :mod:`decimal` module also provides a nice way to "see" the
112exact value that's stored in any particular Python float ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000113
Mark Dickinsond5d32562010-07-30 12:58:44 +0000114 >>> from decimal import Decimal
115 >>> Decimal(2.675)
116 Decimal('2.67499999999999982236431605997495353221893310546875')
Georg Brandl8ec7f652007-08-15 14:28:01 +0000117
Mark Dickinson33e59352010-08-04 21:44:47 +0000118Another consequence is that since 0.1 is not exactly 1/10, summing ten values
119of 0.1 may not yield exactly 1.0, either::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000120
121 >>> sum = 0.0
122 >>> for i in range(10):
123 ... sum += 0.1
124 ...
125 >>> sum
Mark Dickinson6b87f112009-11-24 14:27:02 +0000126 0.9999999999999999
Georg Brandl8ec7f652007-08-15 14:28:01 +0000127
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 `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
131for a more complete account of other common surprises.
132
133As that says near the end, "there are no easy answers." Still, don't be unduly
134wary of floating-point! The errors in Python float operations are inherited
135from the floating-point hardware, and on most machines are on the order of no
136more than 1 part in 2\*\*53 per operation. That's more than adequate for most
137tasks, but you do need to keep in mind that it's not decimal arithmetic, and
138that every float operation can suffer a new rounding error.
139
140While pathological cases do exist, for most casual use of floating-point
141arithmetic you'll see the result you expect in the end if you simply round the
Mark Dickinson33e59352010-08-04 21:44:47 +0000142display of your final results to the number of decimal digits you expect. For
143fine control over how a float is displayed see the :meth:`str.format` method's
144format specifiers in :ref:`formatstrings`.
Georg Brandl8ec7f652007-08-15 14:28:01 +0000145
146
147.. _tut-fp-error:
148
149Representation Error
150====================
151
Mark Dickinson33e59352010-08-04 21:44:47 +0000152This section explains the "0.1" example in detail, and shows how you can
153perform an exact analysis of cases like this yourself. Basic familiarity with
154binary floating-point representation is assumed.
Georg Brandl8ec7f652007-08-15 14:28:01 +0000155
156:dfn:`Representation error` refers to the fact that some (most, actually)
157decimal fractions cannot be represented exactly as binary (base 2) fractions.
158This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
159others) often won't display the exact decimal number you expect::
160
Mark Dickinsond5d32562010-07-30 12:58:44 +0000161 >>> 0.1 + 0.2
162 0.30000000000000004
Georg Brandl8ec7f652007-08-15 14:28:01 +0000163
Mark Dickinsond5d32562010-07-30 12:58:44 +0000164Why is that? 1/10 and 2/10 are not exactly representable as a binary
165fraction. Almost all machines today (July 2010) use IEEE-754 floating point
166arithmetic, and almost all platforms map Python floats to IEEE-754 "double
167precision". 754 doubles contain 53 bits of precision, so on input the computer
168strives to convert 0.1 to the closest fraction it can of the form *J*/2**\ *N*
169where *J* is an integer containing exactly 53 bits. Rewriting ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000170
171 1 / 10 ~= J / (2**N)
172
173as ::
174
175 J ~= 2**N / 10
176
177and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
178the best value for *N* is 56::
179
180 >>> 2**52
Mark Dickinson33e59352010-08-04 21:44:47 +0000181 4503599627370496
Georg Brandl8ec7f652007-08-15 14:28:01 +0000182 >>> 2**53
Mark Dickinson33e59352010-08-04 21:44:47 +0000183 9007199254740992
Georg Brandl8ec7f652007-08-15 14:28:01 +0000184 >>> 2**56/10
Mark Dickinson33e59352010-08-04 21:44:47 +0000185 7205759403792793
Georg Brandl8ec7f652007-08-15 14:28:01 +0000186
Mark Dickinson33e59352010-08-04 21:44:47 +0000187That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.
188The best possible value for *J* is then that quotient rounded::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000189
190 >>> q, r = divmod(2**56, 10)
191 >>> r
Mark Dickinson33e59352010-08-04 21:44:47 +0000192 6
Georg Brandl8ec7f652007-08-15 14:28:01 +0000193
194Since the remainder is more than half of 10, the best approximation is obtained
195by rounding up::
196
197 >>> q+1
Mark Dickinson33e59352010-08-04 21:44:47 +0000198 7205759403792794
Georg Brandl8ec7f652007-08-15 14:28:01 +0000199
200Therefore the best possible approximation to 1/10 in 754 double precision is
201that over 2\*\*56, or ::
202
203 7205759403792794 / 72057594037927936
204
205Note that since we rounded up, this is actually a little bit larger than 1/10;
Mark Dickinson33e59352010-08-04 21:44:47 +0000206if we had not rounded up, the quotient would have been a little bit smaller
207than 1/10. But in no case can it be *exactly* 1/10!
Georg Brandl8ec7f652007-08-15 14:28:01 +0000208
209So the computer never "sees" 1/10: what it sees is the exact fraction given
210above, the best 754 double approximation it can get::
211
212 >>> .1 * 2**56
213 7205759403792794.0
214
215If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
216its 30 most significant decimal digits::
217
Mark Dickinson33e59352010-08-04 21:44:47 +0000218 >>> 7205759403792794 * 10**30 // 2**56
Georg Brandl8ec7f652007-08-15 14:28:01 +0000219 100000000000000005551115123125L
220
221meaning that the exact number stored in the computer is approximately equal to
Mark Dickinsond5d32562010-07-30 12:58:44 +0000222the decimal value 0.100000000000000005551115123125. In versions prior to
223Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits,
Mark Dickinson33e59352010-08-04 21:44:47 +0000224giving '0.10000000000000001'. In current versions, Python displays a value
225based on the shortest decimal fraction that rounds correctly back to the true
226binary value, resulting simply in '0.1'.