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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 Dickinsond5d32562010-07-30 12:58:44 +000051Stop at any finite number of bits, and you get an approximation. On a typical
52machine, there are 53 bits of precision available, so the value stored
53internally is the binary fraction ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000054
Mark Dickinsond5d32562010-07-30 12:58:44 +000055 0.00011001100110011001100110011001100110011001100110011010
Georg Brandl8ec7f652007-08-15 14:28:01 +000056
Mark Dickinsond5d32562010-07-30 12:58:44 +000057which is close to, but not exactly equal to, 1/10.
58
59It's easy to forget that the stored value is an approximation to the original
60decimal fraction, because of the way that floats are displayed at the
61interpreter prompt. Python only prints a decimal approximation to the true
62decimal value of the binary approximation stored by the machine. If Python
63were to print the true decimal value of the binary approximation stored for
640.1, it would have to display ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000065
66 >>> 0.1
67 0.1000000000000000055511151231257827021181583404541015625
68
Mark Dickinsond5d32562010-07-30 12:58:44 +000069That is more digits than most people find useful, so Python keeps the number
70of digits manageable by displaying a rounded value instead ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000071
Mark Dickinsond5d32562010-07-30 12:58:44 +000072 >>> 0.1
73 0.1
Georg Brandl8ec7f652007-08-15 14:28:01 +000074
Mark Dickinsond5d32562010-07-30 12:58:44 +000075It's important to realize that this is, in a real sense, an illusion: the value
76in the machine is not exactly 1/10, you're simply rounding the *display* of the
77true machine value. This fact becomes apparent as soon as you try to do
78arithmetic with these values ::
79
80 >>> 0.1 + 0.2
81 0.30000000000000004
Georg Brandl8ec7f652007-08-15 14:28:01 +000082
83Note that this is in the very nature of binary floating-point: this is not a bug
84in Python, and it is not a bug in your code either. You'll see the same kind of
85thing in all languages that support your hardware's floating-point arithmetic
86(although some languages may not *display* the difference by default, or in all
87output modes).
88
Mark Dickinsond5d32562010-07-30 12:58:44 +000089Other surprises follow from this one. For example, if you try to round the value
902.675 to two decimal places, you get this ::
Georg Brandl8ec7f652007-08-15 14:28:01 +000091
Mark Dickinsond5d32562010-07-30 12:58:44 +000092 >>> round(2.675, 2)
93 2.67
Georg Brandl8ec7f652007-08-15 14:28:01 +000094
Mark Dickinsond5d32562010-07-30 12:58:44 +000095The documentation for the built-in :func:`round` function says that it rounds
96to the nearest value, rounding ties away from zero. Since the decimal fraction
972.675 is exactly halfway between 2.67 and 2.68, you might expect the result
98here to be (a binary approximation to) 2.68. It's not, because when the
99decimal literal ``2.675`` is converted to a binary floating-point number, it's
100again replaced with a binary approximation, whose exact value is ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000101
Mark Dickinsond5d32562010-07-30 12:58:44 +0000102 2.67499999999999982236431605997495353221893310546875
Georg Brandl8ec7f652007-08-15 14:28:01 +0000103
Mark Dickinsond5d32562010-07-30 12:58:44 +0000104Since this approximation is slightly closer to 2.67 than to 2.68, it's rounded
105down.
Georg Brandl8ec7f652007-08-15 14:28:01 +0000106
Mark Dickinsond5d32562010-07-30 12:58:44 +0000107If you're in a situation where you care which way your decimal halfway-cases
108are rounded, you should consider using the :mod:`decimal` module.
109Incidentally, the :mod:`decimal` module also provides a nice way to "see" the
110exact value that's stored in any particular Python float ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000111
Mark Dickinsond5d32562010-07-30 12:58:44 +0000112 >>> from decimal import Decimal
113 >>> Decimal(2.675)
114 Decimal('2.67499999999999982236431605997495353221893310546875')
Georg Brandl8ec7f652007-08-15 14:28:01 +0000115
116Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
1170.1 may not yield exactly 1.0, either::
118
119 >>> sum = 0.0
120 >>> for i in range(10):
121 ... sum += 0.1
122 ...
123 >>> sum
Mark Dickinson6b87f112009-11-24 14:27:02 +0000124 0.9999999999999999
Georg Brandl8ec7f652007-08-15 14:28:01 +0000125
126Binary floating-point arithmetic holds many surprises like this. The problem
127with "0.1" is explained in precise detail below, in the "Representation Error"
128section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
129for a more complete account of other common surprises.
130
131As that says near the end, "there are no easy answers." Still, don't be unduly
132wary of floating-point! The errors in Python float operations are inherited
133from the floating-point hardware, and on most machines are on the order of no
134more than 1 part in 2\*\*53 per operation. That's more than adequate for most
135tasks, but you do need to keep in mind that it's not decimal arithmetic, and
136that every float operation can suffer a new rounding error.
137
138While pathological cases do exist, for most casual use of floating-point
139arithmetic you'll see the result you expect in the end if you simply round the
140display of your final results to the number of decimal digits you expect.
Benjamin Petersonf9ef9882008-05-26 00:54:22 +0000141:func:`str` usually suffices, and for finer control see the :meth:`str.format`
142method's format specifiers in :ref:`formatstrings`.
Georg Brandl8ec7f652007-08-15 14:28:01 +0000143
144
145.. _tut-fp-error:
146
147Representation Error
148====================
149
150This section explains the "0.1" example in detail, and shows how you can perform
151an exact analysis of cases like this yourself. Basic familiarity with binary
152floating-point representation is assumed.
153
154:dfn:`Representation error` refers to the fact that some (most, actually)
155decimal fractions cannot be represented exactly as binary (base 2) fractions.
156This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
157others) often won't display the exact decimal number you expect::
158
Mark Dickinsond5d32562010-07-30 12:58:44 +0000159 >>> 0.1 + 0.2
160 0.30000000000000004
Georg Brandl8ec7f652007-08-15 14:28:01 +0000161
Mark Dickinsond5d32562010-07-30 12:58:44 +0000162Why is that? 1/10 and 2/10 are not exactly representable as a binary
163fraction. Almost all machines today (July 2010) use IEEE-754 floating point
164arithmetic, and almost all platforms map Python floats to IEEE-754 "double
165precision". 754 doubles contain 53 bits of precision, so on input the computer
166strives to convert 0.1 to the closest fraction it can of the form *J*/2**\ *N*
167where *J* is an integer containing exactly 53 bits. Rewriting ::
Georg Brandl8ec7f652007-08-15 14:28:01 +0000168
169 1 / 10 ~= J / (2**N)
170
171as ::
172
173 J ~= 2**N / 10
174
175and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
176the best value for *N* is 56::
177
178 >>> 2**52
179 4503599627370496L
180 >>> 2**53
181 9007199254740992L
182 >>> 2**56/10
183 7205759403792793L
184
185That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
186best possible value for *J* is then that quotient rounded::
187
188 >>> q, r = divmod(2**56, 10)
189 >>> r
190 6L
191
192Since the remainder is more than half of 10, the best approximation is obtained
193by rounding up::
194
195 >>> q+1
196 7205759403792794L
197
198Therefore the best possible approximation to 1/10 in 754 double precision is
199that over 2\*\*56, or ::
200
201 7205759403792794 / 72057594037927936
202
203Note that since we rounded up, this is actually a little bit larger than 1/10;
204if we had not rounded up, the quotient would have been a little bit smaller than
2051/10. But in no case can it be *exactly* 1/10!
206
207So the computer never "sees" 1/10: what it sees is the exact fraction given
208above, the best 754 double approximation it can get::
209
210 >>> .1 * 2**56
211 7205759403792794.0
212
213If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
214its 30 most significant decimal digits::
215
216 >>> 7205759403792794 * 10**30 / 2**56
217 100000000000000005551115123125L
218
219meaning that the exact number stored in the computer is approximately equal to
Mark Dickinsond5d32562010-07-30 12:58:44 +0000220the decimal value 0.100000000000000005551115123125. In versions prior to
221Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits,
222giving '0.10000000000000001'. In current versions, Python displays a value based
223on the shortest decimal fraction that rounds correctly back to the true binary
224value, resulting simply in '0.1'.