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1.. testsetup::
2
3    import math
4
5.. _tut-fp-issues:
6
7**************************************************
8Floating Point Arithmetic:  Issues and Limitations
9**************************************************
10
11.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
12
13
14Floating-point numbers are represented in computer hardware as base 2 (binary)
15fractions.  For example, the decimal fraction ::
16
17   0.125
18
19has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
20
21   0.001
22
23has value 0/2 + 0/4 + 1/8.  These two fractions have identical values, the only
24real difference being that the first is written in base 10 fractional notation,
25and the second in base 2.
26
27Unfortunately, most decimal fractions cannot be represented exactly as binary
28fractions.  A consequence is that, in general, the decimal floating-point
29numbers you enter are only approximated by the binary floating-point numbers
30actually stored in the machine.
31
32The problem is easier to understand at first in base 10.  Consider the fraction
331/3.  You can approximate that as a base 10 fraction::
34
35   0.3
36
37or, better, ::
38
39   0.33
40
41or, better, ::
42
43   0.333
44
45and so on.  No matter how many digits you're willing to write down, the result
46will never be exactly 1/3, but will be an increasingly better approximation of
471/3.
48
49In the same way, no matter how many base 2 digits you're willing to use, the
50decimal value 0.1 cannot be represented exactly as a base 2 fraction.  In base
512, 1/10 is the infinitely repeating fraction ::
52
53   0.0001100110011001100110011001100110011001100110011...
54
55Stop at any finite number of bits, and you get an approximation.  On most
56machines today, floats are approximated using a binary fraction with
57the numerator using the first 53 bits starting with the most significant bit and
58with the denominator as a power of two.  In the case of 1/10, the binary fraction
59is ``3602879701896397 / 2 ** 55`` which is close to but not exactly
60equal to the true value of 1/10.
61
62Many users are not aware of the approximation because of the way values are
63displayed.  Python only prints a decimal approximation to the true decimal
64value of the binary approximation stored by the machine.  On most machines, if
65Python were to print the true decimal value of the binary approximation stored
66for 0.1, it would have to display ::
67
68   >>> 0.1
69   0.1000000000000000055511151231257827021181583404541015625
70
71That is more digits than most people find useful, so Python keeps the number
72of digits manageable by displaying a rounded value instead ::
73
74   >>> 1 / 10
75   0.1
76
77Just remember, even though the printed result looks like the exact value
78of 1/10, the actual stored value is the nearest representable binary fraction.
79
80Interestingly, there are many different decimal numbers that share the same
81nearest approximate binary fraction.  For example, the numbers ``0.1`` and
82``0.10000000000000001`` and
83``0.1000000000000000055511151231257827021181583404541015625`` are all
84approximated by ``3602879701896397 / 2 ** 55``.  Since all of these decimal
85values share the same approximation, any one of them could be displayed
86while still preserving the invariant ``eval(repr(x)) == x``.
87
88Historically, the Python prompt and built-in :func:`repr` function would choose
89the one with 17 significant digits, ``0.10000000000000001``.   Starting with
90Python 3.1, Python (on most systems) is now able to choose the shortest of
91these and simply display ``0.1``.
92
93Note that this is in the very nature of binary floating-point: this is not a bug
94in Python, and it is not a bug in your code either.  You'll see the same kind of
95thing in all languages that support your hardware's floating-point arithmetic
96(although some languages may not *display* the difference by default, or in all
97output modes).
98
99For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits::
100
101   >>> format(math.pi, '.12g')  # give 12 significant digits
102   '3.14159265359'
103
104   >>> format(math.pi, '.2f')   # give 2 digits after the point
105   '3.14'
106
107   >>> repr(math.pi)
108   '3.141592653589793'
109
110
111It's important to realize that this is, in a real sense, an illusion: you're
112simply rounding the *display* of the true machine value.
113
114One illusion may beget another.  For example, since 0.1 is not exactly 1/10,
115summing three values of 0.1 may not yield exactly 0.3, either::
116
117   >>> .1 + .1 + .1 == .3
118   False
119
120Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
1210.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
122:func:`round` function cannot help::
123
124   >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
125   False
126
127Though the numbers cannot be made closer to their intended exact values,
128the :func:`round` function can be useful for post-rounding so that results
129with inexact values become comparable to one another::
130
131    >>> round(.1 + .1 + .1, 10) == round(.3, 10)
132    True
133
134Binary floating-point arithmetic holds many surprises like this.  The problem
135with "0.1" is explained in precise detail below, in the "Representation Error"
136section.  See `The Perils of Floating Point <https://www.lahey.com/float.htm>`_
137for a more complete account of other common surprises.
138
139As that says near the end, "there are no easy answers."  Still, don't be unduly
140wary of floating-point!  The errors in Python float operations are inherited
141from the floating-point hardware, and on most machines are on the order of no
142more than 1 part in 2\*\*53 per operation.  That's more than adequate for most
143tasks, but you do need to keep in mind that it's not decimal arithmetic and
144that every float operation can suffer a new rounding error.
145
146While pathological cases do exist, for most casual use of floating-point
147arithmetic you'll see the result you expect in the end if you simply round the
148display of your final results to the number of decimal digits you expect.
149:func:`str` usually suffices, and for finer control see the :meth:`str.format`
150method's format specifiers in :ref:`formatstrings`.
151
152For use cases which require exact decimal representation, try using the
153:mod:`decimal` module which implements decimal arithmetic suitable for
154accounting applications and high-precision applications.
155
156Another form of exact arithmetic is supported by the :mod:`fractions` module
157which implements arithmetic based on rational numbers (so the numbers like
1581/3 can be represented exactly).
159
160If you are a heavy user of floating point operations you should take a look
161at the NumPy package and many other packages for mathematical and
162statistical operations supplied by the SciPy project. See <https://scipy.org>.
163
164Python provides tools that may help on those rare occasions when you really
165*do* want to know the exact value of a float.  The
166:meth:`float.as_integer_ratio` method expresses the value of a float as a
167fraction::
168
169   >>> x = 3.14159
170   >>> x.as_integer_ratio()
171   (3537115888337719, 1125899906842624)
172
173Since the ratio is exact, it can be used to losslessly recreate the
174original value::
175
176    >>> x == 3537115888337719 / 1125899906842624
177    True
178
179The :meth:`float.hex` method expresses a float in hexadecimal (base
18016), again giving the exact value stored by your computer::
181
182   >>> x.hex()
183   '0x1.921f9f01b866ep+1'
184
185This precise hexadecimal representation can be used to reconstruct
186the float value exactly::
187
188    >>> x == float.fromhex('0x1.921f9f01b866ep+1')
189    True
190
191Since the representation is exact, it is useful for reliably porting values
192across different versions of Python (platform independence) and exchanging
193data with other languages that support the same format (such as Java and C99).
194
195Another helpful tool is the :func:`math.fsum` function which helps mitigate
196loss-of-precision during summation.  It tracks "lost digits" as values are
197added onto a running total.  That can make a difference in overall accuracy
198so that the errors do not accumulate to the point where they affect the
199final total:
200
201   >>> sum([0.1] * 10) == 1.0
202   False
203   >>> math.fsum([0.1] * 10) == 1.0
204   True
205
206.. _tut-fp-error:
207
208Representation Error
209====================
210
211This section explains the "0.1" example in detail, and shows how you can perform
212an exact analysis of cases like this yourself.  Basic familiarity with binary
213floating-point representation is assumed.
214
215:dfn:`Representation error` refers to the fact that some (most, actually)
216decimal fractions cannot be represented exactly as binary (base 2) fractions.
217This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
218others) often won't display the exact decimal number you expect.
219
220Why is that?  1/10 is not exactly representable as a binary fraction. Almost all
221machines today (November 2000) use IEEE-754 floating point arithmetic, and
222almost all platforms map Python floats to IEEE-754 "double precision".  754
223doubles contain 53 bits of precision, so on input the computer strives to
224convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
225an integer containing exactly 53 bits.  Rewriting ::
226
227   1 / 10 ~= J / (2**N)
228
229as ::
230
231   J ~= 2**N / 10
232
233and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
234the best value for *N* is 56::
235
236    >>> 2**52 <=  2**56 // 10  < 2**53
237    True
238
239That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.  The
240best possible value for *J* is then that quotient rounded::
241
242   >>> q, r = divmod(2**56, 10)
243   >>> r
244   6
245
246Since the remainder is more than half of 10, the best approximation is obtained
247by rounding up::
248
249   >>> q+1
250   7205759403792794
251
252Therefore the best possible approximation to 1/10 in 754 double precision is::
253
254   7205759403792794 / 2 ** 56
255
256Dividing both the numerator and denominator by two reduces the fraction to::
257
258   3602879701896397 / 2 ** 55
259
260Note that since we rounded up, this is actually a little bit larger than 1/10;
261if we had not rounded up, the quotient would have been a little bit smaller than
2621/10.  But in no case can it be *exactly* 1/10!
263
264So the computer never "sees" 1/10:  what it sees is the exact fraction given
265above, the best 754 double approximation it can get::
266
267   >>> 0.1 * 2 ** 55
268   3602879701896397.0
269
270If we multiply that fraction by 10\*\*55, we can see the value out to
27155 decimal digits::
272
273   >>> 3602879701896397 * 10 ** 55 // 2 ** 55
274   1000000000000000055511151231257827021181583404541015625
275
276meaning that the exact number stored in the computer is equal to
277the decimal value 0.1000000000000000055511151231257827021181583404541015625.
278Instead of displaying the full decimal value, many languages (including
279older versions of Python), round the result to 17 significant digits::
280
281   >>> format(0.1, '.17f')
282   '0.10000000000000001'
283
284The :mod:`fractions` and :mod:`decimal` modules make these calculations
285easy::
286
287   >>> from decimal import Decimal
288   >>> from fractions import Fraction
289
290   >>> Fraction.from_float(0.1)
291   Fraction(3602879701896397, 36028797018963968)
292
293   >>> (0.1).as_integer_ratio()
294   (3602879701896397, 36028797018963968)
295
296   >>> Decimal.from_float(0.1)
297   Decimal('0.1000000000000000055511151231257827021181583404541015625')
298
299   >>> format(Decimal.from_float(0.1), '.17')
300   '0.10000000000000001'
301