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 ``0.125``
16has value 1/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.001``
17has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only
18real difference being that the first is written in base 10 fractional notation,
19and the second in base 2.
20
21Unfortunately, most decimal fractions cannot be represented exactly as binary
22fractions.  A consequence is that, in general, the decimal floating-point
23numbers you enter are only approximated by the binary floating-point numbers
24actually stored in the machine.
25
26The problem is easier to understand at first in base 10.  Consider the fraction
271/3.  You can approximate that as a base 10 fraction::
28
29   0.3
30
31or, better, ::
32
33   0.33
34
35or, better, ::
36
37   0.333
38
39and so on.  No matter how many digits you're willing to write down, the result
40will never be exactly 1/3, but will be an increasingly better approximation of
411/3.
42
43In the same way, no matter how many base 2 digits you're willing to use, the
44decimal value 0.1 cannot be represented exactly as a base 2 fraction.  In base
452, 1/10 is the infinitely repeating fraction ::
46
47   0.0001100110011001100110011001100110011001100110011...
48
49Stop at any finite number of bits, and you get an approximation.  On most
50machines today, floats are approximated using a binary fraction with
51the numerator using the first 53 bits starting with the most significant bit and
52with the denominator as a power of two.  In the case of 1/10, the binary fraction
53is ``3602879701896397 / 2 ** 55`` which is close to but not exactly
54equal to the true value of 1/10.
55
56Many users are not aware of the approximation because of the way values are
57displayed.  Python only prints a decimal approximation to the true decimal
58value of the binary approximation stored by the machine.  On most machines, if
59Python were to print the true decimal value of the binary approximation stored
60for 0.1, it would have to display ::
61
62   >>> 0.1
63   0.1000000000000000055511151231257827021181583404541015625
64
65That is more digits than most people find useful, so Python keeps the number
66of digits manageable by displaying a rounded value instead ::
67
68   >>> 1 / 10
69   0.1
70
71Just remember, even though the printed result looks like the exact value
72of 1/10, the actual stored value is the nearest representable binary fraction.
73
74Interestingly, there are many different decimal numbers that share the same
75nearest approximate binary fraction.  For example, the numbers ``0.1`` and
76``0.10000000000000001`` and
77``0.1000000000000000055511151231257827021181583404541015625`` are all
78approximated by ``3602879701896397 / 2 ** 55``.  Since all of these decimal
79values share the same approximation, any one of them could be displayed
80while still preserving the invariant ``eval(repr(x)) == x``.
81
82Historically, the Python prompt and built-in :func:`repr` function would choose
83the one with 17 significant digits, ``0.10000000000000001``.   Starting with
84Python 3.1, Python (on most systems) is now able to choose the shortest of
85these and simply display ``0.1``.
86
87Note that this is in the very nature of binary floating-point: this is not a bug
88in Python, and it is not a bug in your code either.  You'll see the same kind of
89thing in all languages that support your hardware's floating-point arithmetic
90(although some languages may not *display* the difference by default, or in all
91output modes).
92
93For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits::
94
95   >>> format(math.pi, '.12g')  # give 12 significant digits
96   '3.14159265359'
97
98   >>> format(math.pi, '.2f')   # give 2 digits after the point
99   '3.14'
100
101   >>> repr(math.pi)
102   '3.141592653589793'
103
104
105It's important to realize that this is, in a real sense, an illusion: you're
106simply rounding the *display* of the true machine value.
107
108One illusion may beget another.  For example, since 0.1 is not exactly 1/10,
109summing three values of 0.1 may not yield exactly 0.3, either::
110
111   >>> .1 + .1 + .1 == .3
112   False
113
114Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
1150.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
116:func:`round` function cannot help::
117
118   >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
119   False
120
121Though the numbers cannot be made closer to their intended exact values,
122the :func:`round` function can be useful for post-rounding so that results
123with inexact values become comparable to one another::
124
125    >>> round(.1 + .1 + .1, 10) == round(.3, 10)
126    True
127
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 `Examples of Floating Point Problems
131<https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/>`_ for
132a pleasant summary of how binary floating-point works and the kinds of
133problems commonly encountered in practice.  Also see
134`The Perils of Floating Point <https://www.lahey.com/float.htm>`_
135for a more complete account of other common surprises.
136
137As that says near the end, "there are no easy answers."  Still, don't be unduly
138wary of floating-point!  The errors in Python float operations are inherited
139from the floating-point hardware, and on most machines are on the order of no
140more than 1 part in 2\*\*53 per operation.  That's more than adequate for most
141tasks, but you do need to keep in mind that it's not decimal arithmetic and
142that every float operation can suffer a new rounding error.
143
144While pathological cases do exist, for most casual use of floating-point
145arithmetic you'll see the result you expect in the end if you simply round the
146display of your final results to the number of decimal digits you expect.
147:func:`str` usually suffices, and for finer control see the :meth:`str.format`
148method's format specifiers in :ref:`formatstrings`.
149
150For use cases which require exact decimal representation, try using the
151:mod:`decimal` module which implements decimal arithmetic suitable for
152accounting applications and high-precision applications.
153
154Another form of exact arithmetic is supported by the :mod:`fractions` module
155which implements arithmetic based on rational numbers (so the numbers like
1561/3 can be represented exactly).
157
158If you are a heavy user of floating-point operations you should take a look
159at the NumPy package and many other packages for mathematical and
160statistical operations supplied by the SciPy project. See <https://scipy.org>.
161
162Python provides tools that may help on those rare occasions when you really
163*do* want to know the exact value of a float.  The
164:meth:`float.as_integer_ratio` method expresses the value of a float as a
165fraction::
166
167   >>> x = 3.14159
168   >>> x.as_integer_ratio()
169   (3537115888337719, 1125899906842624)
170
171Since the ratio is exact, it can be used to losslessly recreate the
172original value::
173
174    >>> x == 3537115888337719 / 1125899906842624
175    True
176
177The :meth:`float.hex` method expresses a float in hexadecimal (base
17816), again giving the exact value stored by your computer::
179
180   >>> x.hex()
181   '0x1.921f9f01b866ep+1'
182
183This precise hexadecimal representation can be used to reconstruct
184the float value exactly::
185
186    >>> x == float.fromhex('0x1.921f9f01b866ep+1')
187    True
188
189Since the representation is exact, it is useful for reliably porting values
190across different versions of Python (platform independence) and exchanging
191data with other languages that support the same format (such as Java and C99).
192
193Another helpful tool is the :func:`math.fsum` function which helps mitigate
194loss-of-precision during summation.  It tracks "lost digits" as values are
195added onto a running total.  That can make a difference in overall accuracy
196so that the errors do not accumulate to the point where they affect the
197final total:
198
199   >>> sum([0.1] * 10) == 1.0
200   False
201   >>> math.fsum([0.1] * 10) == 1.0
202   True
203
204.. _tut-fp-error:
205
206Representation Error
207====================
208
209This section explains the "0.1" example in detail, and shows how you can perform
210an exact analysis of cases like this yourself.  Basic familiarity with binary
211floating-point representation is assumed.
212
213:dfn:`Representation error` refers to the fact that some (most, actually)
214decimal fractions cannot be represented exactly as binary (base 2) fractions.
215This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
216others) often won't display the exact decimal number you expect.
217
218Why is that?  1/10 is not exactly representable as a binary fraction.  Since at
219least 2000, almost all machines use IEEE 754 binary floating-point arithmetic,
220and almost all platforms map Python floats to IEEE 754 binary64 "double
221precision" values.  IEEE 754 binary64 values contain 53 bits of precision, so
222on input the computer strives to convert 0.1 to the closest fraction it can of
223the form *J*/2**\ *N* where *J* is an integer containing exactly 53 bits.
224Rewriting
225::
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 IEEE 754 double precision
253is::
254
255   7205759403792794 / 2 ** 56
256
257Dividing both the numerator and denominator by two reduces the fraction to::
258
259   3602879701896397 / 2 ** 55
260
261Note that since we rounded up, this is actually a little bit larger than 1/10;
262if we had not rounded up, the quotient would have been a little bit smaller than
2631/10.  But in no case can it be *exactly* 1/10!
264
265So the computer never "sees" 1/10:  what it sees is the exact fraction given
266above, the best IEEE 754 double approximation it can get:
267
268   >>> 0.1 * 2 ** 55
269   3602879701896397.0
270
271If we multiply that fraction by 10\*\*55, we can see the value out to
27255 decimal digits::
273
274   >>> 3602879701896397 * 10 ** 55 // 2 ** 55
275   1000000000000000055511151231257827021181583404541015625
276
277meaning that the exact number stored in the computer is equal to
278the decimal value 0.1000000000000000055511151231257827021181583404541015625.
279Instead of displaying the full decimal value, many languages (including
280older versions of Python), round the result to 17 significant digits::
281
282   >>> format(0.1, '.17f')
283   '0.10000000000000001'
284
285The :mod:`fractions` and :mod:`decimal` modules make these calculations
286easy::
287
288   >>> from decimal import Decimal
289   >>> from fractions import Fraction
290
291   >>> Fraction.from_float(0.1)
292   Fraction(3602879701896397, 36028797018963968)
293
294   >>> (0.1).as_integer_ratio()
295   (3602879701896397, 36028797018963968)
296
297   >>> Decimal.from_float(0.1)
298   Decimal('0.1000000000000000055511151231257827021181583404541015625')
299
300   >>> format(Decimal.from_float(0.1), '.17')
301   '0.10000000000000001'
302