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Misc improvements to the float tutorial (GH-102052)

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Raymond Hettinger 2023-02-19 13:21:37 -06:00 committed by GitHub
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Doc/tutorial/floatingpoint.rst
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@ -1,6 +1,7 @@
.. testsetup::
import math
from fractions import Fraction
.. _tut-fp-issues:
@ -9,12 +10,13 @@ Floating Point Arithmetic: Issues and Limitations
**************************************************
.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
.. sectionauthor:: Raymond Hettinger <python at rcn dot com>
Floating-point numbers are represented in computer hardware as base 2 (binary)
fractions. For example, the **decimal** fraction ``0.125``
has value 1/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.001``
has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only
fractions. For example, the **decimal** fraction ``0.625``
has value 6/10 + 2/100 + 5/1000, and in the same way the **binary** fraction ``0.101``
has value 1/2 + 0/4 + 1/8. These two fractions have identical values, the only
real difference being that the first is written in base 10 fractional notation,
and the second in base 2.
@ -57,13 +59,15 @@ Many users are not aware of the approximation because of the way values are
displayed. Python only prints a decimal approximation to the true decimal
value of the binary approximation stored by the machine. On most machines, if
Python were to print the true decimal value of the binary approximation stored
for 0.1, it would have to display ::
for 0.1, it would have to display::
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
That is more digits than most people find useful, so Python keeps the number
of digits manageable by displaying a rounded value instead ::
of digits manageable by displaying a rounded value instead:
.. doctest::
>>> 1 / 10
0.1
@ -90,7 +94,10 @@ thing in all languages that support your hardware's floating-point arithmetic
(although some languages may not *display* the difference by default, or in all
output modes).
For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits::
For more pleasant output, you may wish to use string formatting to produce a
limited number of significant digits:
.. doctest::
>>> format(math.pi, '.12g') # give 12 significant digits
'3.14159265359'
@ -101,33 +108,49 @@ For more pleasant output, you may wish to use string formatting to produce a lim
>>> repr(math.pi)
'3.141592653589793'
It's important to realize that this is, in a real sense, an illusion: you're
simply rounding the *display* of the true machine value.
One illusion may beget another. For example, since 0.1 is not exactly 1/10,
summing three values of 0.1 may not yield exactly 0.3, either::
summing three values of 0.1 may not yield exactly 0.3, either:
>>> .1 + .1 + .1 == .3
.. doctest::
>>> 0.1 + 0.1 + 0.1 == 0.3
False
Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
:func:`round` function cannot help::
:func:`round` function cannot help:
>>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
.. doctest::
>>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1)
False
Though the numbers cannot be made closer to their intended exact values,
the :func:`round` function can be useful for post-rounding so that results
with inexact values become comparable to one another::
the :func:`math.isclose` function can be useful for comparing inexact values:
>>> round(.1 + .1 + .1, 10) == round(.3, 10)
True
.. doctest::
>>> math.isclose(0.1 + 0.1 + 0.1, 0.3)
True
Alternatively, the :func:`round` function can be used to compare rough
approximations::
.. doctest::
>>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2)
True
Binary floating-point arithmetic holds many surprises like this. The problem
with "0.1" is explained in precise detail below, in the "Representation Error"
section. See `The Perils of Floating Point <https://www.lahey.com/float.htm>`_
section. See `Examples of Floating Point Problems
<https://jvns.ca/blog/2023/01/13/examples-of-floating-point-problems/>`_ for
a pleasant summary of how binary floating point works and the kinds of
problems commonly encountered in practice. Also see
`The Perils of Floating Point <https://www.lahey.com/float.htm>`_
for a more complete account of other common surprises.
As that says near the end, "there are no easy answers." Still, don't be unduly
@ -158,26 +181,34 @@ statistical operations supplied by the SciPy project. See <https://scipy.org>.
Python provides tools that may help on those rare occasions when you really
*do* want to know the exact value of a float. The
:meth:`float.as_integer_ratio` method expresses the value of a float as a
fraction::
fraction:
.. doctest::
>>> x = 3.14159
>>> x.as_integer_ratio()
(3537115888337719, 1125899906842624)
Since the ratio is exact, it can be used to losslessly recreate the
original value::
original value:
.. doctest::
>>> x == 3537115888337719 / 1125899906842624
True
The :meth:`float.hex` method expresses a float in hexadecimal (base
16), again giving the exact value stored by your computer::
16), again giving the exact value stored by your computer:
.. doctest::
>>> x.hex()
'0x1.921f9f01b866ep+1'
This precise hexadecimal representation can be used to reconstruct
the float value exactly::
the float value exactly:
.. doctest::
>>> x == float.fromhex('0x1.921f9f01b866ep+1')
True
@ -186,17 +217,43 @@ Since the representation is exact, it is useful for reliably porting values
across different versions of Python (platform independence) and exchanging
data with other languages that support the same format (such as Java and C99).
Another helpful tool is the :func:`math.fsum` function which helps mitigate
loss-of-precision during summation. It tracks "lost digits" as values are
added onto a running total. That can make a difference in overall accuracy
so that the errors do not accumulate to the point where they affect the
final total:
Another helpful tool is the :func:`sum` function which helps mitigate
loss-of-precision during summation. It uses extended precision for
intermediate rounding steps as values are added onto a running total.
That can make a difference in overall accuracy so that the errors do not
accumulate to the point where they affect the final total:
.. doctest::
>>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0
False
>>> math.fsum([0.1] * 10) == 1.0
>>> sum([0.1] * 10) == 1.0
True
The :func:`math.fsum()` goes further and tracks all of the "lost digits"
as values are added onto a running total so that the result has only a
single rounding. This is slower than :func:`sum` but will be more
accurate in uncommon cases where large magnitude inputs mostly cancel
each other out leaving a final sum near zero:
.. doctest::
>>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16,
... -143401161400469.7, 266262841.31058735, -0.003244936839808227]
>>> float(sum(map(Fraction, arr))) # Exact summation with single rounding
8.042173697819788e-13
>>> math.fsum(arr) # Single rounding
8.042173697819788e-13
>>> sum(arr) # Multiple roundings in extended precision
8.042178034628478e-13
>>> total = 0.0
>>> for x in arr:
... total += x # Multiple roundings in standard precision
...
>>> total # Straight addition has no correct digits!
-0.0051575902860057365
.. _tut-fp-error:
Representation Error
@ -225,20 +282,28 @@ as ::
J ~= 2**N / 10
and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
the best value for *N* is 56::
the best value for *N* is 56:
.. doctest::
>>> 2**52 <= 2**56 // 10 < 2**53
True
That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
best possible value for *J* is then that quotient rounded::
best possible value for *J* is then that quotient rounded:
.. doctest::
>>> q, r = divmod(2**56, 10)
>>> r
6
Since the remainder is more than half of 10, the best approximation is obtained
by rounding up::
by rounding up:
.. doctest::
>>> q+1
7205759403792794
@ -256,13 +321,17 @@ if we had not rounded up, the quotient would have been a little bit smaller than
1/10. But in no case can it be *exactly* 1/10!
So the computer never "sees" 1/10: what it sees is the exact fraction given
above, the best 754 double approximation it can get::
above, the best 754 double approximation it can get:
.. doctest::
>>> 0.1 * 2 ** 55
3602879701896397.0
If we multiply that fraction by 10\*\*55, we can see the value out to
55 decimal digits::
55 decimal digits:
.. doctest::
>>> 3602879701896397 * 10 ** 55 // 2 ** 55
1000000000000000055511151231257827021181583404541015625
@ -270,13 +339,17 @@ If we multiply that fraction by 10\*\*55, we can see the value out to
meaning that the exact number stored in the computer is equal to
the decimal value 0.1000000000000000055511151231257827021181583404541015625.
Instead of displaying the full decimal value, many languages (including
older versions of Python), round the result to 17 significant digits::
older versions of Python), round the result to 17 significant digits:
.. doctest::
>>> format(0.1, '.17f')
'0.10000000000000001'
The :mod:`fractions` and :mod:`decimal` modules make these calculations
easy::
easy:
.. doctest::
>>> from decimal import Decimal
>>> from fractions import Fraction