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Floating Point Arithmetic: Issues and Limitations
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.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
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Floating-point numbers are represented in computer hardware as base 2 (binary)
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fractions. For example, the decimal fraction ::
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has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
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has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only
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real difference being that the first is written in base 10 fractional notation,
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and the second in base 2.
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Unfortunately, most decimal fractions cannot be represented exactly as binary
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fractions. A consequence is that, in general, the decimal floating-point
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numbers you enter are only approximated by the binary floating-point numbers
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actually stored in the machine.
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The problem is easier to understand at first in base 10. Consider the fraction
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1/3. You can approximate that as a base 10 fraction::
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and so on. No matter how many digits you're willing to write down, the result
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will never be exactly 1/3, but will be an increasingly better approximation of
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In the same way, no matter how many base 2 digits you're willing to use, the
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decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
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2, 1/10 is the infinitely repeating fraction ::
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0.0001100110011001100110011001100110011001100110011...
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Stop at any finite number of bits, and you get an approximation. On most
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machines today, floats are approximated using a binary fraction with
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the numerator using the first 53 bits starting with the most significant bit and
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with the denominator as a power of two. In the case of 1/10, the binary fraction
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is ``3602879701896397 / 2 ** 55`` which is close to but not exactly
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equal to the true value of 1/10.
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Many users are not aware of the approximation because of the way values are
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displayed. Python only prints a decimal approximation to the true decimal
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value of the binary approximation stored by the machine. On most machines, if
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Python were to print the true decimal value of the binary approximation stored
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for 0.1, it would have to display ::
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0.1000000000000000055511151231257827021181583404541015625
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That is more digits than most people find useful, so Python keeps the number
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of digits manageable by displaying a rounded value instead ::
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Just remember, even though the printed result looks like the exact value
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of 1/10, the actual stored value is the nearest representable binary fraction.
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Interestingly, there are many different decimal numbers that share the same
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nearest approximate binary fraction. For example, the numbers ``0.1`` and
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``0.10000000000000001`` and
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``0.1000000000000000055511151231257827021181583404541015625`` are all
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approximated by ``3602879701896397 / 2 ** 55``. Since all of these decimal
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values share the same approximation, any one of them could be displayed
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while still preserving the invariant ``eval(repr(x)) == x``.
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Historically, the Python prompt and built-in :func:`repr` function would choose
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the one with 17 significant digits, ``0.10000000000000001``. Starting with
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Python 3.1, Python (on most systems) is now able to choose the shortest of
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these and simply display ``0.1``.
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Note that this is in the very nature of binary floating-point: this is not a bug
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in Python, and it is not a bug in your code either. You'll see the same kind of
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thing in all languages that support your hardware's floating-point arithmetic
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(although some languages may not *display* the difference by default, or in all
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For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits::
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>>> format(math.pi, '.12g') # give 12 significant digits
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>>> format(math.pi, '.2f') # give 2 digits after the point
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It's important to realize that this is, in a real sense, an illusion: you're
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simply rounding the *display* of the true machine value.
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One illusion may beget another. For example, since 0.1 is not exactly 1/10,
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summing three values of 0.1 may not yield exactly 0.3, either::
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>>> .1 + .1 + .1 == .3
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Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
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0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
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:func:`round` function cannot help::
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>>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
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Though the numbers cannot be made closer to their intended exact values,
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the :func:`round` function can be useful for post-rounding so that results
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with inexact values become comparable to one another::
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>>> round(.1 + .1 + .1, 10) == round(.3, 10)
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Binary floating-point arithmetic holds many surprises like this. The problem
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with "0.1" is explained in precise detail below, in the "Representation Error"
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section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
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for a more complete account of other common surprises.
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As that says near the end, "there are no easy answers." Still, don't be unduly
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wary of floating-point! The errors in Python float operations are inherited
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from the floating-point hardware, and on most machines are on the order of no
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more than 1 part in 2\*\*53 per operation. That's more than adequate for most
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tasks, but you do need to keep in mind that it's not decimal arithmetic and
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that every float operation can suffer a new rounding error.
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While pathological cases do exist, for most casual use of floating-point
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arithmetic you'll see the result you expect in the end if you simply round the
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display of your final results to the number of decimal digits you expect.
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:func:`str` usually suffices, and for finer control see the :meth:`str.format`
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method's format specifiers in :ref:`formatstrings`.
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For use cases which require exact decimal representation, try using the
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:mod:`decimal` module which implements decimal arithmetic suitable for
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accounting applications and high-precision applications.
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Another form of exact arithmetic is supported by the :mod:`fractions` module
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which implements arithmetic based on rational numbers (so the numbers like
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1/3 can be represented exactly).
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If you are a heavy user of floating point operations you should take a look
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at the Numerical Python package and many other packages for mathematical and
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statistical operations supplied by the SciPy project. See <http://scipy.org>.
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Python provides tools that may help on those rare occasions when you really
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*do* want to know the exact value of a float. The
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:meth:`float.as_integer_ratio` method expresses the value of a float as a
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>>> x.as_integer_ratio()
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(3537115888337719, 1125899906842624)
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Since the ratio is exact, it can be used to losslessly recreate the
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>>> x == 3537115888337719 / 1125899906842624
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The :meth:`float.hex` method expresses a float in hexadecimal (base
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16), again giving the exact value stored by your computer::
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'0x1.921f9f01b866ep+1'
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This precise hexadecimal representation can be used to reconstruct
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the float value exactly::
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>>> x == float.fromhex('0x1.921f9f01b866ep+1')
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Since the representation is exact, it is useful for reliably porting values
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across different versions of Python (platform independence) and exchanging
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data with other languages that support the same format (such as Java and C99).
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Another helpful tool is the :func:`math.fsum` function which helps mitigate
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loss-of-precision during summation. It tracks "lost digits" as values are
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added onto a running total. That can make a difference in overall accuracy
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so that the errors do not accumulate to the point where they affect the
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>>> sum([0.1] * 10) == 1.0
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>>> math.fsum([0.1] * 10) == 1.0
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This section explains the "0.1" example in detail, and shows how you can perform
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an exact analysis of cases like this yourself. Basic familiarity with binary
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floating-point representation is assumed.
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:dfn:`Representation error` refers to the fact that some (most, actually)
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decimal fractions cannot be represented exactly as binary (base 2) fractions.
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This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
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others) often won't display the exact decimal number you expect.
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Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
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machines today (November 2000) use IEEE-754 floating point arithmetic, and
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almost all platforms map Python floats to IEEE-754 "double precision". 754
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doubles contain 53 bits of precision, so on input the computer strives to
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convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
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an integer containing exactly 53 bits. Rewriting ::
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and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
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the best value for *N* is 56::
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>>> 2**52 <= 2**56 // 10 < 2**53
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That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
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best possible value for *J* is then that quotient rounded::
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>>> q, r = divmod(2**56, 10)
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Since the remainder is more than half of 10, the best approximation is obtained
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Therefore the best possible approximation to 1/10 in 754 double precision is::
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7205759403792794 / 2 ** 56
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Dividing both the numerator and denominator by two reduces the fraction to::
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3602879701896397 / 2 ** 55
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Note that since we rounded up, this is actually a little bit larger than 1/10;
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if we had not rounded up, the quotient would have been a little bit smaller than
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1/10. But in no case can it be *exactly* 1/10!
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So the computer never "sees" 1/10: what it sees is the exact fraction given
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above, the best 754 double approximation it can get::
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If we multiply that fraction by 10\*\*55, we can see the value out to
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>>> 3602879701896397 * 10 ** 55 // 2 ** 55
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1000000000000000055511151231257827021181583404541015625
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meaning that the exact number stored in the computer is equal to
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the decimal value 0.1000000000000000055511151231257827021181583404541015625.
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Instead of displaying the full decimal value, many languages (including
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older versions of Python), round the result to 17 significant digits::
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>>> format(0.1, '.17f')
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'0.10000000000000001'
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The :mod:`fractions` and :mod:`decimal` modules make these calculations
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>>> from decimal import Decimal
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>>> from fractions import Fraction
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>>> Fraction.from_float(0.1)
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Fraction(3602879701896397, 36028797018963968)
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>>> (0.1).as_integer_ratio()
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(3602879701896397, 36028797018963968)
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>>> Decimal.from_float(0.1)
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Decimal('0.1000000000000000055511151231257827021181583404541015625')
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>>> format(Decimal.from_float(0.1), '.17')
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'0.10000000000000001'
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