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223 lines
8.1 KiB
ReStructuredText
.. _tut-fp-issues:
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**************************************************
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Floating Point Arithmetic: Issues and Limitations
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**************************************************
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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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0.125
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has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
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0.001
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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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0.3
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or, better, ::
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0.33
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or, better, ::
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0.333
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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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1/3.
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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. This is why
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you see things like::
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>>> 0.1
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0.10000000000000001
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On most machines today, that is what you'll see if you enter 0.1 at a Python
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prompt. You may not, though, because the number of bits used by the hardware to
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store floating-point values can vary across machines, and Python only prints a
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decimal approximation to the true decimal value of the binary approximation
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stored by the machine. On most machines, if Python were to print the true
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decimal value of the binary approximation stored for 0.1, it would have to
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display ::
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>>> 0.1
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0.1000000000000000055511151231257827021181583404541015625
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instead! The Python prompt uses the builtin :func:`repr` function to obtain a
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string version of everything it displays. For floats, ``repr(float)`` rounds
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the true decimal value to 17 significant digits, giving ::
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0.10000000000000001
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``repr(float)`` produces 17 significant digits because it turns out that's
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enough (on most machines) so that ``eval(repr(x)) == x`` exactly for all finite
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floats *x*, but rounding to 16 digits is not enough to make that true.
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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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output modes).
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Python's builtin :func:`str` function produces only 12 significant digits, and
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you may wish to use that instead. It's unusual for ``eval(str(x))`` to
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reproduce *x*, but the output may be more pleasant to look at::
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>>> print(str(0.1))
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0.1
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It's important to realize that this is, in a real sense, an illusion: the value
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in the machine is not exactly 1/10, you're simply rounding the *display* of the
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true machine value.
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Other surprises follow from this one. For example, after seeing ::
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>>> 0.1
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0.10000000000000001
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you may be tempted to use the :func:`round` function to chop it back to the
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single digit you expect. But that makes no difference::
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>>> round(0.1, 1)
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0.10000000000000001
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The problem is that the binary floating-point value stored for "0.1" was already
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the best possible binary approximation to 1/10, so trying to round it again
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can't make it better: it was already as good as it gets.
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Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
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0.1 may not yield exactly 1.0, either::
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>>> sum = 0.0
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>>> for i in range(10):
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... sum += 0.1
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...
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>>> sum
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0.99999999999999989
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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 discussion of
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Python's ``%`` format operator: the ``%g``, ``%f`` and ``%e`` format codes
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supply flexible and easy ways to round float results for display.
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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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.. _tut-fp-error:
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Representation Error
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====================
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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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>>> 0.1
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0.10000000000000001
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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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1 / 10 ~= J / (2**N)
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as ::
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J ~= 2**N / 10
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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
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4503599627370496L
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>>> 2**53
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9007199254740992L
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>>> 2**56/10
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7205759403792793L
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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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>>> r
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6L
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Since the remainder is more than half of 10, the best approximation is obtained
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by rounding up::
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>>> q+1
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7205759403792794L
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Therefore the best possible approximation to 1/10 in 754 double precision is
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that over 2\*\*56, or ::
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7205759403792794 / 72057594037927936
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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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>>> .1 * 2**56
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7205759403792794.0
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If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
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its 30 most significant decimal digits::
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>>> 7205759403792794 * 10**30 / 2**56
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100000000000000005551115123125L
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meaning that the exact number stored in the computer is approximately equal to
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the decimal value 0.100000000000000005551115123125. Rounding that to 17
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significant digits gives the 0.10000000000000001 that Python displays (well,
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will display on any 754-conforming platform that does best-possible input and
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output conversions in its C library --- yours may not!).
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