1 .. _tut-fp-issues:

     2 

     3 **************************************************

     4 Floating Point Arithmetic:  Issues and Limitations

     5 **************************************************

     6 

     7 .. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>

     8 

     9 

    10 Floating-point numbers are represented in computer hardware as base 2 (binary)

    11 fractions.  For example, the decimal fraction ::

    12 

    13    0.125

    14 

    15 has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::

    16 

    17    0.001

    18 

    19 has value 0/2 + 0/4 + 1/8.  These two fractions have identical values, the only

    20 real difference being that the first is written in base 10 fractional notation,

    21 and the second in base 2.

    22 

    23 Unfortunately, most decimal fractions cannot be represented exactly as binary

    24 fractions.  A consequence is that, in general, the decimal floating-point

    25 numbers you enter are only approximated by the binary floating-point numbers

    26 actually stored in the machine.

    27 

    28 The problem is easier to understand at first in base 10.  Consider the fraction

    29 1/3.  You can approximate that as a base 10 fraction::

    30 

    31    0.3

    32 

    33 or, better, ::

    34 

    35    0.33

    36 

    37 or, better, ::

    38 

    39    0.333

    40 

    41 and so on.  No matter how many digits you're willing to write down, the result

    42 will never be exactly 1/3, but will be an increasingly better approximation of

    43 1/3.

    44 

    45 In the same way, no matter how many base 2 digits you're willing to use, the

    46 decimal value 0.1 cannot be represented exactly as a base 2 fraction.  In base

    47 2, 1/10 is the infinitely repeating fraction ::

    48 

    49    0.0001100110011001100110011001100110011001100110011...

    50 

    51 Stop at any finite number of bits, and you get an approximation.  This is why

    52 you see things like::

    53 

    54    >>> 0.1

    55    0.10000000000000001

    56 

    57 On most machines today, that is what you'll see if you enter 0.1 at a Python

    58 prompt.  You may not, though, because the number of bits used by the hardware to

    59 store floating-point values can vary across machines, and Python only prints a

    60 decimal approximation to the true decimal value of the binary approximation

    61 stored by the machine.  On most machines, if Python were to print the true

    62 decimal value of the binary approximation stored for 0.1, it would have to

    63 display ::

    64 

    65    >>> 0.1

    66    0.1000000000000000055511151231257827021181583404541015625

    67 

    68 instead!  The Python prompt uses the builtin :func:`repr` function to obtain a

    69 string version of everything it displays.  For floats, ``repr(float)`` rounds

    70 the true decimal value to 17 significant digits, giving ::

    71 

    72    0.10000000000000001

    73 

    74 ``repr(float)`` produces 17 significant digits because it turns out that's

    75 enough (on most machines) so that ``eval(repr(x)) == x`` exactly for all finite

    76 floats *x*, but rounding to 16 digits is not enough to make that true.

    77 

    78 Note that this is in the very nature of binary floating-point: this is not a bug

    79 in Python, and it is not a bug in your code either.  You'll see the same kind of

    80 thing in all languages that support your hardware's floating-point arithmetic

    81 (although some languages may not *display* the difference by default, or in all

    82 output modes).

    83 

    84 Python's builtin :func:`str` function produces only 12 significant digits, and

    85 you may wish to use that instead.  It's unusual for ``eval(str(x))`` to

    86 reproduce *x*, but the output may be more pleasant to look at::

    87 

    88    >>> print str(0.1)

    89    0.1

    90 

    91 It's important to realize that this is, in a real sense, an illusion: the value

    92 in the machine is not exactly 1/10, you're simply rounding the *display* of the

    93 true machine value.

    94 

    95 Other surprises follow from this one.  For example, after seeing ::

    96 

    97    >>> 0.1

    98    0.10000000000000001

    99 

   100 you may be tempted to use the :func:`round` function to chop it back to the

   101 single digit you expect.  But that makes no difference::

   102 

   103    >>> round(0.1, 1)

   104    0.10000000000000001

   105 

   106 The problem is that the binary floating-point value stored for "0.1" was already

   107 the best possible binary approximation to 1/10, so trying to round it again

   108 can't make it better:  it was already as good as it gets.

   109 

   110 Another consequence is that since 0.1 is not exactly 1/10, summing ten values of

   111 0.1 may not yield exactly 1.0, either::

   112 

   113    >>> sum = 0.0

   114    >>> for i in range(10):

   115    ...     sum += 0.1

   116    ...

   117    >>> sum

   118    0.99999999999999989

   119 

   120 Binary floating-point arithmetic holds many surprises like this.  The problem

   121 with "0.1" is explained in precise detail below, in the "Representation Error"

   122 section.  See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_

   123 for a more complete account of other common surprises.

   124 

   125 As that says near the end, "there are no easy answers."  Still, don't be unduly

   126 wary of floating-point!  The errors in Python float operations are inherited

   127 from the floating-point hardware, and on most machines are on the order of no

   128 more than 1 part in 2\*\*53 per operation.  That's more than adequate for most

   129 tasks, but you do need to keep in mind that it's not decimal arithmetic, and

   130 that every float operation can suffer a new rounding error.

   131 

   132 While pathological cases do exist, for most casual use of floating-point

   133 arithmetic you'll see the result you expect in the end if you simply round the

   134 display of your final results to the number of decimal digits you expect.

   135 :func:`str` usually suffices, and for finer control see the :meth:`str.format`

   136 method's format specifiers in :ref:`formatstrings`.

   137 

   138 

   139 .. _tut-fp-error:

   140 

   141 Representation Error

   142 ====================

   143 

   144 This section explains the "0.1" example in detail, and shows how you can perform

   145 an exact analysis of cases like this yourself.  Basic familiarity with binary

   146 floating-point representation is assumed.

   147 

   148 :dfn:`Representation error` refers to the fact that some (most, actually)

   149 decimal fractions cannot be represented exactly as binary (base 2) fractions.

   150 This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many

   151 others) often won't display the exact decimal number you expect::

   152 

   153    >>> 0.1

   154    0.10000000000000001

   155 

   156 Why is that?  1/10 is not exactly representable as a binary fraction. Almost all

   157 machines today (November 2000) use IEEE-754 floating point arithmetic, and

   158 almost all platforms map Python floats to IEEE-754 "double precision".  754

   159 doubles contain 53 bits of precision, so on input the computer strives to

   160 convert 0.1 to the closest fraction it can of the form *J*/2\*\**N* where *J* is

   161 an integer containing exactly 53 bits.  Rewriting ::

   162 

   163    1 / 10 ~= J / (2**N)

   164 

   165 as ::

   166 

   167    J ~= 2**N / 10

   168 

   169 and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),

   170 the best value for *N* is 56::

   171 

   172    >>> 2**52

   173    4503599627370496L

   174    >>> 2**53

   175    9007199254740992L

   176    >>> 2**56/10

   177    7205759403792793L

   178 

   179 That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.  The

   180 best possible value for *J* is then that quotient rounded::

   181 

   182    >>> q, r = divmod(2**56, 10)

   183    >>> r

   184    6L

   185 

   186 Since the remainder is more than half of 10, the best approximation is obtained

   187 by rounding up::

   188 

   189    >>> q+1

   190    7205759403792794L

   191 

   192 Therefore the best possible approximation to 1/10 in 754 double precision is

   193 that over 2\*\*56, or ::

   194 

   195    7205759403792794 / 72057594037927936

   196 

   197 Note that since we rounded up, this is actually a little bit larger than 1/10;

   198 if we had not rounded up, the quotient would have been a little bit smaller than

   199 1/10.  But in no case can it be *exactly* 1/10!

   200 

   201 So the computer never "sees" 1/10:  what it sees is the exact fraction given

   202 above, the best 754 double approximation it can get::

   203 

   204    >>> .1 * 2**56

   205    7205759403792794.0

   206 

   207 If we multiply that fraction by 10\*\*30, we can see the (truncated) value of

   208 its 30 most significant decimal digits::

   209 

   210    >>> 7205759403792794 * 10**30 / 2**56

   211    100000000000000005551115123125L

   212 

   213 meaning that the exact number stored in the computer is approximately equal to

   214 the decimal value 0.100000000000000005551115123125.  Rounding that to 17

   215 significant digits gives the 0.10000000000000001 that Python displays (well,

   216 will display on any 754-conforming platform that does best-possible input and

   217 output conversions in its C library --- yours may not!).

   218 

   219