/Doc/tutorial/floatingpoint.rst

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  1. .. _tut-fp-issues:
  2. **************************************************
  3. Floating Point Arithmetic: Issues and Limitations
  4. **************************************************
  5. .. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
  6. Floating-point numbers are represented in computer hardware as base 2 (binary)
  7. fractions. For example, the decimal fraction ::
  8. 0.125
  9. has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
  10. 0.001
  11. has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only
  12. real difference being that the first is written in base 10 fractional notation,
  13. and the second in base 2.
  14. Unfortunately, most decimal fractions cannot be represented exactly as binary
  15. fractions. A consequence is that, in general, the decimal floating-point
  16. numbers you enter are only approximated by the binary floating-point numbers
  17. actually stored in the machine.
  18. The problem is easier to understand at first in base 10. Consider the fraction
  19. 1/3. You can approximate that as a base 10 fraction::
  20. 0.3
  21. or, better, ::
  22. 0.33
  23. or, better, ::
  24. 0.333
  25. and so on. No matter how many digits you're willing to write down, the result
  26. will never be exactly 1/3, but will be an increasingly better approximation of
  27. 1/3.
  28. In the same way, no matter how many base 2 digits you're willing to use, the
  29. decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
  30. 2, 1/10 is the infinitely repeating fraction ::
  31. 0.0001100110011001100110011001100110011001100110011...
  32. Stop at any finite number of bits, and you get an approximation. This is why
  33. you see things like::
  34. >>> 0.1
  35. 0.10000000000000001
  36. On most machines today, that is what you'll see if you enter 0.1 at a Python
  37. prompt. You may not, though, because the number of bits used by the hardware to
  38. store floating-point values can vary across machines, and Python only prints a
  39. decimal approximation to the true decimal value of the binary approximation
  40. stored by the machine. On most machines, if Python were to print the true
  41. decimal value of the binary approximation stored for 0.1, it would have to
  42. display ::
  43. >>> 0.1
  44. 0.1000000000000000055511151231257827021181583404541015625
  45. instead! The Python prompt uses the built-in :func:`repr` function to obtain a
  46. string version of everything it displays. For floats, ``repr(float)`` rounds
  47. the true decimal value to 17 significant digits, giving ::
  48. 0.10000000000000001
  49. ``repr(float)`` produces 17 significant digits because it turns out that's
  50. enough (on most machines) so that ``eval(repr(x)) == x`` exactly for all finite
  51. floats *x*, but rounding to 16 digits is not enough to make that true.
  52. Note that this is in the very nature of binary floating-point: this is not a bug
  53. in Python, and it is not a bug in your code either. You'll see the same kind of
  54. thing in all languages that support your hardware's floating-point arithmetic
  55. (although some languages may not *display* the difference by default, or in all
  56. output modes).
  57. Python's built-in :func:`str` function produces only 12 significant digits, and
  58. you may wish to use that instead. It's unusual for ``eval(str(x))`` to
  59. reproduce *x*, but the output may be more pleasant to look at::
  60. >>> print str(0.1)
  61. 0.1
  62. It's important to realize that this is, in a real sense, an illusion: the value
  63. in the machine is not exactly 1/10, you're simply rounding the *display* of the
  64. true machine value.
  65. Other surprises follow from this one. For example, after seeing ::
  66. >>> 0.1
  67. 0.10000000000000001
  68. you may be tempted to use the :func:`round` function to chop it back to the
  69. single digit you expect. But that makes no difference::
  70. >>> round(0.1, 1)
  71. 0.10000000000000001
  72. The problem is that the binary floating-point value stored for "0.1" was already
  73. the best possible binary approximation to 1/10, so trying to round it again
  74. can't make it better: it was already as good as it gets.
  75. Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
  76. 0.1 may not yield exactly 1.0, either::
  77. >>> sum = 0.0
  78. >>> for i in range(10):
  79. ... sum += 0.1
  80. ...
  81. >>> sum
  82. 0.99999999999999989
  83. Binary floating-point arithmetic holds many surprises like this. The problem
  84. with "0.1" is explained in precise detail below, in the "Representation Error"
  85. section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
  86. for a more complete account of other common surprises.
  87. As that says near the end, "there are no easy answers." Still, don't be unduly
  88. wary of floating-point! The errors in Python float operations are inherited
  89. from the floating-point hardware, and on most machines are on the order of no
  90. more than 1 part in 2\*\*53 per operation. That's more than adequate for most
  91. tasks, but you do need to keep in mind that it's not decimal arithmetic, and
  92. that every float operation can suffer a new rounding error.
  93. While pathological cases do exist, for most casual use of floating-point
  94. arithmetic you'll see the result you expect in the end if you simply round the
  95. display of your final results to the number of decimal digits you expect.
  96. :func:`str` usually suffices, and for finer control see the :meth:`str.format`
  97. method's format specifiers in :ref:`formatstrings`.
  98. .. _tut-fp-error:
  99. Representation Error
  100. ====================
  101. This section explains the "0.1" example in detail, and shows how you can perform
  102. an exact analysis of cases like this yourself. Basic familiarity with binary
  103. floating-point representation is assumed.
  104. :dfn:`Representation error` refers to the fact that some (most, actually)
  105. decimal fractions cannot be represented exactly as binary (base 2) fractions.
  106. This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
  107. others) often won't display the exact decimal number you expect::
  108. >>> 0.1
  109. 0.10000000000000001
  110. Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
  111. machines today (November 2000) use IEEE-754 floating point arithmetic, and
  112. almost all platforms map Python floats to IEEE-754 "double precision". 754
  113. doubles contain 53 bits of precision, so on input the computer strives to
  114. convert 0.1 to the closest fraction it can of the form *J*/2\*\**N* where *J* is
  115. an integer containing exactly 53 bits. Rewriting ::
  116. 1 / 10 ~= J / (2**N)
  117. as ::
  118. J ~= 2**N / 10
  119. and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
  120. the best value for *N* is 56::
  121. >>> 2**52
  122. 4503599627370496L
  123. >>> 2**53
  124. 9007199254740992L
  125. >>> 2**56/10
  126. 7205759403792793L
  127. That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
  128. best possible value for *J* is then that quotient rounded::
  129. >>> q, r = divmod(2**56, 10)
  130. >>> r
  131. 6L
  132. Since the remainder is more than half of 10, the best approximation is obtained
  133. by rounding up::
  134. >>> q+1
  135. 7205759403792794L
  136. Therefore the best possible approximation to 1/10 in 754 double precision is
  137. that over 2\*\*56, or ::
  138. 7205759403792794 / 72057594037927936
  139. Note that since we rounded up, this is actually a little bit larger than 1/10;
  140. if we had not rounded up, the quotient would have been a little bit smaller than
  141. 1/10. But in no case can it be *exactly* 1/10!
  142. So the computer never "sees" 1/10: what it sees is the exact fraction given
  143. above, the best 754 double approximation it can get::
  144. >>> .1 * 2**56
  145. 7205759403792794.0
  146. If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
  147. its 30 most significant decimal digits::
  148. >>> 7205759403792794 * 10**30 / 2**56
  149. 100000000000000005551115123125L
  150. meaning that the exact number stored in the computer is approximately equal to
  151. the decimal value 0.100000000000000005551115123125. Rounding that to 17
  152. significant digits gives the 0.10000000000000001 that Python displays (well,
  153. will display on any 754-conforming platform that does best-possible input and
  154. output conversions in its C library --- yours may not!).