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annotate Doc/tutorial/floatingpoint.rst @ 80661:94de77bd0b4b 2.7

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Issue #16194: document imp.load_dynamic problems when read several extension modules from the same binary object file.
author Andrew Svetlov <andrew.svetlov@gmail.com>
date Thu, 29 Nov 2012 19:51:28 +0200
parents b36cb4602e21
children
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1 .. _tut-fp-issues:
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2
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3 **************************************************
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4 Floating Point Arithmetic: Issues and Limitations
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5 **************************************************
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6
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7 .. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
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8
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9
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10 Floating-point numbers are represented in computer hardware as base 2 (binary)
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11 fractions. For example, the decimal fraction ::
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12
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13 0.125
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14
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15 has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
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16
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17 0.001
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18
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19 has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only
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20 real difference being that the first is written in base 10 fractional notation,
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21 and the second in base 2.
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22
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23 Unfortunately, most decimal fractions cannot be represented exactly as binary
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24 fractions. A consequence is that, in general, the decimal floating-point
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25 numbers you enter are only approximated by the binary floating-point numbers
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26 actually stored in the machine.
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27
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28 The problem is easier to understand at first in base 10. Consider the fraction
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29 1/3. You can approximate that as a base 10 fraction::
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30
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31 0.3
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32
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33 or, better, ::
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34
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35 0.33
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36
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37 or, better, ::
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38
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39 0.333
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40
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41 and so on. No matter how many digits you're willing to write down, the result
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42 will never be exactly 1/3, but will be an increasingly better approximation of
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43 1/3.
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44
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45 In the same way, no matter how many base 2 digits you're willing to use, the
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46 decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
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47 2, 1/10 is the infinitely repeating fraction ::
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48
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49 0.0001100110011001100110011001100110011001100110011...
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50
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51 Stop at any finite number of bits, and you get an approximation.
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52
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53 On a typical machine running Python, there are 53 bits of precision available
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54 for a Python float, so the value stored internally when you enter the decimal
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55 number ``0.1`` is the binary fraction ::
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56
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57 0.00011001100110011001100110011001100110011001100110011010
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58
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59 which is close to, but not exactly equal to, 1/10.
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60
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61 It's easy to forget that the stored value is an approximation to the original
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62 decimal fraction, because of the way that floats are displayed at the
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63 interpreter prompt. Python only prints a decimal approximation to the true
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64 decimal value of the binary approximation stored by the machine. If Python
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65 were to print the true decimal value of the binary approximation stored for
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66 0.1, it would have to display ::
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67
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68 >>> 0.1
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69 0.1000000000000000055511151231257827021181583404541015625
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70
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71 That is more digits than most people find useful, so Python keeps the number
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72 of digits manageable by displaying a rounded value instead ::
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73
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74 >>> 0.1
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75 0.1
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76
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77 It's important to realize that this is, in a real sense, an illusion: the value
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78 in the machine is not exactly 1/10, you're simply rounding the *display* of the
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79 true machine value. This fact becomes apparent as soon as you try to do
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80 arithmetic with these values ::
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81
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82 >>> 0.1 + 0.2
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83 0.30000000000000004
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84
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85 Note that this is in the very nature of binary floating-point: this is not a
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86 bug in Python, and it is not a bug in your code either. You'll see the same
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87 kind of thing in all languages that support your hardware's floating-point
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88 arithmetic (although some languages may not *display* the difference by
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89 default, or in all output modes).
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90
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91 Other surprises follow from this one. For example, if you try to round the
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92 value 2.675 to two decimal places, you get this ::
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93
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94 >>> round(2.675, 2)
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95 2.67
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96
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97 The documentation for the built-in :func:`round` function says that it rounds
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98 to the nearest value, rounding ties away from zero. Since the decimal fraction
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99 2.675 is exactly halfway between 2.67 and 2.68, you might expect the result
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100 here to be (a binary approximation to) 2.68. It's not, because when the
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101 decimal string ``2.675`` is converted to a binary floating-point number, it's
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102 again replaced with a binary approximation, whose exact value is ::
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103
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104 2.67499999999999982236431605997495353221893310546875
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105
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106 Since this approximation is slightly closer to 2.67 than to 2.68, it's rounded
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107 down.
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108
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109 If you're in a situation where you care which way your decimal halfway-cases
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110 are rounded, you should consider using the :mod:`decimal` module.
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111 Incidentally, the :mod:`decimal` module also provides a nice way to "see" the
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112 exact value that's stored in any particular Python float ::
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113
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114 >>> from decimal import Decimal
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115 >>> Decimal(2.675)
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116 Decimal('2.67499999999999982236431605997495353221893310546875')
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117
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118 Another consequence is that since 0.1 is not exactly 1/10, summing ten values
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119 of 0.1 may not yield exactly 1.0, either::
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120
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121 >>> sum = 0.0
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122 >>> for i in range(10):
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123 ... sum += 0.1
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124 ...
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125 >>> sum
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126 0.9999999999999999
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127
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128 Binary floating-point arithmetic holds many surprises like this. The problem
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129 with "0.1" is explained in precise detail below, in the "Representation Error"
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130 section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
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131 for a more complete account of other common surprises.
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132
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133 As that says near the end, "there are no easy answers." Still, don't be unduly
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134 wary of floating-point! The errors in Python float operations are inherited
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135 from the floating-point hardware, and on most machines are on the order of no
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136 more than 1 part in 2\*\*53 per operation. That's more than adequate for most
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137 tasks, but you do need to keep in mind that it's not decimal arithmetic, and
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138 that every float operation can suffer a new rounding error.
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139
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140 While pathological cases do exist, for most casual use of floating-point
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141 arithmetic you'll see the result you expect in the end if you simply round the
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142 display of your final results to the number of decimal digits you expect. For
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143 fine control over how a float is displayed see the :meth:`str.format` method's
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144 format specifiers in :ref:`formatstrings`.
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145
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146
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147 .. _tut-fp-error:
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148
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149 Representation Error
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150 ====================
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151
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152 This section explains the "0.1" example in detail, and shows how you can
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153 perform an exact analysis of cases like this yourself. Basic familiarity with
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154 binary floating-point representation is assumed.
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155
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156 :dfn:`Representation error` refers to the fact that some (most, actually)
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157 decimal fractions cannot be represented exactly as binary (base 2) fractions.
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158 This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
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159 others) often won't display the exact decimal number you expect::
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160
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161 >>> 0.1 + 0.2
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162 0.30000000000000004
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163
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164 Why is that? 1/10 and 2/10 are not exactly representable as a binary
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165 fraction. Almost all machines today (July 2010) use IEEE-754 floating point
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166 arithmetic, and almost all platforms map Python floats to IEEE-754 "double
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167 precision". 754 doubles contain 53 bits of precision, so on input the computer
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168 strives to convert 0.1 to the closest fraction it can of the form *J*/2**\ *N*
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169 where *J* is an integer containing exactly 53 bits. Rewriting ::
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170
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171 1 / 10 ~= J / (2**N)
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172
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173 as ::
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174
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175 J ~= 2**N / 10
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176
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177 and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
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178 the best value for *N* is 56::
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179
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180 >>> 2**52
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181 4503599627370496
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182 >>> 2**53
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183 9007199254740992
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184 >>> 2**56/10
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185 7205759403792793
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186
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187 That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.
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188 The best possible value for *J* is then that quotient rounded::
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189
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190 >>> q, r = divmod(2**56, 10)
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191 >>> r
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192 6
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193
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194 Since the remainder is more than half of 10, the best approximation is obtained
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195 by rounding up::
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196
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197 >>> q+1
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198 7205759403792794
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199
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200 Therefore the best possible approximation to 1/10 in 754 double precision is
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201 that over 2\*\*56, or ::
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202
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203 7205759403792794 / 72057594037927936
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204
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205 Note that since we rounded up, this is actually a little bit larger than 1/10;
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206 if we had not rounded up, the quotient would have been a little bit smaller
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207 than 1/10. But in no case can it be *exactly* 1/10!
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208
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209 So the computer never "sees" 1/10: what it sees is the exact fraction given
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210 above, the best 754 double approximation it can get::
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211
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212 >>> .1 * 2**56
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213 7205759403792794.0
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214
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215 If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
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216 its 30 most significant decimal digits::
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217
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218 >>> 7205759403792794 * 10**30 // 2**56
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219 100000000000000005551115123125L
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220
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221 meaning that the exact number stored in the computer is approximately equal to
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222 the decimal value 0.100000000000000005551115123125. In versions prior to
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223 Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits,
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224 giving '0.10000000000000001'. In current versions, Python displays a value
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225 based on the shortest decimal fraction that rounds correctly back to the true
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226 binary value, resulting simply in '0.1'.