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Fixed a wrong assumption in configure.in and Include/pyport.h. The is finite function is not called isfinite() but finite(). Sorry, my fault. :)
author Christian Heimes <christian@cheimes.de>
date Sun, 20 Jan 2008 14:28:28 +0000
parents 9e1529bf0442
children 70f9a5db279c 95d6fe2e8e0e
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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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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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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. This is why
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52 you see things like::
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53
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54 >>> 0.1
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55 0.10000000000000001
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56
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57 On most machines today, that is what you'll see if you enter 0.1 at a Python
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58 prompt. You may not, though, because the number of bits used by the hardware to
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59 store floating-point values can vary across machines, and Python only prints a
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60 decimal approximation to the true decimal value of the binary approximation
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61 stored by the machine. On most machines, if Python were to print the true
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62 decimal value of the binary approximation stored for 0.1, it would have to
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63 display ::
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64
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65 >>> 0.1
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66 0.1000000000000000055511151231257827021181583404541015625
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67
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68 instead! The Python prompt uses the builtin :func:`repr` function to obtain a
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69 string version of everything it displays. For floats, ``repr(float)`` rounds
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70 the true decimal value to 17 significant digits, giving ::
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71
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72 0.10000000000000001
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73
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74 ``repr(float)`` produces 17 significant digits because it turns out that's
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75 enough (on most machines) so that ``eval(repr(x)) == x`` exactly for all finite
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76 floats *x*, but rounding to 16 digits is not enough to make that true.
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77
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78 Note that this is in the very nature of binary floating-point: this is not a bug
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79 in Python, and it is not a bug in your code either. You'll see the same kind of
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80 thing in all languages that support your hardware's floating-point arithmetic
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81 (although some languages may not *display* the difference by default, or in all
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82 output modes).
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83
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84 Python's builtin :func:`str` function produces only 12 significant digits, and
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85 you may wish to use that instead. It's unusual for ``eval(str(x))`` to
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86 reproduce *x*, but the output may be more pleasant to look at::
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87
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88 >>> print str(0.1)
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89 0.1
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90
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91 It's important to realize that this is, in a real sense, an illusion: the value
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92 in the machine is not exactly 1/10, you're simply rounding the *display* of the
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93 true machine value.
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94
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95 Other surprises follow from this one. For example, after seeing ::
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96
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97 >>> 0.1
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98 0.10000000000000001
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99
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100 you may be tempted to use the :func:`round` function to chop it back to the
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101 single digit you expect. But that makes no difference::
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102
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103 >>> round(0.1, 1)
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104 0.10000000000000001
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105
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106 The problem is that the binary floating-point value stored for "0.1" was already
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107 the best possible binary approximation to 1/10, so trying to round it again
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108 can't make it better: it was already as good as it gets.
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109
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110 Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
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111 0.1 may not yield exactly 1.0, either::
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112
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113 >>> sum = 0.0
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114 >>> for i in range(10):
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115 ... sum += 0.1
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116 ...
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117 >>> sum
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118 0.99999999999999989
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119
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120 Binary floating-point arithmetic holds many surprises like this. The problem
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121 with "0.1" is explained in precise detail below, in the "Representation Error"
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122 section. See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
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123 for a more complete account of other common surprises.
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124
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125 As that says near the end, "there are no easy answers." Still, don't be unduly
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126 wary of floating-point! The errors in Python float operations are inherited
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127 from the floating-point hardware, and on most machines are on the order of no
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128 more than 1 part in 2\*\*53 per operation. That's more than adequate for most
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129 tasks, but you do need to keep in mind that it's not decimal arithmetic, and
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130 that every float operation can suffer a new rounding error.
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131
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132 While pathological cases do exist, for most casual use of floating-point
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133 arithmetic you'll see the result you expect in the end if you simply round the
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134 display of your final results to the number of decimal digits you expect.
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135 :func:`str` usually suffices, and for finer control see the discussion of
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136 Python's ``%`` format operator: the ``%g``, ``%f`` and ``%e`` format codes
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137 supply flexible and easy ways to round float results for display.
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138
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139
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140 .. _tut-fp-error:
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141
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142 Representation Error
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143 ====================
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144
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145 This section explains the "0.1" example in detail, and shows how you can perform
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146 an exact analysis of cases like this yourself. Basic familiarity with binary
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147 floating-point representation is assumed.
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148
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149 :dfn:`Representation error` refers to the fact that some (most, actually)
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150 decimal fractions cannot be represented exactly as binary (base 2) fractions.
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151 This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
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152 others) often won't display the exact decimal number you expect::
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153
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154 >>> 0.1
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155 0.10000000000000001
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156
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157 Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
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158 machines today (November 2000) use IEEE-754 floating point arithmetic, and
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159 almost all platforms map Python floats to IEEE-754 "double precision". 754
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160 doubles contain 53 bits of precision, so on input the computer strives to
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161 convert 0.1 to the closest fraction it can of the form *J*/2\*\**N* where *J* is
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162 an integer containing exactly 53 bits. Rewriting ::
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163
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164 1 / 10 ~= J / (2**N)
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165
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166 as ::
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167
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168 J ~= 2**N / 10
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169
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170 and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
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171 the best value for *N* is 56::
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172
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173 >>> 2**52
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174 4503599627370496L
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175 >>> 2**53
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176 9007199254740992L
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177 >>> 2**56/10
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178 7205759403792793L
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179
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180 That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
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181 best possible value for *J* is then that quotient rounded::
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182
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183 >>> q, r = divmod(2**56, 10)
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184 >>> r
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185 6L
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186
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187 Since the remainder is more than half of 10, the best approximation is obtained
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188 by rounding up::
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189
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190 >>> q+1
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191 7205759403792794L
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192
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193 Therefore the best possible approximation to 1/10 in 754 double precision is
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194 that over 2\*\*56, or ::
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195
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196 7205759403792794 / 72057594037927936
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197
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198 Note that since we rounded up, this is actually a little bit larger than 1/10;
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199 if we had not rounded up, the quotient would have been a little bit smaller than
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200 1/10. But in no case can it be *exactly* 1/10!
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201
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202 So the computer never "sees" 1/10: what it sees is the exact fraction given
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203 above, the best 754 double approximation it can get::
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204
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205 >>> .1 * 2**56
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206 7205759403792794.0
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207
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208 If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
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209 its 30 most significant decimal digits::
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210
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211 >>> 7205759403792794 * 10**30 / 2**56
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212 100000000000000005551115123125L
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213
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214 meaning that the exact number stored in the computer is approximately equal to
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215 the decimal value 0.100000000000000005551115123125. Rounding that to 17
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216 significant digits gives the 0.10000000000000001 that Python displays (well,
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217 will display on any 754-conforming platform that does best-possible input and
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218 output conversions in its C library --- yours may not!).
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219
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220