Clean-up floating point tutorial
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@ -50,7 +50,7 @@ decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
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Stop at any finite number of bits, and you get an approximation. On most
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machines today, floats are approximated using a binary fraction with
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the numerator using the first 53 bits following the most significant bit and
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the numerator using the first 53 bits starting with the most significant bit and
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with the denominator as a power of two. In the case of 1/10, the binary fraction
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is ``3602879701896397 / 2 ** 55`` which is close to but not exactly
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equal to the true value of 1/10.
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@ -230,12 +230,8 @@ as ::
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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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4503599627370496
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>>> 2**53
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9007199254740992
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>>> 2**56/10
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7205759403792794.0
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>>> 2**52 <= 2**56 // 10 < 2**53
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True
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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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@ -250,14 +246,13 @@ by rounding up::
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>>> q+1
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7205759403792794
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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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Therefore the best possible approximation to 1/10 in 754 double precision is::
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7205759403792794 / 72057594037927936
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7205759403792794 / 2 ** 56
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Dividing both the numerator and denominator by two reduces the fraction to::
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3602879701896397 / 36028797018963968
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3602879701896397 / 2 ** 55
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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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@ -269,24 +264,34 @@ above, the best 754 double approximation it can get::
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>>> 0.1 * 2 ** 55
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3602879701896397.0
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If we multiply that fraction by 10\*\*60, we can see the value of out to
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60 decimal digits::
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If we multiply that fraction by 10\*\*55, we can see the value out to
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55 decimal digits::
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>>> 3602879701896397 * 10 ** 60 // 2 ** 55
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>>> 3602879701896397 * 10 ** 55 // 2 ** 55
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1000000000000000055511151231257827021181583404541015625
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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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meaning that the exact number stored in the computer is equal to
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the decimal value 0.1000000000000000055511151231257827021181583404541015625.
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Instead of displaying the full decimal value, many languages (including
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older versions of Python), round the result to 17 significant digits::
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>>> format(0.1, '.17f')
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'0.10000000000000001'
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The :mod:`fractions` and :mod:`decimal` modules make these calculations
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easy::
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>>> from decimal import Decimal
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>>> from fractions import Fraction
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>>> print(Fraction.from_float(0.1))
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3602879701896397/36028797018963968
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>>> print(Decimal.from_float(0.1))
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0.1000000000000000055511151231257827021181583404541015625
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>>> Fraction.from_float(0.1)
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Fraction(3602879701896397, 36028797018963968)
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>>> (0.1).as_integer_ratio()
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(3602879701896397, 36028797018963968)
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>>> Decimal.from_float(0.1)
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Decimal('0.1000000000000000055511151231257827021181583404541015625')
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>>> format(Decimal.from_float(0.1), '.17')
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'0.10000000000000001'
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