More tweaks to floating-point section of the tutorial.

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Mark Dickinson 2010-08-04 21:44:47 +00:00
parent dd1d8f72f9
commit 33e5935b53
1 changed files with 34 additions and 32 deletions

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@ -48,9 +48,11 @@ decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base
0.0001100110011001100110011001100110011001100110011... 0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation. On a typical Stop at any finite number of bits, and you get an approximation.
machine, there are 53 bits of precision available, so the value stored
internally is the binary fraction :: On a typical machine running Python, there are 53 bits of precision available
for a Python float, so the value stored internally when you enter the decimal
number ``0.1`` is the binary fraction ::
0.00011001100110011001100110011001100110011001100110011010 0.00011001100110011001100110011001100110011001100110011010
@ -80,14 +82,14 @@ arithmetic with these values ::
>>> 0.1 + 0.2 >>> 0.1 + 0.2
0.30000000000000004 0.30000000000000004
Note that this is in the very nature of binary floating-point: this is not a bug Note that this is in the very nature of binary floating-point: this is not a
in Python, and it is not a bug in your code either. You'll see the same kind of bug in Python, and it is not a bug in your code either. You'll see the same
thing in all languages that support your hardware's floating-point arithmetic kind of thing in all languages that support your hardware's floating-point
(although some languages may not *display* the difference by default, or in all arithmetic (although some languages may not *display* the difference by
output modes). default, or in all output modes).
Other surprises follow from this one. For example, if you try to round the value Other surprises follow from this one. For example, if you try to round the
2.675 to two decimal places, you get this :: value 2.675 to two decimal places, you get this ::
>>> round(2.675, 2) >>> round(2.675, 2)
2.67 2.67
@ -96,7 +98,7 @@ The documentation for the built-in :func:`round` function says that it rounds
to the nearest value, rounding ties away from zero. Since the decimal fraction to the nearest value, rounding ties away from zero. Since the decimal fraction
2.675 is exactly halfway between 2.67 and 2.68, you might expect the result 2.675 is exactly halfway between 2.67 and 2.68, you might expect the result
here to be (a binary approximation to) 2.68. It's not, because when the here to be (a binary approximation to) 2.68. It's not, because when the
decimal literal ``2.675`` is converted to a binary floating-point number, it's decimal string ``2.675`` is converted to a binary floating-point number, it's
again replaced with a binary approximation, whose exact value is :: again replaced with a binary approximation, whose exact value is ::
2.67499999999999982236431605997495353221893310546875 2.67499999999999982236431605997495353221893310546875
@ -113,8 +115,8 @@ exact value that's stored in any particular Python float ::
>>> Decimal(2.675) >>> Decimal(2.675)
Decimal('2.67499999999999982236431605997495353221893310546875') Decimal('2.67499999999999982236431605997495353221893310546875')
Another consequence is that since 0.1 is not exactly 1/10, summing ten values of Another consequence is that since 0.1 is not exactly 1/10, summing ten values
0.1 may not yield exactly 1.0, either:: of 0.1 may not yield exactly 1.0, either::
>>> sum = 0.0 >>> sum = 0.0
>>> for i in range(10): >>> for i in range(10):
@ -137,9 +139,9 @@ that every float operation can suffer a new rounding error.
While pathological cases do exist, for most casual use of floating-point While pathological cases do exist, for most casual use of floating-point
arithmetic you'll see the result you expect in the end if you simply round the arithmetic you'll see the result you expect in the end if you simply round the
display of your final results to the number of decimal digits you expect. display of your final results to the number of decimal digits you expect. For
:func:`str` usually suffices, and for finer control see the :meth:`str.format` fine control over how a float is displayed see the :meth:`str.format` method's
method's format specifiers in :ref:`formatstrings`. format specifiers in :ref:`formatstrings`.
.. _tut-fp-error: .. _tut-fp-error:
@ -147,9 +149,9 @@ method's format specifiers in :ref:`formatstrings`.
Representation Error Representation Error
==================== ====================
This section explains the "0.1" example in detail, and shows how you can perform This section explains the "0.1" example in detail, and shows how you can
an exact analysis of cases like this yourself. Basic familiarity with binary perform an exact analysis of cases like this yourself. Basic familiarity with
floating-point representation is assumed. binary floating-point representation is assumed.
:dfn:`Representation error` refers to the fact that some (most, actually) :dfn:`Representation error` refers to the fact that some (most, actually)
decimal fractions cannot be represented exactly as binary (base 2) fractions. decimal fractions cannot be represented exactly as binary (base 2) fractions.
@ -176,24 +178,24 @@ and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
the best value for *N* is 56:: the best value for *N* is 56::
>>> 2**52 >>> 2**52
4503599627370496L 4503599627370496
>>> 2**53 >>> 2**53
9007199254740992L 9007199254740992
>>> 2**56/10 >>> 2**56/10
7205759403792793L 7205759403792793
That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.
best possible value for *J* is then that quotient rounded:: The best possible value for *J* is then that quotient rounded::
>>> q, r = divmod(2**56, 10) >>> q, r = divmod(2**56, 10)
>>> r >>> r
6L 6
Since the remainder is more than half of 10, the best approximation is obtained Since the remainder is more than half of 10, the best approximation is obtained
by rounding up:: by rounding up::
>>> q+1 >>> q+1
7205759403792794L 7205759403792794
Therefore the best possible approximation to 1/10 in 754 double precision is Therefore the best possible approximation to 1/10 in 754 double precision is
that over 2\*\*56, or :: that over 2\*\*56, or ::
@ -201,8 +203,8 @@ that over 2\*\*56, or ::
7205759403792794 / 72057594037927936 7205759403792794 / 72057594037927936
Note that since we rounded up, this is actually a little bit larger than 1/10; Note that since we rounded up, this is actually a little bit larger than 1/10;
if we had not rounded up, the quotient would have been a little bit smaller than if we had not rounded up, the quotient would have been a little bit smaller
1/10. But in no case can it be *exactly* 1/10! than 1/10. But in no case can it be *exactly* 1/10!
So the computer never "sees" 1/10: what it sees is the exact fraction given So the computer never "sees" 1/10: what it sees is the exact fraction given
above, the best 754 double approximation it can get:: above, the best 754 double approximation it can get::
@ -213,12 +215,12 @@ above, the best 754 double approximation it can get::
If we multiply that fraction by 10\*\*30, we can see the (truncated) value of If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
its 30 most significant decimal digits:: its 30 most significant decimal digits::
>>> 7205759403792794 * 10**30 / 2**56 >>> 7205759403792794 * 10**30 // 2**56
100000000000000005551115123125L 100000000000000005551115123125L
meaning that the exact number stored in the computer is approximately equal to meaning that the exact number stored in the computer is approximately equal to
the decimal value 0.100000000000000005551115123125. In versions prior to the decimal value 0.100000000000000005551115123125. In versions prior to
Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits, Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits,
giving '0.10000000000000001'. In current versions, Python displays a value based giving '0.10000000000000001'. In current versions, Python displays a value
on the shortest decimal fraction that rounds correctly back to the true binary based on the shortest decimal fraction that rounds correctly back to the true
value, resulting simply in '0.1'. binary value, resulting simply in '0.1'.