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...
Stop at any finite number of bits, and you get an approximation. On a typical
machine, there are 53 bits of precision available, so the value stored
internally is the binary fraction ::
Stop at any finite number of bits, and you get an approximation.
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
@ -80,14 +82,14 @@ arithmetic with these values ::
>>> 0.1 + 0.2
0.30000000000000004
Note that this is in the very nature of binary floating-point: this is not a bug
in Python, and it is not a bug in your code either. You'll see the same kind of
thing in all languages that support your hardware's floating-point arithmetic
(although some languages may not *display* the difference by default, or in all
output modes).
Note that this is in the very nature of binary floating-point: this is not a
bug in Python, and it is not a bug in your code either. You'll see the same
kind of thing in all languages that support your hardware's floating-point
arithmetic (although some languages may not *display* the difference by
default, or in all output modes).
Other surprises follow from this one. For example, if you try to round the value
2.675 to two decimal places, you get this ::
Other surprises follow from this one. For example, if you try to round the
value 2.675 to two decimal places, you get this ::
>>> round(2.675, 2)
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
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
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 ::
2.67499999999999982236431605997495353221893310546875
@ -113,8 +115,8 @@ exact value that's stored in any particular Python float ::
>>> Decimal(2.675)
Decimal('2.67499999999999982236431605997495353221893310546875')
Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
0.1 may not yield exactly 1.0, either::
Another consequence is that since 0.1 is not exactly 1/10, summing ten values
of 0.1 may not yield exactly 1.0, either::
>>> sum = 0.0
>>> 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
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.
:func:`str` usually suffices, and for finer control see the :meth:`str.format`
method's format specifiers in :ref:`formatstrings`.
display of your final results to the number of decimal digits you expect. For
fine control over how a float is displayed see the :meth:`str.format` method's
format specifiers in :ref:`formatstrings`.
.. _tut-fp-error:
@ -147,9 +149,9 @@ method's format specifiers in :ref:`formatstrings`.
Representation Error
====================
This section explains the "0.1" example in detail, and shows how you can perform
an exact analysis of cases like this yourself. Basic familiarity with binary
floating-point representation is assumed.
This section explains the "0.1" example in detail, and shows how you can
perform an exact analysis of cases like this yourself. Basic familiarity with
binary floating-point representation is assumed.
:dfn:`Representation error` refers to the fact that some (most, actually)
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::
>>> 2**52
4503599627370496L
4503599627370496
>>> 2**53
9007199254740992L
9007199254740992
>>> 2**56/10
7205759403792793L
7205759403792793
That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The
best possible value for *J* is then that quotient rounded::
That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.
The best possible value for *J* is then that quotient rounded::
>>> q, r = divmod(2**56, 10)
>>> r
6L
6
Since the remainder is more than half of 10, the best approximation is obtained
by rounding up::
>>> q+1
7205759403792794L
7205759403792794
Therefore the best possible approximation to 1/10 in 754 double precision is
that over 2\*\*56, or ::
@ -201,8 +203,8 @@ that over 2\*\*56, or ::
7205759403792794 / 72057594037927936
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
1/10. But in no case can it be *exactly* 1/10!
if we had not rounded up, the quotient would have been a little bit smaller
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
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
its 30 most significant decimal digits::
>>> 7205759403792794 * 10**30 / 2**56
>>> 7205759403792794 * 10**30 // 2**56
100000000000000005551115123125L
meaning that the exact number stored in the computer is approximately equal 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,
giving '0.10000000000000001'. In current versions, Python displays a value based
on the shortest decimal fraction that rounds correctly back to the true binary
value, resulting simply in '0.1'.
giving '0.10000000000000001'. In current versions, Python displays a value
based on the shortest decimal fraction that rounds correctly back to the true
binary value, resulting simply in '0.1'.