B. Floating Point Arithmetic: Issues and Limitations
Floatingpoint numbers are represented in computer hardware as
base 2 (binary) fractions. For example, the decimal fraction
has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction
has value 0/2 + 0/4 + 1/8. These two fractions have identical values,
the only real difference being that the first is written in base 10
fractional notation, and the second in base 2.
Unfortunately, most decimal fractions cannot be represented exactly as
binary fractions. A consequence is that, in general, the decimal
floatingpoint numbers you enter are only approximated by the binary
floatingpoint numbers actually stored in the machine.
The problem is easier to understand at first in base 10. Consider the
fraction 1/3. You can approximate that as a base 10 fraction:
or, better,
or, better,
and so on. No matter how many digits you're willing to write down, the
result will never be exactly 1/3, but will be an increasingly better
approximation of 1/3.
In the same way, no matter how many base 2 digits you're willing to
use, the decimal value 0.1 cannot be represented exactly as a base 2
fraction. In base 2, 1/10 is the infinitely repeating fraction
0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation. This
is why you see things like:
>>> 0.1
0.10000000000000001
On most machines today, that is what you'll see if you enter 0.1 at
a Python prompt. You may not, though, because the number of bits
used by the hardware to store floatingpoint values can vary across
machines, and Python only prints a decimal approximation to the true
decimal value of the binary approximation stored by the machine. On
most machines, if Python were to print the true decimal value of
the binary approximation stored for 0.1, it would have to display
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
instead! The Python prompt uses the builtin
repr() function to obtain a string version of everything it
displays. For floats, repr(float) rounds the true
decimal value to 17 significant digits, giving
repr(float) produces 17 significant digits because it
turns out that's enough (on most machines) so that
eval(repr(x)) == x exactly for all finite floats
x, but rounding to 16 digits is not enough to make that true.
Note that this is in the very nature of binary floatingpoint: 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 floatingpoint arithmetic (although some languages may
not display the difference by default, or in all output modes).
Python's builtin str() function produces only 12
significant digits, and you may wish to use that instead. It's
unusual for eval(str(x)) to reproduce x, but the
output may be more pleasant to look at:
It's important to realize that this is, in a real sense, an illusion:
the value in the machine is not exactly 1/10, you're simply rounding
the display of the true machine value.
Other surprises follow from this one. For example, after seeing
>>> 0.1
0.10000000000000001
you may be tempted to use the round() function to chop it
back to the single digit you expect. But that makes no difference:
>>> round(0.1, 1)
0.10000000000000001
The problem is that the binary floatingpoint value stored for "0.1"
was already the best possible binary approximation to 1/10, so trying
to round it again can't make it better: it was already as good as it
gets.
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):
... sum += 0.1
...
>>> sum
0.99999999999999989
Binary floatingpoint arithmetic holds many surprises like this. The
problem with "0.1" is explained in precise detail below, in the
"Representation Error" section. See
The Perils of Floating
Point for a more complete account of other common surprises.
As that says near the end, ``there are no easy answers.'' Still,
don't be unduly wary of floatingpoint! The errors in Python float
operations are inherited from the floatingpoint hardware, and on most
machines are on the order of no more than 1 part in 2**53 per
operation. That's more than adequate for most tasks, but you do need
to keep in mind that it's not decimal arithmetic, and that every float
operation can suffer a new rounding error.
While pathological cases do exist, for most casual use of
floatingpoint 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. str() usually suffices, and for
finer control see the discussion of Python's % format
operator: the %g , %f and %e format codes
supply flexible and easy ways to round float results for display.
B.1 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 floatingpoint representation is assumed.
Representation error refers to the fact that some (most, actually)
decimal fractions cannot be represented exactly as binary (base 2)
fractions. This is the chief reason why Python (or Perl, C, C++,
Java, Fortran, and many others) often won't display the exact decimal
number you expect:
>>> 0.1
0.10000000000000001
Why is that? 1/10 is not exactly representable as a binary fraction.
Almost all machines today (November 2000) use IEEE754 floating point
arithmetic, and almost all platforms map Python floats to IEEE754
"double precision". 754 doubles contain 53 bits of precision, so on
input the computer strives to convert 0.1 to the closest fraction it can
of the form J/2**N where J is an integer containing
exactly 53 bits. Rewriting
as
and recalling that J has exactly 53 bits (is >= 2**52 but
< 2**53 ), the best value for N is 56:
>>> 2**52
4503599627370496L
>>> 2**53
9007199254740992L
>>> 2**56/10
7205759403792793L
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
Since the remainder is more than half of 10, the best approximation is
obtained by rounding up:
>>> q+1
7205759403792794L
Therefore the best possible approximation to 1/10 in 754 double
precision is 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!
So the computer never ``sees'' 1/10: what it sees is the exact
fraction given above, the best 754 double approximation it can get:
>>> .1 * 2**56
7205759403792794.0
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
100000000000000005551115123125L
meaning that the exact number stored in the computer is approximately
equal to the decimal value 0.100000000000000005551115123125. Rounding
that to 17 significant digits gives the 0.10000000000000001 that Python
displays (well, will display on any 754conforming platform that does
bestpossible input and output conversions in its C library  yours may
not!).
Release 2.5.2, documentation updated on 21st February, 2008.
See About this document... for information on suggesting changes.
