round(22.47,2) gives 22.469999999999999?

[Eric van Riet Paap]

Hi,

On python2.1.3, python2.2.1 and python2.2.3 round(22.47,2) gives
22.469999999999999 . Does anyone know if this is a bug or some weird, yet
unexpected, behaviour?

P.S. I haven't tested python2.3, but python1.5.2 doesn't have this problem.

 

[Damien Wyart]

On python2.1.3, python2.2.1 and python2.2.3 round(22.47,2) gives
22.469999999999999 .

This is explained in detail at
<http://www.python.org/doc/current/tut/node14.html>.

 

[Duncan Booth]

Hi,

On python2.1.3, python2.2.1 and python2.2.3 round(22.47,2) gives
22.469999999999999 . Does anyone know if this is a bug or some weird,
yet unexpected, behaviour?

P.S. I haven't tested python2.3, but python1.5.2 doesn't have this
problem.

Please read the FAQ, particularly entry 4.98 Why are floating point
calculations so inaccurate?

http://www.python.org/cgi-bin/faqw.py?req=show&file=faq04.098.htp

 

[Thomas Andrews]

Hi,

On python2.1.3, python2.2.1 and python2.2.3 round(22.47,2) gives
22.469999999999999 . Does anyone know if this is a bug or some weird, yet
unexpected, behaviour?

P.S. I haven't tested python2.3, but python1.5.2 doesn't have this problem.

In IDLE, just typing:
22.469999999999999

That seems to indicate that 22.47 is best approximated by the floating
number 22.499999999999.

Rounding should probably not be used for gettting "formatable" values.
Instead, you can do something like:
'22.47'

if you want a string version of the number rounded to 2 decimal places.

=thomas

 

[John Machin]

Wouldn't

int(22.47 + .5)

get you to where you want to go?

Sure would, if 22 was where you wanted to go. One trusts that you
would be lonely on arrival, though. Perhaps you might then notice a
bunch of folks over at 22.47 ...

 

[Helmut Jarausch]

Hi,

On python2.1.3, python2.2.1 and python2.2.3 round(22.47,2) gives
22.469999999999999 . Does anyone know if this is a bug or some weird, yet
unexpected, behaviour?

You're asking too much. There simply is no number 22.47 representable in
binary arithmetic with a finite number of bits. Remember most current
CPUs use binary and not decimal arithmetic.
So 22.469999999999999 is the best you can get and it doesn't make sense
to print a rounded number with many more digits after the decimal point.

If you printed

print '%6.2f' % 22.47

you got '22.47'
so this limited precision (nearly 16 decimal digits)
doesn't hurt.
You still could write a fix-point class which internally
store the value multiplied by 10**d with a given 'd'.
You'd have to overload the standard math operators (and perhaps
functions) and especially the '__str__' method.



--
Helmut Jarausch

Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany

 

[Bernhard Herzog]

Sure would, if 22 was where you wanted to go.

Even if he would want to round to an int, using int(x + 0.5) would fail
for negative values:

-1

but

-2.0


Bernhard

 

[John Machin]

Even if he would want to round to an int, using int(x + 0.5) would fail
for negative values:

-2.0


Bernhard

"Fail" depends on your/his definition of "round to an int" -- up?
down? towards zero? away from zero? closest in either direction [ties
broken how?]?