working of round()

[Krishna.K.1900@gmail.com]

Does round() always perfectly return the output expected or are there
some artifacts which don't allow perfect functionality

Using python 2.5:

>>round(12.23 4, 2)

12.23

>>round(12.23 4, 3)

12.234

>>round(12.23 4, 1)

12.199999999999 999

>>>

but was expecting 12.2

Also, for round(x,n), can't 'x' be an expression

round(5.25/2, 2)

was expecting 2.62 , but

>>round(5.25/2, 2)

2.6299999999999 999

[Robert Kern]

Re: working of round()

Krishna.K.1900@ gmail.com wrote:

Does round() always perfectly return the output expected or are there
some artifacts which don't allow perfect functionality
>
Using python 2.5:

>>>round(12.234 , 2)

12.23

>>>round(12.234 , 3)

12.234

>>>round(12.234 , 1)

12.199999999999 999
>
but was expecting 12.2

http://docs.python.org/tut/node16.html

Also, for round(x,n), can't 'x' be an expression
>
round(5.25/2, 2)
>
was expecting 2.62 , but
>

>>>round(5.25/2, 2)

2.6299999999999 999

round(x, n) essentially does the following:

math.floor(x * 10**n + 0.5) / 10**n

Since (5.25/2)*100 == 262.5, adding 0.5 gives 263.0 and ultimately 2.63 as the
rounded number. round() does not do anything more complicated like round-to-even.

Use the decimal module if you want decimal arithmetic rather than binary
floating point.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

[Steven Bethard]

Re: working of round()

Krishna.K.1900@ gmail.com wrote:

Does round() always perfectly return the output expected or are there
some artifacts which don't allow perfect functionality
>
Using python 2.5:

>>>round(12.234 , 2)

12.23

>>>round(12.234 , 3)

12.234

>>>round(12.234 , 1)

12.199999999999 999
>
but was expecting 12.2
>
Also, for round(x,n), can't 'x' be an expression
>
round(5.25/2, 2)
>
was expecting 2.62 , but
>

>>>round(5.25/2, 2)

2.6299999999999 999

You're running into floating-point issues (e.g. it's impossible to
represent 2.63 perfectly in binary). What are you really trying to do?
If you just want to format these with only two decimal places, use
string formatting::

>>'%.2f' % 12.234

'12.23'

>>'%.2f' % (5.25 / 2)

'2.63'

I'm not sure why you would have expected 2.62 for the latter when::

>>5.25 / 2

2.625

STeVe

[subscriber123]

Re: working of round()

On Apr 15, 8:06 pm, Krishna.K.1...@ gmail.com wrote:

Does round() always perfectly return the output expected or are there
some artifacts which don't allow perfect functionality
>
Using python 2.5:
>

>round(12.234 , 2)

12.23

>round(12.234 , 3)

12.234

>round(12.234 , 1)

12.199999999999 999
>
but was expecting 12.2
>
Also, for round(x,n), can't 'x' be an expression
>
round(5.25/2, 2)
>
was expecting 2.62 , but
>

>round(5.25/2, 2)

>
2.6299999999999 999

The problem is that floats are encoded as fractions where the
denominator is an exponent of 2.
2.63 is not representable as such a fraction.
2.6299999999999 999999999999999 99999999... is the closest fraction.
Rounding this number will only give you the same thing.
If you want decimals to act as expected, use the Decimal class in
module decimal. It works as expected, but is much slower.