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Where is the correct round() method?

P: n/a
Hello,

I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:
>>round(0.5)
0
>>round(1.5)
2
>>round(2.5)
2
I would think this is a common need, but I cannot find a function in
the Python library to do it. I wrote my own, but did I miss such a
method in my search of the Python library?

Thanks
Jul 27 '08 #1
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9 Replies


P: n/a
josh logan wrote:
Hello,

I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:

>>>round(0.5)
0
>>>round(1.5)
2
>>>round(2.5)
2

Huh?
>>round(2.5)
3.0
Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo
CPU". What system are you on?
It could be that 2.5 is really 2.49999... which would round down to 2,
but on any modern CPU (using IEEE floating point), 2.5 should be
representable exactly.

However, as with any floating point calculations, if you expect exact
representation or calculations with any numbers, then you are misusing
floating points.

Gary Herron
I would think this is a common need, but I cannot find a function in
the Python library to do it. I wrote my own, but did I miss such a
method in my search of the Python library?

Thanks
--
http://mail.python.org/mailman/listinfo/python-list
Jul 28 '08 #2

P: n/a
it could be that 3.0 is using "banker's rounding" --- rounding to the
even digit. the idea behind it behind it being to reduce error
accumulation when working with large sets of values.

Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo
CPU". What system are you on?
It could be that 2.5 is really 2.49999... which would round down to 2,
but on any modern CPU (using IEEE floating point), 2.5 should be
representable exactly.

Jul 28 '08 #3

P: n/a
On Jul 27, 7:58*pm, Gary Herron <gher...@islandtraining.comwrote:
josh logan wrote:
Hello,
I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:
>>round(0.5)
0
>>round(1.5)
2
>>round(2.5)
2

Huh?

*>>round(2.5)
3.0

Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo
CPU". *What system are you on?

It could be that 2.5 is really 2.49999... which would round down to 2,
but on any modern CPU (using IEEE floating point), 2.5 should be
representable exactly.

However, as with any floating point calculations, if you expect exact
representation or calculations with any numbers, then you are misusing
floating points.

Gary Herron
I would think this is a common need, but I cannot find a function in
the Python library to do it. I wrote my own, but did I miss such a
method in my search of the Python library?
Thanks
--
http://mail.python.org/mailman/listinfo/python-list

I should reiterate that I am using Python 3.0 and not Python 2.x.
It looks like the behavior round() has changed between these two
versions.
Here is the documentation for round() in Python 3.0:
http://docs.python.org/dev/3.0/libra...ons.html#round

Of interest in this discussion is the second paragraph, which explains
the change:

Does anyone know the reason behind this change, and what replacement
method I can use to get the original behavior?
Jul 28 '08 #4

P: n/a
On Jul 27, 8:45*pm, pigmartian <scottp...@comcast.netwrote:
it could be that 3.0 is using "banker's rounding" --- rounding to the
even digit. *the idea behind it behind it being to reduce error
accumulation when working with large sets of values.
Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo
CPU". *What system are you on?
It could be that 2.5 is really 2.49999... which would round down to 2,
but on any modern CPU (using IEEE floating point), 2.5 should be
representable exactly.

That's exactly what's happening, pigmartian. Thank you for explaining
the reasoning behind this change.
So am I relegated to building my own round() function that behaves
like the original function? Or did they move the functionality to a
new method somewhere for backwards-compatibility?
Jul 28 '08 #5

P: n/a


Gary Herron wrote:
josh logan wrote:
>I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:
>>>>round(2.5)
>
2
Huh?
>>round(2.5)
3.0
As the OP said, PY 3.0, where statisticians' unbiased round to even was
explicitly adopted. (I think before it was maybe left to the C library?)
>I would think this is a common need,
If you need any particular rounding for legal/finance-rule reasons, you
probably need the decimal module, which has several rounding modes and
other features catering to money calculation rules.

For general data analysis with floats, round to even is better.

tjr

Jul 28 '08 #6

P: n/a
josh logan wrote:
Hello,

I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:
>>>round(0.5)
0
>>>round(1.5)
2
>>>round(2.5)
2
I would think this is a common need, but I cannot find a function in
the Python library to do it. I wrote my own, but did I miss such a
method in my search of the Python library?

Thanks
I think what you want is something like:

math.ceil(x-0.4999999999999)

-Larry
Jul 28 '08 #7

P: n/a
On Jul 27, 8:55*pm, Larry Bates <larry.ba...@websafe.com`wrote:
josh logan wrote:
Hello,
I need a round function that _always_ rounds to the higher integer if
the argument is equidistant between two integers. In Python 3.0, this
is not the advertised behavior of the built-in function round() as
seen below:
>>round(0.5)
0
>>round(1.5)
2
>>round(2.5)
2
I would think this is a common need, but I cannot find a function in
the Python library to do it. I wrote my own, but did I miss such a
method in my search of the Python library?
Thanks

I think what you want is something like:

math.ceil(x-0.4999999999999)

-Larry- Hide quoted text -

- Show quoted text -
The version I learned back in my FORTRAN days was:

int(x + 0.5)

-- Paul
Jul 28 '08 #8

P: n/a
josh logan wrote:
On Jul 27, 8:45 pm, pigmartian <scottp...@comcast.netwrote:
>it could be that 3.0 is using "banker's rounding" --- rounding to the
even digit. the idea behind it behind it being to reduce error
accumulation when working with large sets of values.

>>Works for me on Python 2.5 on Linux running on "Intel(R) Core(TM)2 Duo
CPU". What system are you on?

It could be that 2.5 is really 2.49999... which would round down to 2,
but on any modern CPU (using IEEE floating point), 2.5 should be
representable exactly.

That's exactly what's happening, pigmartian. Thank you for explaining
the reasoning behind this change.
So am I relegated to building my own round() function that behaves
like the original function? Or did they move the functionality to a
new method somewhere for backwards-compatibility?
This will work as you wish:
math.floor(x+0.5)

Gary Herron

--
http://mail.python.org/mailman/listinfo/python-list
Jul 28 '08 #9

P: n/a
On Jul 28, 12:34*am, Gary Herron <gher...@islandtraining.comwrote:
This will work as you wish:
* math.floor(x+0.5)
This works fine for positive x but what about negative:
>>round(2.5)
3.0
>>floor(2.5 + 0.5)
3.0
>>round(-2.5)
-3.0
>>floor(-2.5 + 0.5)
-2.0

Maybe:

def round2(x):
return math.floor(x + (0.5 if x >= 0 else -0.5))
Jul 28 '08 #10

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