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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 builtin 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  
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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 builtin 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/pythonlist  
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.
 
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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 builtin 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/pythonlist
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?  
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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 backwardscompatibility?  
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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 builtin 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/financerule 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  
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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 builtin 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(x0.4999999999999)
Larry  
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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 builtin 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(x0.4999999999999)
Larry Hide quoted text 
 Show quoted text 
The version I learned back in my FORTRAN days was:
int(x + 0.5)
 Paul  
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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 backwardscompatibility?
This will work as you wish:
math.floor(x+0.5)
Gary Herron
 http://mail.python.org/mailman/listinfo/pythonlist  
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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))   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 5919
 replies: 9
 date asked: Jul 27 '08
