The fact that the numbers are stored in binary is significant:
In [16]: nums = [eval('1.%d5' % n) for n in range(10)]
In [17]: nums
Out[17]: [1.05, 1.15, 1.25, 1.35, 1.45, 1.55, 1.65, 1.75, 1.85, 1.95]
In [18]: [round(n, 1) for n in nums]
Out[18]: [1.1, 1.1, 1.2, 1.4, 1.4, 1.6, 1.6, 1.8, 1.9, 1.9]
Proper decimal rounding might look like:
In [20]: from decimal import Decimal
In [21]: nums = [Decimal('1.%d5' % n) for n in range(10)]
In [22]: nums
Out[22]:
[Decimal('1.05'),
Decimal('1.15'),
Decimal('1.25'),
Decimal('1.35'),
Decimal('1.45'),
Decimal('1.55'),
Decimal('1.65'),
Decimal('1.75'),
Decimal('1.85'),
Decimal('1.95')
In [23]: [round(n, 1) for n in nums]
Out[23]:
[Decimal('1.1'),
Decimal('1.2'),
Decimal('1.3'),
Decimal('1.4'),
Decimal('1.5'),
Decimal('1.6'),
Decimal('1.7'),
Decimal('1.8'),
Decimal('1.9'),
Decimal('2.0')]
Or with half-even rounding:
In [24]: from decimal import getcontext
In [25]: getcontext().rounding = 'ROUND_HALF_EVEN'
In [26]: [round(n, 1) for n in nums]
Out[26]:
[Decimal('1.0'),
Decimal('1.2'),
Decimal('1.2'),
Decimal('1.4'),
Decimal('1.4'),
Decimal('1.6'),
Decimal('1.6'),
Decimal('1.8'),
Decimal('1.8'),
Decimal('2.0')]
The binary floats don't work out correct because some are above and
some are below the number suggested by the original float literal.
On Thu, 11 Apr 2019 at 02:03, Chris Smith <smi...@gmail.com> wrote:
>
> That floats are stored in binary is an implementation detail which need not
> prevent base-10 rounding to still work. The 2nd argument to round is intended
> to tell at which base-10 digit the rounding is to take place. By shifting
> that digit to the ones position and then doing the rounding one can easily
> detect whether what follows is above or below a half.
>
> >>> .345.as_integer_ratio()
> (1553741871442821, 4503599627370496)
> >>> (_*100).as_integer_ratio()
> (69, 2)
>
> "round to even" means "if what follows the digit to which you are rounding is
> 5 then round so your digit is even". As is shown in the integer ratios above,
> what follows the 4 is exactly 1/2 so we can round to the even 4 (as in .34)
> instead of the odd 5 (as in .35) just like round(0.75,1) rounds to 0.8 while
> round(0.25) rounds to 0.2.
>
> On Wednesday, April 10, 2019 at 7:29:18 PM UTC-5, Aaron Meurer wrote:
>>
>> Doesn't Python do rounding based on the binary representation of the float?
>>
>> I'm a little confused what "round to even" means in that case.
>>
>> Aaron Meurer
>>
>> On Wed, Apr 10, 2019 at 6:16 PM Chris Smith <smi...@gmail.com> wrote:
>> >
>> > Python 3 implements "round to even on tie" logic so `round(12.5)` -> 12,
>> > not 13. I have updated, in #16608, the round function but there is a
>> > difference in how ties are detected. I shift the desired position to the
>> > ones position and then check for a tie so 12.345 is shifted to 1234.5 and
>> > rounded to 1234 then is divided by 100 to give 12.34. Python doesn't do
>> > this. I suspect it adds 0.05 and then detects that12.395 > 12395/1000 and
>> > rounds up to 12.35
>> >
>> >
>> > >>> Rational(*.345.as_integer_ratio())-Rational(345,1000) # .345 <
>> > >>> 345/1000
>> > -3/112589990684262400
>> > >>> Rational(*.395.as_integer_ratio())-Rational(395,1000) # .395 >
>> > >>> 395/1000
>> > 1/56294995342131200
>> >
>> >
>> > >>> round(12.345,2) # 12.345 rounds up b/c a tie is not detected
>> > 12.35
>> >
>> >
>> >
>> > Does anyone have objections to the proposed rounding?
>> >
>> > /c
>> >
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