I think that Python's float.__round__ is correct. AIUI it rounds
correctly based on the true value represented by the float:
In [4]: round(1.05, 1)
Out[4]: 1.1
In [5]: import decimal
In [6]: decimal.Decimal(1.1)
Out[6]: Decimal('1.100000000000000088817841970012523233890533447265625')
That's surprising just because the true exact value isn't what we
thought it should be. There is a rounding error but it happens in the
literal 1.1 that doesn't give us the float we think it should. It's
extra confusing because str(1.1) gives '1.1' making it seem like the
exact number we requested. The behaviour of round is mathematically
correct here though given the input received: it is computing the true
exact result correctly rounded (where "correctly rounded" refers to
the decimal to binary rounding at the end).
For decimals that can be exactly represented in binary the results are
as we would intuitively expect:
In [7]: round(1.25, 1)
Out[7]: 1.2
In [8]: round(1.75, 1)
Out[8]: 1.8
(Of course 1.2 and 1.8 are not exactly representable in binary
floating point but we have the floats that result from correctly
rounding the true numbers 1.2 and 1.8.)
--
Oscar
On Thu, 11 Apr 2019 at 05:29, Chris Smith <smi...@gmail.com> wrote:
>
> I am aware of how the numbers are stored but was overly optimistic that the
> shift could resolve this in all cases. It can't (and thanks for the
> correction).
> But my suggested alternative makes a significant difference in how often the
> problem arises:
>
> >>> bad=[]
> >>> for i in range(1,1000):
> ... n = str(i)+'5'
> ... if int(str(round(int(n)/10**len(n),len(n)-1))[-1])%2!=0:bad.append(i) #
> e.g. round(0.1235, 3)
> ...
> >>> len(bad)
> 546
>
>
> >>> bad=[]
> >>> for i in range(1,1000):
> ... n = str(i)+'5'
> ... if round(int(n)/10**len(n)*10**(len(n)-1))%2!=0:bad.append(i) # e.g.
> round(0.1235*1000)
> ...
> >>> len(bad)
> 8
> >>> bad # e.g. 0.545*100 != 54.5
> [54, 57, 501, 503, 505, 507, 509, 511]
>
>> So the question is whether we want to do better and keep the SymPy algorithm.
>
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