Question about Decimal and rounding

Frank Millman frank at chagford.com
Fri Apr 27 09:51:14 EDT 2018


"Thomas Jollans"  wrote in message 
news:19223891-2006-d496-bdfe-32776834e318 at tjol.eu...

> On 27/04/18 10:21, Frank Millman wrote:
>
> > I have an object which represents a Decimal type.
> >
> > It can receive input from various sources. It has to round the value to
> > a particular scale factor before storing it. The scale factor can vary,
> > so it has to be looked up every time, which is a slight overhead. I
> > thought I could speed it up a bit by checking first to see if the value
> > has any decimal places. If not, I can skip the scaling routine.
> >
> > This is how I do it -
> >
> >    s = str(value)
> >    if '.' in s:
> >        int_portion, dec_portion = s.split('.')
> >        is_integer = (int(int_portion) == value)
> >    else:
> >        is_integer = True
> >
> > It assumes that the value is in the form iii.ddd or just iii. Today I
> > found the following value -
> >
> >    -1.4210854715202004e-14
> >
> > which does not match my assumption.
> >
> > It happens to work, but I don't understand enough about the notation to
> > know if this is reliable, or if there are any corner cases where my test
> > would fail.
>
> This is not reliable. Decimal is happy to give you integers in
> scientific notation if this is needed to keep track of the number of
> significant digits.
>
> >>> str(Decimal('13.89e2'))
> '1389'
> >>> str(Decimal('13.89e3'))
> '1.389E+4'
> >>> Decimal('13.89e3') == 13890
> True
>
> It appears to me that the "obvious" way to check whether a Decimal
> number is an integer is simply:
>
> >>> d1 = Decimal('1.1')
> >>> d2 = Decimal('3')
> >>> int(d1) == d1
> False
> >>> int(d2) == d2
> True
>

Thanks, Thomas - makes perfect sense.

I have been burned before by -

>>> int('1.1')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '1.1'

so I did not think of trying -

>>> int(Decimal('1.1'))
1

but it works just fine.

Frank




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