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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Thu, 16 Feb 2012 10:34:18 -0700
Subject: Re: format a measurement result and its error in "scientific" way
To: Daniel Fetchinson <fetchinson@googlemail.com>
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On Thu, Feb 16, 2012 at 1:36 AM, Daniel Fetchinson
<fetchinson@googlemail.com> wrote:
>>> Hi folks, often times in science one expresses a value (say
>>> 1.03789291) and its error (say 0.00089) in a short way by parentheses
>>> like so: 1.0379(9)
>>>
>>> One can vary things a bit, but let's take the simplest case when we
>>> only keep 1 digit of the error (and round it of course) and round the
>>> value correspondingly. I've been searching around for a simple
>>> function that would take 2 float arguments and would return a string
>>> but didn't find anything although something tells me it's been done a
>>> gazillion times.
>>>
>>> What would be the simplest such function?
>>
>> Well, this basically works:
>>
>>>>> def format_error(value, error):
>> ... =A0 =A0 precision =3D int(math.floor(math.log(error, 10)))
>> ... =A0 =A0 format =3D "%%.%df(%%d)" % max(-precision, 0)
>> ... =A0 =A0 return format % (round(value, -precision),
>> ... =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0int(round(error / 10 ** p=
recision)))
>> ...
>>>>> format_error(1.03789291, 0.00089)
>> '1.0379(9)'
>>
>> Note that "math.floor(math.log(error, 10))" may return the wrong
>> decimal precision due to binary floating point rounding error, which
>> could produce some strange results:
>>
>>>>> format_error(10378929, 1000)
>> '10378900(10)'
>>
>> So you'll probably want to use decimals instead:
>>
>> def format_error(value, error):
>> =A0 =A0 value =3D decimal.Decimal(value)
>> =A0 =A0 error =3D decimal.Decimal(error)
>> =A0 =A0 value_scale =3D value.log10().to_integral(decimal.ROUND_FLOOR)
>> =A0 =A0 error_scale =3D error.log10().to_integral(decimal.ROUND_FLOOR)
>> =A0 =A0 precision =3D value_scale - error_scale
>> =A0 =A0 if error_scale > 0:
>> =A0 =A0 =A0 =A0 format =3D "%%.%dE" % max(precision, 0)
>> =A0 =A0 else:
>> =A0 =A0 =A0 =A0 format =3D "%%.%dG" % (max(precision, 0) + 1)
>> =A0 =A0 value_str =3D format % value.quantize(decimal.Decimal("10") **
>> error_scale)
>> =A0 =A0 error_str =3D '(%d)' % error.scaleb(-error_scale).to_integral()
>> =A0 =A0 if 'E' in value_str:
>> =A0 =A0 =A0 =A0 index =3D value_str.index('E')
>> =A0 =A0 =A0 =A0 return value_str[:index] + error_str + value_str[index:]
>> =A0 =A0 else:
>> =A0 =A0 =A0 =A0 return value_str + error_str
>>
>>>>> format_error(1.03789291, 0.00089)
>> '1.0379(9)'
>>>>> format_error(103789291, 1000)
>> '1.03789(1)E+08'
>>
>> I haven't tested this thoroughly, so use at your own risk. :-)
>
> Thanks a lot, this indeed mostly works, except for cases when the
> error needs to be rounded up and becomes two digits:
>
>>>> format_error( '1.34883', '0.0098' )
> '1.349(10)'
>
> But in this case I'd like to see 1.35(1)

A small adjustment to the scale fixes that.  Also tidied up the string
formatting part:

import decimal

def format_error(value, error):
    value =3D decimal.Decimal(value)
    error =3D decimal.Decimal(error)
    error_scale =3D error.adjusted()
    error_scale +=3D error.scaleb(-error_scale).to_integral().adjusted()
    value_str =3D str(value.quantize(decimal.Decimal("1E%d" % error_scale))=
)
    error_str =3D '(%d)' % error.scaleb(-error_scale).to_integral()
    if 'E' in value_str:
        index =3D value_str.index('E')
        return value_str[:index] + error_str + value_str[index:]
    else:
        return value_str + error_str

Cheers,
Ian