keeping same formatting for floating point values

I have a python program that reads floating point values using the following regular expression

 (-?\d+\.\d+)

once I extract the value using float(match.group(1)), I get the actual floating point number. However, I am not able to distinguish if the number was 1.2345678 or 1.234 or 1.2340000.

The problem I am facing is to print out the floating point value again, with the exact same formatting. An easy solution is to "split and count" the floating point value when still a string, eg splitting at the decimal point, and counting the integer part length and the fractional part length, then create the formatter as

print "%"+str(total_len)+"."+str(fractional_len)+"f" % value

but maybe you know a standard way to achieve the same result ?

From stackoverflow

• If you want to keep a fixed precision, avoid using floats and use Decimal instead:

  >>> from decimal import Decimal
  >>> d = Decimal('-1.2345')
  >>> str(d)
  '-1.2345'
  >>> float(d)
  -1.2344999999999999
  

  Stefano Borini : Cool, I didn't know that!

  S.Lott : Better still, keep it as string as long as possible.

• >>> from decimal import Decimal as d
  >>> d('1.13200000')
  Decimal('1.13200000')
  >>> print d('1.13200000')
  1.13200000
  

• You method is basically correct. String formatting has a less often used * operator you can put for the formatting sizes, here's some code:

  import re
  
  def parse_float(str):
    re_float = re.compile(r'(-?)(\d+)\.(\d+)')
    grps = re_float.search(str)
    sign, decimal, fraction = grps.groups()
    float_val = float('%s%s.%s' % (sign, decimal, fraction))
    total_len = len(grps.group(0))
    print '%*.*f'  % (total_len, len(fraction), float_val)
  
  parse_float('1.2345678')
  parse_float('1.234')
  parse_float('1.2340000')
  

  and it outputs

  1.2345678
  1.234
  1.2340000