Convert a floating point number to a decimal with no loss of information in Python.

``````import decimal

"""
Nicely representing a floating-point number in python

QUESTION
--------

I want to represent a floating-point number as a string rounded to some number
of significant digits, and never using the exponential format. Essentially, I
want to display any floating-point number and make sure it "looks nice".

There are several parts to this problem:

- I need to be able to specify the number of significant digits.
- The number of significant digits needs to be variable, which can't be
done with with the string formatting operator. I've been corrected;
the string formatting operator can do this.
- I need it to be rounded the way a person would expect, not something like
`1.999999999999`

I've figured out one way of doing this, though it looks like a work-round
and it's not quite perfect. (The maximum precision is 15 significant digits.)

>>> def f(number, sigfig):
return ("%.15f" % (round(number, int(-1 * floor(log10(number)) + (sigfig - 1))))).rstrip("0").rstrip(".")

>>> print f(0.1, 1)
0.1
>>> print f(0.0000000000368568, 2)
0.000000000037
>>> print f(756867, 3)
757000

Is there a better way to do this? Why doesn't Python have a built-in function
for this?

------

It appears there is no built-in string formatting trick which allows you to
(1) print floats whose first significant digit appears after the 15th decimal
place and (2) not in scientific notation. So that leaves manual string
manipulation.

Below I use the `decimal` module to extract the decimal digits from the float.
The `float_to_decimal` function is used to convert the float to a `Decimal`
object. The obvious way `decimal.Decimal(str(f))` is wrong because `str(f)`
can lose significant digits.

`float_to_decimal` was lifted from the decimal module's documentation.

Once the decimal digits are obtained as a tuple of ints, the code below does
the obvious thing: chop off the desired number of sigificant digits, round up
if necessary, join the digits together into a string, tack on a sign, place a
decimal point and zeros to the left or right as appropriate.

At the bottom you'll find a few cases I used to test the `f` function.

http://stackoverflow.com/questions/2663612/nicely-representing-a-floating-point-number-in-python
"""

def float_to_decimal(f):
# http://docs.python.org/library/decimal.html#decimal-faq
"Convert a floating point number to a Decimal with no loss of information"
n, d = f.as_integer_ratio()
numerator, denominator = decimal.Decimal(n), decimal.Decimal(d)
ctx = decimal.Context(prec=60)
result = ctx.divide(numerator, denominator)
while ctx.flags[decimal.Inexact]:
ctx.flags[decimal.Inexact] = False
ctx.prec *= 2
result = ctx.divide(numerator, denominator)
return result

def f(number, sigfig):
# http://stackoverflow.com/questions/2663612/nicely-representing-a-floating-point-number-in-python/2663623#2663623
assert(sigfig > 0)
try:
d = decimal.Decimal(number)
except TypeError:
d = float_to_decimal(float(number))

sign, digits, exponent = d.as_tuple()
if len(digits) < sigfig:
digits = list(digits)
digits.extend( * (sigfig - len(digits)))
result = int(''.join(map(str, digits[:sigfig])))

# Round the result
if len(digits) > sigfig and digits[sigfig] >= 5:
result += 1
result = list(str(result))

# Rounding can change the length of result
shift += len(result) - sigfig

# reset len of result to sigfig
result = result[:sigfig]
if shift >= sigfig - 1:
# Tack more zeros on the end
result += ['0'] * (shift - sigfig + 1)
elif 0 <= shift:
# Place the decimal point in between digits
result.insert(shift + 1, '.')
else:
# Tack zeros on the front
assert(shift < 0)
result = ['0.'] + ['0'] * (-shift - 1) + result
if sign:
result.insert(0, '-')
return ''.join(result)

if __name__ == '__main__':
tests = [
(0.1, 1, '0.1'),
(0.0000000000368568, 2, '0.000000000037'),
(0.00000000000000000000368568, 2, '0.0000000000000000000037'),
(756867, 3, '757000'),
(-756867, 3, '-757000'),
(-756867, 1, '-800000'),
(0.0999999999999, 1, '0.1'),
(0.00999999999999, 1, '0.01'),
(0.00999999999999, 2, '0.010'),
(0.0099, 2, '0.0099'),
(1.999999999999, 1, '2'),
(1.999999999999, 2, '2.0'),
(34500000000000000000000, 17, '34500000000000000000000'),
('34500000000000000000000', 17, '34500000000000000000000'),
(756867, 7, '756867.0'),
]

for number, sigfig, answer in tests:
try:
result = f(number, sigfig)