python - Convert float to rounded decimal equivalent

Question

When you convert a float to Decimal, the Decimal will contain as accurate a representation of the binary number that it can. It's nice to be accurate, but it isn't always what you want. Since many decimal numbers can't be represented exactly in binary, the resulting Decimal will be a little off - sometimes a little high, sometimes a little low.

>>> from decimal import Decimal
>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
    print Decimal(f)

0.1000000000000000055511151231257827021181583404541015625
0.299999999999999988897769753748434595763683319091796875
10000000000000000905969664
9999999999999999583119736832
1.000000000000099920072216264088638126850128173828125

Ideally we'd like the Decimal to be rounded to the most likely decimal equivalent.

I tried converting to str since a Decimal created from a string will be exact. Unfortunately str rounds a little too much.

>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
    print Decimal(str(f))

0.1
0.3
1E+25
1E+28
1.0

Is there a way of getting a nicely rounded Decimal from a float?

Answer 1

It turns out that repr does a better job of converting a float to a string than str does. It's the quick-and-easy way to do the conversion.

>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
    print Decimal(repr(f))

0.1
0.3
1E+25
1E+28
1.0000000000001

Before I discovered that, I came up with a brute-force way of doing the rounding. It has the advantage of recognizing that large numbers are accurate to 15 digits - the repr method above only recognizes one significant digit for the 1e25 and 1e28 examples.

from decimal import Decimal,DecimalTuple

def _increment(digits, exponent):
    new_digits = [0] + list(digits)
    new_digits[-1] += 1
    for i in range(len(new_digits)-1, 0, -1):
        if new_digits[i] > 9:
            new_digits[i] -= 10
            new_digits[i-1] += 1
    if new_digits[0]:
        return tuple(new_digits[:-1]), exponent + 1
    return tuple(new_digits[1:]), exponent

def nearest_decimal(f):
    sign, digits, exponent = Decimal(f).as_tuple()
    if len(digits) > 15:
        round_up = digits[15] >= 5
        exponent += len(digits) - 15
        digits = digits[:15]
        if round_up:
            digits, exponent = _increment(digits, exponent)
    while digits and digits[-1] == 0 and exponent < 0:
        digits = digits[:-1]
        exponent += 1
    return Decimal(DecimalTuple(sign, digits, exponent))

>>> for f in (0.1, 0.3, 1e25, 1e28, 1.0000000000001):
    print nearest_decimal(f)

0.1
0.3
1.00000000000000E+25
1.00000000000000E+28
1.0000000000001

Edit: I discovered one more reason to use the brute-force rounding. repr tries to return a string that uniquely identifies the underlying float bit representation, but it doesn't necessarily ensure the accuracy of the last digit. By using one less digit, my rounding function will more often be the number you would expect.

>>> print Decimal(repr(2.0/3.0))
0.6666666666666666
>>> print dec.nearest_decimal(2.0/3.0)
0.666666666666667

The decimal created with repr is actually more accurate, but it implies a level of precision that doesn't exist. The nearest_decimal function delivers a better match between precision and accuracy.